Advanced Engineering

Mathematics

Magdi Salah El-Azab

 Contents

Advanced Engineering Mathematics-Fourth Edition

2626/2001 ‫رقم اإليداع بدار الكتب والوثائق القومية‬

ISBN 977-5272-93-9

© Magdi Salah El-Azab, 2003

All Rights Reserved. No part of this book may be reproduced by any

means without a written permission of the author.
 Contents

Contents

Chapter 1 Fourier Analysis 1

1.1 Fourier Series 1
1.1.1 Definitions and notations 1
1.1.2 Trigonometric Fourier series 9
1.1.3 Convergence of Fourier series 15
1.1.4 Fourier series for even and odd functions (Half-interval
expansion) 17
1.1.5 Fourier cosine and sine series 19
1.1.6 Exponential (Complex) Fourier series 27
1.1.7 Parseval’s identity 29
1.2 Fourier Integrals 33
1.2.1 Introduction 33
1.2.2 Fourier cosine integrals 37
1.2.3 Fourier sine integrals 37
Chapter 2 Functions of a complex variable 43
2.1 Complex functions 43
2.2 Limits 48
2.3. Continuity 51
2.4 Derivative of complex functions 52
2.5 Cauchy-Riemann equations 55
2.6 Cauchy-Riemann equations in polar form 59
2.7 Harmonic functions 62
Chapter 3 Elementary Functions 68
3.1 The complex exponential function 68
3.2 The complex logarithmic function 70

i
 Contents

3.3 General powers 76

3.3.1 The root of complex numbers 76
3.3.2 Expansion of and 80
3.3.3 Complex powers 82
3.4 Trigonometric functions 85
3.5 Hyperbolic functions 91
3.6 Inverse trigonometric and hyperbolic functions 96
Chapter 4 Complex integrations 102
4.1 Contour integrals 102
4.2 Definition of complex line integration 102
4.3 Properties of complex line integrals 103
4.4 Evaluation of complex line integrals (Direct method) 104
4.5 Cauchy’s integral theorem 106
4.5.1 Cauchy’s integral theorem for simply connected domains
107
4.5.2 Cauchy’s integral theorem for multiply connected
domains 109
4.6 Cauchy’s integral formula 111
4.7 The fundamental theorem for complex integral 115
4.8 Integral of elementary functions 119
4.9 Derivatives of analytic functions 121
Chapter 5 The Residue theorem 128
5.1 Partial fractions 128
5.2 Residues 132
5.3 The residue theorem 133
5.4 Definite integrals 142
Chapter 6 Multiple integrals 161
6.1 Double integrals 161
6.1.1 Interpretation of double integrals 162
6.1.2 Basic properties of double integrals 162
6.1.3 Evaluation of double integrals 163
6.1.4 Applications of double integrals 168
6.1.5 Interchange of the order of integration 173
6.1.6 Change of variables in double integrals 175
6.1.7 Double integrals in polar coordinates 177
 Contents

6.2 Triple integrals 186

6.2.1 Applications of triple integrals 189
6.2.2 Triple integrals in spherical Co-ordinates 196
6.2.3 Triple integrals in cylindrical Co-ordinates 199
6.3 Line Integrals 203
6.3.1 Line integrals of scalar functions in the plane 203
6.3.2 Line integrals of scalar functions in space 207
6.3.3 Line integrals of vector fields 208
6.3.4 Independence of path 212
6.3.5 The fundamental theorem for line integrals 213
6.3.6 Green’s theorem 219
6.4 Surface Integrals 228
6.4.1 Surface integrals of scalar functions 228
6.4.2 Surface integrals of vector fields 234
6.4.3 The Divergence theorem 238
6.4.4 Stokes theorem 246
References 255

iii
1. Fourier Analysis

The subject of Fourier analysis is one of the oldest subjects in

mathematical analysis and is of great importance to mathematicians and
engineers alike. From a practical point of view, when one thinks of
Fourier analysis, one usually refers to Fourier series, Fourier integrals
and Fourier transforms. The importance of Fourier analysis stems not
only from the significance of their physical interpretation, such as time-
frequency analysis of signals, but also from the fact that the Fourier
analytic techniques are extremely powerful. They constitute a very
important tool in solving problems that involve ordinary and partial
differential equations.
1.1 Fourier Series
1.1.1. Definitions and notations

Throughout this chapter, we shall need the following definitions and

notations. It is convenient to use x  x 0
𝑥 → 𝑥0+  𝑥 → 𝑥0 , 𝑥 > 𝑥0 ,
𝑥 → 𝑥0−  𝑥 → 𝑥0 , 𝑥 < 𝑥0 , x0  x

𝑓(𝑥0+ ) lim+ 𝑓(𝑥) x  x 0

𝑥 → 𝑥0

𝑓(𝑥0− ) lim− 𝑓(𝑥) x  x0

𝑥 → 𝑥0
Fig. 2.1

From now on we shall use the following identities. For every positive
integer 𝑛 and real numbers 𝑎 and 𝑏, we have
𝟏. sin 𝑛𝜋 = 0
𝟐. cos 𝑛𝜋 = (−1)𝑛
1
𝟑. sin 𝑎 cos 𝑏 = (sin(𝑎 − 𝑏) + sin(𝑎 + 𝑏))
2
 2 Chapter 1 Fourier Analysis

1
𝟒. cos 𝑎 cos 𝑏 = (cos(𝑎 − 𝑏) + cos(𝑎 + 𝑏))
2
1
𝟓. sin 𝑎 sin 𝑏 = (cos(𝑎 − 𝑏) − cos(𝑎 + 𝑏))
2
𝑎𝑥
𝑒 𝑎𝑥
𝟔. ∫ 𝑒 sin 𝑏𝑥 𝑑𝑥 = 2 (𝑎 sin 𝑏𝑥 − 𝑏 cos 𝑏𝑥) + 𝑐
𝑎 + 𝑏2
𝑎𝑥
𝑒 𝑎𝑥
𝟕. ∫ 𝑒 cos 𝑏𝑥 𝑑𝑥 = 2 (𝑎 cos 𝑏𝑥 + 𝑏 sin 𝑏𝑥) + 𝑐
𝑎 + 𝑏2
𝜋⁄2 𝜋⁄2
𝟖. ∫ cos 𝑛 𝑥 𝑑𝑥 = ∫ sin𝑛 𝑥 𝑑𝑥
0 0
(𝑛 − 1)(𝑛 − 3)(𝑛 − 5) ⋯ 1 𝑓𝑜𝑟 𝑛 𝑜𝑑𝑑
= {𝜋
𝑛(𝑛 − 2)(𝑛 − 4) ⋯ 𝑓𝑜𝑟 𝑛 𝑒𝑣𝑒𝑛
2
2𝜋 2𝜋
𝟗. ∫ cos 𝑥 𝑑𝑥 = ∫ sin𝑛 𝑥 𝑑𝑥
𝑛
0 0
0 𝑓𝑜𝑟 𝑛 𝑜𝑑𝑑
𝜋⁄2
={
4∫ cos 𝑛 𝑥 𝑑𝑥 𝑓𝑜𝑟 𝑛 𝑒𝑣𝑒𝑛
0

Definition 1.1 (Periodic functions). A function 𝑓(𝑥)is said to be periodic

of period 𝑝 > 0 if and only if
𝑓(𝑥 + 𝑝) = 𝑓(𝑥) 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 (1)

Example 1 The function 𝑓(𝑥) = sin 𝜔𝑥 is periodic with period 2𝜋⁄𝜔

since
sin 𝜔𝑥 = sin(𝜔𝑥 + 2𝜋) = sin 𝜔(𝑥 + 2𝜋⁄𝜔)

Theorem 1.1 Let 𝑓(𝑥) be a periodic function with period 2𝐿. Then
𝑎+2𝐿 2𝐿 𝐿
∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥
𝑎 0 −𝐿
 Chapter 1 Fourier Analysis 3

This theorem says that when we integrate a periodic function over an

interval of one period, the starting point is immaterial (see Fig. 1.2)

g( x )

 0  2 3 x

Fig. 1.2

Example 2 Evaluate the integral of the function 𝑔 shown below over

two different intervals of length 2𝜋 and show that the integrals are
equal. Let
0, (2𝑛 − 1)𝜋 < 𝑥 < 2𝑛𝜋
𝑔(𝑥) = { , 𝑛 = 0, ±1, ±2, ⋯
𝑥 − 2𝑛𝜋, 2𝑛𝜋 < 𝑥 < (2𝑛 + 1)𝜋

Solution For the interval −𝜋 to 𝜋, we have

𝜋 0 𝜋 𝜋
𝑥2 𝜋2
∫ 𝑔(𝑥)𝑑𝑥 = ∫ 0𝑑𝑥 + ∫ 𝑥𝑑𝑥 = | = ,
−𝜋 −𝜋 0 2 0 2
𝜋 5𝜋
whereas, for to , we can write
2 2
5𝜋⁄2 𝜋 2𝜋 5𝜋⁄2
∫ 𝑔(𝑥)𝑑𝑥 = ∫ 𝑥𝑑𝑥 + ∫ 0 𝑑𝑥 + ∫ (𝑥 − 2𝜋)𝑑𝑥 =
𝜋⁄2 𝜋⁄2 𝜋 2𝜋
𝜋 5𝜋⁄2
𝑥2 𝑥2
= | + ( − 2𝜋𝑥)|
2 𝜋⁄2 2 2𝜋

3𝜋 2 −15𝜋 2
=( ) + (( ) − (−2𝜋 2 ))
8 8

3𝜋 2 15𝜋 2 3𝜋 2 𝜋 2 𝜋 2
= + (− + 2𝜋 2 ) = + =
8 8 8 8 2
 4 Chapter 1 Fourier Analysis

Definition 1.2 (Piecewise continuous functions). A function 𝑓(𝑥) is said

to be piecewise continuous function on the interval [𝑎, 𝑏] if
 The interval [𝑎, 𝑏] can be divided into a finite number of
subintervals in each of which 𝑓(𝑥) is continuous, and
 𝑓(𝑥) has right and left hand limits at each point of discontinuity.

As an example of a piecewise continuous function is shown in Fig. 2.3.

Fig. 1.3

Definition 1.3 (Inner product of functions). The inner or scalar product

of two functions 𝑓(𝑥) and 𝑔(𝑥) defined on the interval [𝑎, 𝑏] is given
by
𝑏
(𝑓, 𝑔) = ∫ 𝑓(𝑥)𝑔(𝑥)𝑑𝑥 , (2)
𝑎

where 𝑔(𝑥) denotes the complex conjugate of 𝑔(𝑥).

Definition 1.4 (Norm of functions). The norm of the function 𝑓(𝑥) over
the interval [𝑎, 𝑏], denoted by ‖𝑓(𝑥)‖, is defined by
𝑏 1⁄2
‖𝑓(𝑥)‖ = (∫ |𝑓(𝑥)|2 𝑑𝑥) (3)
𝑎
 Chapter 1 Fourier Analysis 5

Definition 1.5 (Orthogonal functions). Two functions 𝑓(𝑥) and 𝑔(𝑥) are
called orthogonal on an interval [𝑎, 𝑏] if
(𝑓, 𝑔) = 0 (4)

Example 3 Prove that the functions sin 𝑚𝑥 and sin 𝑛𝑥 are orthogonal
on [0, 2𝜋] for all integers 𝑚, 𝑛 such that 𝑚 ≠ 𝑛.

Solution To prove the orthogonal property of the given functions we

compute the inner product of them;
2𝜋
(sin 𝑚𝑥 , sin 𝑛𝑥) = ∫ sin 𝑚𝑥 sin 𝑛𝑥 𝑑𝑥
0

1 2𝜋
= ∫ [cos(𝑚 − 𝑛)𝑥 − cos(𝑚 + 𝑛)𝑥]𝑑𝑥
2 0
2𝜋
1 sin(𝑚 − 𝑛)𝑥 sin(𝑚 + 𝑛)𝑥
= [ − ] = 0,
2 (𝑚 − 𝑛) (𝑚 + 𝑛)
0

which implies that for 𝑚 ≠ 𝑛, the functions sin 𝑚𝑥 and sin 𝑛𝑥 are
orthogonal on [0, 2𝜋].

Example 4 Are the functions 𝑒 𝑖𝑚𝑥 and 𝑒 𝑖𝑛𝑥 orthogonal over [0, 2𝜋]
for all integers 𝑚, 𝑛 such that 𝑚 ≠ 𝑛?

Solution The inner product of the functions 𝑒 𝑖𝑚𝑥 and 𝑒 𝑖𝑛𝑥 is

2𝜋 2𝜋
(𝑒 𝑖𝑚𝑥 , 𝑒 𝑖𝑛𝑥 ) = ∫ 𝑒 𝑖𝑚𝑥 ̅̅̅̅̅
𝑒 𝑖𝑛𝑥 𝑑𝑥 = ∫ 𝑒 𝑖𝑚𝑥 𝑒 −𝑖𝑛𝑥 𝑑𝑥
0 0

2𝜋 2𝜋
𝑖(𝑚−𝑛)𝑥
𝑒 𝑖(𝑚−𝑛)𝑥 𝑒 𝑖2(𝑚−𝑛)𝜋 − 1
=∫ 𝑒 𝑑𝑥 = [ ] =
0 𝑖(𝑚 − 𝑛) 0 𝑖(𝑚 − 𝑛)
 6 Chapter 1 Fourier Analysis

𝑒 𝑖2(𝑚−𝑛)𝜋 − 1
= =
𝑖(𝑚 − 𝑛)
1
= (cos 2(𝑚 − 𝑛)𝜋 + 𝑖 sin 2(𝑚 − 𝑛)𝜋 − 1)
𝑖(𝑚 − 𝑛)
1
= ((−1)2(𝑚−𝑛) − 1) = 0
𝑖(𝑚 − 𝑛)
Therefore, the functions 𝑒 𝑖𝑚𝑥 and 𝑒 𝑖𝑛𝑥 are orthogonal over [0, 2𝜋]
for all integers 𝑚, 𝑛 such that 𝑚 ≠ 𝑛.

Definition 1.6 (Orthogonal and orthonormal systems). If {φ𝑛 }𝑘𝑛=1 is a

set of functions such that these functions are mutually orthogonal , i.e.,
(φ𝑚 , φ𝑛 ) = 0, 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑚 ≠ 𝑛 (5)
Then {𝜑𝑛 }𝑘𝑛=1 is called an orthogonal system. Moreover, if
‖φ𝑛 ‖ = 1, 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛 (6)
Then the system {φ𝑛 }𝑘𝑛=1 is called an orthonormal system.

That is to say, the system {φ𝑛 }𝑘𝑛=1 is called an orthonormal system if:
 The system is orthogonal (its functions satisfy (5)),
 Each function of this system is normalized (satisfies (6)).
However, if a function 𝑓(𝑥) is not normalized over [𝑎, 𝑏], it is easy to
show that 𝑓(𝑥)⁄‖𝑓(𝑥)‖ is a normalized function over [𝑎, 𝑏].

Theorem 2.2 The set of functions {1, cos 𝑛𝜔𝑥 , sin 𝑚𝜔𝑥} forms an
𝜋
orthogonal set over the interval [−𝐿, 𝐿]; where 𝜔 = 𝐿 , that is we have,

for all 𝑚, 𝑛 = 0,1,2, ⋯

(sin 𝑚𝜔𝑥 , cos 𝑛𝜔𝑥) = 0,
0, 𝑚≠𝑛
(cos 𝑛𝜔𝑥 , cos 𝑚𝜔𝑥) = { ,
𝐿, 𝑚=𝑛
0, 𝑚≠𝑛
(sin 𝑛𝜔𝑥 , sin 𝑚𝜔𝑥) = {
𝐿, 𝑚=𝑛
 Chapter 1 Fourier Analysis 7

Proof. ■ Case 1 For 𝑚 ≠ 𝑛

𝐿
(sin 𝑚𝜔𝑥 , cos 𝑛𝜔𝑥) = ∫ sin 𝑚𝜔𝑥 cos 𝑛𝜔𝑥 𝑑𝑥
−𝐿
1 𝐿
= ∫ [sin(𝑚 − 𝑛)𝜔𝑥 + sin(𝑚 + 𝑛)𝜔𝑥]𝑑𝑥
2 −𝐿
𝐿
1 − cos(𝑚 − 𝑛)𝜔𝑥 − cos(𝑚 + 𝑛)𝜔𝑥
= [ + ]
2 (𝑚 − 𝑛)𝜔 (𝑚 + 𝑛)𝜔
−𝐿

1 cos(𝑚 − 𝑛)𝜋 cos(𝑚 − 𝑛)(−𝜋)

= ( − )
2 (𝑚 − 𝑛)𝜔 (𝑚 − 𝑛)𝜔
1 cos(𝑚 + 𝑛)𝜋 cos(𝑚 + 𝑛)(−𝜋)
− ( − )=0
2 (𝑚 + 𝑛)𝜔 (𝑚 + 𝑛)𝜔
■ Case 2. For 𝑚 = 𝑛
𝐿
(sin 𝑛𝜔𝑥 , cos 𝑛𝜔𝑥) = ∫ sin 𝑛𝜔𝑥 cos 𝑛𝜔𝑥 𝑑𝑥
−𝐿
𝐿
= ∫ sin 𝑛𝜔𝑥 cos 𝑛𝜔𝑥 𝑑𝑥
−𝐿
1 1
= [sin2 𝑛𝜔𝑥]𝐿−𝐿 = (sin2 𝑛𝜋 − sin2 (−𝑛𝜋)) = 0
2𝑛𝜔 2𝑛𝜔
Similarly, it can be shown that, for all 𝑚, 𝑛 = 0,1,2, ⋯
0, 𝑚≠𝑛
(cos 𝑚𝜔𝑥 , cos 𝑛𝜔𝑥) = {
𝐿, 𝑚=𝑛

Remark As a special case if 𝐿 = 𝜋  𝜔 = 1, then the set

{1, cos 𝑛𝑥 , sin 𝑚𝑥} is orthogonal system over the
interval [−𝜋, 𝜋], that is,

For all 𝑚, 𝑛 = 0,1,2, ⋯, we have

(sin 𝑚𝑥 , cos 𝑛𝑥) = 0,
0, 𝑚≠𝑛
(cos 𝑛𝜔𝑥 , cos 𝑚𝜔𝑥) = { ,
𝜋, 𝑚=𝑛
0, 𝑚≠𝑛
(sin 𝑛𝜔𝑥 , sin 𝑚𝜔𝑥) = {
𝜋, 𝑚=𝑛
 8 Chapter 1 Fourier Analysis

Example 5 Normalize the functions {1, cos 𝑛𝑥 , sin 𝑛𝑥} over the
interval [−𝜋, 𝜋].

Solution
𝜋
‖1‖2 = ∫ 12 𝑑𝑥 = 2𝜋
−𝜋
𝜋
1 𝜋
‖cos 𝑛𝑥‖2 = ∫ cos 2 𝑛𝑥 𝑑𝑥 = ∫ (1 + cos 2𝑛𝑥)𝑑𝑥
−𝜋 2 −𝜋
1 sin 2𝑛𝑥 𝜋
= [𝑥 + ] =𝜋
2 2𝑛 −𝜋
𝜋
1 𝜋
‖sin 𝑛𝑥‖2 = ∫ sin2 𝑛𝑥 𝑑𝑥 = ∫ (1 − cos 2𝑛𝑥)𝑑𝑥
−𝜋 2 −𝜋
1 sin 2𝑛𝑥 𝜋
= [𝑥 − ] =𝜋
2 2𝑛 −𝜋
Therefore, the functions {1, cos 𝑛𝑥 , sin 𝑛𝑥} are normalized by
multiplying them by
1 1 1
, ,
√2𝜋 √𝜋 √𝜋
respectively. Therefore, the system
1 cos 𝑛𝑥 sin 𝑛𝑥
{ , , }, 𝑛 = 1,2,3, ⋯,
√2𝜋 √𝜋 √𝜋
is an orthonormal system.

𝑬𝒙𝒆𝒓𝒄𝒊𝒔𝒆𝒔 𝟏. 𝟏

In problems 1-5, show that the given functions are orthogonal on the
indicated interval.
𝟏. 𝑓1 (𝑥) = 𝑥, 𝑓2 (𝑥) = 𝑥 2 , [−2,2]
𝟐. 𝑓1 (𝑥) = 𝑥 2 , 𝑓2 (𝑥) = 𝑥 3 + 𝑥 , [−1,1]
 Chapter 1 Fourier Analysis 9

𝟑. 𝑓1 (𝑥) = 𝑒 𝑥 , 𝑓2 (𝑥) = 𝑥𝑒 −𝑥 − 𝑒 −𝑥 , [0,2]

𝟒. 𝑓1 (𝑥) = 𝑥, 𝑓2 (𝑥) = cos 𝑥 , [− 𝜋⁄2 , 𝜋⁄2]
𝟓. 𝑓1 (𝑥) = cos 𝑥 , 𝑓2 (𝑥) = sin2 𝑥 , [0, 𝜋]
In problems 𝟓 − 𝟏𝟎 , normalize the following functions over the
indicated interval.
𝟔. {sin 𝑥 , sin 3𝑥 , sin 5𝑥 , ⋯ }, [0, 𝜋⁄2]
𝟕. {cos 𝑥 , cos 3𝑥 , cos 5𝑥 , ⋯ }, [0, 𝜋⁄2]
𝟖. {sin 𝑛𝑥}, 𝑛 = 1,2,3, ⋯, [0, 𝜋]
𝑛𝜋𝑥
𝟗. {sin } , 𝑛 = 1,2,3, ⋯, [0, 𝐿]
𝐿
𝑛𝜋𝑥 𝑛𝜋𝑥
𝟏𝟎. {1, sin , sin } , 𝑛 = 1,2,3, ⋯, [−𝐿, 𝐿]
𝐿 𝐿

1.1.2 Trigonometric Fourier series

Representation of functions over a certain interval by linear

combination of mutually orthogonal functions is called Fourier series
representation of a function. There exist, however, a large number of
sets of orthogonal functions such as trigonometric functions,
exponential functions, Bessel functions and Legendre polynomials.
This section is devoted to the representation of functions in terms of the
trigonometric sine and cosine functions. We begin with the following
theorem.

Theorem 1.3 Let 𝑓(𝑥) be a periodic function with period 2𝐿. Then the
Fourier series of 𝑓(𝑥) is given by
∞
𝑎0 𝜋
𝑓(𝑥) = + ∑(𝑎𝑛 cos 𝑛𝜔𝑥 + 𝑏𝑛 sin 𝑛𝜔𝑥) , 𝜔 = (1)
2 𝐿
𝑛=1

provided the coefficients

 10 Chapter 1 Fourier Analysis

1 𝐿
𝑎0 = ∫ 𝑓(𝑥)𝑑𝑥,
𝐿 −𝐿
1 𝐿
𝑎𝑛 = ∫ 𝑓(𝑥) cos 𝑛𝜔𝑥 𝑑𝑥 ,
𝐿 −𝐿
1 𝐿
𝑏𝑛 = ∫ 𝑓(𝑥) sin 𝑛𝜔𝑥 𝑑𝑥,
𝐿 −𝐿
all exist.

The coefficients 𝑎0 , 𝑎𝑛 , and 𝑏𝑛 are called Fourier coefficients.

Proof. We want to compute the coefficients 𝑎𝑛 , 𝑛 = 0,1,2, ⋯, and the
coefficients 𝑏𝑛 , 𝑛 = 0,1,2, ⋯, of (1). Consider the orthogonality
𝜋
properties of the functions cos 𝑛𝜔𝑥 and sin 𝑛𝜔𝑥 , 𝜔 = in the
𝐿

interval [−𝐿, 𝐿].

(𝒊) To find 𝑎𝑛 , multiply both sides of (1) by cos 𝑛𝜔𝑥 and then integrate
−𝐿 to 𝐿.
𝑎0
(𝑓(𝑥), cos 𝑚𝜔𝑥) = (1, cos 𝑚𝜔𝑥)
⏟
2 =0

(cos 𝑛𝜔𝑥 , cos 𝑚𝜔𝑥) + 𝑏𝑛 ⏟

+ ∑ (𝑎𝑛 ⏟ (sin 𝑛𝜔𝑥 , cos 𝑚𝜔𝑥))
𝑛=1 0 𝑚≠𝑛
= {𝜋 =0
𝑚=𝑛

From theorem 2.2 we have

𝑎0
(𝑓(𝑥), cos 𝑚𝜔𝑥) = ∙ 0 + 𝑎𝑚 ∙ 𝐿 + 0 = 𝑎 𝑚 ∙ 𝐿
2
Solving for 𝑎𝑚
1
𝑎𝑚 = (𝑓(𝑥), cos 𝑚𝜔𝑥)
𝐿
Since 𝑚 is a dummy variable, we can write
1 1 𝐿
𝑎𝑛 = (𝑓(𝑥), cos 𝑚𝜔𝑥) = ∫ 𝑓(𝑥) cos 𝑛𝜔𝑥 𝑑𝑥 , 𝑛 = 1,2, ⋯
𝐿 𝐿 −𝐿
 Chapter 1 Fourier Analysis 11

(𝒊𝒊) To find 𝑎0 we multiply (1) by 1 (cos 0𝑥) and integrate both sides
from −𝐿 to 𝐿.
∞
𝑎0
(𝑓(𝑥), 1) = ⏟ (1, 1) + ∑ (𝑎𝑛 ⏟
(cos 𝑛𝜔𝑥 , 1) + 𝑏𝑛 ⏟
(sin 𝑛𝜔𝑥 , 1))
2 =2𝐿
𝑛=1 = 0 =0
𝑎0
(𝑓(𝑥), 1) = ∙ 2𝐿 + 𝑎𝑚 ∙ 0 + 𝑏𝑛 ∙ 0 = 𝑎0 ∙ 𝐿
2
Solving (3) we find
1 1 𝐿
𝑎0 = (𝑓(𝑥), 1) = ∫ 𝑓(𝑥)𝑑𝑥
𝐿 𝐿 −𝐿
(𝑖𝑖𝑖) Finally, to compute 𝑏𝑛 for 𝑛 = 1,2, ⋯, multiply both sides of (1)
by sin 𝑚𝜔𝑥 and then integrate −𝐿 to 𝐿. Using theorem 2.2 we have
𝑎0
(𝑓(𝑥), sin 𝑚𝜔𝑥) = (1, ⏟ sin 𝑚𝜔𝑥)
2 =0

+ ∑ (𝑎𝑛 (cos
⏟ 𝑛𝜔𝑥 , sin 𝑚𝜔𝑥) + 𝑏𝑛 (sin
⏟ 𝑛𝜔𝑥 , sin 𝑚𝜔𝑥))
𝑛=1 = 0 0 𝑚≠𝑛
= {𝜋 𝑚=𝑛

Thus, replacing 𝑚 by 𝑛, we find

1 1 𝐿
𝑏𝑛 = (𝑓(𝑥), sin 𝑛𝜔𝑥) = ∫ 𝑓(𝑥) sin 𝑛𝜔𝑥 𝑑𝑥 , 𝑛 = 1,2, ⋯
𝐿 𝐿 −𝐿
However, the questions arise now is does the series in the right hand
side of (1) converge? and if it is convergent does it converge to 𝑓(𝑥) as
we had written already in equation (1)? Briefly, these questions can be
stated in the form: Under what conditions can the function 𝑓(𝑥) can be
represented by a series of the form (1). These questions will be
answered in the following subsection.

Example 1 Find the Fourier series of the function 𝑓(𝑥) = 𝑥 on the

interval [0, 2𝜋], with period 2𝜋.
 12 Chapter 1 Fourier Analysis

Solution Since,
2𝐿 = 2𝜋  𝐿=𝜋  ω=1
Then from (1), we can write
∞
𝑎0
𝑓(𝑥) = + ∑(𝑎𝑛 cos 𝑛𝑥 + 𝑏𝑛 sin 𝑛𝑥),
2
𝑛=1

where
2𝜋
1 2𝜋 1 2𝜋 1 𝑥2
𝑎0 = ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑥𝑑𝑥 = [ ] = 2𝜋,
𝜋 0 𝜋 0 𝜋 2 0

𝑓(𝑥)

−4𝜋 −2𝜋 0 2𝜋 4𝜋 x

Fig. 2.4
and
1 2𝜋
𝑎𝑛 = ∫ 𝑓(𝑥) cos 𝑛𝑥 𝑑𝑥
𝜋 0
𝒙 + 𝐜𝐨𝐬 𝒏𝒙
2𝜋
1
= ∫ 𝑥 cos 𝑛𝑥 𝑑𝑥 𝐬𝐢𝐧 𝒏𝒙
𝜋 0 1
− 𝒏
1 sin 𝑛𝑥 cos 𝑛𝑥 2𝜋 𝐜𝐨𝐬 𝒏𝒙
= [(𝑥) ( ) − (1) (− )] 0 −
𝜋 𝑛 𝑛2 0 𝒏𝟐

1 cos 2𝑛𝜋 − 1 1 (−1)2𝑛 − 1

= (0 + ( )) = ( ) = 0,
𝜋 𝑛2 𝜋 𝑛2
and the coefficients 𝒙 + 𝐬𝐢𝐧 𝒏𝒙
1 2𝜋 𝐜𝐨𝐬 𝒏𝒙
𝑏𝑛 = ∫ 𝑓(𝑥) sin 𝑛𝑥 𝑑𝑥 1 − −
𝜋 0 𝒏
1 2𝜋 𝐬𝐢𝐧 𝒏𝒙
= ∫ 𝑥 sin 𝑛𝑥 𝑑𝑥 0 −
𝜋 0 𝒏𝟐
 Chapter 1 Fourier Analysis 13

1 cos 𝑛𝑥 sin 𝑛𝑥 2𝜋
= [(𝑥) (− ) − (1) (− )]
𝜋 𝑛 𝑛2 0
1 cos 2𝑛𝜋
= [(2𝜋) (− )]
𝜋 𝑛
2 2
= − [(−1)2𝑛 ] = −
𝑛 𝑛
The trigonometric Fourier series is given by
∞
𝑎0
𝑓(𝑥) = + ∑(𝑎𝑛 cos 𝑛𝑥 + 𝑏𝑛 sin 𝑛𝑥)
2
𝑛=1
∞
2𝜋 2
= + ∑ (0 ∙ cos 𝑛𝑥 − sin 𝑛𝑥)
2 𝑛
𝑛=1
∞
sin 𝑛𝑥 sin 𝑥 sin 2𝑥 sin 3𝑥
= 𝜋−2∑ = 𝜋 − 2( + + +⋯)
𝑛 1 2 3
𝑛=1

Example 2 Find the Fourier series of the function 𝑓(𝑥) = 𝑒 𝑥 in the

interval − 𝜋 < 𝑥 < 𝜋, with period 2𝜋.

Solution Since,
2𝐿 = 2𝜋  𝐿=𝜋  ω = 1,
we can write
∞
𝑎0
𝑓(𝑥) = + ∑(𝑎𝑛 cos 𝑛𝑥 + 𝑏𝑛 sin 𝑛𝑥),
2
𝑛=1

where
1 𝜋 1 𝜋 𝑥 1
𝑎0 = ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑒 𝑑𝑥 = [𝑒 𝑥 ]𝜋−𝜋
𝜋 −𝜋 𝜋 −𝜋 𝜋
𝑒 𝜋 − 𝑒 −𝜋 2
= = sinh 𝜋,
𝜋 𝜋
 14 Chapter 1 Fourier Analysis

1 𝜋 1 𝜋 𝑥
𝑎𝑛 = ∫ 𝑓(𝑥) cos 𝑛𝑥 𝑑𝑥 = ∫ 𝑒 cos 𝑛𝑥 𝑑𝑥
𝜋 −𝜋 𝜋 −𝜋
Using the integral relation
𝑒 𝑎𝑥
∫ 𝑒 𝑎𝑥 cos 𝑏𝑥 𝑑𝑥 = (𝑎 cos 𝑏𝑥 + 𝑏 sin 𝑏𝑥) + 𝑐,
𝑎2 + 𝑏 2
we get
𝜋
1 𝑒𝑥
𝑎𝑛 = [ (cos 𝑛𝑥 + 𝑛 sin 𝑛𝑥)]
𝜋 1 + 𝑛2 −𝜋
1
= [𝑒 𝜋 (cos 𝑛𝜋) − 𝑒 −𝜋 (cos(−𝑛𝜋))]
𝜋(1 + 𝑛2 )
(−1)𝑛 𝜋 −𝜋 ] 𝑛
2
= [𝑒 − 𝑒 = (−1) sinh 𝜋
𝜋(1 + 𝑛2 ) 𝜋(1 + 𝑛2 )
Similarly we compute the coefficients 𝑏𝑛 :
1 𝜋 1 𝜋
𝑏𝑛 = ∫ 𝑓(𝑥) sin 𝑛𝑥 𝑑𝑥 = ∫ 𝑒 𝑥 sin 𝑛𝑥 𝑑𝑥
𝜋 −𝜋 𝜋 −𝜋
𝜋
1 𝑒𝑥
= [ (sin 𝑛𝑥 − 𝑛 cos 𝑛𝑥)]
𝜋 1+𝑛 −𝜋
−𝑛 𝜋 −𝜋
= [𝑒 cos 𝑛𝜋 − 𝑒 (cos(−𝑛𝜋))]
𝜋(1 + 𝑛2 )
−𝑛(−1)𝑛 𝜋 −2𝑛
= 2
[𝑒 − 𝑒 −𝜋 ] = (−1)𝑛 sinh 𝜋
𝜋(1 + 𝑛 ) 𝜋(1 + 𝑛2 )
𝑒 𝑎𝑥
[𝐻𝑖𝑛𝑡: ∫ 𝑒 𝑎𝑥 sin 𝑏𝑥 𝑑𝑥 = (𝑎 sin 𝑏𝑥 − 𝑏 cos 𝑏𝑥) + 𝑐]
𝑎2 + 𝑏2
Hence in − 𝜋 < 𝑥 < 𝜋
2 ∞
sinh 𝜋 2
𝑓(𝑥) = 𝜋 + ∑ [(−1)𝑛 sinh 𝜋 cos 𝑛𝑥
2 𝜋(1 + 𝑛2 )
𝑛=1
−2𝑛
+(−1)𝑛 sinh 𝜋 sin 𝑛𝑥],
𝜋(1 + 𝑛2 )
or
∞
sinh 𝜋 2 sinh 𝜋 cos 𝑛𝑥 𝑛 sin 𝑛𝑥
𝑓(𝑥) = + ∑(−1)𝑛 [ − ]
𝜋 𝜋 (1 + 𝑛 ) (1 + 𝑛2 )
2
𝑛=1
 Chapter 1 Fourier Analysis 15

1.1.3. Convergence of Fourier series

We are now in a position to state a classic theorem, which tells us, for
a certain set of functions, that our method of constructing a Fourier
Series is valid.

Theorem 2.4 Let 𝑓(𝑥) be a periodic function with period 2𝐿, and let
𝑓(𝑥) and 𝑓 ′ (𝑥) be piecewise continuous in [−𝐿, 𝐿]. Then the Fourier
series of 𝑓(𝑥)
∞
𝑎0 𝜋
𝑓(𝑥) = + ∑(𝑎𝑛 cos 𝑛𝜔𝑥 + 𝑏𝑛 sin 𝑛𝜔𝑥) , 𝜔 =
2 𝐿
𝑛=1
converges to
1
[𝑓(𝑥 + ) + 𝑓(𝑥 − )]
2
for all 𝑥.

It is convenient from theorem 2.4 that for sufficiently large 𝑛 the

Fourier series of 𝑓(𝑥) converges to 𝑓(𝑥) at the points of continuity,
since at these points we have
𝑓(𝑥 + ) + 𝑓(𝑥 − ) = 𝑓(𝑥)
On the other hand when 𝑓(𝑥) has a jump discontinuity, the Fourier
series of 𝑓(𝑥) converges to a value halfway between 𝑓(𝑥 + ) and 𝑓(𝑥 − ).
The hypotheses of the above theorem are known as by the name
Dirichlet conditions. Hence, if 𝑓(𝑥) satisfies the Dirichlet conditions,
then
∞
𝑎0
+ ∑(𝑎𝑛 cos 𝑛𝜔𝑥 + 𝑏𝑛 sin 𝑛𝜔𝑥) =
2
𝑛=1

𝑓(𝑥) 𝑖𝑓 𝑥 𝑖𝑠 𝑎 𝑝𝑡 𝑜𝑓 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑡𝑦
= {𝑓(𝑥 + ) + 𝑓(𝑥 − )
𝑖𝑓 𝑥 𝑖𝑠 𝑎 𝑝𝑡 𝑜𝑓 𝑑𝑖𝑠𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑡𝑦
2
 16 Chapter 1 Fourier Analysis

For instance, in Example 1 of the previous section, we note that at

𝑥 = 0, the Fourier series converges to
1 1
[𝑓(0+ ) + 𝑓(0− )] = [0 + 2𝜋] = 𝜋
2 2

2

    

4  2 0 2 4 x

Fig. 1.5

Therefore, the graph of the function to which the series converges is

shown in Fig. 1.5.

𝑬𝒙𝒆𝒓𝒄𝒊𝒔𝒆𝒔 𝟏. 𝟐

In problems (𝟏) − (𝟑), answer true or false

1
𝟏. The function 𝑓(𝑥) = 𝑥 is piecewise continuous in the interval

− 𝜋 ≤ 𝑥 ≤ 𝜋.
𝟐. The Fourier series of the function 𝑓(𝑥) = |𝑥|, −𝜋 ≤ 𝑥 ≤ 𝜋,
𝑓(𝑥 + 2𝜋) = 𝑓(𝑥) converges to 𝑓(𝑥).
𝟑. The Fourier series of the function 𝑓(𝑥) = 𝑥 2 , − 𝜋 ≤ 𝑥 ≤ 𝜋,
𝑓(𝑥 + 2𝜋) = 𝑓(𝑥) does not involve any sine terms.
In problems (𝟒) − (𝟐𝟎), find the Fourier series representation of the
functions
𝟒. 𝑓(𝑥) = 𝑥 2 , 0 < 𝑥 < 1, 𝑓(𝑥 + 1) = 𝑓(𝑥)
𝟓. 𝑓(𝑥) = 𝑒 𝑥 , 0 < 𝑥 < 1, 𝑓(𝑥 + 1) = 𝑓(𝑥)
 Chapter 1 Fourier Analysis 17

𝑥2
𝟔. 𝑓(𝑥) = , 0 < 𝑥 < 2, 𝑓(𝑥 + 2) = 𝑓(𝑥)
4
1 − 𝜋⁄2 < 𝑥 < 𝜋⁄2
𝟕. 𝑓(𝑥) = { , 𝑓(𝑥 + 2𝜋) = 𝑓(𝑥)
−1 𝜋⁄2 < 𝑥 < 3𝜋⁄2
0 −2<𝑥 <0
𝟖. 𝑓(𝑥) = { , 𝑓(𝑥 + 4) = 𝑓(𝑥)
2 0<𝑥<2
𝑥 0<𝑥<1
𝟗. 𝑓(𝑥) = { , 𝑓(𝑥 + 2) = 𝑓(𝑥)
1−𝑥 1<𝑥 <2
−1 𝑖𝑓 −𝜋 < 𝑥 < − 𝜋⁄2
𝟏𝟎. 𝑓(𝑥) = { 0 𝑖𝑓 − 𝜋⁄2 < 𝑥 < 𝜋⁄2, 𝑓(𝑥 + 2𝜋) = 𝑓(𝑥)
1 𝑖𝑓 𝜋⁄2 < 𝑥 < 𝜋
1 1
+𝑥 − <𝑥<0
𝟏𝟏. 𝑓(𝑥) = { 2 2 , 𝑓(𝑥 + 1) = 𝑓(𝑥)
1 1
+𝑥 0<𝑥<
2 2
𝑥2 − 𝜋⁄2 < 𝑥 < 𝜋⁄2
𝟏𝟐. 𝑓(𝑥) = { 𝜋 2 , 𝑓(𝑥 + 2𝜋) = 𝑓(𝑥)
𝜋⁄2 < 𝑥 < 3𝜋⁄2
4
1 0<𝑥<1
𝟏𝟑. 𝑓(𝑥) = { , 𝑓(𝑥 + 4) = 𝑓(𝑥)
0 1 < |𝑥| < 2
𝟏𝟒. 𝑓(𝑥) = 𝑥 + |𝑥| , − 𝜋 < 𝑥 < 𝜋, 𝑓(𝑥 + 2𝜋) = 𝑓(𝑥)
𝟏𝟓. 𝑓(𝑥) = 3𝑥 2 , − 1 < 𝑥 < 1, 𝑓(𝑥 + 2) = 𝑓(𝑥)
𝟏𝟔. 𝑓(𝑥) = 𝜋 sin 𝜋𝑥 , 0 < 𝑥 < 1, 𝑓(𝑥 + 1) = 𝑓(𝑥)
𝟏𝟕. 𝑓(𝑥) = 𝑥 4 , − 𝜋 < 𝑥 < 𝜋, 𝑓(𝑥 + 2𝜋) = 𝑓(𝑥)
𝟏𝟖. 𝑓(𝑥) = 𝑥 2 , − 𝜋⁄2 < 𝑥 < 𝜋⁄2 , 𝑓(𝑥 + 𝜋) = 𝑓(𝑥)
𝟏𝟗. 𝑓(𝑥) = sin 𝑥 , 0 < 𝑥 < 2𝜋, 𝑓(𝑥 + 2𝜋) = 𝑓(𝑥)
𝟐𝟎. 𝑓(𝑥) = cosh 𝑥 , − 𝜋 < 𝑥 < 𝜋, 𝑓(𝑥 + 2𝜋) = 𝑓(𝑥)

1.1.4. Fourier series for even and odd functions (Halfinterval expansion)

First, we begin this section by the following definitions of even and odd
functions.
 18 Chapter 1 Fourier Analysis

Definition 1.7 (Even and Odd functions). A function 𝑓(𝑥) on the interval
(−𝑎, 𝑎) is said to be even if
𝑓(−𝑥) = 𝑓(𝑥) (1)
This means that this function is symmetric about the 𝑦𝑎𝑥𝑖𝑠 in the 𝑥𝑦 −
𝑝𝑙𝑎𝑛𝑒.
A function 𝑓(𝑥) is said to be odd on the interval (−𝑎, 𝑎) if
𝑓(−𝑥) = − 𝑓(𝑥) (2)
This function is symmetric with respect to the origin in the 𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒.

For example, the functions cos ω𝑥 , 1, 𝑥 2 , 𝑥 4 , |𝑥| are even and

sin ω𝑥 , 𝑥, 𝑥 3 , 𝑥 5 are odd on the interval (−𝑎, 𝑎).
With respect to the operations of addition and multiplication, even and
odd functions have the following properties:
𝒆𝒗𝒆𝒏 + 𝒆𝒗𝒆𝒏 ≡ 𝒆𝒗𝒆𝒏 𝒐𝒅𝒅 + 𝒐𝒅𝒅 ≡ 𝒐𝒅𝒅
𝒆𝒗𝒆𝒏 × 𝒆𝒗𝒆𝒏 ≡ 𝒆𝒗𝒆𝒏 𝒐𝒅𝒅 × 𝒐𝒅𝒅 ≡ 𝒆𝒗𝒆𝒏
𝒆𝒗𝒆𝒏 × 𝒐𝒅𝒅 ≡ 𝒐𝒅𝒅

All of the above properties are obtained if we denote for the even
function by the 𝑝𝑙𝑢𝑠 𝑠𝑖𝑔𝑛 (+), and denote by 𝑚𝑖𝑛𝑢𝑠 𝑠𝑖𝑔𝑛 (−) for the
odd functions. For instance,
𝒐𝒅𝒅 × 𝒐𝒅𝒅 ≡ (−) × (−) = (+) ≡ 𝒆𝒗𝒆𝒏
On computing the Fourier coefficients, we noticed that the limits of the
integrals extend from − 𝐿 to 𝐿. But we know from the properties of
definite integrals that
𝐿 𝐿
∫ (𝑒𝑣𝑒𝑛)𝑑𝑥 = 2 ∫ (𝑒𝑣𝑒𝑛)𝑑𝑥 (3)
−𝐿 0
𝐿
∫ (𝑜𝑑𝑑)𝑑𝑥 = 0 (4)
−𝐿
 Chapter 1 Fourier Analysis 19

Moreover, very often a function 𝑓(𝑥) is defined on the ℎ𝑎𝑙𝑓 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙

(0, 𝐿) rather than the 𝑓𝑢𝑙𝑙 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 (−𝐿, 𝐿), and we desire a Fourier
representation of 𝑓(𝑥) over [0, 𝐿]. All we have to do is to extend the
function 𝑓(𝑥) into the interval [− 𝐿, 0] in such a way that we can write
the Fourier series over [−𝐿, 𝐿] and then periodically on the rest of 𝑅.
The present section is devoted to study the following cases.
 On one hand, the use of property (3) of integrals to give some
simplification in computing the Fourier coefficients.
 On the other hand, the achievement of desired extension of
functions in such a way that they satisfy some given boundary
conditions.

However, we have three options to extend functions as shown in Fig.

2.6

𝒇(𝒙) 𝒇(𝒙) 𝒇(𝒙)

L o L x L o L x L o L x

Even function Odd function Not even Not odd

function

Fig. 1.6

1.1.5 Fourier cosine and sine series

If 𝑓(𝑥) is an 𝑒𝑣𝑒𝑛 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 over [−𝐿, 𝐿] or we have to extend it as an

even function, we use the following 𝐹𝑜𝑢𝑟𝑖𝑒𝑟 𝑐𝑜𝑠𝑖𝑛𝑒 𝑠𝑒𝑟𝑖𝑒𝑠.
 20 Chapter 1 Fourier Analysis

∞
𝑎0 𝜋
𝑓(𝑥) = + ∑ 𝑎𝑛 cos 𝑛𝜔𝑥 , 𝜔 = (1)
2 𝐿
𝑛=1

𝑏𝑛 = 0 (2)
2 𝐿
𝑎0 = ∫ 𝑓(𝑥)𝑑𝑥 (3)
𝐿 −𝐿
2 𝐿
𝑎𝑛 = ∫ 𝑓(𝑥) cos 𝑛𝜔𝑥 𝑑𝑥 (4)
𝐿 −𝐿

Similarly, if 𝑓(𝑥) is an 𝑜𝑑𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 over [−𝐿, 𝐿] or we have to

extend it as an odd function, we use the following
𝐹𝑜𝑢𝑟𝑖𝑒𝑟 𝑠𝑖𝑛𝑒 𝑠𝑒𝑟𝑖𝑒𝑠.

∞
𝜋
𝑓(𝑥) = ∑ 𝑏𝑛 sin 𝑛𝜔𝑥 , 𝜔 = (5)
𝐿
𝑛=1

𝑎0 = 𝑎𝑛 = 0 (6)
2 𝐿
𝑏𝑛 = ∫ 𝑓(𝑥) sin 𝑛𝜔𝑥 𝑑𝑥 (7)
𝐿 −𝐿

Remark To obtain Fourier series of functions in correct and an

easy way, we have to take the following comments into
considerations.
 𝑭𝒊𝒓𝒔𝒕, we must note the extension of the period of the function,
whether extended from −𝐿 𝑡𝑜 𝐿, or from 0 𝑡𝑜 2𝐿.
 Second, we study the symmetry of functions, i.e., are these
functions symmetric about the y-axis or with respect to the
origin, or there is no symmetry at all? This will be done by the
aid of 𝒔𝒌𝒆𝒕𝒄𝒉𝒊𝒏𝒈 𝒕𝒉𝒆 𝒈𝒓𝒂𝒑𝒉 of 𝑓(𝑥).
 Chapter 1 Fourier Analysis 21

 Third, sometimes the required assertion of the problem

determines the kind of symmetry of the function to be dealt
with. If the required assertion is given in the form:
𝐸𝑥𝑝𝑎𝑛𝑑 𝑓(𝑥) 𝑖𝑛 𝑐𝑜𝑠𝑖𝑛𝑒 𝐹𝑜𝑢𝑟𝑖𝑒𝑟 𝑠𝑒𝑟𝑖𝑒𝑠,

or

𝑆ℎ𝑜𝑤 𝑡ℎ𝑎𝑡 𝑓(𝑥) ≡ 𝐶𝑜𝑠𝑖𝑛𝑒 𝑡𝑒𝑟𝑚𝑠 𝑜𝑛𝑙𝑦 𝑜𝑛 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡 ℎ𝑎𝑛𝑑 𝑠𝑖𝑑𝑒.

This means that we will deal with the function 𝑓(𝑥) as an even function
and we shall use (1) − (4). In this case the given interval of definition
of 𝑓(𝑥) will be considered as half-period of 𝑓(𝑥).
Similarly, if the required assertion is given in the form:

𝐸𝑥𝑝𝑎𝑛𝑑 𝑓(𝑥) 𝑖𝑛 𝑠𝑖𝑛𝑒 𝐹𝑜𝑢𝑟𝑖𝑒𝑟 𝑠𝑒𝑟𝑖𝑒𝑠,

or

𝑆ℎ𝑜𝑤 𝑡ℎ𝑎𝑡 𝑓(𝑥) ≡ 𝑠𝑖𝑛𝑒 𝑡𝑒𝑟𝑚𝑠 𝑜𝑛𝑙𝑦 𝑜𝑛 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡 ℎ𝑎𝑛𝑑 𝑠𝑖𝑑𝑒.

This means that we will deal with the function 𝑓(𝑥) as an odd function
and we shall use (5) − (7). In this case the given interval of definition
of 𝑓(𝑥) will be considered as half-period of 𝑓(𝑥).
The above remark will be illustrated by the following examples.

Example 1 Find the Fourier series of the function 𝑓(𝑥) = 𝑥 2 on the

interval [0,2𝜋], 𝑓(𝑥) = 𝑓(𝑥 + 2𝜋). Hence, prove that
𝜋2 1 1 1 1
= 2 + 2 + 2 + ⋯+ 2 + ⋯
6 1 2 3 𝑛

Solution From Fig. 1.7(a), the function 𝑓(𝑥) is neither odd nor even
and we use the general laws of Fourier series with ω = 1:
22 Chapter 1 Fourier Analysis

∞
𝑎0
𝑓(𝑥) = + ∑(𝑎𝑛 cos 𝑛𝑥 + 𝑏𝑛 sin 𝑛𝑥),
2
𝑛=1

where
y

 4 2 0 2 4 6 x

𝑦 = 𝑥 2, 0 < 𝑥 < 2𝜋
𝑵𝒐𝒕 𝒆𝒗𝒆𝒏 𝒏𝒐𝒕 𝒐𝒅𝒅 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏

Fig. 1.7(a)

1 2𝜋 𝒙𝟐 + 𝐜𝐨𝐬 𝒏𝒙
𝑎0 = ∫ 𝑓(𝑥)𝑑𝑥
𝜋 0
𝐬𝐢𝐧 𝒏𝒙
2x −
2𝜋 3 2𝜋 2 𝒏
1 1 𝑥 8𝜋
= ∫ 𝑥 2 𝑑𝑥 = [ ] = ,
𝜋 0 𝜋 3 0 3 𝐜𝐨𝐬 𝒏𝒙
2 + −
𝒏𝟐
1 2𝜋
𝑎𝑛 = ∫ 𝑓(𝑥) cos 𝑛𝑥 𝑑𝑥 𝐬𝐢𝐧 𝒏𝒙
𝜋 0 0 −
𝑛3
1 2𝜋 2
= ∫ 𝑥 cos 𝑛𝑥 𝑑𝑥 =
𝜋 0
1 2)
sin 𝑛𝑥 cos 𝑛𝑥 sin 𝑛𝑥 2𝜋
= [(𝑥 ( ) − (2𝑥) (− ) + (2) (− )]
𝜋 𝑛 𝑛2 𝑛3 0
1 cos 2𝑛𝜋 4
= (0 + (4𝜋) ( 2
) + 0) = 2 ,
𝜋 𝑛 𝑛
 Chapter 1 Fourier Analysis 23

𝒙𝟐 + 𝐬𝐢𝐧 𝒏𝒙
1 2𝜋
𝑏𝑛 = ∫ 𝑓(𝑥) sin 𝑛𝑥 𝑑𝑥 𝐜𝐨𝐬 𝒏𝒙
𝜋 0 2x − −
𝒏

1 2𝜋 2 2 + −
𝐬𝐢𝐧 𝒏𝒙
= ∫ 𝑥 sin 𝑛𝑥 𝑑𝑥 𝒏𝟐
𝜋 0
𝐜𝐨𝐬 𝒏𝒙
0
1 cos 𝑛𝑥 𝑛3
= [(𝑥 2 ) (− )
𝜋 𝑛
sin 𝑛𝑥 cosn 𝑛𝑥 2𝜋
−(2𝑥) (− ) + (2) (− )]
𝑛2 𝑛3 0

1 cos 2𝑛𝜋 cos 2𝑛𝜋 − 1

= ((4𝜋 2 ) (− ) − 0 + (2) (− ))
𝜋 𝑛 𝑛3

4𝜋
=−
𝑛

Hence,
∞
2
4𝜋 2 4 4𝜋
𝑥 = + ∑ ( 2 cos 𝑛𝑥 − sin 𝑛𝑥)
3 𝑛 𝑛
𝑛=1

4𝜋 2 cos 𝑥 cos 2𝑥 cos 3𝑥

= + 4( 2 + + +⋯)
3 1 22 32
sin 𝑥 sin 2𝑥 sin 3𝑥
− 4𝜋 ( + + +⋯)
1 2 3
This expansion is valid for all 𝑥. To get the required assertion put 𝑥 =
0 or 𝑥 = 2𝜋 in both sides of this series. At this discontinuous point we
have
1
𝑥 = 𝑓(2𝜋) = (4𝜋 2 + 0) = 2𝜋 2 ,
2
hence,
4𝜋 2 1 1 1
2𝜋 2 = + 4 ( 2 + 2 + 2 + ⋯ ),
3 1 2 3
 24 Chapter 1 Fourier Analysis

or
2𝜋 2 1 1 1
= 4 ( 2 + 2 + 2 + ⋯ ),
3 1 2 3
from which we obtain
𝜋2 1 1 1
= ( 2 + 2 + 2 +⋯)
6 1 2 3

Example 2 Find the Fourier series of the function 𝑓(𝑥) = 𝑥 2 on the

interval [−𝜋, 𝜋], 𝑓(𝑥) = 𝑓(𝑥 + 2𝜋).

Solution From Fig. 1.7(b) the function 𝑓(𝑥) = 𝑥 2 on the interval

[−𝜋, 𝜋], with period 2𝜋 is an 𝑒𝑣𝑒𝑛 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 and (1) − (4) will be
𝜋
used with ω = = 1. Thus
𝐿
∞
𝑎0
𝑓(𝑥) = + ∑ 𝑎𝑛 cos 𝑛𝑥 ,
2
𝑛=1

where
𝜋
2 𝜋 2 𝜋 2 𝑥3 2𝜋 2
𝑎0 = ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑥 2 𝑑𝑥 = [ ] = ,
𝜋 0 𝜋 0 𝜋 3 0 3
2 𝜋 2 𝜋 2
𝑎𝑛 = ∫ 𝑓(𝑥) cos 𝑛𝑥 𝑑𝑥 = ∫ 𝑥 cos 𝑛𝑥 𝑑𝑥
𝜋 0 𝜋 0
Integrating by parts we get

2 sin 𝑛𝑥 cos 𝑛𝑥 sin 𝑛𝑥 𝜋

𝑎𝑛 = [(𝑥 2 ) ( ) − (2𝑥) (− ) + (2) (− )]
𝜋 𝑛 𝑛2 𝑛3 0

2 cos 𝑛𝜋 (−1)𝑛
= [0 + (2𝜋) ( 2 ) + 0] = 4
𝜋 𝑛 𝑛2
Substituting with 𝑎0 and 𝑎𝑛 we get
 Chapter 1 Fourier Analysis 25

∞
2
𝜋2 (−1)𝑛
𝑥 = + ∑ (4 ) cos 𝑛𝑥
3 𝑛2
𝑛=1
∞
𝜋2 cos 𝑛𝑥
= + 4 ∑(−1)𝑛
3 𝑛2
𝑛=1

𝜋2 cos 𝑥 cos 2𝑥 cos 3𝑥

= − 4( 2 − + −⋯)
3 1 22 32

  x

𝑦 = 𝑥 2 , −𝜋 < 𝑥 < 𝜋
𝒆𝒗𝒆𝒏 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏

Fig. 1.7 (b)

Example 3 Expand in 𝑠𝑖𝑛𝑒 𝐹𝑜𝑢𝑟𝑖𝑒𝑟 𝑠𝑒𝑟𝑖𝑒𝑠 the function

𝑓(𝑥) = 𝑥 2 , 𝑥 ∈ [0, 𝜋], 𝑓(𝑥) = 𝑓(𝑥 + 2𝜋).

Solution In this case, we extend 𝑓(𝑥) so that the new function defined
over [−𝜋, 𝜋] is 𝑜𝑑𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 as shown in Fig. 1.7(c). So that we have
𝑎0 = 𝑎𝑛 = 0,
and from (5)
∞

𝑓(𝑥) = ∑ 𝑏𝑛 sin 𝑛𝑥 ,
𝑛=1

with the coefficients 𝑏𝑛 calculated from (7) as follows.

 26 Chapter 1 Fourier Analysis

  x

𝑦 = 𝑥 2 , −𝜋 < 𝑥 < 𝜋
𝑶𝒅𝒅 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏

Fig. 1.7 (c)

2 𝜋 2 𝜋 2
𝑏𝑛 = ∫ 𝑓(𝑥) sin 𝑛𝑥 𝑑𝑥 = ∫ 𝑥 sin 𝑛𝑥 𝑑𝑥
𝜋 0 𝜋 0
2 cos 𝑛𝑥 sin 𝑛𝑥 cosn 𝑛𝑥 𝜋
= [(𝑥 2 ) (− ) −(2𝑥) (− ) + (2) ( )]
𝜋 𝑛 𝑛2 𝑛3 0

2 (−1)𝑛 (−1)𝑛 − 1
= ((𝜋 2 ) (− ) − 0 + (2) ( ))
𝜋 𝑛 𝑛3

(−1)𝑛+1 4 (−1)𝑛 − 1
= 2𝜋 ( )+ ( )
𝑛 𝜋 𝑛3
Therefore,
∞ ∞
2
(−1)𝑛+1 sin 𝑛𝑥 4 sin 𝑛𝑥
𝑥 = 2𝜋 ∑ ( ) + ∑[(−1)𝑛 − 1] ( 3 )
𝑛 𝜋 𝑛
𝑛=1 𝑛=1
∞ ∞
sin 𝑛𝑥 8 sin 𝑛𝑥
= 2𝜋 ∑(−1)𝑛+1 − ∑
𝑛 𝜋 𝑛3
𝑛=1 𝑛=𝑜𝑑𝑑
 Chapter 1 Fourier Analysis 27

∞ ∞
sin 𝑛𝑥 8 sin(2𝑛 − 1)𝑥
= 2𝜋 ∑(−1)𝑛+1 − ∑
𝑛 𝜋 (2𝑛 − 1)3
𝑛=1 𝑛=1
sin 𝑥 sin 2𝑥 sin 3𝑥
= 2𝜋 ( + + +⋯)
1 2 3
8 sin 𝑥 sin 3𝑥 sin 5𝑥
− ( 3 + + +⋯)
𝜋 1 33 53

1.1.6 Exponential (Complex) Fourier series

The set of functions

{𝑒 𝑖𝑛ω𝑥 }, 𝑛 = ±1, ±2, ⋯,

forms a complete orthogonal set over the interval [−𝐿, 𝐿], where
𝜋
𝜔 = 𝐿 . So it is possible to represent an arbitrary function by a linear

combination of these functions. In doing so we get

∞
𝜋
𝑓(𝑥) = ∑ 𝑐𝑛 𝑒𝑖𝑛ω𝑥 , 𝜔 =
𝐿
𝑛=−∞

Multiplying by 𝑒 𝑖𝑚ω𝑥 and integrate from −𝐿 to 𝐿, we have

∞

(𝑓(𝑥), 𝑒𝑖𝑚ω𝑥 ) = ∑ 𝑐𝑛 (𝑒𝑖𝑛ω𝑥 , 𝑒𝑖𝑚ω𝑥 )

𝑛=−∞

It is easy to prove that the inner product of the exponential functions is

given by
0 𝑚≠𝑛
(𝑒 𝑖𝑛ω𝑥 , 𝑒 𝑖𝑚ω𝑥 ) = {
2𝐿 𝑚=𝑛
So we get
∞

(𝑓(𝑥), 𝑒𝑖𝑚𝑥 ) = ∑ 𝑐𝑛 ⏟
(𝑒𝑖𝑛ω𝑥 , 𝑒𝑖𝑚ω𝑥 ),
𝑛=−∞ ={ 0 𝑚≠𝑛
2𝐿 𝑚=𝑛

from which we have

 28 Chapter 1 Fourier Analysis

(𝑓(𝑥), 𝑒𝑖𝑚ω𝑥 ) = 𝑐𝑚 ∙ 2𝐿,

or, by renaming the dummy variable 𝑚,
1 1 𝐿
𝑐𝑛 = (𝑓(𝑥), 𝑒𝑖𝑛ω𝑥 ) = ∫ 𝑓(𝑥)𝑒−𝑖𝑛ω𝑥 𝑑𝑥
2𝐿 2𝐿 −𝐿
Summarizing, the above in the following theorem
Theorem 1.5 The exponential (complex) Fourier series of the periodic
function 𝑓(𝑥), with period 2𝐿, is given by
∞
𝜋
𝑓(𝑥) = ∑ 𝑐𝑛 𝑒𝑖𝑛ω𝑥 , 𝜔 = , (1)
𝐿
𝑛=−∞

where
1 𝑖𝑛ω𝑥 1 𝐿
𝑐𝑛 = (𝑓(𝑥), 𝑒 )= ∫ 𝑓(𝑥)𝑒−𝑖𝑛ω𝑥 𝑑𝑥 (2)
2𝐿 2𝐿 −𝐿

Example 1 Find the exponential Fourier series of the function

𝑓(𝑥) = 𝑒 𝑥 in the interval − 𝜋 < 𝑥 < 𝜋, with period 2𝜋.

Solution Since 2𝐿 = 2𝜋  𝐿 = 𝜋  𝜔 = 1. Then

∞

𝑓(𝑥) = ∑ 𝑐𝑛 𝑒 𝑖𝑛𝑥 ,
𝑛=−∞
𝜋
1 1
𝑐𝑛 = (𝑓(𝑥), 𝑒 −𝑖𝑛𝑥 ) = ∫ 𝑒 𝑥 𝑒 −𝑖𝑛𝑥 𝑑𝑥
2𝜋 2𝜋 −𝜋
𝜋
1 𝜋 (1−𝑖𝑛)𝑥 1 𝑒 (1−𝑖𝑛)𝑥
= ∫ 𝑒 𝑑𝑥 = [ ]
2𝜋 −𝜋 2𝜋 (1 − 𝑖𝑛) −𝜋

1 𝑒 (1−𝑖𝑛)𝜋 − 𝑒 − (1−𝑖𝑛)𝜋
=
2𝜋 (1 − 𝑖𝑛)
1
= (𝑒 𝜋 𝑒 −𝑖𝑛𝜋 − 𝑒 −𝜋 𝑒 𝑖𝑛𝜋 )
2𝜋(1 − 𝑖𝑛)
 Chapter 1 Fourier Analysis 29

1
= (𝑒 𝜋 (cos 𝑛𝜋 − 𝑖 sin 𝑛𝜋) − 𝑒 −𝜋 (cos 𝑛𝜋 + 𝑖 sin 𝑛𝜋))
2𝜋(1 − 𝑖𝑛)
cos 𝑛𝜋 𝑒 𝜋 − 𝑒 −𝜋 cos 𝑛𝜋
= ( )= sinh 𝜋
𝜋(1 − 𝑖𝑛) 2 𝜋(1 − 𝑖𝑛)
cos 𝑛𝜋 1 (1 + 𝑖𝑛)
= ∙ sinh 𝜋
𝜋 (1 − 𝑖𝑛) (1 + 𝑖𝑛)
(1 + 𝑖𝑛) sinh 𝜋
= (−1)𝑛 ( )
(1 + 𝑛2 ) 𝜋
Thus, the exponential Fourier series of 𝑓(𝑥) = 𝑒 𝑥 is
∞
sinh 𝜋 (1 + 𝑖𝑛) 𝑖𝑛𝑥
𝑓(𝑥) = ∑ (−1)𝑛 𝑒
𝜋 (1 + 𝑛2 )
𝑛=−∞

Remark It should be noted that the Fourier trigonometric series

and Fourier complex series are the same series and the
coefficients of these two series are related by
𝑎0
= 𝑐0 ,
2
𝑎𝑛 = 𝑐𝑛 + 𝑐−𝑛
𝑏𝑛 = 𝑖(𝑐𝑛 − 𝑐−𝑛 )

1.1.7 Parseval’s identity

Theorem 2.6 Let 𝑓(𝑥) be a periodic function with period 2𝐿 and whose
Fourier series is given by
∞
𝑎0 𝜋
𝑓(𝑥) = + ∑(𝑎𝑛 cos 𝑛𝜔𝑥 + 𝑏𝑛 sin 𝑛𝜔𝑥) , 𝜔 = (1)
2 𝐿
𝑛=1

Then
∞
1 𝑎2
‖𝑓‖2 = 0 + ∑(𝑎𝑛2 + 𝑏𝑛2 ) (2)
𝐿 2
𝑛=1
 30 Chapter 1 Fourier Analysis

Proof. On Multiplying (1) by 𝑓(𝑥) and integrate term by term from

−𝐿 to 𝐿, we get
𝑎0
(𝑓(𝑥), 𝑓(𝑥)) = (1, 𝑓(𝑥))
2
∞

+ ∑(𝑎𝑛 (𝑓(𝑥), cos 𝑛𝜔𝑥) + 𝑏𝑛 (𝑓(𝑥), sin 𝑛𝜔𝑥)) (3)

𝑛=1

From the definition of Fourier coefficients, we obtain

∞
𝑎0
‖𝑓‖2 = (𝐿𝑎0 ) + ∑(𝑎𝑛 (𝐿𝑎𝑛 ) + (𝐿𝑏𝑛 ))
2
𝑛=1

Therefore,
∞
1 𝑎2
‖𝑓‖2 = 0 + ∑(𝑎𝑛2 + 𝑏𝑛2 ) , (4)
𝐿 2
𝑛=1

which is the required proof. ∎

Example 1 Write Parseval’s identity corresponding to the Fourier

series of the function 𝑓(𝑥) = 𝑥 2 , 𝑥 ∈ [−𝜋, 𝜋], 𝑓(𝑥) = 𝑓(𝑥 + 2𝜋),
given in Example 2 in section 2.1.5.

Solution Here, we have

2𝜋 2 (−1)𝑛
𝑎0 = , 𝑎𝑛 = 4 , 𝑏𝑛 = 0
3 𝑛2
and
1 1 1
‖𝑓‖2 = (𝑓(𝑥), 𝑓(𝑥)) = (𝑥 2 , 𝑥 2 )
𝐿 𝜋 𝜋
𝜋
2 𝜋 4 2 𝑥4 2𝜋 4
= ∫ 𝑥 𝑑𝑥 = [ ] =
𝜋 0 𝜋 5 0 5
Hence, we have from (2) that
 Chapter 1 Fourier Analysis 31

2
2𝜋 2 ∞
(
2𝜋 4
3 ) 4 2
= + ∑ (((−1)𝑛 2 ) + 0),
5 2 𝑛
𝑛=1

or
∞
2𝜋 4 2𝜋 4 1
− = 16 ∑ 4
5 9 𝑛
𝑛=1

Thus
∞
𝜋4 1 1 1 1
= ∑ 4 = ( 4 + 4 + 4 +⋯)
90 𝑛 1 2 3
𝑛=1

𝑬𝒙𝒆𝒓𝒄𝒊𝒔𝒆𝒔 𝟏. 𝟑

In problems 1-2, find the exponential Fourier series representation of

the given functions
− cos 𝑥 − 𝜋 < 𝑥 < 0
𝟏. 𝑓(𝑥) = { , 𝑓(𝑥) = 𝑓(𝑥 + 2𝜋)
cos 𝑥 0<𝑥<𝜋
−1 − 𝜋 < 𝑥 < 0
𝟐. 𝑓(𝑥) = { 0 𝑥=0 , 𝑓(𝑥) = 𝑓(𝑥 + 2𝜋)
1 0<𝑥<𝜋
𝟑. Show that
∞
2 4 cos 2𝑛𝑥
|sin 𝑥| = − ∑ 2
𝜋 𝜋 4𝑛 − 1
𝑛=1

𝟒. Show that the Fourier expansion of

1 + (𝑥⁄𝜋) − 𝜋 ≤ 𝑥 ≤ 0
𝑓(𝑥) = { ,
1 − (𝑥⁄𝜋) 0≤𝑥≤𝜋
in the range −𝜋 ≤ 𝑥 ≤ 𝜋 is
1 4 cos 3𝑥 cos 5𝑥
𝑓(𝑥) = + 2 (cos 𝑥 + + +⋯)
2 𝜋 32 52
Hence deduce that
 32 Chapter 1 Fourier Analysis

∞
𝜋2 1 1 1
= 1+ 2+ 2+⋯= ∑
8 3 5 (2𝑛 − 1)2
𝑛=1

𝟓. Expand the function

𝑥 0 ≤ 𝑥 ≤ 𝜋⁄2
𝑓(𝑥) = {
𝜋⁄2 𝜋⁄2 ≤ 𝑥 ≤ 𝜋
in (𝒊) a 𝑠𝑖𝑛𝑒 𝑠𝑒𝑟𝑖𝑒𝑠, and (𝒊𝒊) a 𝑐𝑜𝑠𝑖𝑛𝑒 𝑠𝑒𝑟𝑖𝑒𝑠, valid in 0 ≤ 𝑥 < 𝜋.
Sketch the graphs of the functions represented by these series in the
range −2𝜋 ≤ 𝑥 ≤ 2𝜋.
𝟔. If 𝑓(𝑥) is an odd function of 𝑥, of period 2𝜋, and
𝑥 0 ≤ 𝑥 ≤ 𝜋⁄3
𝑓(𝑥) = { 𝜋⁄3 𝜋⁄3 ≤ 𝑥 ≤ 2𝜋⁄3
𝜋−𝑥 2𝜋⁄3 ≤ 𝑥 ≤ 𝜋
show that the Fourier series of 𝑓(𝑥) may be expressed in the form
∞
4 1 𝑛𝜋
∑ 2 sin sin 𝑛𝑥
𝜋 𝑛 3
𝑛=1

where 𝑛 is an odd positive integer, and determine the first three non-
zero terms.
𝟕. Prove that for 0 ≤ 𝑥 ≤ 𝜋
∞
𝜋2 cos 2𝑛𝑥
(𝒊) 𝑥(𝜋 − 𝑥) = −∑ ,
6 𝑛2
𝑛=1
∞
8 sin(2𝑛 − 1)𝑥
(𝒊𝒊) 𝑥(𝜋 − 𝑥) = ∑ ,
𝜋 (2𝑛 − 1)3
𝑛=1

and hence, show that

∞ ∞
1 𝜋2 (−1)𝑛 𝜋 2
(𝒂) ∑ 2 = , (𝒃) ∑ = ,
𝑛 6 𝑛2 12
𝑛=1 𝑛=1
∞
(−1)𝑛−1 𝜋3
(𝒄) ∑ =
(2𝑛 − 1)3 32
𝑛=1
 Chapter 1 Fourier Analysis 33

𝟖. Use problem 7 to find the sum of each of the following series

∞ ∞ ∞ ∞
1 1 1 1
∑ 4, ∑ 6, ∑ , ∑
𝑛 𝑛 (2𝑛 − 1)4 (2𝑛 − 1)6
𝑛=1 𝑛=1 𝑛=1 𝑛=1

𝟗. (𝒂) Expand the function 𝑓(𝑥) = |𝑥| , − 1 ≤ 𝑥 ≤ 1 , which is

periodic with period 2, in a Fourier series.
(𝒃) By giving a special value to x in part (𝒂) above, find the sum
1 1 1
of the series (12 + 32 + 52 + ⋯ ).

In problems 𝟏𝟎 − 𝟏𝟓, find the Fourier exponential series of the

following periodic functions
𝟏𝟎. 𝑓(𝑥) = 𝑥, − 𝜋<𝑥<𝜋
𝟏𝟏. 𝑓(𝑥) = 𝑥, 0 < 𝑥 < 2𝜋
𝟏𝟐. 𝑓(𝑥) = 𝑥 2 , 0 < 𝑥 < 2𝜋
1 − (𝑥⁄𝜋) − 𝜋 ≤ 𝑥 ≤ 0
𝟏𝟑. 𝑓(𝑥) = {
1 + (𝑥⁄𝜋) 0≤𝑥≤𝜋
−1 − 2 < 𝑥 < 0
𝟏𝟒. 𝑓(𝑥) = {
1 0<𝑥<2
0 − 1⁄2 < 𝑥 < 0
𝟏𝟓. 𝑓(𝑥) = {1 0 < 𝑥 ≤ 1⁄4
0 1⁄4 ≤ 𝑥 ≤ 1⁄2

1.2. Fourier Integrals

1.2.1. Introduction

Since many practical problems involve non-periodic functions, then we

cannot use the Fourier series in dealing with these functions. However,
the non-periodic functions may be considered as periodic functions
with period 2𝐿 = ∞ . In other words we shall construct a periodic
function 𝑓 of period 2𝐿 and in the limit we let 2𝐿 → ∞ so that the
function f has only one cycle in the interval (−∞, ∞).
 34 Chapter 1 Fourier Analysis

Let 𝑓𝐿 (𝑥) be a periodic function with period 2𝐿. Then the Fourier series
of this function will be in the form
∞
𝑎0 𝜋
𝑓𝐿 (𝑥) = + ∑(𝑎𝑛 cos 𝑛𝜔𝑥 + 𝑏𝑛 sin 𝑛𝜔𝑥) , ω = (1)
2 𝐿
𝑛=1

Inserting 𝑎0 , 𝑎𝑛 and 𝑏𝑛 in this equation

∞
1 𝐿 1 𝐿
𝑓𝐿 (𝑥) = ∫ 𝑓𝐿 (𝑧)𝑑𝑧 + ∑ [cos 𝑛ω𝑥 ∫ 𝑓𝐿 (𝑧) cos 𝑛ω𝑧 𝑑𝑧
2𝐿 −𝐿 𝐿 −𝐿
𝑛=1
𝐿
+ sin 𝑛ω𝑥 ∫ 𝑓𝐿 (𝑧) sin 𝑛ω𝑧 𝑑𝑧] (2)
−𝐿

Put
ω𝑛 = 𝑛ω,
then
𝜋 𝜋
∆ω𝑛 = (𝑛 + 1)ω − 𝑛ω = ω = ω =  𝐿=
𝐿 ∆ω𝑛
Substitute in (2) we have
∞
1 𝐿 1 𝐿
𝑓𝐿 (𝑥) = ∫ 𝑓 (𝑧)𝑑𝑧 + ∑ [cos ω𝑛 𝑥 ∙ ∆ω𝑛 ∫ 𝑓𝐿 (𝑧) cos ω𝑛 𝑧 𝑑𝑧
2𝐿 −𝐿 𝐿 𝜋 −𝐿
𝑛=1
𝐿
+ sin ω𝑛 𝑥 ∙ ∆ω𝑛 ∫ 𝑓𝐿 (𝑧) sin ω𝑛 𝑧 𝑑𝑧] (3)
−𝐿

Taking the limit 𝐿 → ∞  ω𝑛 → 0 and assume that

𝑓(𝑥) = lim 𝑓𝐿 (𝑥) (4)
𝑛→∞

Thus we have
1 ∞ ∞
𝑓(𝑥) = ∫ [cos ω𝑛 𝑥 ∫ 𝑓(𝑧) cos ω𝑛 𝑧 𝑑𝑧
𝜋 0 −∞
∞
+ sin ω𝑛 𝑥 ∫ 𝑓(𝑧) sin ω𝑛 𝑧 𝑑𝑧] 𝑑ω𝑛 (5)
−∞

Renaming the dummy variables, equation (5) may be rewritten as

 Chapter 1 Fourier Analysis 35

1 ∞ ∞
𝑓(𝑥) = ∫ [cos ω𝑥 ∫ 𝑓(𝑥) cos ω𝑥 𝑑𝑥
𝜋 0 −∞
∞
+ sin ω𝑥 ∫ 𝑓(𝑥) sin ω𝑥 𝑑𝑥] 𝑑ω (6)
−∞
On using the notation
∞
𝐴(ω) = ∫ 𝑓(𝑥) cos ω𝑥 𝑑𝑥 (7)
−∞
∞
𝐵(ω) = ∫ 𝑓(𝑥) sin ω𝑥 𝑑𝑥 (8)
−∞

We can write (6) in the form

1 ∞
𝑓(𝑥) = ∫ (𝐴(ω) cos ω𝑥 + 𝐵(ω) sin ω𝑥) 𝑑ω (9)
𝜋 0
The integral representation of 𝑓(𝑥) (Equation (9)) is called the
𝐹𝑜𝑢𝑟𝑖𝑒𝑟 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 of 𝑓(𝑥) , and 𝐴(ω) and 𝐵(ω) are called the
𝐹𝑜𝑢𝑟𝑖𝑒𝑟 𝑐𝑜𝑠𝑖𝑛𝑒 𝑎𝑛𝑑 𝑠𝑖𝑛𝑒 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚 , respectively, of the function
𝑓(𝑥). In certain situations 𝐴(ω) and 𝐵(ω) are called the 𝑠𝑝𝑒𝑐𝑡𝑟𝑢𝑚 of
𝑓(𝑥).

Remark Equation (4) is valid if and only if 𝑓(𝑥) is absolutely

integrable on the 𝑥 − 𝑎𝑥𝑖𝑠; that is, the following finite
limit exists
∞
∫ ‖𝑓(𝑥)‖𝑑𝑥 (10)
−∞
Summarize the above discussion in the following theorem.

Theorem 1.7 If 𝑓(𝑥) is a piecewise continuous in every finite interval

and has a piecewise differentiable for all 𝑥 and if the integral (10) exists,
then 𝑓(𝑥)can be represented by a Fourier integral (9). At a discontinuity
point 𝑥0 the value of Fourier integral converges to
1
[𝑓(𝑥0+ ) + 𝑓(𝑥0− )]
2
 36 Chapter 1 Fourier Analysis

Example 1 Find the Fourier integral representation of the gate function

f (x )
(single pulse) defined by
1
1 𝑖𝑓 |𝑥| < δ
𝑓(𝑥) = {   x
0 𝑖𝑓 |𝑥| > δ

Fig. 1.8
Solution
1 ∞
𝑓(𝑥) = ∫ (𝐴(ω) cos ω𝑥 + 𝐵(ω) sin ω𝑥) 𝑑ω,
𝜋 0
where
∞ δ
𝐴(ω) = ∫ 𝑓(𝑥) cos ω𝑥 𝑑𝑥 = ∫ cos ω𝑥 𝑑𝑥
−∞ −δ
δ
sin ω𝑥 2 sin ωδ
=[ ] =
ω −δ ω
and
∞ δ
𝐵(ω) = ∫ 𝑓(𝑥) sin ω𝑥 𝑑𝑥 = ∫ sin ω𝑥 𝑑𝑥
−∞ −δ

cos ω𝑥 δ
= [− ] =0
ω −δ
Then
2 ∞ sin ωδ cos ω𝑥
𝑓(𝑥) = ∫ 𝑑ω (11)
𝜋 0 ω
At 𝑥 = δ, we have
1 1
𝑓(δ) = [𝑓(δ+ ) + 𝑓(δ− )] =
2 2
Therefore, we can write from (11)
𝜋
𝑖𝑓 |𝑥| < δ
∞
sin ωδ cos ω𝑥 𝜋 2
∫ 𝑑ω = 𝑓(𝑥) = 𝜋 𝑖𝑓 |𝑥| = δ
0 ω 2 4
{ 0 𝑖𝑓 |𝑥| > δ
 Chapter 1 Fourier Analysis 37

From this last integral we can estimate the integral

∞ ∞ ∞
sin ωδ sin ωδ sin x 𝜋
∫ 𝑑ω = ∫ 𝑑δω = ∫ 𝑑𝑥 =
0 ω 0 ωδ 0 𝑥 2

1.2.2. Fourier cosine integrals

Using the properties of integrals, the Fourier integral becomes simpler.

The simplifications arise if the function is either even or odd. Here we
consider the Fourier integrals of even functions.
If 𝑓(𝑥) is an even function (symmetric about the 𝑦 − 𝑎𝑥𝑖𝑠), then from
(8) we have
∞
𝐵(ω) = ∫ 𝑓(𝑥) sin ω𝑥 𝑑𝑥 = 0, (11)
−∞

and (7) gives

∞
𝐴(ω) = 2 ∫ 𝑓(𝑥) cos ω𝑥 𝑑𝑥 (12)
0

and the Fourier integral (9) reduces to the so-called

𝐹𝑜𝑢𝑟𝑖𝑒𝑟 𝑐𝑜𝑠𝑖𝑛𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙
1 ∞
𝑓(𝑥) = ∫ 𝐴(ω) cos ω𝑥 𝑑ω (13)
𝜋 0
1.2.3. Fourier sine integrals
Similarly, if 𝑓(𝑥) is an odd function (symmetric with respect to the
origin), then from (7) we have
∞
𝐴(ω) = ∫ 𝑓(𝑥) cos ω𝑥 𝑑𝑥 = 0, (14)
−∞

and from(8) we get

∞
𝐵(ω) = 2 ∫ 𝑓(𝑥) sin ω𝑥 𝑑𝑥 , (15)
0
 38 Chapter 1 Fourier Analysis

and the Fourier integral (9) reduces to the so-called

𝐹𝑜𝑢𝑟𝑖𝑒𝑟 𝑠𝑖𝑛𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙
1 ∞
𝑓(𝑥) = ∫ 𝐵(ω) sin ω𝑥 𝑑ω (16)
𝜋 0

Example 1 Using the Fourier integral, show that

∞
ω3 sin ω𝑥 𝜋
∫ 4
𝑑ω = 𝑒 −𝑥 cos 𝑥 , 𝑥 > 0
0 ω +4 2

Solution. From the left hand side of the required assertion, we note that
this is an expansion of some function in terms of Fourier sine integral.
Therefore, we use (15) with 𝑓(𝑥) = 𝑒 −𝑥 cos 𝑥, 𝑥 > 0 to obtain

∞ ∞
𝐵(ω) = 2 ∫ 𝑓(𝑥) sin ω𝑥 𝑑𝑥 = 2 ∫ 𝑒 −𝑥 cos 𝑥 sin ω𝑥 𝑑𝑥
0 0
∞
1
= 2 × ∫ 𝑒 −𝑥 [sin(ω − 1)𝑥 + sin(ω + 1)𝑥]𝑑𝑥
2 0
∞ ∞
= ∫ 𝑒 −𝑥 sin(ω − 1)𝑥 𝑑𝑥 + ∫ 𝑒 −𝑥 sin(ω + 1)𝑥 𝑑𝑥
0 0

Since,

𝑎𝑥
𝑒 𝑎𝑥
∫𝑒 sin 𝑏𝑥 𝑑𝑥 = 2 (𝑎 sin 𝑏𝑥 − 𝑏 cos 𝑏𝑥) + 𝑐,
𝑎 + 𝑏2
then
∞ ∞
𝐵(ω) = ∫ 𝑒 −𝑥 sin(ω − 1)𝑥 𝑑𝑥 + ∫ 𝑒 −𝑥 sin(ω + 1)𝑥 𝑑𝑥
0 0
∞
𝑒 −𝑥
=[ (− sin(ω − 1)𝑥 − (ω − 1) cos(ω − 1)𝑥)]
1 + (ω − 1)2 0
 Chapter 1 Fourier Analysis 39

∞
𝑒 −𝑥
+[ (− sin(ω + 1)𝑥 − (ω + 1) cos(ω + 1)𝑥)]
1 + (ω + 1)2 0

(ω − 1) (ω + 1)
= +
1 + (ω − 1)2 1 + (ω + 1)2
(ω − 1) (ω + 1)
= + 2
ω2 + 2 − 2ω ω + 2 + 2ω
(ω − 1)(ω2 + 2 + 2ω) + (ω + 1)(ω2 + 2 − 2ω)
=
(ω2 + 2 − 2ω)(ω2 + 2 + 2ω)
2ω3 2ω3 2ω3
= = =
(ω2 + 2)2 − 4ω2 ω4 + 4ω2 + 4 − 4ω2 ω4 + 4
Then the Fourier sine integral can be written as
1 ∞ 2ω3
𝑓(𝑥) = 𝑒 −𝑥 cos 𝑥 = ∫ sin ω𝑥 𝑑ω,
𝜋 0 ω4 + 4
which implies that
2 ∞ ω3
𝑒 −𝑥 cos 𝑥 = ∫ sin ω𝑥 𝑑ω,
𝜋 0 ω4 + 4
or
∞
ω3 sin ω𝑥 𝜋 −𝑥
∫ 𝑑ω = 𝑒 cos 𝑥 , 𝑥 > 0
0 ω4 + 4 2
Hint: On evaluating the integrals we can use the Laplace transform
concept as following
∞ ∞
−𝑥
𝐵(ω) = ∫ 𝑒 sin(ω − 1)𝑥 𝑑𝑥 + ∫ 𝑒 −𝑥 sin(ω + 1)𝑥 𝑑𝑥
0 0

= ℒ(sin(ω − 1)𝑥)⌋𝑠=1 + ℒ(sin(ω + 1)𝑥)⌋𝑠=1

(ω − 1) (ω + 1)
= ⌋ + ⌋
𝑠 2 + (ω − 1)2 𝑠=1 𝑠 2 + (ω + 1)2 𝑠=1
(ω − 1) (ω + 1)
= +
1 + (ω − 1)2 1 + (ω + 1)2
 40 Chapter 1 Fourier Analysis

(ω − 1) (ω + 1)
= +
ω2 + 2 − 2ω ω2 + 2 + 2ω
(ω − 1)(ω2 + 2 + 2ω) + (ω + 1)(ω2 + 2 − 2ω)
=
(ω2 + 2 − 2ω)(ω2 + 2 + 2ω)
2ω3 2ω3 2ω3
= = =
(ω2 + 2)2 − 4ω2 ω4 + 4ω2 + 4 − 4ω2 ω4 + 4

𝑬𝒙𝒆𝒓𝒄𝒊𝒔𝒆𝒔 𝟏. 𝟒

Find the Fourier integral representations of the functions given in

problems 𝟏 − 𝟓.
1 −4<𝑥 <4
𝟏. 𝑓(𝑥) = {
0 𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒
1−𝑥 0<𝑥<1
𝟐. 𝑓(𝑥) = { 1 + 𝑥 −1<𝑥 <0
0 𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒
−𝑥
𝑒 𝑥>0
𝟑. 𝑓(𝑥) = { 𝑥
𝑒 𝑥<0
sin 𝑥 −𝜋 <𝑥 <𝜋
𝟒. 𝑓(𝑥) = {
0 𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒
𝑥 0<𝑥<1
𝟓. 𝑓(𝑥) = {2 − 𝑥 1 < 𝑥 < 2
0 𝑥>2
In problems 6 − 11, by the use of Fourier integral representation, show
that
𝜋𝑒 −𝑥 𝑖𝑓 𝑥 > 0
∞
cos ω𝑥 + ω sin ω𝑥 𝜋
𝟔. ∫ 2
𝑑ω = { 𝑖𝑓 𝑥 = 0
0 ω +1 2
0 𝑖𝑓 𝑥 < 0
∞
cos ω𝑥 𝜋 −𝑥
𝟕. ∫ 2
𝑑ω = 𝑒 , 𝑥>0
0 ω +1 2
∞ 𝜋
sin πω sin ω𝑥 sin 𝑥 0≤𝑥≤𝜋
𝟖. ∫ 2
𝑑ω = {2
0 1−ω 0 𝑥>𝜋
 Chapter 1 Fourier Analysis 41

∞
ω3 sin ω𝑥 𝜋
𝟗. ∫ 4
𝑑ω = 𝑒 −𝑥 cos 𝑥 , 𝑥 > 0
0 ω +4 2
𝜋
0≤𝑥<1
∞
cos ω𝑥 + ω sin ω𝑥 2
𝟏𝟎. ∫ 𝑑ω = 𝜋
0 ω 2+1 𝑥=1
4
{0 𝑥>1
𝜋 𝜋
∞
cos(πω⁄2) cos ω𝑥 cos 𝑥 𝑥<
𝟏𝟏. ∫ 2
𝑑ω = {2 2
𝜋
0 1 − ω 0 𝑥>
2
In problems 12 − 15, find the Fourier cosine integral of the functions

𝟏𝟐. 𝑓(𝑥) = 𝑒 −𝑥 , 𝑥 > 0

𝟏𝟑. 𝑓(𝑥) = 𝑎2 − 𝑥 2 , 𝑖𝑓 0 < 𝑥 < 𝑎, 𝑓(𝑥) = 0 𝑖𝑓 𝑥 > 𝑎
𝑥 0<𝑥<𝑎
𝟏𝟒. 𝑓(𝑥) = {
0 𝑥>𝑎
sin 𝑥 0<𝑥<𝜋
𝟏𝟓. 𝑓(𝑥) = {
0 𝑥>𝜋
2. Functions of a Complex Variable

This chapter focuses primarily on functions of a complex

variable, their representations, and properties associated with functions
such as limits, continuity and derivatives.
2.1 Complex functions
The idea of a function of a complex variable is a particular case of the
general mathematical notation of a function. Recall that the real valued
function 𝑓: 𝑋 → 𝑌, 𝑋, 𝑌 ∈ 𝑅, is a relation between the two sets 𝑋
and 𝑌 such that to each element 𝑥 of 𝑋 there corresponds exactly one
element 𝑦 of 𝑌. In this chapter we consider functions in which the
independent and dependent variables are complex numbers. These
functions are defined in a similar way to functions of real numbers that
we studied in calculus; the only difference is that they operate on
complex numbers rather than real numbers. Now we begin by giving
the following definition to functions of a complex variable.

Definition 2.1 (A complex function) A 𝑐𝑜𝑚𝑝𝑙𝑒𝑥 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑓 on a set

𝐷  𝐶 of complex numbers to a set 𝑅  𝐶 is a rule which assigns to
every 𝑧 ∈ 𝐷 a complex number 𝑤 = 𝑓(𝑧) ∈ 𝑅, i.e.,
𝑓: 𝐷 → 𝑅,
such that
𝑤 = 𝑓 (𝑧 )

The set 𝐷 is called the 𝑑𝑜𝑚𝑎𝑖𝑛 of the function and the set 𝑅 is called
the 𝑟𝑎𝑛𝑔𝑒 of 𝑓 . For any complex function, both the independent
variable and the dependent variable may be separated into real and
imaginary parts. So that the image 𝑤 of a complex number 𝑧 = 𝑥 + 𝑖𝑦
will be some complex number 𝑤 = 𝑢 + 𝑖𝑣, that is,
 44 Chapter 2 Functions of a Complex Variable

𝑤 = 𝑓(𝑧) = 𝑓(𝑥 + 𝑖𝑦) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦) (1)

where 𝑢 and 𝑣 are the real and imaginary parts of 𝑤 and they are real
valued functions.

𝒛 − 𝒑𝒍𝒂𝒏𝒆 𝒘 − 𝒑𝒍𝒂𝒏𝒆
𝑦 𝑣

𝐷𝑜𝑚𝑎𝑖𝑛 𝑓(𝑧) 𝑅𝑎𝑛𝑔𝑒

𝐷  𝑅

𝑥 𝑢

Fig. 2.1 The mapping 𝑤 = 𝑓(𝑧) = 𝑓(𝑥 + 𝑖𝑦) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦)

Graphically, we say that the complex function (1) is a mapping

(transformation) from the 𝑥𝑦 − plane ( 𝑧 − plane) to the 𝑢𝑣 − plane
(𝑤 −plane) that maps 𝐷 into 𝑅, as shown in Fig. 2.1. Some examples
of functions of a complex variable are:
𝑓(𝑧) = 𝑧 2 − 4𝑧,
𝑧
𝑓(𝑧) = 2 , 𝑧 ≠ ±2𝑖
𝑧 +4
𝑓(𝑧) = 𝑧 + Im(𝑧)
Each of these functions can be expressed in form (1).
If to each value of 𝑧 there corresponds only one value of 𝑤, we say that
the function 𝑤 = 𝑓(𝑧) is 𝑠𝑖𝑛𝑔𝑙𝑒 𝑣𝑎𝑙𝑢𝑒𝑑; but if some values of 𝑧 have
corresponding more than one value, the function is said to be
𝑚𝑢𝑙𝑡𝑖𝑣𝑎𝑙𝑢𝑒𝑑 𝑜𝑟 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑣𝑎𝑙𝑢𝑒𝑑. For instance, 𝑤 = 𝑧 𝑛 , (𝑛 is a
positive integer) 𝑤 = |𝑧|, 𝑎𝑛𝑑 𝑤 = 𝑧 are single valued functions;
𝑛
while 𝑤 = √𝑧, 𝑎𝑛𝑑 𝑤 = arg 𝑧, are multivalued functions.
 Chapter 2 Functions of a Complex Variable 45

Example 1 Express 𝑓(𝑧) = 𝑧 2 − 4𝑧 in the form 𝑓(𝑧) = 𝑢(𝑥, 𝑦) +

𝑖𝑣(𝑥, 𝑦).

Solution
𝑓(𝑧) = 𝑧 2 − 4𝑧 = (𝑥 + 𝑖𝑦 )2 − 4(𝑥 + 𝑖𝑦 )
= (𝑥 2 − 𝑦 2 + 2𝑖𝑥𝑦 ) − 4(𝑥 + 𝑖𝑦 )
= (𝑥 2 − 𝑦 2 − 4𝑥 ) + 𝑖(2𝑥𝑦 − 4𝑦 )
= 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦)
Hence,
𝑢(𝑥, 𝑦) = (𝑥 2 − 𝑦 2 − 4𝑥 ), 𝑣(𝑥, 𝑦) = (2𝑥𝑦 − 4𝑦 )

Example 2 Express 𝑓(𝑧) = 𝑧 Re 𝑧 + 𝑧 2 + Im 𝑧 in the form 𝑓(𝑧) =

𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦).

Solution
𝑓(𝑧) = 𝑧 Re 𝑧 + 𝑧 2 + Im 𝑧 = (𝑥 − 𝑖𝑦) (𝑥) + (𝑥 + 𝑖𝑦)2 + (𝑦)
= (𝑥 2 − 𝑖𝑥𝑦) + (𝑥 2 − 𝑦 2 + 2𝑖𝑥𝑦) + 𝑦
= (2𝑥 2 − 𝑦 2 + 𝑦) + 𝑖𝑥𝑦
Thus,
𝑢(𝑥, 𝑦) = (2𝑥 2 − 𝑦 2 + 𝑦), 𝑣(𝑥, 𝑦) = 𝑥𝑦

Example 3 Write the complex function 𝑓(𝑧) = 𝑧 4 − 2𝑧 3 in the form

𝑓(𝑧) = 𝑢(𝑟, θ) + 𝑖𝑣(𝑟, θ), where 𝑧 = 𝑟𝑒 𝑖θ .

Solution From the polar form of 𝑧, we get

4 3
𝑓(𝑧) = 𝑓(𝑟𝑒 𝑖θ ) = (𝑟𝑒 𝑖θ ) − 2(𝑟𝑒 𝑖θ ) = 𝑟 4 𝑒 𝑖4θ − 2𝑟 3 𝑒 𝑖3θ
 46 Chapter 2 Functions of a Complex Variable

= 𝑟 4 (cos 4θ + 𝑖 sin 4θ) − 2𝑟 3 (cos 3θ + 𝑖 sin 3θ)

= (𝑟 4 cos 4θ − 2𝑟 3 cos 3θ) + 𝑖(𝑟 4 sin 4θ − 2𝑟 3 sin 3θ)
From which we obtain
𝑢(𝑟, θ) = (𝑟 4 cos 4θ − 2𝑟 3 cos 3θ),
𝑣(𝑟, θ) = (𝑟 4 sin 4θ − 2𝑟 3 sin 3θ)

Remark As we see from Example 3, it is convenient to use the

polar form of 𝑧 in case of power function and we shall
see later that this form is convenient also for logarithmic
functions.
However, we shall keep to the notation
𝑧 = 𝑥 + 𝑖𝑦 = 𝑟𝑒 𝑖θ = 𝑟(cos θ + 𝑖 sin θ), (2)
𝑤 = 𝑢 + 𝑖𝑣 = ρ𝑒 𝑖φ = ρ(cos φ + 𝑖 sin φ), (3)
1 1
Re 𝑧 = (𝑧 + 𝑧), Im 𝑧 = (𝑧 − 𝑧), (4)
2 2𝑖
1 1
Re 𝑤 = (𝑤 + 𝑤), Im 𝑤 = (𝑤 − 𝑤), (5)
2 2𝑖
Examples 1, 2 and 3 show how to find 𝑢(𝑥, 𝑦) and 𝑣(𝑥, 𝑦) when a rule
for computing 𝑓 is given. Conversely, if 𝑢(𝑥, 𝑦) and 𝑣(𝑥, 𝑦) are two
real-valued functions of the real variables 𝑥 and 𝑦, they determine a
complex-valued function 𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦) and we can use
formula (4) to find a formula for 𝑓 involving the variables 𝑧 and 𝑧.

Example 4 If 𝑢 = 𝑥 2 − 𝑦 2 + 𝑥 and 𝑣 = 2𝑥𝑦 − 𝑦, express 𝑤 as a

function of the complex number 𝑧.

Solution Since
 Chapter 2 Functions of a Complex Variable 47

1
𝑥 = Re 𝑧 = (𝑧 + 𝑧),
2
1
𝑦 = Im 𝑧 = (𝑧 − 𝑧),
2𝑖

then the complex function

𝑤 = (𝑥 2 − 𝑦 2 + 𝑥) + 𝑖(2𝑥𝑦 − 𝑦),

can be expressed in the form

2 2
𝑧+𝑧 𝑧−𝑧 𝑧+𝑧
𝑤 = (( ) −( ) +( ))
2 2𝑖 2

𝑧+𝑧 𝑧−𝑧 𝑧−𝑧

+𝑖 (2 ( )( )−( )),
2 2𝑖 2𝑖

which is written by simplification as

2 2
𝑧 2 + 2𝑧𝑧 + 𝑧 𝑧 2 − 2𝑧𝑧 + 𝑧 𝑧+𝑧
𝑤 = (( )−( )+( ))
4 −4 2

2
𝑧2 − 𝑧 𝑧−𝑧
+ (( )−( )),
2 2

2 2
2𝑧 2 + 2𝑧 𝑧+𝑧 𝑧2 − 𝑧 𝑧−𝑧
=( + )+( − )
4 2 2 2
2 2
𝑧2 + 𝑧 𝑧+𝑧 𝑧2 − 𝑧 𝑧−𝑧
=( + )+( − )
2 2 2 2
= (𝑧 2 + 𝑧)
 48 Chapter 2 Functions of a Complex Variable

2.2 Limits

Definition 2.2 (The limit of a complex function) A function 𝑤 = 𝑓(𝑧) is

said to have the limit 𝑤0 as 𝑧 approaches a point 𝑧0 , written
lim 𝑓(𝑧) = 𝑤0 ,
𝑧 → 𝑧0

if for every number ε > 0 there is a corresponding number δ > 0

such that for all 𝑧 ≠ 𝑧0 in the disk |𝑧 − 𝑧0 | < δ we have
|𝑓(𝑧) − 𝑤0 | < ε.

The above ε − δ definition of the limit is illustrated in Fig. 2.2.

𝒛 − 𝒑𝒍𝒂𝒏𝒆 𝒘 − 𝒑𝒍𝒂𝒏𝒆
𝑦 𝑣

 𝑤2 ε
δ  𝑧2
 𝑤0
 𝑧0
 𝑧1  𝑤1

|z − 𝑧0 | < δ |w − 𝑤0 | < ε
𝑥 𝑢
 𝑧1

Fig. 2.2 The limit lim 𝑓(𝑧) = 𝑤0

𝑧 → 𝑧0

That is, the values of 𝑓 are “close” to 𝐿 for all 𝑧 “close” to 𝑧0 . Let
𝑤 = 𝑓(𝑧) = 𝑓(𝑥 + 𝑖𝑦) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦),
𝑧0 = 𝑥0 + 𝑖𝑦0 ,
𝑤0 = 𝑢0 + 𝑖𝑣0 = 𝑢(𝑥0 , 𝑦0 ) + 𝑖𝑣(𝑥0 , 𝑦0 )
Then
lim 𝑓(𝑧) = 𝑤0 ,
𝑧 → 𝑧0
if and only if
lim 𝑢(𝑥, 𝑦) = 𝑢0 ,
𝑥 → 𝑥0
lim 𝑣(𝑥, 𝑦) = 𝑣0
𝑥 → 𝑥0
𝑦 → 𝑦0 𝑦 → 𝑦0
 Chapter 2 Functions of a Complex Variable 49

Remark As we had studied in the course of the partial

differentiation, we note that z may approach 𝑧0 in
infinitely many ways as shown in Fig. 2.3 and Fig. 2.4..
However, if a limit exists, it is unique.

𝒚 𝒚
𝒛 𝒛

 𝑧0  𝑧0

𝒙 𝒙
Fig. 2.3. Fig. 2.4.

The following theorem gives the main properties of limits of complex

functions.

Theorem 2.1 If 𝑓(𝑧) and 𝑔(𝑧) are single valued functions, and 𝑎 and
𝑏 are constants, then
 lim [𝑎𝑓(𝑧) ± 𝑏𝑔(𝑧)] = 𝑎 lim 𝑓(𝑧) ± 𝑏 lim 𝑔(𝑧)
𝑧 → 𝑧0 𝑧 → 𝑧0 𝑧 → 𝑧0

 lim 𝑓(𝑧)𝑔(𝑧) = lim 𝑓(𝑧) ∙ lim 𝑔(𝑧)

𝑧 → 𝑧0 𝑧 → 𝑧0 𝑧 → 𝑧0
lim 𝑓(𝑧)
𝑓(𝑧)

𝑧 → 𝑧0
lim = , lim 𝑔(𝑧) ≠ 0
𝑧 → 𝑧0 𝑔(𝑧) lim 𝑔(𝑧) 𝑧 → 𝑧0
𝑧 → 𝑧0

 lim |𝑓(𝑧)| = | lim 𝑓(𝑧)|

𝑧 → 𝑧0 𝑧 → 𝑧0

Example 1 Evaluate each of the following limits

(𝑧 + 3)(𝑧 − 1)
(𝒊) lim (𝑧 2 − 𝑧 + 4), (𝒊𝒊) lim
𝑧 → 1−𝑖 𝑧 → −2𝑖 (𝑧 2 − 2𝑧 + 4)
 50 Chapter 2 Functions of a Complex Variable

Solution
(𝒊) lim (𝑧 2 − 𝑧 + 4) = (1 − 𝑖)2 − (1 − 𝑖) + 4
𝑧 → 1−𝑖

= 1 − 1 − 2𝑖 − 1 + 𝑖 + 4 = 3 − 𝑖

(𝑧 + 3)(𝑧 − 1) 𝑧 lim (𝑧 + 3)(𝑧 − 1)

→ −2𝑖
(𝒊𝒊) lim =
𝑧 → −2𝑖 (𝑧 2 − 2𝑧 + 4) lim (𝑧 2 − 2𝑧 + 4)
𝑧 → −2𝑖

(−2𝑖 + 3)(−2𝑖 − 1) −7 − 4𝑖
= =
(−4 + 4𝑖 + 4) 4𝑖
−7 − 4𝑖 7
= −𝑖 = −1 + 𝑖
4 4

𝑧
𝐄𝐱𝐚𝐦𝐩𝐥𝐞 𝟐 Show that lim does not exist.
𝑧→0 𝑧

Solution
𝑧 𝑥 − 𝑖𝑦
lim = lim
𝑧→0 𝑧 𝑥 → 0 𝑥 + 𝑖𝑦
𝑦→0

To prove that the limit does not exist, we try to choose two different
paths and show that the result of the limit on these paths takes different
values. Let us take first the path 𝑥 = 0 (we move to the origin along
the 𝑦 −axis)
𝑧 𝑥 − 𝑖𝑦
lim = lim = −1,
𝑧→0 𝑧 𝑥= 0 𝑥 + 𝑖𝑦
𝑦→0

and on the path 𝑦 = 0 (we move to the origin along the 𝑥 −axis)
𝑧 𝑥 − 𝑖𝑦
lim = lim = 1,
𝑧→0 𝑧 𝑥 → 0 𝑥 + 𝑖𝑦
𝑦= 0

Therefore, the given limit does not exist.

 Chapter 2 Functions of a Complex Variable 51

2.3. Continuity

Definition 2.3 (A continuous function) A function 𝑤 = 𝑓(𝑧) is said to

be continuous at a point 𝑧 = 𝑧0 if 𝑓(𝑧0 ) is defined and
lim 𝑓(𝑧) = 𝑓(𝑧0 )
𝑧 → 𝑧0

𝑓(𝑧) is said to be 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑖𝑛 𝑎 𝑑𝑜𝑚𝑎𝑖𝑛 if it is continuous at each

point of the domain. Moreover, a function, which is not continuous at
𝑧0 is called discontinuous at 𝑧0 .

Remark Observe that if 𝑓 is continuous at 𝑧0 , if

 𝑓(𝑧0 ) exists,
 lim 𝑓(𝑧) exists,
𝑧 → 𝑧0

 lim 𝑓(𝑧) = 𝑓(𝑧0 )

𝑧 → 𝑧0

Example 1 Discuss the continuity of the following functions at z=0.

|𝑧|2 𝑧
(𝒊) 𝑓(𝑧) = , (𝒊𝒊) 𝑓(𝑧) = { 𝑧 if 𝑧 ≠ 0
𝑧
0 if 𝑧 = 0
|𝑧|2
(𝒊𝒊𝒊) 𝑓(𝑧) = { 𝑧 if 𝑧 ≠ 0
0 if 𝑧 = 0

Solution
|𝑧|2
(𝒊) The function 𝑓(𝑧) = is not defined at 𝑧 = 0 so it is
𝑧

discontinuous at this point.

(𝒊𝒊) For the function
𝑧
𝑓(𝑧) = { 𝑧 if 𝑧 ≠ 0 ,
0 if 𝑧 = 0
 52 Chapter 2 Functions of a Complex Variable

we have 𝑓(0) = 0, but the limit of this function does not exist so it is
discontinuous at 𝑧 = 0.
(𝒊𝒊𝒊) The third function is defined at 𝑧 = 0 and 𝑓(0) = 0. The limit of
this function at 𝑧 = 0 is given by
|𝑧|2 𝑧𝑧
lim 𝑓(𝑧) = lim = lim = lim 𝑧 = 0
𝑧→0 𝑧→0 𝑧 𝑧→0 𝑧 𝑧→0

Thus we have
lim 𝑓(𝑧) = 𝑓(0),
𝑧→0

which implies that

|𝑧|2
𝑓(𝑧) = { 𝑧 if 𝑧 ≠ 0 ,
0 if 𝑧 = 0
is continuous at 𝑧 = 0.

2.4. Derivative of complex functions

We are familiar with derivatives of real valued functions. For complex

functions, the definition is formally the same as the definition for
derivatives of real functions as given in the following definition.

Definition 2.4 (Derivative of a complex function) Let 𝑤 = 𝑓(𝑧) be a

complex function and 𝑧0 ∈ 𝐷. If the limit of the quotient
𝑓(𝑧0 + ∆𝑧) − 𝑓(𝑧0 )
,
∆𝑧
exists as ∆𝑧 → 0, then 𝑓 is said to be 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑏𝑙𝑒 at 𝑧0 , and the
limit is called the derivative of the function 𝑓 at 𝑧0 and is denoted
by 𝑓 ′ (𝑧0 ), i.e.,
𝑓(𝑧0 + ∆𝑧) − 𝑓(𝑧0 )
𝑓 ′ (𝑧0 ) = lim
∆𝑧 → 0 ∆𝑧
 Chapter 2 Functions of a Complex Variable 53

As in real variables, 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑡𝑦: If 𝑓(𝑧)

is differentiable at 𝑧0 , then 𝑓 is continuous at 𝑧0 . Moreover, the rules of
differentiation are the same as in the calculus of real variables. If 𝑓 and
𝑔 are differentiable at a point 𝑧 and 𝑘 is a complex constant, then
𝑑
 𝑘 = 0,
𝑑𝑧
𝑑
 𝑘𝑓(𝑧) = 𝑘𝑓 ′ (𝑧),
𝑑𝑧
𝑑
 [𝑓(𝑧) ± 𝑔(𝑧)] = 𝑓 ′ (𝑧) ± 𝑔′ (𝑧),
𝑑𝑧
𝑑
 [𝑓(𝑧)𝑔(𝑧)] = 𝑓(𝑧)𝑔′ (𝑧) ± 𝑓 ′ (𝑧)𝑔(𝑧),
𝑑𝑧
𝑑 𝑓(𝑧) 𝑔(𝑧)𝑓 ′ (𝑧)−𝑓(𝑧)𝑔′ (𝑧)
 ( )= , 𝑔(𝑧) ≠ 0
𝑑𝑧 𝑔(𝑧) 𝑔2 (𝑧)
𝑑
 𝑓(𝑔(𝑧)) = 𝑓 ′ (𝑔(𝑧))𝑔′ (𝑧)
𝑑𝑧

Example 1 Find the derivative of the complex function 𝑓(𝑧) = 𝑧 2 .

Solution

′ (𝑧)
𝑓(𝑧 + ∆𝑧) − 𝑓(𝑧) (𝑧 + ∆𝑧)2 − 𝑧 2
𝑓 = lim = lim
∆𝑧 → 0 ∆𝑧 ∆𝑧 → 0 ∆𝑧
2
2𝑧∆𝑧 + (∆𝑧)
= lim = lim (2𝑧 + ∆𝑧) = 2𝑧
∆𝑧 → 0 ∆𝑧 ∆𝑧 → 0

𝑑
Example 2 Show that 𝑑𝑧
𝑧 does not exist anywhere.

Solution
𝑓(𝑧 + ∆𝑧) − 𝑓(𝑧) (𝑧 + ∆𝑧) − 𝑧
𝑓 ′ (𝑧) = lim = lim
∆𝑧 → 0 ∆𝑧 ∆𝑧 → 0 ∆𝑧
∆𝑧 ∆𝑥 − 𝑖∆𝑦
= lim = lim
∆𝑧 → 0 ∆𝑧 𝑥 → 0 ∆𝑥 + 𝑖∆𝑦
𝑦= 0
 54 Chapter 2 Functions of a Complex Variable

(𝒂) If ∆𝑦 = 0, the required limit equals 1,

(𝒃) If ∆𝑥 = 0, the required limit equals – 1.
Hence the limit depends on the manner in which ∆𝑧 → 0, therefore the
derivative does not exist.

Definition 2.5 (Regular functions) A function 𝑓(𝑧) is said to be regular

in any domain 𝐷 if the function is single-valued and differentiable at
every point of the domain 𝐷.

Definition 2.6 (Analytic functions) A function 𝑓(𝑧) is said to be analytic

at a point 𝑧0 ∈ 𝐶 if it is defined and differentiable not only at 𝑧0 but
at every point in some neighbourhood of 𝑧0 .

A function 𝑓 is said to be analytic in a domain 𝐷  𝐶 if 𝑓(𝑧) is analytic

at all points of 𝐷. The term holomorphic function is sometimes used
interchangeably with "analytic function". Many mathematicians prefer
the term holomorphic function to "analytic function", while "analytic"
appears to be in widespread use among physicists, engineers, and in
some older texts.

Definition 2.7 (Entire functions) An entire function is one that is analytic

at every point of the 𝑧 −plane, that is, throughout the entire plane.

Remark Analyticity at a point is a neighborhood property.

Therefore, analyticity at a point is not the same as
differentiability at a point. However, analyticity implies
differentiability but not vice versa.
 Chapter 2 Functions of a Complex Variable 55

Exercise 2.1
𝟏. Suppose 𝑓(𝑧) = 3𝑥𝑦 + 𝑖(𝑥 − 𝑦)2. Find lim 𝑓(𝑧).
𝑧 → 3+2𝑖

𝟐. Prove that if 𝑓 has a derivative at 𝑧, then 𝑓 is continuous at 𝑧.

𝟑. Find all points at which 𝑓(𝑧) = 𝑧𝑧 has a derivative.
𝟒. Find all points at which the function

𝑓(𝑧) = (2 + 𝑖)𝑧 3 − 𝑖𝑧 2 + 4𝑧 − (1 + 7𝑖),

has a derivative.
(𝑧)2
𝟓. Is the function 𝑓(𝑧) = { if 𝑧 ≠ 0 differentiable at 𝑧 = 0?
𝑧
0 if 𝑧 = 0
Explain.
2.5 Cauchy-Riemann equations
In the preceding section we saw that a function 𝑓 of a complex variable
𝑧 is analytic in a domain 𝐷 if 𝑓(𝑧) is differentiable at all points in 𝐷.
We shall now derive a very important test for the analyticity of a
complex function 𝑓 based on the partial derivatives of its real and
imaginary parts.

Theorem 2.2 For a function 𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦), defined on a

domain 𝐷 to be differentiable at a point 𝑧 ∈ 𝐷 as a function of a
complex variable, it is necessary and sufficient that:
 The functions 𝑢(𝑥, 𝑦) and 𝑣(𝑥, 𝑦) be differentiable at the same
point (as functions of two real variables) and;
 Their partial derivatives satisfy the Cauchy-Riemann Equations:
𝑢𝑥 = 𝑣𝑦 , 𝑢𝑦 = −𝑣𝑥 (1)
′ (𝑧)
If the conditions of the theorem are fulfilled, the derivative 𝑓 may
be found from the following form:
𝑑𝑓(𝑧)
𝑓 ′ (𝑧) = = 𝑢𝑥 + 𝑖𝑣𝑥 = 𝑣𝑦 − 𝑖𝑢𝑦 (2)
𝑑𝑧
 56 Chapter 2 Functions of a Complex Variable

Proof. By assumption, the derivative 𝑓 ′ (𝑧) at 𝑧 exists. It is given by

𝑓(𝑧 + ∆𝑧) − 𝑓(𝑧)
𝑓 ′ (𝑧) = lim (3)
∆𝑧 → 0 ∆𝑧
Since ∆𝑧 = ∆𝑥 + 𝑖∆𝑦, then we can write (3) in terms of 𝑢 and 𝑣 as
𝑓 ′ (𝑧) =
[𝑢(𝑥 + ∆𝑥, 𝑦 + ∆𝑦) + 𝑖𝑣(𝑥 + ∆𝑥, 𝑦 + ∆𝑦)] − [𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦)]
= lim
∆𝑥 → 0 ∆𝑥 + 𝑖∆𝑦
∆𝑦 →0

[𝑢(𝑥 + ∆𝑥, 𝑦 + ∆𝑦) − 𝑢(𝑥, 𝑦)] + 𝑖[𝑣(𝑥 + ∆𝑥, 𝑦 + ∆𝑦) − 𝑣(𝑥, 𝑦)]
= lim
∆𝑥 → 0 ∆𝑥 + 𝑖∆𝑦
∆𝑦 →0

 Case 1 when ∆𝑦 = 0, and ∆𝑥 → 0, we get

[𝑢(𝑥 + ∆𝑥, 𝑦) − 𝑢(𝑥, 𝑦)] + 𝑖[𝑣(𝑥 + ∆𝑥, 𝑦) − 𝑣(𝑥, 𝑦)]
𝑓 ′ (𝑧) = lim
∆𝑥 → 0 ∆𝑥
∆𝑦=0

𝜕𝑢 𝜕𝑣
= +𝑖 (4)
𝜕𝑥 𝜕𝑥
 Case 2 when ∆𝑥 = 0, and ∆𝑦 → 0, we get
[𝑢(𝑥, 𝑦 + ∆𝑦) − 𝑢(𝑥, 𝑦)] + 𝑖[𝑣(𝑥, 𝑦 + ∆𝑦) − 𝑣(𝑥, 𝑦)]
𝑓 ′ (𝑧) = lim
∆𝑥= 0 𝑖∆𝑦
∆𝑦→0

𝜕𝑣 𝜕𝑢
= −𝑖 (5)
𝜕𝑦 𝜕𝑦
The existence of the derivative 𝑓 ′ (𝑧) implies the existence of the four
partial derivatives in (4) and (5). By equating the real and imaginary
parts in (4) and (5) we get the Cauchy-Riemann equations (1).
Furthermore, from these equations we obtain the required assertion in
(2).

Remark Cauchy-Riemann equations are fundamental because

they are 𝑛𝑜𝑡 𝑜𝑛𝑙𝑦 𝑛𝑒𝑐𝑒𝑠𝑠𝑎𝑟𝑦 but also 𝑠𝑢𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 for a
function to be analytic. However, when we add the
 Chapter 2 Functions of a Complex Variable 57

condition of continuity to 𝑢 and 𝑣 and the four partial

derivatives, the Cauchy-Riemann equations can be
shown to imply analyticity.

𝑑
Example 1 Show that (𝑅𝑒 𝑧) does not exist anywhere.
𝑑𝑧

Solution Since
𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦) = 𝑅𝑒 𝑧 = 𝑥,
then
𝑢(𝑥, 𝑦) = 𝑥, 𝑣(𝑥, 𝑦) = 0,
and this implies that
𝑢𝑥 = 1, 𝑢𝑦 = 0, 𝑣𝑥 = 𝑣𝑦 = 0,
and the Cauchy-Riemann equations are not satisfied anywhere.
Therefore, the given function is not differentiable.

Example 2 Show that the function 𝑓(𝑧) = |𝑧|2 is differentiable but not
analytic at the point 𝑧 = 0.

Solution
𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦) = 𝑥 2 + 𝑦 2 ,
so that
𝑢(𝑥, 𝑦) = 𝑥 2 + 𝑦 2 , 𝑣(𝑥, 𝑦) = 0
Therefore, we have
𝑢𝑥 = 2𝑥, 𝑢𝑦 = 2𝑦, 𝑣𝑥 = 𝑣𝑦 = 0,
 For 𝑧 ≠ 0, these derivatives are continuous but the Cauchy-Riemann
Equations are not satisfied, therefore our function is not differentiable
for all 𝑧 ≠ 0.
 58 Chapter 2 Functions of a Complex Variable

 At the point 𝑧 = 0, Cauchy-Riemann Equations are satisfied which

mean that the function 𝑓(𝑧) = |𝑧|2 is differentiable at this point, but
there is no neighbourhood of 𝑧 = 0 through which the function and its
derivatives exist. Therefore the function is not analytic at the point 𝑧 =
0.

Example 3 Is 𝑓(𝑧) = 𝑧 2 analytic?

Solution.
𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦) = (𝑥 + 𝑖𝑦) = 𝑥 2 − 𝑦 2 + 2𝑖𝑥𝑦,
Hence,
𝑢(𝑥, 𝑦) = 𝑥 2 − 𝑦 2 , 𝑣(𝑥, 𝑦) = 2𝑥𝑦
Therefore,
𝑢𝑥 = 2𝑥, 𝑢𝑦 = −2𝑦, 𝑣𝑥 = 2𝑦, 𝑣𝑦 = 2𝑥
The partial derivatives satisfy the Cauchy-Riemann equations for every
𝑧. Hence 𝑓(𝑧) = 𝑧 2 is analytic. Moreover,
𝑑𝑓(𝑧)
𝑓 ′ (𝑧) = = 𝑢𝑥 + 𝑖𝑣𝑥 = 2𝑥 + 𝑖2𝑦 = 2(𝑥 + 𝑖𝑦) = 2𝑧
𝑑𝑧

Example 4 Find the points at which the function

𝑓(𝑧) = 𝑥 3 − 𝑖(1 − 𝑦)3 ,
is differentiable.

Solution Here we have

𝑢(𝑥, 𝑦) = 𝑥 3 , 𝑣(𝑥, 𝑦) = −(1 − 𝑦)3 (6)
Differentiating equations (6) with respect to 𝑥 and 𝑦, respectively we
get
 Chapter 2 Functions of a Complex Variable 59

𝑢𝑥 = 3𝑥 2 , 𝑣𝑥 = 0, (7)
𝑢𝑦 = 0, 𝑣𝑦 = 3(1 − 𝑦)2 (8)
The Cauchy-Riemann equations yield
3𝑥 2 = 3(1 − 𝑦)2 ,
or
𝑥 = ±(1 − 𝑦) 𝑦
 𝑥 = (1 − 𝑦), 𝑥 = −(1 − 𝑦)
That is to say the given function is
differentiable on the set of all points
𝑥
on the cross formed by the two
straight lines shown in Fig. 2.5. Fig. 2.5

2.6 Cauchy-Riemann equations in polar form

In many cases it is necessary to have a formula for differentiability at a
point 𝑧 ≠ 0 of a function of a complex variable expressed in terms of
the polar coordinates 𝑧 = 𝑟(cos θ + 𝑖 sin θ).

Theorem 1.3 The necessary and sufficient conditions of

differentiability of 𝑓(𝑧) = 𝑢(𝑟, θ) + 𝑖𝑣(𝑟, θ) are:
 𝑢 and 𝑣 are differentiable functions of 𝑟 and θ.
 Their partial derivatives are related by the Cauchy-Riemann
Equations in its polar form:
1 1
𝑢𝑟 = 𝑣θ , 𝑣𝑟 = − 𝑢θ (1)
𝑟 𝑟
′ (𝑧)
Moreover, the derivative 𝑓 is expressed in the form:
𝑓 ′ (𝑧) = (𝑢𝑟 + 𝑖𝑣𝑟 )(cos θ − 𝑖 sin θ) = (𝑢𝑟 + 𝑖𝑣𝑟 )𝑒 −𝑖θ (2)

Proof. From the well known relations

 60 Chapter 2 Functions of a Complex Variable

𝑥 = 𝑟 cos θ , 𝑦 = 𝑟 sin θ,
𝑦
𝑟 = √𝑥 2 + 𝑦 2 , θ = tan−1 ( ),
𝑥
we get
𝜕𝑟 𝑥 𝜕θ −𝑦 sin θ
= = cos θ, = 2 2
=−
𝜕𝑥 𝑟 𝜕𝑥 𝑥 + 𝑦 𝑟
𝜕𝑟 𝑦 𝜕θ 𝑥 cos θ
= = sin θ, = 2 =
𝜕𝑦 𝑟 𝜕𝑦 𝑥 + 𝑦 2 𝑟
Using the Chain Rule to obtain the following partial derivatives of
𝑢(𝑟, θ) and 𝑣(𝑟, θ).

𝑥
𝜕𝑟 𝜕𝑟
𝑟 = cos θ, = sin θ
𝑦 𝜕𝑥 𝜕𝑦
𝑢, 𝑣
𝑥 𝜕θ sin θ 𝜕θ cos θ
θ =− , =
𝑦 𝜕𝑥 𝑟 𝜕𝑦 𝑟

𝜕𝑢 𝜕𝑟 𝜕θ sin θ
= 𝑢𝑟 + 𝑢θ = cos θ 𝑢𝑟 − 𝑢θ (3)
𝜕𝑥 𝜕𝑥 𝜕𝑥 𝑟
𝜕𝑣 𝜕𝑟 𝜕θ cos θ
= 𝑣𝑟 + 𝑣θ = sin θ 𝑣𝑟 + 𝑣θ (4)
𝜕𝑦 𝜕𝑦 𝜕𝑦 𝑟
But we have from the Cauchy-Riemann equations in the Cartesian:
𝑢𝑥 = 𝑣𝑦
Substitute from (3) and (4) we find
sin θ cos θ
cos θ 𝑢𝑟 − 𝑢θ = sin θ 𝑣𝑟 + 𝑣θ (5)
𝑟 𝑟
This equation is valid for all θ so that
1
θ=0  𝑢𝑟 = 𝑣θ ,
𝑟
π 1
θ=  − 𝑢θ = 𝑣𝑟 ,
2 𝑟
 Chapter 2 Functions of a Complex Variable 61

which are the Cauchy-Riemann Equations in its polar form given by

(1). To get (2) we need the partial derivative
𝜕𝑣 𝜕𝑟 𝜕θ sin θ
= 𝑣𝑟 + 𝑣θ = cos θ 𝑣𝑟 − 𝑣 , (6)
𝜕𝑥 𝜕𝑥 𝜕𝑥 𝑟 θ
Therefore from (3) and (6)
𝑑𝑓(𝑧) 𝜕𝑢 𝜕𝑣
𝑓 ′ (𝑧) = = +𝑖
𝑑𝑧 𝜕𝑥 𝜕𝑥
sin θ sin θ
= (cos θ 𝑢𝑟 − 𝑢θ ) + 𝑖 (cos θ 𝑣𝑟 − 𝑣 )
𝑟 𝑟 θ
𝑢θ 𝑣θ
= cos θ (𝑢𝑟 + 𝑖𝑣𝑟 ) − sin θ (( ) + 𝑖 ( ))
𝑟 𝑟
Substitute from equation (1), we get
𝑓 ′ (𝑧) = cos θ (𝑢𝑟 + 𝑖𝑣𝑟 ) − sin θ (−𝑣r + 𝑖𝑢𝑟 )
= cos θ (𝑢𝑟 + 𝑖𝑣𝑟 ) − 𝑖 sin θ (𝑢𝑟 + 𝑖𝑣r )
= (𝑢𝑟 + 𝑖𝑣𝑟 )(cos θ − 𝑖 sin θ)
= (𝑢𝑟 + 𝑖𝑣𝑟 )𝑒 −𝑖θ
This completes the proof of the theorem. ∎

𝑑
Example 1 Prove that for all 𝑛, 𝑑𝑧 𝑧 𝑛 = 𝑛𝑧 𝑛−1.

Solution The given function is written in terms of 𝑟 and θ in the form

𝑛
𝑓(𝑧) = 𝑢(𝑟, θ) + 𝑖𝑣(𝑟, θ) = 𝑧 𝑛 = (𝑟𝑒 𝑖θ ) = 𝑟 𝑛 (cos 𝑛θ + 𝑖 sin 𝑛θ)
Hence,
𝑢(𝑟, θ) = 𝑟 𝑛 cos 𝑛θ , 𝑣(𝑟, θ) = 𝑟 𝑛 sin 𝑛θ
Differentiating these equations with respect to 𝑟 and θ, we get
𝑢𝑟 = 𝑛𝑟 𝑛−1 cos 𝑛θ, 𝑢θ = −𝑛𝑟 𝑛 sin 𝑛θ,
𝑣𝑟 = 𝑛𝑟 𝑛−1 sin 𝑛θ, 𝑣θ = 𝑛𝑟 𝑛 cos 𝑛θ,
 62 Chapter 2 Functions of a Complex Variable

which shows that Cauchy-Riemann equations are satisfied. Then the

substitution in (2) yields
𝑓 ′ (𝑧) = (𝑢𝑟 + 𝑖𝑣𝑟 )𝑒 −𝑖θ
= [(𝑛𝑟 𝑛−1 cos 𝑛θ) + 𝑖(𝑛𝑟 𝑛−1 sin 𝑛θ)]𝑒 −𝑖θ
= 𝑛𝑟 𝑛−1 (cos 𝑛θ + 𝑖 sin 𝑛θ)𝑒 −𝑖θ
= 𝑛𝑟 𝑛−1 𝑒 𝑖𝑛θ 𝑒 −𝑖θ = 𝑛𝑟 𝑛−1 𝑒 𝑖(𝑛−1)θ
𝑛−1
= 𝑛(𝑟𝑒 𝑖θ ) = 𝑛𝑧 𝑛−1

2.7 Harmonic functions

One of the main reasons for the great practical importance of complex
analysis in engineering mathematics results from the fact that both the
real and imaginary parts of an analytic function satisfy the most
important differential equation of physics, Laplace equation, which
occurs in gravitation, electrostatics, heat conduction and fluid flow.

Theorem 1.4 If 𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦) is analytic in some domain

of the complex plane, then 𝑢(𝑥, 𝑦) and 𝑣(𝑥, 𝑦) satisfy Laplace’s
equation
∇2 𝑢 = 𝑢𝑥𝑥 + 𝑢𝑦𝑦 = 0, (1)
and
∇2 𝑣 = 𝑣𝑥𝑥 + 𝑣𝑦𝑦 = 0, (2)

Proof. Since 𝑓(𝑧) is analytic, then the real and imaginary parts satisfy
the Cauchy-Riemann equations
𝑢𝑥 = 𝑣𝑦 , 𝑢𝑦 = −𝑣𝑥 (3)
Differentiating the first part of this equation with respect to 𝑥 and the
second part with respect to 𝑦 we obtain (provided the second derivatives
exist)
𝑢𝑥𝑥 = 𝑣𝑦𝑥 , 𝑢𝑦𝑦 = −𝑣𝑥𝑦 ,
 Chapter 2 Functions of a Complex Variable 63

which implies that the real part 𝑢(𝑥, 𝑦) satisfies Laplace’s equation.
Similarly, equation (2) is obtained by differentiating the first part of
equation (3) with respect to 𝑦 and differentiating the second part with
respect to 𝑥 and subtracting the result equations.

Definition 2.7 (Harmonic functions) Any function that has continuous

partial derivatives of the second order and satisfies Laplace’s
equation is called a ℎ𝑎𝑟𝑚𝑜𝑛𝑖𝑐 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛. Hence the real and
imaginary parts of an analytic function are harmonic functions.

If two harmonic functions 𝑢 and 𝑣 satisfy the Cauchy-Riemann

equations, then they are the real and imaginary parts of an analytic
function 𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦). In this case the functions 𝑢 and 𝑣
are called 𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 ℎ𝑎𝑟𝑚𝑜𝑛𝑖𝑐 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠. (Of course, this has
nothing to do with the use of “conjugate” for 𝑧). Given one of two
conjugate harmonic functions, the Cauchy-Riemann equations can be
used to find the other.

Example 1 Verify that 𝑢 = 𝑥 2 − 𝑦 2 − 𝑦 is harmonic in the whole

complex plane and find a conjugate harmonic 𝑣 of 𝑢.

Solution By direct calculation

𝑢𝑥 = 2𝑥, 𝑢𝑥𝑥 = 2
𝑢𝑦 = −2𝑦 − 1, 𝑢𝑦𝑦 = −2,
therefore
∇2 𝑢 = 𝑢𝑥𝑥 + 𝑢𝑦𝑦 = 0,
and so 𝑢 is a harmonic function. From Cauchy-Riemann equations
𝑢𝑥 = 𝑣𝑦  𝑣𝑦 = 2𝑥, 𝑣𝑥 = −𝑢𝑦 = 2𝑦 + 1 (4)
Thus we have
 64 Chapter 2 Functions of a Complex Variable

𝑑𝑣 = 𝑣𝑥 𝑑𝑥 + 𝑣𝑦 𝑑𝑦 = (2𝑦 + 1)𝑑𝑥 + (2𝑥)𝑑𝑦

We can get 𝑣 by one of the following two methods:

 First method:
𝑥 𝑦
𝑣 = ∫ 𝑣𝑥 (𝑥, 𝑦)𝑑𝑥 + ∫ 𝑣𝑦 (0, 𝑦)𝑑𝑦
0 0
𝑥 𝑦
= ∫ (2𝑦 + 1)𝑑𝑥 + ∫ (0)𝑑𝑦 = 2𝑥𝑦 + 𝑥 + 𝑐
0 0

 Second method:

Integrate both sides of the two parts of equation (4) without limits and
take the repeated terms in the result once. Since from (4)
𝑣𝑥 = 2𝑦 + 1  𝑣 = 2𝑥𝑦 + 𝑥 + 𝑐1 (5)
Also
𝑣𝑦 = 2𝑥  𝑣 = 2𝑥𝑦 + 𝑐2 (6)
Adding (5) and (6) and taking the repeated terms once, we obtain
𝑣 = 2𝑥𝑦 + 𝑥 + 𝑐
Therefore, the analytic function is

𝑤 = 𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦)

= (𝑥 2 − 𝑦 2 − 𝑦) + 𝑖(2𝑥𝑦 + 𝑥 + 𝑐)

= (𝑥 2 − 𝑦 2 + 𝑖2𝑥𝑦) + 𝑖(𝑥 + 𝑖𝑦 + 𝑐)

= 𝑧 2 + 𝑖𝑧 + 𝑖𝑐

Remark 1 To express 𝑤 as a function of 𝑧, as often required, in

addition to the method used in the above example, we
may use one of the following two methods:
 Chapter 2 Functions of a Complex Variable 65

 First method:
𝑤 = 𝑓(𝑧) = 𝑓(𝑥 + 𝑖𝑦) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦)
Put 𝑦 = 0,
𝑓(𝑥) = 𝑢(𝑥, 0) + 𝑖𝑣(𝑥, 0)
Replacing 𝑥 by 𝑧, we get the formula
𝑓(𝑧) = 𝑢(𝑧, 0) + 𝑖𝑣(𝑧, 0)
That is to say, in both of the real and imaginary parts we replace 𝑥 by 𝑧
and 𝑦 by 0. On doing this in Example 1, we get
𝑤 = (𝑥 2 − 𝑦 2 − 𝑦) + 𝑖(2𝑥𝑦 + 𝑥 + 𝑐)

= (𝑧 2 − 0 − 0) + 𝑖(0 + 𝑧 + 𝑐)
= 𝑧 2 + 𝑖𝑧 + 𝑖𝑐
 Second method:
In the real and imaginary parts of 𝑤, we replace 𝑥 and 𝑦 by
𝑧+𝑧 𝑧−𝑧
𝑥= , 𝑦= ,
2 2𝑖
and then simplify the resulting expression of 𝑤. Thus, in Example 1
we find
2 2
𝑧+𝑧 𝑧−𝑧 𝑧−𝑧
𝑤 = [( ) −( ) −( )]
2 2𝑖 2𝑖
𝑧+𝑧 𝑧−𝑧 𝑧+𝑧
+𝑖 [2 ( )( )+( ) + 𝑐],
2 2𝑖 2
2 2
𝑧 2 + 𝑧 + 2𝑧𝑧 𝑧 2 + 𝑧 − 2𝑧𝑧 𝑧−𝑧
=( )−( )+𝑖( )
4 −4 2
2
𝑧2 − 𝑧 𝑧+𝑧
+( )+𝑖( ) + 𝑖𝑐
2 2
2 2
𝑧2 + 𝑧 2𝑧 𝑧2 − 𝑧
=( )+𝑖( )+( ) + 𝑖𝑐
2 2 2
= 𝑧 2 + 𝑖𝑧 + 𝑖𝑐
 66 Chapter 2 Functions of a Complex Variable

Remark 2 Suppose that 𝑢 and 𝑣 are the real and imaginary parts of
an analytic function, then they are harmonic functions.
The level curves 𝑢(𝑥, 𝑦) = 𝑐1 and 𝑣(𝑥, 𝑦) = 𝑐2 defined
by these functions form two orthogonal families of
curves. For example, the level curves generated by the
simple analytic function 𝑓(𝑧) = 𝑧 = 𝑥 + 𝑖𝑦 are 𝑥 = 𝑐1
and 𝑦 = 𝑐2 . The family of vertical lines defined by 𝑥 =
𝑐1 is clearly orthogonal to the family of horizontal lines
defined by 𝑦 = 𝑐2 .

Exercise 2.2

Are the functions in problems 1 − 4 analytic?

𝟏. 𝑓(𝑧) = Re (𝑧 2 ) 𝟐. 𝑓(𝑧) = 𝑧 − 𝑧
𝑦 𝑥 2
𝟑. 𝑓(𝑧) = 2 2
+𝑖 2 𝟒. 𝑓(𝑧) = 𝑧
𝑥 +𝑦 𝑥 + 𝑦2
𝟓. Find all points at which 𝑓(𝑧) = 2𝑦 − 𝑖𝑥 is differentiable.
𝑦 𝑦
𝟔. Find all points at which 𝑓(𝑧) = 𝑥 2 +𝑦 2 − 𝑖 𝑥 2 +𝑦 2 is differentiable. At

what points is 𝑓 analytic? Explain.

𝟕. If 𝑓(𝑧) = 𝑒 𝑥 (cos 𝑦 + 𝑖 sin 𝑦), show that 𝑓 ′ (𝑧) exists everywhere.
Find 𝑓 ′ (𝑧).
𝟖. (𝒂) Prove that 𝑒 −𝑥 (𝑥sin 𝑦 − 𝑦 cos 𝑦) is harmonic.
(𝒃) Find v such that 𝑓(𝑧) = 𝑢 + 𝑖𝑣 will be analytic function.
In problems 9 − 12, are the given functions harmonic? If so, find the
conjugate harmonic function.
𝟗. 3𝑥 2 𝑦 + 2𝑥 2 − 𝑦 3 − 2𝑦 2 𝟏𝟎. 𝑒 −𝑥 cos 2𝑦
𝟏𝟏. 𝑒 𝑥 (𝑥 cos 𝑦 − 𝑦 sin 𝑦) 𝟏𝟐. ln(𝑥 2 + 𝑦 2 )
 Chapter 2 Functions of a Complex Variable 67

In problems 13 − 14, find the derivative of the following functions at

the indicated points
(𝑧 + 2𝑖)(𝑖 − 𝑧)
𝟏𝟑. ,𝑧 = 𝑖 𝟏𝟒. [𝑧 + (𝑧 2 + 1)2 ]2 , 𝑧 = 1 + 𝑖
(2𝑧 − 1)
In problems 15 − 16, show that each of the following functions is
everywhere continuous but nowhere analytic
𝟏𝟓. 𝑓(𝑧) = (𝑥 3 + 𝑥𝑦 2 ) − 𝑖(𝑥 2 𝑦 + 𝑦 3 ) 1𝟔. 𝑓(𝑧) = 3𝑥 − 𝑖𝑦
In problems 17 − 19, prove that each of the following functions
𝟏𝟕. 𝑣(𝑥, 𝑦) = 𝑥 3 − 3𝑥𝑦 2 𝟏𝟖. 𝑢(𝑥, 𝑦) = 𝑥𝑦
𝟏𝟗. 𝑢(𝑥, 𝑦) = 𝑒 −𝑥 (𝑥 sin 𝑦 − 𝑦 cos 𝑦)
is harmonic. Find its harmonic conjugate then express the analytic
function 𝑓(𝑧) = 𝑢 + 𝑖𝑣 as a function of 𝑧.
𝟐𝟎. Starting from the Cauchy-Riemann equations in the Cartesian form,
prove that the Cauchy-Riemann equations is expressed in the complex
form
𝜕𝑓(𝑧, 𝑧) 𝜕𝑓(𝑧, 𝑧)
= 0, 𝑓 ′ (𝑧) =
𝜕𝑧 𝜕𝑧
3. Elementary Functions

In this chapter, we shall examine the exponential, logarithmic,

trigonometric, and hyperbolic functions of a complex variable 𝑧.
Although the definitions of these complex functions are motivated by
their real variable analogues, the properties of these complex functions
will yield some surprises.

3.1. The complex exponential function

We define the exponential function of complex variable by
𝑒 𝑧 = 𝑒 (𝑥+𝑖𝑦) = 𝑒 𝑥 𝑒 𝑖𝑦 = 𝑒 𝑥 (cos 𝑦 + 𝑖 sin 𝑦) (1)
Properties of 𝒆𝒛
(𝒂) In case 𝑦 = 0, 𝑒 𝑧 reduces to the real exponential function 𝑒 𝑥
(𝒃) In case 𝑥 = 0, from equation (1) 𝑒 𝑧 becomes
𝑒 𝑖𝑦 = cos 𝑦 + 𝑖 sin 𝑦 , (2)
which is Euler formula.
(𝒄) 𝑒 𝑧 is an entire function and
𝑑 𝑧
(𝑒 ) = 𝑒 𝑧 (3)
𝑑𝑧
(𝒅) Writing 𝑒 𝑧 in its polar form, equation (1) becomes
𝜌𝑒 𝑖φ = 𝑒 𝑥 𝑒 𝑖𝑦 ,
and this yields
|𝑒 𝑧 | = 𝑒 𝑥 , (4)
arg 𝑒 𝑧 = 𝑦 + 2𝑘𝜋, (𝑘 = 0, ±1, ±2, ⋯ ) (5)
(𝒆) 𝑒 𝑧 ≠ 0 for any complex number 𝑧, since 𝑒 𝑥 is never zero, and the
cosine and sine functions are never zero at the same point.
𝑒 𝑧1
(𝒇) 𝑒 𝑧1 𝑒 𝑧2 = 𝑒 𝑧1 +𝑧2 , 𝑒 𝑧2
= 𝑒 𝑧1 −𝑧2 ,

(𝒈) 𝑒 0 = 1,
 Chapter 3 Elementary Functions 69

(𝒉) 𝑒 𝑧 = 𝑒 𝑧 ,
(𝒊) 𝑒 𝑧 is a 𝑝𝑒𝑟𝑖𝑜𝑑𝑖𝑐 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ 𝑝𝑒𝑟𝑖𝑜𝑑 2𝜋𝑖. The periodicity of
cos 𝑦 and sin 𝑦 yields
𝑒 𝑧 = 𝑒 𝑥 (cos 𝑦 + 𝑖 sin 𝑦) = 𝑒 𝑥 [cos(𝑦 + 2𝑘𝜋) + 𝑖 sin(𝑦 + 2𝑘𝜋)]
= 𝑒 𝑥 𝑒 𝑖(𝑦+2𝜋) = 𝑒 (𝑥+𝑖𝑦)+2𝜋𝑖 = 𝑒 𝑧+2𝜋𝑖
That is,
𝑒 𝑧 = 𝑒 𝑧+2𝜋𝑖 for all 𝑧
Because of this complex
𝑦
periodicity, all possible functional
𝑦=𝜋
values of 𝑓(𝑧) = 𝑒 𝑧 are assumed
in any infinite horizontal strip of 𝑥
width 2𝜋. The strip −𝜋 < 𝑦 < 𝜋
𝑦 = −𝜋
is called the fundamental region
Fig. 3.1 fundamental region of 𝑒 𝑧
for the exponential 𝑓(𝑧) = 𝑒 𝑧 .

Example 1 Solve the equation 𝑒 𝑧 = 1 − √3𝑖.

𝑦
Solution with the aid of Fig. 3.2, we have
1 − √3𝑖 = 2𝑒 −𝑖𝜋⁄3 1
𝝅⁄ 𝟑 𝑥
Thus the given equation is written as
√3
𝑒 = 1 − √3𝑖  𝑒 𝑒
𝑧 𝑥 𝑖𝑦
= 2𝑒 −𝑖𝜋⁄3
,
 1 − √3𝑖
from which we get Fig. 3.2 𝑧 = 1 − √3𝑖
𝑒 = 2  𝑥 = ln 2 , 𝑦 = − 𝜋⁄3
𝑥

So we have the solution of our problem as

𝜋
𝑧 = 𝑥 + 𝑖𝑦 = ln 2 + 𝑖 (− + 2𝑘𝜋) , 𝑘 = 0, ±1, ±2, ⋯
3
 70 Chapter 3 Elementary Functions

𝜋
Example 2 Compute 𝑒 𝑧 and |𝑒 𝑧 | if 𝑧 = 2 − 𝑖 6 .

Solution
𝜋 𝜋 𝜋 𝜋
𝑒 𝑧 = 𝑒 (2−𝑖 6 ) = 𝑒 2 𝑒 −𝑖 6 = 𝑒 2 (cos (− ) + 𝑖 sin (− ))
6 6
√3 1 𝑒2
= 𝑒 2 ( − 𝑖 ) = (√3 − 𝑖),
2 2 2
and
|𝑒 𝑧 | = 𝑒 2

3.2. The complex logarithmic function

The 𝑐𝑜𝑚𝑝𝑙𝑒𝑥 𝑛𝑎𝑡𝑢𝑟𝑎𝑙 𝑙𝑜𝑔𝑎𝑟𝑖𝑡ℎ𝑚 of a complex variable 𝑧,
where 𝑧 = 𝑟𝑒 𝑖θ is defined as the inverse of the exponential function-
that is,
𝑤 = ln 𝑧 if 𝑧 = 𝑒𝑤
To find the real and imaginary parts of 𝑤 = ln 𝑧, we write
ln 𝑧 = ln(𝑟𝑒 𝑖θ ) = ln 𝑟 + 𝑖θ (1)
From the periodicity of the exponential function 𝑒 𝑖θ we may write
𝑧 = 𝑟𝑒 𝑖(θ+2kπ) , 𝑘 = 0, ±1, ±2, ⋯ (2)
Thus formula (1) can be written
ln 𝑧 = ln(𝑟𝑒 𝑖(θ+2kπ) ) = ln 𝑟 + 𝑖(θ + 2kπ), 𝑘 = 0, ±1, ±2, ⋯ (3)
That is, the function ln 𝑧 is a 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 − 𝑣𝑎𝑙𝑢𝑒𝑑 with infinity many
values. The value of ln 𝑧 corresponding to 𝑘 = 0 in (3) is denoted by
Ln 𝑧 and is called 𝑡ℎ𝑒 𝑝𝑟𝑖𝑛𝑐𝑖𝑝𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 of ln 𝑧. Thus
Ln 𝑧 = ln 𝑟 + 𝑖θ (4)
 Chapter 3 Elementary Functions 71

The uniqueness of θ implies that Ln 𝑧 is 𝑠𝑖𝑛𝑔𝑙𝑒 − 𝑣𝑎𝑙𝑢𝑒𝑑, that is, a

function in the usual sense. Since the other values of θ differ by 2π
integer multiples, the other values of ln 𝑧 are given by
ln 𝑧 = Ln 𝑧 + 2kπ𝑖, 𝑘 = 0, ±1, ±2, ⋯ (5)
They all have the same real part, and their imaginary parts differ by
integer multiples of 2π.

Remark 1 If 𝑧 is positive real, then θ = 0, and Ln 𝑧 becomes

identical with the real natural logarithm known from
calculus. If 𝑧 is negative real (so that the natural
logarithm of calculus is not defined!), then
θ = π,
and
ln 𝑧 = ln 𝑟 + 𝑖π

Example 1 Find Ln(−1).

Solution From Fig. 3.3, we may write

−1 = 𝑒 𝑖(𝜋+2kπ) , 𝑘 = 0, ±1, ±2, ⋯ 𝑦
Thus, for 𝑘 = 0, ±1, ±2, ⋯
ln(−1) = ln 𝑒 𝑖(𝜋+2kπ) = 𝑖(𝜋 + 2kπ) 𝒛 = −𝟏
𝝅
𝑥
or
Fig. 3.3 𝑧 = −1
ln(−1) = ±𝜋𝑖, ±3𝜋𝑖, ±5𝜋𝑖, ⋯,
However, the principal value is

Ln(−1) = 𝜋𝑖
 72 Chapter 3 Elementary Functions

Properties of 𝐥𝐧 𝒛
(𝒂) The single-values function
Ln 𝑧 = ln 𝑟 + 𝑖θ, (𝑟 > 0, − 𝜋 < θ < 𝜋)
is analytic except at 𝑧 = 0 and except on the negative real axis (where
the imaginary part of such a function is not continuous) and
𝑑 1
(ln 𝑧) = , (𝑟 > 0, − 𝜋 < θ < 𝜋) (6)
𝑑𝑧 𝑧
Proof.
𝑤 = 𝑢 + 𝑖𝑣 = ln(𝑟𝑒 𝑖θ ) = ln 𝑟 + 𝑖θ
Hence,
𝑢(𝑟, θ) = ln 𝑟 , 𝑣(𝑟, θ) = θ
Moreover, the derivatives
1
𝑢𝑟 = , 𝑢θ = 0, 𝑣𝑟 = 0, 𝑣θ = 1,
𝑟
are continuous functions in the domain (𝑟 > 0, −𝜋 < θ < 𝜋), and
1
satisfy the Cauchy-Riemann equations in polar form (𝑢𝑟 = 𝑟 𝑣θ , 𝑣𝑟 =
1
− 𝑟 𝑢θ ). Therefore,
𝑑𝑤 1 −1 1
= (𝑢𝑟 + 𝑖𝑣𝑟 )𝑒 −𝑖θ = ( + 0) 𝑒 −𝑖θ = (𝑟𝑒 𝑖θ ) =
𝑑𝑧 𝑟 𝑧
(𝒃) Since 𝑒 ln 𝑟 = 𝑟, then from (1) and for positive real 𝑟, we get
𝑒 ln 𝑟+𝑖θ = 𝑒 ln 𝑟 𝑒 𝑖θ = 𝑟𝑒 𝑖θ = 𝑧
That is,
𝑒 ln 𝑧 = 𝑧 (7)
But since
arg 𝑧 = 𝑦 + 2𝑘𝜋, (𝑘 = 0, ±1, ±2, ⋯ ),
is multivalued, then
ln(𝑒 𝑧 ) = ln(𝑒 𝑥+𝑖𝑦 ) = ln(𝑒 𝑥 𝑒 𝑖(𝑦+2𝑘𝜋) ) = ln(𝑒 𝑥+𝑖𝑦+𝑖2𝑘𝜋 )
= 𝑧 + 𝑖2𝑘𝜋, (𝑘 = 0, ±1, ±2, ⋯ ) (8)
 Chapter 3 Elementary Functions 73

Example 2 Find the following complex logarithmic functions

(𝒊) ln(3 − 4𝑖) (𝒊𝒊) ln 7𝑖

𝑦
Solution
(𝒊) From Fig. 3.4, 3
θ 𝑥
𝑖 tan−1 (−4⁄3) −0.9273𝑖
3 − 4𝑖 = 5𝑒 = 5𝑒 ,
4
therefore
 3 − 4𝑖
ln(3 − 4𝑖) = ln 5 − 𝑖(0.9273 + 2𝑘𝜋)
Fig. 3.4 𝑧 = 3 − 4𝑖
The principal value is
Ln(3 − 4𝑖) = ln 5 − 0.9273 𝑖

(𝒊𝒊) From Fig. 3.5, writing 7𝑖 = 7𝑒 𝑖𝜋⁄2 , then 𝑦

𝑖(𝜋⁄2+2𝑘𝜋)  7𝑖
ln 7𝑖 = ln(7𝑒 )
7
= ln 7 + 𝑖(𝜋⁄2 + 2𝑘𝜋)
𝑥
The principal value is
𝜋 Fig. 3.5 𝑧 = 7𝑖
Ln 7𝑖 = ln 7 + 𝑖
2

Example 3 Solve the following equation for 𝑧

ln 𝑧 = (4 + 3𝑖) 𝑦

Solution  4 + 3𝑖

 First method 3
θ
From Fig. 3.6, 4 𝑥

ln 𝑧 = ln(𝑟𝑒 𝑖θ ) = ln 𝑟 + 𝑖θ = 4 + 3𝑖 Fig. 3.6 𝑧 = 4 + 3𝑖

Equating the real and imaginary parts,

ln 𝑟 = 4  𝑟 = 𝑒 4 , θ = 3,
 74 Chapter 3 Elementary Functions

Hence,
𝑧 = 𝑟𝑒 𝑖θ = 𝑒 4 𝑒 𝑖3 = 𝑒 4 (cos 3 + 𝑖 sin 3)
= −54.052 + 7.705𝑖
 Second method
ln 𝑧 = (4 + 3𝑖)  𝑧 = 𝑒 (4+3𝑖)
Therefore,
𝑧 = 𝑒 4 𝑒 3𝑖 = 𝑒 4 (cos 3 + 𝑖 sin 3)
= −54.052 + 7.705𝑖

Example 4 Prove that

𝑥 + 𝑖𝑦 𝑦
ln = 2𝑖 tan−1 ( )
𝑥 − 𝑖𝑦 𝑥

Solution Using the polar form of complex numbers, we can write

𝑥 + 𝑖𝑦 = 𝑟𝑒 𝑖θ  𝑥 − 𝑖𝑦 = 𝑟𝑒 −𝑖θ ,
where
𝑦
𝑟 = √𝑥 2 + 𝑦 2 , θ = tan−1 ( )
𝑥
Thus,
𝑥 + 𝑖𝑦 𝑟𝑒 𝑖θ 𝑦
ln = ln −𝑖θ = ln 𝑒 𝑖2θ = 2𝑖θ = 2𝑖 tan−1 ( )
𝑥 − 𝑖𝑦 𝑟𝑒 𝑥

Example 5 Solve for 𝑧, the equation 𝑒 𝑧 = − 8𝑖.

𝑦
Solution
𝑥
 First method (see Fig. 3.7) 8
𝑧 𝑥 𝑖𝑦 −𝑖𝜋⁄2  − 8𝑖
𝑒 =𝑒 𝑒 = − 8𝑖 = 8𝑒 ,
Fig. 3.7 𝑧 = − 8𝑖
from which we get
 Chapter 3 Elementary Functions 75

𝜋
𝑒 𝑥 = 8  𝑥 = ln 8 , 𝑦=−
2
Thus
𝜋
𝑧 = ln 8 − 𝑖
2
 Second method The given problem
𝑒 𝑧 = − 8𝑖  𝑧 = ln(−8𝑖)
Therefore,
𝜋
𝑧 = ln(−8𝑖) = ln(8𝑒 −𝑖𝜋⁄2 ) = ln 8 − 𝑖
2

Exercise 3.1

In problems 1 − 10, express 𝑒 𝑧 in the form 𝑢 + 𝑖𝑣 if

𝜋 3𝜋
𝟏. 𝑧 = −1 + 𝑖 𝟐. 𝑧 = − 𝜋 + 𝑖
4 2
𝟑. 𝑧 = (1 + 𝑖)𝜋 𝟒. 𝑧 = − 0.3 + 0.5𝑖
𝜋
𝟓. 𝑧 = 1.5 + 2𝑖 𝟔. 𝑧 = 𝑖
6
𝜋 𝜋
𝟕. 𝑧 = 2 − 𝑖 𝟖. 𝑧 = 5 + 𝑖
2 4
𝜋 𝜋
𝟗. 𝑧 = −𝑖 𝟏𝟎. 𝑧 = − 2 + 𝑖
2 3
In problems 11 − 16, express ln 𝑧 in the form 𝑢 + 𝑖𝑣 if
𝟏𝟏. 𝑧 = √2 + 𝑖√6 𝟏𝟐. 𝑧 = − 𝑒𝑖
5
𝟏𝟑. 𝑧 = 1 + 𝑖 𝟏𝟒. 𝑧 = (1 + 𝑖√3)
𝟏𝟓. 𝑧 = −100 𝟏𝟔. 𝑧 = (1 + 𝑖)4
In problems 17 − 21, show that
2
𝟏𝟕. 𝑓(𝑧) = 𝑒 𝑧 is entire function.
2
𝟏𝟖. 𝑢(𝑥, 𝑦) = Re (𝑒 𝑧 ) is a harmonic function.
𝟏𝟗. 𝑢(𝑥, 𝑦) = ln(𝑥 2 + 𝑦 2 ) is a harmonic function.
 76 Chapter 3 Elementary Functions

𝟐𝟎. 𝑓(𝑧) = 𝑒 𝑧 is nowhere analytic function.

𝑒2
𝟐𝟏. 𝑒 (2+𝑖𝜋⁄4) = (1 + 𝑖).
√2

In problems 22 − 31, find all values of 𝑧 such that

𝟐𝟐. 𝑒 𝑧 = −2 𝟐𝟑. 𝑒 𝑧 = 1 + 𝑖√3
𝟐𝟒. 𝑒 (2𝑧−1) = 1 𝟐𝟓. 𝑒 (𝑧−1) = −𝑖𝑒 2
𝟐𝟔. 𝑒 2𝑧 + 𝑒 𝑧 + 1 = 0 𝟐𝟕. 𝑒 1⁄𝑧 = −1
𝟐𝟖. 𝑒 𝑧 = 𝑒 𝑧 𝟐𝟗. |𝑒 −𝑧 | < 1
𝟑𝟎. 𝑒 𝑧 is real 𝟑𝟏. Re (𝑒 2𝑧 ) = 0
In problems 32 − 40, solve the following equations for 𝑧
𝟑𝟐. ln 𝑧 = √2 + 𝜋𝑖 𝟑𝟑. ln 𝑧 = 4 − 3𝑖
𝟑𝟒. ln 𝑧 = 0.3 + 0.7𝑖 𝟑𝟓. ln 𝑧 = −4 + 𝑖
𝜋
𝟑𝟔. ln 𝑧 = −5 + 0.01𝑖 𝟑𝟕. ln 𝑧 = 2 + 𝑖
4
𝟑𝟖. 𝑧 8 − 2𝑧 4 + 1 = 0 𝟑𝟗. 𝑧 4 + 1 = 0
𝟒𝟎. 𝑧 2 + 𝑧 + 1 − 𝑖 = 0

3.3 General powers

The integer power of 𝑧 is given by

𝑛
𝑧 𝑛 = (𝑟𝑒 𝑖θ ) = 𝑟 𝑛 𝑒 𝑖𝑛θ = 𝑟 𝑛 (cos 𝑛θ + 𝑖 sin 𝑛θ) (1)
Putting |𝑧| = 1 in equation (1), we get 𝐷𝑒 𝑀𝑜𝑖𝑣𝑟𝑒 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 which
states that, for any integer 𝑛, the following identity holds
(cos θ + 𝑖 sin θ)𝑛 = (cos 𝑛θ + 𝑖 sin 𝑛θ) (2)

3.3.1 The 𝒏𝒕𝒉 root of complex numbers

If
𝑤 𝑛 = 𝑧, (3)
 Chapter 3 Elementary Functions 77

then each value of 𝑤 is called an 𝑛𝑡ℎ root of the complex number 𝑧,

and we write
𝑛
𝑤 = √𝑧 , (4)
Hence 𝑤 has the 𝑛 values:
𝑧 1⁄𝑛 = [𝑟(cos(θ + 2kπ) + 𝑖 sin(θ + 2kπ))]1⁄𝑛
θ + 2kπ θ + 2kπ
= 𝑟 1⁄𝑛 (cos ( ) + 𝑖 sin ( )) , 𝑘 = 0,1,2, ⋯ , 𝑛 − 1 (5)
𝑛 𝑛
𝑛
These 𝑛 values of 𝑤 = √𝑧 lie on a circle of radius 𝑟 1⁄𝑛 with centre
at the origin and constitute the vertices of a regular polygon of 𝑛 sides.

Example 1 Find 𝑧 3 if 𝑧 = 1 + 𝑖√3.

Solution. Since  1 + √3𝑖

1 + 𝑖√3 = 2𝑒 𝑖𝜋⁄3 , 2
√3
then, we have 𝝅⁄𝟑
1 𝑥
3 𝑖𝜋⁄3 3 𝑖𝜋 Fig. 3.8 𝑧 = 1 + 𝑖√3
𝑧 = (2𝑒 ) = 8𝑒
= 8(cos 𝜋 + 𝑖 sin 𝜋) = −8

Example 2 (𝑛𝑡ℎ 𝑟𝑜𝑜𝑡 𝑜𝑓 𝑢𝑛𝑖𝑡𝑦) Solve the equation

𝑧𝑛 = 1 𝑦

Solution Since  𝑥
1
𝑖0 𝑖(0+2𝑘𝜋)
1=𝑒 =𝑒 ,
Fig. 3.9 𝑧 = 1
Thus
𝑛
√1 = 𝑒 𝑖(2𝑘𝜋⁄𝑛) , 𝑘 = 0,1,2, ⋯ , 𝑛 − 1,
or
𝑛
√1 = (cos(2𝑘𝜋⁄𝑛) + 𝑖 sin(2𝑘𝜋⁄𝑛)), 𝑘 = 0,1,2, ⋯ , 𝑛 − 1
 78 Chapter 3 Elementary Functions

For example, the three roots of the equation 𝑧 3 = 1 are

𝑧1 = cos 0 + 𝑖 sin 0 = 1
1 √3
𝑧2 = cos(2𝜋⁄3) + 𝑖 sin(2𝜋⁄3) = +𝑖
2 2
1 √3
𝑧3 = cos(4𝜋⁄3) + 𝑖 sin(4𝜋⁄3) = −𝑖
2 2
These three roots of the unity are denoted by 1, ω, ω2 .
Similarly, the four roots of the unit are
𝑧1 = 𝑒 𝑖0 = cos 0 + 𝑖 sin 0 = 1
𝑧2 = 𝑒 𝑖2𝜋⁄4 = cos(𝜋⁄2) + 𝑖 sin(𝜋⁄2) = 𝑖
𝑧3 = 𝑒 𝑖4𝜋⁄4 = cos 𝜋 + 𝑖 sin 𝜋 = −1
𝑧4 = 𝑒 𝑖6𝜋⁄4 = cos(3𝜋⁄2) + 𝑖 sin(3𝜋⁄2) = −𝑖
If we denote by ω = 𝑒 𝑖𝜋⁄2 , then the four roots of the unit may be
denoted by 1, ω, ω2 , ω3 .
We note that from Fig. 3.10 that

1 + ω + ω2 + ω3 = 0

𝑦 𝑦
𝒛𝟐
𝒛𝟐  

𝒛 𝑥 𝒛𝟑  𝒛 𝑥
𝟏 𝟏

𝒛𝟑 𝒛
𝟒

Fig. 3.10 Roots of 𝑧 3 = 1 Fig. 3.11 Roots of 𝑧 4 = 1

Example 3 Solve the equation 𝑧 2 = 𝑖.

 Chapter 3 Elementary Functions 79

Solution From Fig. 3.12, we have

𝑖 = 𝑒 𝑖𝜋⁄2 ,
then for 𝑘 = 0,1, De Moivre theorem yields 𝑦
𝑖
𝜋⁄2 + 2kπ 𝜋⁄2 + 2kπ
𝑧 = cos ( ) + 𝑖 sin ( ) 1
2 2
thus 𝑥

𝑧1 = cos(𝜋⁄4) + 𝑖 sin(𝜋⁄4) Fig. 3.12 𝑧 = 𝑖

1 1 𝑦
= +𝑖
√2 √2
 𝒛𝟏
𝑧2 = cos(5 𝜋⁄4) + 𝑖 sin(5 𝜋⁄4)
𝑥

1 1 𝒛𝟐
= −𝑖
√2 √2 Fig. 3.13 roots of 𝑧 2 = 𝑖

Example 4 Find the roots of

(𝒊) 𝑧 4 = −2√3 − 2𝑖 (𝒊𝒊) 𝑧 3 = −1 + 𝑖

Solution
−2√3
(𝒊) From Fig. 3.12, we have −5 𝜋⁄6 𝑥
4
−2√3 − 2𝑖 = 4𝑒 𝑖(−5𝜋⁄6+2𝑘𝜋)  −2
−2√3 − 2𝑖
Therefore, for 𝑘 = 0,1,2,3 the four roots
are computed from the relations Fig. 3.14 𝑧 = −2√3 − 2𝑖
1⁄4
(−2√3 − 2𝑖) = 41⁄4 𝑒 𝑖(−5𝜋⁄6+2𝑘𝜋⁄4) ,
−5 𝜋⁄6 + 2𝑘𝜋 −5 𝜋⁄6 + 2𝑘𝜋
= √2 cos ( ) + 𝑖 sin ( )
4 4
Hence
 80 Chapter 3 Elementary Functions

𝑧1 = √2 cos(−5 𝜋⁄24) + 𝑖 sin(−5 𝜋⁄24) 𝑦

𝑧2 = √2 cos(7 𝜋⁄24) + 𝑖 sin(7 𝜋⁄24)
𝒛𝟑  𝒛𝟐

𝑧3 = √2 cos(19 𝜋⁄24) + 𝑖 sin(19 𝜋⁄24)
𝑥
𝑧4 = √2 cos(31 𝜋⁄24) + 𝑖 sin(31 𝜋⁄24)  𝒛𝟏
𝒛𝟒

(𝒊𝒊) Since Fig. 3.15

−1 + 𝑖 = √2𝑒 𝑖(3𝜋⁄4+2𝑘𝜋) , 𝑦
then −1 + 𝑖
 1
3𝜋⁄4+2𝑘𝜋 √2 3 𝜋⁄4
𝑖( )
(−1 + 𝑖)1⁄3 = 21⁄6 𝑒 3 , 𝑘 = 0,1,2
−1 𝑥
Hence the three roots are Fig. 3.16 𝑧 = −1 + 𝑖
𝑦
𝑧1 = 21⁄6 cos(𝜋⁄4) + 𝑖 sin(𝜋⁄4)
𝑧2 = 21⁄6 cos(11 𝜋⁄12) + 𝑖 sin(11 𝜋⁄12) 𝒛𝟑  𝒛𝟐

𝑧3 = 21⁄6 cos(19 𝜋⁄12) + 𝑖 sin(19 𝜋⁄12) 𝑥
 𝒛𝟏
𝒛𝟒

3.3.2 Expansion of 𝐬𝐢𝐧𝒏 𝛉 and 𝐜𝐨𝐬 𝒏 𝛉 Fig. 3.17

By the use of the following facts we can expand 𝐬𝐢𝐧𝒏 𝛉 and 𝐜𝐨𝐬𝒏 𝛉 in
a series of cosines and sines of a multiple of 𝛉: Namely, if
𝑧 = cos θ + 𝑖 sin θ,
then
1
= cos θ − 𝑖 sin θ,
𝑧
𝑧 𝑛 = cos 𝑛θ + 𝑖 sin 𝑛θ,
and
1 1
𝑧+ = 2 cos θ , 𝑧− = 2𝑖 sin θ
𝑧 𝑧
1 1
𝑧𝑛 + = 2 cos 𝑛θ, 𝑧 𝑛 − 𝑛 = 2𝑖 sin 𝑛θ
𝑧𝑛 𝑧
 Chapter 3 Elementary Functions 81

Therefore, for any integer 𝑛 we can get the required expansion of sin𝑛 θ
and cos𝑛 θ as illustrated by the following example.

1
Example 1 Expand cos6 θ 1
1
1 2 1
1 3 3 1
Solution Using the above relations and the 1 4 6 4 1
1 5 10 10 5 1
binomial triangle we obtain 1 6 15 20 15 6 1
1 6 ⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯
(2 cos θ)6 = (𝑧 + ) The binomial triangle
𝑧
1 1 1
= 𝑧 6 + 6(𝑧 5 ) ( ) + 15(𝑧 4 ) ( 2 ) + 20(𝑧 3 ) ( 3 )
𝑧 𝑧 𝑧
1 1 1
+15(𝑧 2 ) ( 4 ) + 6(𝑧) ( 5 ) + ( 6 )
𝑧 𝑧 𝑧
1 1 1
= (𝑧 6 + 6 ) + 6 (𝑧 4 + 4 ) + 15 (𝑧 2 + 2 ) + 20
𝑧 𝑧 𝑧
= (2 cos 6θ) + 6(2 cos 4θ) + 15(2 cos 2θ) + 20
Hence,
(2 cos θ)6 = (2 cos 6θ) + 6(2 cos 4θ) + 15(2 cos 2θ) + 20
or
1
cos 6 θ = (cos 6θ + 6 cos 4θ + 15 cos 2θ + 10)
32

Example 2 Expand sin5 θ in terms of cosine and sine of .

Solution
5
1 5
(2𝑖 sin θ) = (𝑧 − )
𝑧
 82 Chapter 3 Elementary Functions

1 1 1
= 𝑧 5 + 5(𝑧 4 ) (− ) + 10(𝑧 3 ) ( 2 ) + 10(𝑧 2 ) (− 3 )
𝑧 𝑧 𝑧
1 1
+5(𝑧) ( 4 ) + (− 5 )
𝑧 𝑧
1 1 1
= (𝑧 5 − 5 ) − 5 (𝑧 3 − 3 ) + 10 (𝑧 − )
𝑧 𝑧 𝑧
= (2𝑖 sin 5θ) − 5(2𝑖 sin 3θ) + 10(2𝑖 sin θ),
or
1
sin5 θ = (sin 5θ − 5 sin 3θ + 10 sin θ)
16

3.3.3 Complex powers

The general power of a complex number 𝒛 is defined by

α
𝑧 α = 𝑒 ln 𝑧 = 𝑒 α ln 𝑧 , (α ≡ complex, z ≠ 0) (1)
Since ln 𝑧 is multiple-valued, 𝑧 α will be multiple-valued. The principal
value is
𝑧 α = 𝑒 α Ln 𝑧 (2)
The single-valued function 𝑧 α = 𝑒 α Ln 𝑧 is analytic in the domain
(𝑟 > 0, −𝜋 < θ < 𝜋), and
𝑑 α
(𝑧 ) = α𝑧 α−1 (3)
𝑑𝑧

Remark Equation (3) is proved for α = 𝑛 where 𝑛 is an integer in

§2.5. Namely;
𝑑 𝑛
(𝑧 ) = 𝑛𝑧 𝑛−1 , 𝑛 is an integer
𝑑𝑧
According to the definition (1) the exponential function
with base 𝑎, where 𝑎 is a complex number can be written

𝑎 𝑧 = 𝑒 𝑧 ln 𝑎 (α ≡ complex, z ≠ 0) (4)
 Chapter 3 Elementary Functions 83

Example 1 Find the principal value of

(𝒊) (−𝑖)𝑖 (𝒊𝒊) (5 − 2𝑖)3+𝜋𝑖

Solution
𝑦
(𝒊) (−𝑖)𝑖 ln(−𝑖)𝑖 𝑖 ln(−𝑖)
=𝑒 =𝑒
𝑥
From Fig. 3.18, 1
−𝑖(𝜋⁄2)
−𝑖 = 𝑒 ,  −𝑖
Fig. 3.18
thus
(𝜋⁄2)
𝑖 ln(𝑒 𝑖 ) (𝜋⁄2)
𝑖 ln(𝑒 −𝑖(𝜋⁄2) )
(−𝑖)𝑖 = 𝑒 =𝑒 = 𝑒𝑖 𝑖 = 𝑒 (𝜋⁄2)
3+𝜋𝑖
(𝒊𝒊) (5 − 2𝑖)3+𝜋𝑖 = 𝑒 ln(5−2𝑖) = 𝑒 (3+𝜋𝑖) ln(5−2𝑖) ,
with
𝑦
−2
𝑖 tan−1 ( )
(5 − 2𝑖) = √29𝑒 5 ,
5
then θ 𝑥
−𝑖 tan−1 (0.4) √29
ln(5 − 2𝑖) = ln(√29𝑒 ) −2− 
5 − 2𝑖
= ln √29 − 𝑖 tan−1(0.4)
Fig. 3.19 𝑧 = 5 − 2𝑖
Therefore, we have
(3 + 𝜋𝑖) ln(5 − 2𝑖) = (3 + 𝜋𝑖)(ln √29 − 𝑖 tan−1 (0.4))

= (3 ln √29 + 𝜋 tan−1(0.4)) + 𝑖(𝜋 ln √29 − 3 tan−1(0.4))

= 6.246 + 4.147𝑖
Finally,
(5 − 2𝑖)3+𝜋𝑖 = 𝑒 (6.246+4.147𝑖) = 𝑒 6.246 𝑒 4.147𝑖
= 𝑒 6.246 (cos 4.147 + 𝑖 sin 4.147)
≅ − (276.152 + 436.027𝑖)
 84 Chapter 3 Elementary Functions

Exercise 3.2

In problems 1 − 7, prove that

𝟏. cos 5θ = 5 cos θ − 20 cos3 θ + 16 cos 5 θ
sin 5θ
𝟐. = 1 − 12 cos 2 θ + 16 cos4 θ, sin θ ≠ 0
sin θ
2π 4π 6π 8π
𝟑. 1 + cos + cos + cos + cos =0
5 5 5 5
2π 4π 6π 8π
𝟒. sin + sin + sin + sin =0
5 5 5 5
𝟓. cos 4θ = cos4 θ − 6 cos 2 θ sin2 θ + sin4 θ
𝟔. sin 4θ = 4 sin θ cos θ (cos 2 θ − sin2 θ)
1
𝟕. cos4 θ sin2 θ = (2 + cos 2θ − 2 cos 4θ − cos 6θ)
32
𝟖. Expand cos5 θ, sin6 θ, and sin7 θ cos θ in terms of cos θ and sin θ.
In problems 9 − 18, compute the principal values of

𝟗. (−𝑖)4𝑖 𝟏𝟎. 3𝑖⁄𝜋

3𝑖
𝟏𝟏. (1 + 𝑖)(1+𝑖) 𝟏𝟐. (1 + 𝑖√3)
𝟏𝟑. (5 − 2𝑖)(3+𝜋𝑖) 𝟏𝟒. (−5)(2−4𝑖)
𝟏𝟓. (−1)(−2𝑖⁄𝜋) 𝟏𝟔. (1 − 𝑖)2𝑖
𝑖⁄2
𝟏𝟕. 4(3−𝑖) 𝟏𝟖. (2 + 𝑖√21)
𝟏𝟗. Find the complex number 𝑧 which satisfies the equation
5(1 − 𝑖)
(1 + 𝑖)𝑧 + (3 + 𝑖)𝑧 =
(2 + 𝑖)
In problems 20 − 25, find all values of the given quantity
𝟐𝟎. (2 − 𝑖)(1+𝑖) 𝟐𝟏. 3(3−𝑖)
𝟐𝟐. |𝑖 𝑖 | 𝟐𝟑. 𝑖 𝑖⁄2
5
𝟐𝟒. (1 + 𝑖)4 𝟐𝟓. (1 + 𝑖√3)
 Chapter 3 Elementary Functions 85

𝟐𝟔. At what points is the function given by Ln(𝑧 2 + 1) analytic?

Explain.
In problems 27 − 32 , are the functions harmonic? If so, find a
conjugate harmonic.
𝟐𝟕. 𝑥 2 𝑦 2 𝟐𝟖. 𝑥 3 − 3𝑥𝑦 2
𝟐𝟗. 𝑥 2 − 𝑦 2 𝟑𝟎. 4𝑥 2 𝑦 + 5𝑦 2
𝟑𝟏. 𝑥𝑦(𝑥 3 − 3𝑥𝑦 2 ) 𝟑𝟐. 𝑥 2 𝑦 2 (𝑥 2 + 3𝑥𝑦 2 )
𝟑𝟑. Determine the fourth roots of
(−2 + 𝑖)3 (−2 − 3𝑖)3
𝑧= ,
(4 − 3𝑖)
expressing them in the Cartesian form.

𝟑𝟒. Is the collection of all values of ln(𝑖 2 ) the same as the collection
of all values of 2 ln 𝑖 ? Explain.
35. Using the fact that the 𝑛𝑡ℎ root of 1 are spaced equally round the
unit circle with one root of 𝑧 = 1, verify that the fourth roots of 1 are
1, 𝑖, −1, −𝑖. Find the fifth roots of 1. Label the first root anti-clockwise
from 𝑧 = 1 as γ. Do the other roots have a simple relationship to γ? Are
there any relationship akin to 1 + ω + ω2 = 0? Can you say anything
in general about the nature of the 𝑛𝑡ℎ root of 1? Consider the cases 𝑛
even and 𝑛 odd.

3.4. Trigonometric functions

From the Euler formula
𝑒 𝑖𝑦 = cos 𝑦 + 𝑖 sin 𝑦,
𝑒 −𝑖𝑦 = cos 𝑦 − 𝑖 sin 𝑦
By addition and subtraction, we obtain
 86 Chapter 3 Elementary Functions

1 𝑖𝑦
cos 𝑦 = (𝑒 + 𝑒 −𝑖𝑦 ),
2
1
sin 𝑦 = (𝑒 𝑖𝑦 − 𝑒 −𝑖𝑦 )
2𝑖
This suggests the following definitions for complex values 𝑧:
𝑒 𝑖𝑧 + 𝑒 −𝑖𝑧
cos 𝑧 = (1)
2
𝑒 𝑖𝑧 − 𝑒 −𝑖𝑧
sin 𝑧 = (2)
2𝑖
With these definitions in place, it is now easy to create the other
complex trigonometric functions, provided the denominators in the
following expressions do not equal zero.
sin 𝑧 cos 𝑧
tan 𝑧 = , cot 𝑧 = ,
cos 𝑧 sin 𝑧
1 1
sec 𝑧 = , csc 𝑧 = ,
cos 𝑧 sin 𝑧
Properties of trigonometric functions
(𝒂) If we replace 𝑧 by 𝑖𝑧 everywhere in (1) and (2), we obtain
𝑒 −𝑧 + 𝑒 𝑧
cos 𝑖𝑧 = = cosh 𝑧 (3)
2
𝑒 −𝑧 − 𝑒 𝑧
sin 𝑖𝑧 = = 𝑖 sinh 𝑧 (4)
2𝑖
Here we use the definitions of the complex hyperbolic cosine and sine
functions:
𝑒 𝑧 + 𝑒 −𝑧 𝑒 𝑧 − 𝑒 −𝑧
cosh 𝑧 = , sinh 𝑧 =
2 2
These functions will be studied in detail later.
(𝒃) By the aid of equations (3) and (4), we deduce that
cos 𝑧 = cos(𝑥 + 𝑖𝑦) = cos 𝑥 cos 𝑖𝑦 − sin 𝑥 sin 𝑖𝑦
= cos 𝑥 cosh 𝑦 − 𝑖 sin 𝑥 sinh 𝑦 (5)
 Chapter 3 Elementary Functions 87

Therefore,
|cos 𝑧|2 = cos2 𝑥 cosh2 𝑦 + sin2 𝑥 sinh2 𝑦
= cos2 𝑥 (1 + sinh2 𝑦) + sin2 𝑥 sinh2 𝑦
= cos2 𝑥 + (cos 2 𝑥 + sin2 𝑥) sinh2 𝑦
= cos2 𝑥 + sinh2 𝑦, (6)
and in a similar way, we have
sin 𝑧 = sin 𝑥 cosh 𝑦 + 𝑖 cos 𝑥 sinh 𝑦 , (7)
|sin 𝑧|2 = sin2 𝑥 + sinh2 𝑦 , (8)

Remark 1 In ordinary trigonometry we know that |sin 𝑥| ≤ 1 and

|cos 𝑥| ≤ 1. But in complex analysis equations (6) and
(8) show that these inequalities do not hold for complex
sine and cosine, since sinh 𝑦 can range from −∞ to ∞.

(𝒄) Both cos 𝑧 and sin 𝑧 are entire functions because they are linear
combinations of the entire functions 𝑒 𝑖𝑧 and 𝑒 −𝑖𝑧 . Moreover,
𝑑
sin 𝑧 = cos 𝑧,
𝑑𝑧
𝑑
cos 𝑧 = − sin 𝑧
𝑑𝑧

𝑑
Example 1 Show that sin 𝑧 = cos 𝑧.
𝑑𝑧

Solution  First method Since

𝑑 𝑧
𝑒 = 𝑒 𝑧,
𝑑𝑧
it follows from the chain rule that
𝑑 𝑖𝑧 𝑑 −𝑖𝑧
𝑒 = 𝑖𝑒 𝑖𝑧 , 𝑒 = −𝑖𝑒 −𝑖𝑧
𝑑𝑧 𝑑𝑧
 88 Chapter 3 Elementary Functions

Hence,
𝑑 𝑑 𝑒 𝑖𝑧 − 𝑒 −𝑖𝑧 𝑖𝑒 𝑖𝑧 + 𝑖𝑒 −𝑖𝑧 𝑒 𝑖𝑧 + 𝑒 −𝑖𝑧
sin 𝑧 = ( )= = = cos 𝑧
𝑑𝑧 𝑑𝑧 2𝑖 2𝑖 2
 Second method

𝑓(𝑧) = 𝑢 + 𝑖𝑣 = sin 𝑧 = sin(𝑥 + 𝑖𝑦)

= sin 𝑥 cos 𝑖𝑦 + cos 𝑥 sin 𝑖𝑦
= sin 𝑥 cosh 𝑦 + 𝑖 cos 𝑥 sinh 𝑦
Thus
𝑢 = sin 𝑥 cosh 𝑦 , 𝑣 = cos 𝑥 sinh 𝑦
𝑢𝑥 = cos 𝑥 cosh 𝑦 , 𝑣𝑥 = − sin 𝑥 sinh 𝑦
𝑢𝑦 = sin 𝑥 sinh 𝑦, 𝑣𝑦 = cos 𝑥 cosh 𝑦
So we see that the Cauchy-Riemann equations are satisfied for every 𝑧.
Therefore
𝑓 ′ (𝑧) = 𝑢𝑥 + 𝑖𝑣𝑥
= cos 𝑥 cosh 𝑦 − 𝑖 sin 𝑥 sinh 𝑦
= cos 𝑥 cos 𝑖𝑦 − sin 𝑥 sin 𝑖𝑦 = cos(𝑥 + 𝑖𝑦) = cos 𝑧

(𝒅) Zeros of 𝐜𝐨𝐬 𝒛 and 𝐬𝐢𝐧 𝒛. The zeros of cos 𝑧 and sin 𝑧 are
evaluated as illustrated by the following example.

Example 2 Solve the equations

(𝒊) cos 𝑧 = 0, (𝒊𝒊) sin 𝑧 = 0

Solution Let us begin by the first function

cos 𝑧 = cos(𝑥 + 𝑖𝑦)

= cos 𝑥 cos 𝑖𝑦 − sin 𝑥 sin 𝑖𝑦
= cos 𝑥 cosh 𝑦 − 𝑖 sin 𝑥 sinh 𝑦 = 0
 Chapter 3 Elementary Functions 89

Thus
cos 𝑥 cosh 𝑦 = 0, sin 𝑥 sinh 𝑦 = 0
The equation cos 𝑥 cosh 𝑦 = 0 implies cos 𝑥 = 0 (cosh 𝑦 ≠ 0), then
𝜋
𝑥 = (2𝑛 + 1) 2 , (𝑛 = 0, ±1, ±2, ⋯ ). Substitute in the second

equation we get sinh 𝑦 = 0 and so 𝑦 = 0. Therefore, we have

𝜋
𝑧 = (2𝑛 + 1) , (𝑛 = 0, ±1, ±2, ⋯ ) (9)
2
(𝒊𝒊) Another way to find the zeros of sin 𝑧 is to start from equation (7),
which yields
sin 𝑧 = 0  sin 𝑥 = 0, sinh 𝑦 = 0
Hence,
𝑥 = 𝑛𝜋, 𝑦 = 0,
or alternatively we write
𝑧 = 𝑛𝜋, (𝑛 = 0, ±1, ±2, ⋯ ) (10)

Remark 2 Equations (9) and (10) imply that the only zeros of cos 𝑧
and sin 𝑧 are those of the real cosine and sine functions.

(𝒆) tan 𝑧 and sec 𝑧 are analytic in any domain where cos 𝑧 ≠ 0, i.e.
𝜋
they are analytic functions for all 𝑧 except at the 𝑧 = (2𝑛 + 1) 2 ,
(𝑛 = 0, ±1, ±2, ⋯ ) and
𝑑
tan 𝑧 = sec 2 𝑧 (11)
𝑑𝑧
𝑑
sec 𝑧 = sec 𝑧 tan 𝑧 (12)
𝑑𝑧
Also cot 𝑧 and csc 𝑧 are analytic in any domain where sin 𝑧 ≠ 0, i.e.
they are analytic functions for all 𝑧 except at the points 𝑧 = 𝑛𝜋,
(𝑛 = 0, ±1, ±2, ⋯ ), and
 90 Chapter 3 Elementary Functions

𝑑
cot 𝑧 = −csc 2 𝑧, (13)
𝑑𝑧
𝑑
csc 𝑧 = − csc 𝑧 cot 𝑧 (14)
𝑑𝑧

5
Example 3 Solve the equation cos 𝑧 = 3.

Solution According to Remark 1, it is possible to have such equations

in complex analysis
𝑒 𝑖𝑧 + 𝑒 −𝑖𝑧 5
cos 𝑧 = =
2 3
Multiply both sides by 𝑒 𝑖𝑧 , we get
𝑒 𝑖𝑧 − 3
3𝑒 𝑖2𝑧 + 3 = 10𝑒 𝑖𝑧 ,
or −
3𝑒 𝑖𝑧 1
3𝑒 𝑖2𝑧 − 10𝑒 𝑖𝑧 + 3 = 0,
which is a quadratic equation in 𝑒 𝑖𝑧 , hence
(𝑒 𝑖𝑧 − 3)(3𝑒 𝑖𝑧 − 1) = 0
 (𝑒 𝑖𝑧 − 3) = 0  𝑒 𝑖𝑧 = 3,
1
 (3𝑒 𝑖𝑧 − 1) = 0  𝑒 𝑖𝑧 =
3
from which we get
𝑒 𝑖𝑥−𝑦 = 3  𝑒 𝑖𝑥 𝑒 −𝑦 = 3𝑒 𝑖(0+2𝑘𝜋) , (15)
or
1 1
𝑒 𝑖𝑥−𝑦 =  𝑒 𝑖𝑥 𝑒 −𝑦 = 𝑒 𝑖(0+2𝑘𝜋) (16)
3 3
Equating the real and imaginary parts of (15), we get
𝑒 −𝑦 = 3  𝑦 = − ln 3 , 𝑥 = 2𝑘𝜋
Similarly, equation (16) implies
 Chapter 3 Elementary Functions 91

1 1
𝑒 −𝑦 =  𝑦 = − ln = ln 3 , 𝑥 = 2𝑘𝜋
3 3
Therefore, the solution of the given equation is
𝑧 = 2𝑘𝜋 ± 𝑖 ln 3 (𝑘 = 0, ±1, ±2, ⋯ )
(f) The familiar trigonometric identities are also the same in the
complex case.
1. cos(−𝑧) = cos 𝑧 , sin(−𝑧) = − sin 𝑧
𝜋 𝜋
2. sin (2 − 𝑧) = sin ( 2 + 𝑧) cos 𝑧

3. sin2 𝑧 + cos 2 𝑧 = 1
4. sin(𝑧1 ± 𝑧2 ) = sin 𝑧1 cos 𝑧2 ± cos 𝑧1 sin 𝑧2
5. cos(𝑧1 ± 𝑧2 ) = cos 𝑧1 cos 𝑧2 ∓ sin 𝑧1 sin 𝑧2
6. sin 2𝑧 = 2 sin 𝑧 cos 𝑧
7. cos 2𝑧 = cos2 𝑧 − sin2 𝑧
8. sin(𝑧 + 2𝜋) = sin 𝑧, (𝑃𝑒𝑟𝑖𝑜𝑑𝑖𝑐 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦)
9. cos(𝑧 + 2𝜋) = cos 𝑧, (𝑃𝑒𝑟𝑖𝑜𝑑𝑖𝑐 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦)
10. tan(𝑧 + 𝜋) = tan 𝑧, (𝑃𝑒𝑟𝑖𝑜𝑑𝑖𝑐 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦)
The last three identities indicate that sin 𝑧 and cos 𝑧 are periodic
functions with period 2𝜋 and tan 𝑧 is periodic with period 𝜋.

3.5 Hyperbolic functions

The complex hyperbolic cosine and sine are defined by the formulas
𝑒 𝑧 + 𝑒 −𝑧
cosh 𝑧 = , (1)
2
𝑒 𝑧 − 𝑒 −𝑧
sinh 𝑧 = , (2)
2
It is now easy to create the other complex trigonometric functions,
provided the denominators in the following expressions do not equal
zero.
 92 Chapter 3 Elementary Functions

sinh 𝑧 cosh 𝑧
tanh 𝑧 = , coth 𝑧 =
cosh 𝑧 sinh 𝑧
1 1
sech 𝑧 = , csch 𝑧 =
cosh 𝑧 sinh 𝑧

Properties of hyperbolic functions

(𝒂) The relations between the complex hyperbolic and trigonometric

functions follow from the definition of the hyperbolic functions if we
replace 𝑧 by 𝑖𝑧; thus
cosh 𝑖𝑧 = cos 𝑧 (3)
sinh 𝑖𝑧 = 𝑖 sin 𝑧 (4)
(𝒃) Both sinh 𝑧 and cosh 𝑧 are entire functions and
𝑑
sinh 𝑧 = cosh 𝑧,
𝑑𝑧
𝑑
cosh 𝑧 = sinh 𝑧
𝑑𝑧
(𝒄) Zeros of 𝐜𝐨𝐬𝐡 𝒛 and 𝐬𝐢𝐧𝐡 𝒛. The zeros of cosh 𝑧 and sinh 𝑧 are
evaluated by the same way used for cos 𝑧 and sin 𝑧. This is illustrated
by the following example.

Example 1 Solve the equations

(𝒊) cosh 𝑧 = 0, (𝒊𝒊) sinh 𝑧 = 0

Solution
(𝒊) cosh 𝑧 = cosh(𝑥 + 𝑖𝑦)
= cosh 𝑥 cosh 𝑖𝑦 + sinh 𝑥 sinh 𝑖𝑦
= cosh 𝑥 cos 𝑦 + 𝑖 sinh 𝑥 sin 𝑦 = 0
Thus
cosh 𝑥 cos 𝑦 = 0, sinh 𝑥 sin 𝑦 = 0
 Chapter 3 Elementary Functions 93

Since (cosh 𝑥 > 0), then cos 𝑦 = 0, i.e.,

𝜋
𝑦 = (2𝑘 + 1) , (𝑘 = 0, ±1, ±2, ⋯ )
2
But for these values of 𝑦, from the second equation, sin 𝑦 ≠ 0, and thus
sinh 𝑥 = 0, i.e., 𝑥 = 0. Therefore the zeros of cosh 𝑧 are
𝜋
𝑧 = 𝑖(2𝑘 + 1) , (𝑘 = 0, ±1, ±2, ⋯ ) (5)
2
(ii) In like manner we find that the zeros of sinh 𝑧 are
𝑧 = 𝑖𝑘𝜋, (𝑘 = 0, ±1, ±2, ⋯ ) (6)
(d) tanh 𝑧 and sech 𝑧 are analytic in any domain where cosh 𝑧 ≠ 0,
i.e. they are analytic functions for all 𝑧 except at the points 𝑧 =
𝜋
𝑖(2𝑘 + 1) 2 , (𝑘 = 0, ±1, ±2, ⋯ ), and
𝑑
tanh 𝑧 = sech2 𝑧 (7)
𝑑𝑧
𝑑
sech 𝑧 = − sech 𝑧 tanh 𝑧 (8)
𝑑𝑧
Also coth 𝑧 and csch 𝑧 are analytic in any domain where sinh 𝑧 ≠ 0,
i.e. they are analytic functions for all 𝑧 except at the points 𝑧 =
𝑖𝑘𝜋, (𝑘 = 0, ±1, ±2, ⋯ ), and
𝑑
coth 𝑧 = − csch2 𝑧, (9)
𝑑𝑧
𝑑
csch 𝑧 = − csch 𝑧 coth 𝑧 (10)
𝑑𝑧
(𝒆) Periodicity of 𝐜𝐨𝐬𝐡 𝐳 and 𝐬𝐢𝐧𝐡 𝐳.
The functions sinh 𝑧 and cosh 𝑧 are periodic with period 2𝜋𝑖; but their
ratio tanh 𝑧 is periodic with period 𝜋𝑖.
(𝒇) The familiar hyperbolic identities are also the same in the complex
case.
1. cosh(−𝑧) = cosh 𝑧, sinh(−𝑧) = − sinh 𝑧
 94 Chapter 3 Elementary Functions

2. cosh2 𝑧 − sinh2 𝑧 = 1
3. sinh(𝑧1 ± 𝑧2 ) = sinh 𝑧1 cosh 𝑧2 ± cosh 𝑧1 sinh 𝑧2
4. cosh(𝑧1 ± 𝑧2 ) = cosh 𝑧1 cosh 𝑧2 ± sinh 𝑧1 sinh 𝑧2
5. sinh 2𝑧 = 2 sinh 𝑧 cosh 𝑧
6. cosh 2𝑧 = cosh2 𝑧 + sinh2 𝑧
7. sinh(𝑧 + 2𝜋𝑖) = sinh 𝑧, (𝑃𝑒𝑟𝑖𝑜𝑑𝑖𝑐 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦)
8. cosh(𝑧 + 2𝜋𝑖) = cosh 𝑧, (𝑃𝑒𝑟𝑖𝑜𝑑𝑖𝑐 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦)
9. tanh(𝑧 + 𝜋𝑖) = tanh 𝑧, (𝑃𝑒𝑟𝑖𝑜𝑑𝑖𝑐 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦)

Example 2 Compute sinh(4 − 3𝑖) in the form 𝑢 + 𝑖𝑣.

Solution
sinh(4 − 3𝑖) = sinh 4 cosh 3𝑖 − cosh 4 sinh 3𝑖
= sinh 4 cos 3 − 𝑖 cosh 4 sin 3

Example 3 Solve the equations

1
(𝒊) cosh 𝑧 = , (𝒊𝒊) sinh 𝑧 = 𝑖
2

1
Solution (𝒊) The equation cosh 𝑧 = 2 gives
𝑒 𝑧 + 𝑒 −𝑧 1
=  𝑒 𝑧 + 𝑒 −𝑧 = 1 (× 𝑒 𝑧 )  𝑒 2𝑧 − 𝑒 𝑧 + 1 = 0
2 2
Thus
1 ± √1 − 4 1 √3
𝑒𝑧 = = ±𝑖 ,
2 2 2
from which we obtain
 Chapter 3 Elementary Functions 95

1 √3
𝑒𝑧 = +𝑖 = 𝑒 𝑖(𝜋⁄3+2𝑘𝜋) , (𝑘 = 0, ±1, ±2, ⋯ ) (11)
2 2
or
1 √3
𝑒𝑧 = −𝑖 = 𝑒 𝑖(−𝜋⁄3+2𝑘𝜋) , (𝑘 = 0, ±1, ±2, ⋯ ) (12)
2 2

𝑦 𝑦
1 √3 1⁄2
 +𝑖 − 𝜋⁄3 𝑥
2 2

√3⁄2 √3⁄2

𝜋⁄3 1 √3
1⁄2  −𝑖
𝑥 2 2

1
− 𝑖 23
√
1
+ 𝑖 23
√
Fig. 3.20 z = Fig. 3.21 z =
2 2

The solution of equation (11) is given by

𝑧 = 𝑖(𝜋⁄3 + 2𝑘𝜋), (𝑘 = 0, ±1, ±2, ⋯ )
and that of (12) is
𝑧 = −𝑖(𝜋⁄3 + 2𝑘𝜋), (𝑘 = 0, ±1, ±2, ⋯ )
1
Finally, the solution of cosh 𝑧 = 2 put in the form

𝑧 = 𝑖(± 𝜋⁄3 + 2𝑘𝜋), (𝑘 = 0, ±1, ±2, ⋯ )

(𝒊𝒊) From the equation sinh 𝑧 = 𝑖, we get
𝑒 𝑧 − 𝑒 −𝑧
= 𝑖  𝑒 𝑧 − 𝑒 −𝑧 = 2𝑖 (× 𝑒 𝑧 )  𝑒 2𝑧 − 2𝑖𝑒 𝑧 − 1 = 0
2
Hence,
𝑦
2𝑖 ± √(2𝑖)2 + 4 𝑖
𝑒𝑧 = =𝑖
2 1
𝑖(𝜋⁄2+2𝑘𝜋)
=𝑒 , (𝑘 = 0, ±1, ±2, ⋯ ),
𝑥
Therefore, the solution of the equation
Fig. 3.22 𝑧 = 𝑖
sinh 𝑧 = 𝑖 is
𝑧 = 𝑖(𝜋⁄2 + 2𝑘𝜋), (𝑘 = 0, ±1, ±2, ⋯ )
 96 Chapter 3 Elementary Functions

Example 4 Determine the constant 𝑎 such that the function 𝑢(𝑥, 𝑦) =

cos 𝑎𝑥 cosh 𝑦 is harmonic and find its conjugate harmonic.

Solution
𝑢𝑥 (𝑥, 𝑦) = −𝑎 sin 𝑎𝑥 cosh 𝑦 , 𝑢𝑥𝑥 (𝑥, 𝑦) = −𝑎2 cos 𝑎𝑥 cosh 𝑦
𝑢𝑦 (𝑥, 𝑦) = cos 𝑎𝑥 sinh 𝑦 , 𝑢𝑦𝑦 (𝑥, 𝑦) = cos 𝑎𝑥 cosh 𝑦
Since 𝑢(𝑥, 𝑦) is harmonic by assumption, then
∇2 𝑢 = 𝑢𝑥𝑥 + 𝑢𝑦𝑦 = 0
 −𝑎2 cos 𝑎𝑥 cosh 𝑦 + cos 𝑎𝑥 cosh 𝑦 = 0  𝑎 = ±1
Using the Cauchy-Riemann equations
𝑣𝑦 = 𝑢𝑥 = − sin 𝑥 cosh 𝑦,
𝑣𝑥 = − 𝑢𝑦 = − cos 𝑥 sinh 𝑦
Hence
𝑥 𝑦
𝑣 = ∫ 𝑣𝑥 (𝑥, 𝑦)𝑑𝑥 + ∫ 𝑣𝑦 (0, 𝑦)𝑑𝑦
0 0
𝑥 𝑦
= ∫ − cos 𝑥 sinh 𝑦 𝑑𝑥 + ∫ 0𝑑𝑦
0 0

= − sin 𝑥 sinh 𝑦 + 𝑐
Finally, the analytic function is
𝑓(𝑧) = 𝑢(𝑧, 0) + 𝑖𝑣(𝑧, 0)
= cos 𝑧 cosh 0 + 𝑖(− sin 𝑧 sinh 0 + 𝑐)
= cos 𝑧 + 𝑖𝑐
3.6 Inverse trigonometric and hyperbolic functions
Inverses of the trigonometric and hyperbolic functions can be described
in terms of logarithms. We define the inverse cosine 𝑤 = cos −1 𝑧 such
that:
 Chapter 3 Elementary Functions 97

𝑤 = cos−1 𝑧  𝑧 = cos 𝑤
In terms of the exponential function, we get
𝑒 𝑖𝑤 + 𝑒 −𝑖𝑤
𝑧 = cos 𝑤 = ,
2
or
𝑒 𝑖𝑤 + 𝑒 −𝑖𝑤 = 2𝑧
Multiply both sides of the last equation by 𝑒 𝑖𝑤 we get

𝑒 2𝑖𝑤 − 2𝑧𝑒 𝑖𝑤 + 1 = 0,
which is quadratic in 𝑒 𝑖𝑤 . We find that
2𝑧 + √4𝑧 2 − 4
𝑒 𝑖𝑤 = = 𝑧 + √𝑧 2 − 1
2
where √𝑧 2 − 1 is a double-valued function of 𝑧. So we have

𝑖𝑤 = ln (𝑧 + √𝑧 2 − 1),

which gives

cos−1 𝑧 = −𝑖 ln (𝑧 + √𝑧 2 − 1) (1)

Similarly, the inverse sine and inverse tangent functions are found to be

sin−1 𝑧 = −𝑖 ln (𝑖𝑧 + √1 − 𝑧 2 ) (2)

𝑖 𝑖+𝑧
tan−1 𝑧 = ln (3)
2 𝑖−𝑧
The inverse of the hyperbolic functions can be written in the form:

cosh−1 𝑧 = ln (𝑧 + √𝑧 2 − 1) (4)

sinh−1 𝑧 = ln (𝑧 + √𝑧 2 + 1) (5)
1 1+𝑧
tanh−1 𝑧 = ln (6)
2 1−𝑧
 98 Chapter 3 Elementary Functions

Example 1 Find the values of

(𝒊) tan−1(1 + 𝑖) (𝒊𝒊) cosh−1 (−1)

Solution (𝒊) The direct substitution in (𝟑) gives

𝑖 𝑖+𝑧 𝑖 𝑖+1+𝑖
tan−1 𝑧 = ln  tan−1(1 + 𝑖) = ln
2 𝑖−𝑧 2 𝑖−1−𝑖
Hence, 𝑦
𝑖 1 + 2𝑖
tan−1(1 + 𝑖) = ln ( ) −𝟏 𝛉
2 −1 𝑥
𝑖
= ln(−1 − 2𝑖)
2 −𝟏 − 𝟐𝒊  −𝟐 −

𝑖 −1
= ln(√5𝑒 𝑖(𝜋+tan 2+2𝑘𝜋) ) Fig. 3.23 𝐳 = −𝟏 − 𝟐𝐢
2
𝑖
= (ln √5 + 𝑖(𝜋 + tan−1 2 + 2𝑘𝜋)), (𝑘 = 0, ±1, ±2, ⋯ )
2

(𝒊𝒊) Substitute in equation (4), we get for 𝑦

every, for all 𝑘 = 0, ±1, ±2, ⋯
cosh−1 (−1) = ln(−1) 𝝅
𝒛 = −𝟏 𝑥
𝑖(𝜋+2𝑘𝜋)
= ln(𝑒 ) = 𝑖(𝜋 + 2𝑘𝜋)
Fig. 3.24 𝑧 = −1

Example 2 Solve the equation sin 𝑧 = 2.

𝑦
 (𝟐 + √𝟑)𝑖
Solution 𝒛 = 𝐬𝐢𝐧−𝟏 𝟐 = −𝒊 𝐥𝐧(𝟐𝒊 + √𝟏 − 𝟒)

= −𝑖 ln[(2 + √3)𝑖]
𝑥
= −𝑖 ln ((2 + √3)𝑒 𝑖(𝜋⁄2+2𝑘𝜋) )
Fig. 3.25 𝒛 = (𝟐 + √𝟑)𝒊
= −𝑖[ln(2 + √3) + 𝑖(𝜋⁄2 + 2𝑘𝜋)]
= (𝜋⁄2 + 2𝑘𝜋) − 𝑖 ln(2 + √3) , 𝑘 = 0, ±1, ±2, ⋯
 Chapter 3 Elementary Functions 99

 Derivatives of inverse functions

The derivatives of the inverse functions considered above can be found
by implicit differentiation. For example, to find the derivative of the
inverse cosine function 𝑤 = cos −1 𝑧, we begin by differentiating 𝑧 =
cos 𝑤:
𝑑𝑤 1 1
= =
𝑑𝑧 𝑑𝑧⁄𝑑𝑤 − sin 𝑤
Using the identity sin2 𝑤 + cos 2 𝑤 = 1, we get
𝑑𝑤 𝑑 −1
= cos −1 𝑧 =
𝑑𝑧 𝑑𝑧 √1 − 𝑧 2
Similarly, we can prove the formulas of the derivatives of the other
inverse trigonometric functions as shown in the following table:

𝒅 −𝟏 𝒅 𝟏
𝐜𝐨𝐬 −𝟏 𝒛 = 𝐜𝐨𝐬𝐡−𝟏 𝒛 =
𝒅𝒛 √𝟏 − 𝒛𝟐 𝒅𝒛 √𝒛𝟐 − 𝟏

𝒅 𝟏 𝑑 1
𝐬𝐢𝐧−𝟏 𝒛 = sinh−1 𝑧 =
𝒅𝒛 √𝟏 − 𝒛𝟐 𝑑𝑧 √1 + 𝑧 2

𝒅 𝟏 𝑑 1
𝐭𝐚𝐧−𝟏 𝒛 = tanh−1 𝑧 =
𝒅𝒛 𝟏 + 𝒛𝟐 𝑑𝑧 1 − 𝑧2

𝒅 −𝟏 𝑑 −1
𝐜𝐨𝐭 −𝟏 𝒛 = coth−1 𝑧 =
𝒅𝒛 𝟏 + 𝒛𝟐 𝑑𝑧 1 − 𝑧2

𝒅 𝟏 𝑑 −1
𝐬𝐞𝐜 −𝟏 𝒛 = sech−1 𝑧 =
𝒅𝒛 𝒛√𝒛𝟐 − 𝟏 𝑑𝑧 𝑧√1 − 𝑧 2

𝒅 −𝟏 𝑑 −1
𝐜𝐬𝐜 −𝟏 𝒛 = csch−1 𝑧 =
𝒅𝒛 𝒛√𝒛𝟐 − 𝟏 𝑑𝑧 𝑧√𝑧 2 + 1

Table 1 Derivatives of inverse functions

 100 Chapter 3 Elementary Functions

Exercise 3.3
In problems 1 − 8, express the given quantity in the form 𝑢 + 𝑖𝑣.
𝟏. sin(−2𝑖) 𝟐. cos(3𝑖)
𝟑. cos(2 − 4𝑖) 𝟒. sinh(2 + 𝑖)
𝟓. cos(−2 + 3𝑖) 𝟔. tan 5𝑖
𝜋
𝟕. csc(1 + 𝑖) 𝟖. cot ( + 5𝑖)
2
In problems 9 − 14, find all values of 𝑧 such that
𝟗. cosh 𝑧 = 0 𝟏𝟎. cos 𝑧 = −3𝑖
𝟏𝟏. cos 𝑧 = sin 𝑧 𝟏𝟐. cos 𝑧 = 𝑖 sin 𝑧
𝟏𝟑. sinh 𝑧 = 𝑖 𝟏𝟒. cosh 𝑧 = 0.5
In problems 15 − 21, prove that
𝟏𝟓. cos 𝑧 = cos 𝑧 𝟏𝟔. sin 𝑧 = sin 𝑧
𝟏𝟕. sin 𝑖𝑧 = sin(𝑖𝑧) 𝟏𝟖. cos 𝑖𝑧 = cos(𝑖𝑧)

𝟏𝟗. tan 𝑖𝑧 = − tan(𝑖𝑧)

𝟐𝟎. sinh 𝑧 and cosh 𝑧 are periodic functions of period 2𝜋𝑖.
𝟐𝟏. tanh 𝑧 is a periodic function of period 𝜋𝑖.
𝟐𝟐. Show that
𝑓(𝑧) = −(2𝑥𝑦 + 5𝑥) + 𝑖(𝑥 2 − 5𝑦 − 𝑦 2 )
is analytic for all 𝑧, while neither cos 𝑧 nor sin 𝑧 is analytic function of
𝑧 anywhere.
In problems 23 − 26, find an analytic function 𝑓(𝑧) = 𝑢 + 𝑖𝑣 such
that
𝑦
𝟐𝟑. 𝑢 = sin 𝑥 cosh 𝑦 𝟐𝟒. 𝑣 =
𝑥2 + 𝑦2
𝟐𝟓. 𝑢 = 𝑒 𝑥⁄3 cos(𝑦⁄3) 𝟐𝟔. 𝑣 = cos 𝑥 cosh 𝑦
In problems 27 − 34, find all values of each of the given quantity
 Chapter 3 Elementary Functions 101

𝟐𝟕. cosh−1 𝑖 𝟐𝟖. tan−1(1 + 𝑖)

𝟐𝟗. cosh−1 (−2) 𝟑𝟎. cos−1 (1⁄2)
𝟑𝟏. cos−1 (2𝑖) 𝟑𝟐. sinh−1 (4⁄3)
𝟑𝟑. tanh−1(1 + 2𝑖) 𝟑𝟒. cos−1 √2
𝟑𝟓. Solve the equation sin 𝑧 = 3 for 𝑧
(𝒊) by identifying real and imaginary component of its members;
(𝒊𝒊) by using the logarithmic form of sin−1 𝑧.
In problems 36 − 40, prove that
𝟑𝟔. cosh−1 𝑧 = ln(𝑧 + √𝑧 2 − 1).

𝟑𝟕. sinh−1 𝑧 = ln (𝑧 + √𝑧 2 + 1)

𝟑𝟖 sin−1 𝑧 = −𝑖 ln (𝑖𝑧 + √1 − 𝑧 2 )
𝑖 𝑖+𝑧
𝟑𝟗 tan−1 𝑧 = ln
2 𝑖−𝑧
1 1+𝑧
𝟒𝟎 tanh−1 𝑧 = ln
2 1−𝑧
4. Complex Integrations

In chapter 3 we saw how to differentiate a complex function. We

now turn our attention to the problem of integrating complex functions.
We will find that integrals of analytic functions are well behaved and
that many properties from calculus carry over to the complex case.

4.1 Contour integrals

We begin by introducing the definition of the complex line integral. To
do that we need the following notations. The definition of definite
integrals, or line integrals, of complex function 𝑓(𝑧) of 𝑧 = 𝑥 + 𝑖𝑦
along a curve γ in the complex plane needs a short preparation on curves
in the complex plane, as follows.

Definition 4.1 (Path or contour of integration) In the complex integration

we integrate a complex function 𝑓(𝑧) along a curve γ, called the
𝑝𝑎𝑡ℎ 𝑜𝑓 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛. These curves are represented in the
parametric form
γ: 𝑧(𝑡) = 𝑥(𝑡) + 𝑖𝑦(𝑡), α ≤ 𝑡 ≤ β,
where 𝑡 is a real parameter.

Definition 4.2 (Smooth curve) A smooth curve is the curve which has a
continuous and nonzero derivative
𝑧̇ (𝑡) = 𝑥̇ (𝑡) + 𝑖𝑦̇ (𝑡)

4.2 Definition of complex line integration

Let 𝑓(𝑧) be a regular function which is defined at each point of a
smooth curve γ: 𝑧 = 𝑥 + 𝑖𝑦, joining between two points 𝑎 and 𝑏. We
subdivide the interval 𝑎 ≤ 𝑧 ≤ 𝑏 by points
 Chapter 4 Complex Integration 103

𝑧0 (= 𝑎), 𝑧1 , 𝑧2 , ⋯ , 𝑧𝑛−1 , 𝑧𝑛 (= 𝑏)
On each portion of subdivision of γ we choose an arbitrary point, say,
a point 1 between 𝑧0 and 𝑧1 , a point 2 between 𝑧1 and 𝑧2 , etc. (See
Fig. 4.1). Then we form the sum
𝑛 𝑛

𝑆𝑛 = ∑ 𝑓(𝑗 ) (𝑧𝑗 − 𝑧𝑗−1 ) = ∑ 𝑓(𝑗 ) ∆𝑧𝑗

𝑗=1 𝑗=1

𝑦
 𝒛 =𝒃
𝑧𝑛−1
 𝑧𝑗

 𝑧𝑗−1 𝑗

 𝑧2
 𝑧1
 𝒛=𝒂

𝑥
Fig. 4.1. Contour of integration

But |∆𝑧𝑗 | = |𝑧𝑗 − 𝑧𝑗−1 | → 0 as 𝑛 → ∞ and in this case the limit of

𝑆𝑛 is called the 𝑙𝑖𝑛𝑒 or 𝑐𝑜𝑛𝑡𝑜𝑢𝑟 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 of 𝑓(𝑧) over the curve γ and
is denoted by

∫ 𝑓(𝑧) 𝑑𝑧 (1)
γ

If γ is a closed path, we use the standard notation

∮ 𝑓(𝑧) 𝑑𝑧 (2)
γ

4.3 Properties of complex line integrals

Complex integrals have properties that are similar to those of real
integrals. If (𝑎 + 𝑖𝑏) denotes a constant complex number, then
 104 Chapter 4 Complex Integration

(𝒂) ∫ (𝑎 + 𝑖𝑏)𝑓(𝑧) 𝑑𝑧 = (𝑎 + 𝑖𝑏) ∫ 𝑓(𝑧) 𝑑𝑧

γ γ

(𝒃) ∫ (𝑓1 (𝑧) + 𝑓2 (𝑧)) 𝑑𝑧 = ∫ 𝑓1 (𝑧) 𝑑𝑧 + ∫ 𝑓2 (𝑧) 𝑑𝑧

γ γ γ

(𝒄) ∫ 𝑓(𝑧) 𝑑𝑧 = ∫ 𝑓(𝑧) 𝑑𝑧 + ∫ 𝑓(𝑧) 𝑑𝑧 , (γ = γ1 ∪ γ2 )

γ γ1 γ2
𝑏 𝑎
(𝒅) ∫ 𝑓(𝑧) 𝑑𝑧 = − ∫ 𝑓(𝑧) 𝑑𝑧
𝑎 𝑏

4.4 Evaluation of complex line integrals (Direct method)

We turn now to the question of evaluating the complex line integral (1).
To compute this integral, we use the direct method which takes one of
the following two forms:
 First form
Let the contour γ is given by: γ: 𝑧 = 𝑥 + 𝑖𝑦, then
𝑑𝑧 = 𝑑𝑥 + 𝑖𝑑𝑦
𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦)
So we have

∫ 𝑓(𝑧) 𝑑𝑧 = ∫ (𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦)) (𝑑𝑥 + 𝑖𝑑𝑦)

γ γ

= ∫ (𝑢𝑑𝑥 − 𝑣𝑑𝑦) + 𝑖 ∫ (𝑣𝑑𝑥 + 𝑢𝑑𝑦) (3)

γ γ

 Second form
Let the contour γ is given by the parametric form:
γ: 𝑧(𝑡) = 𝑥(𝑡) + 𝑖𝑦(𝑡), α ≤ 𝑡 ≤ β,
then we obtain the line integral in the form
 Chapter 4 Complex Integration 105

β
∫ 𝑓(𝑧) 𝑑𝑧 = ∫ 𝑓(𝑧(𝑡)) 𝑧̇ (𝑡)𝑑𝑡 (4)
γ α

Note that the "smooth" condition guarantees that 𝑧̇ (𝑡) is continuous

and, hence, that the integral exists.

Example 1 Evaluate ∫γ 𝑥 𝑑𝑧, from 𝑧 = 0 to 𝑧 = 1 + 𝑖,

(𝒂) along the path γ = γ∗ in Fig. 4.2,

(𝒃) along γ consisting of γ1 and γ2 as shown in Fig. 4.2.

Solution 𝑦
∗ 𝑧 =1+𝑖
(𝒂) Along the path γ = γ 
γ∗ can be represented in the parametric form as
𝑧 = 𝑡 + 𝑖𝑡, (0 ≤ 𝑡 ≤ 1) γ2

𝑧̇ (𝑡) = 1 + 𝑖 γ1 𝑥

𝑓(𝑧(𝑡)) = 𝑡 Fig. 4.2 contour of Ex 1

𝑓(𝑧(𝑡))𝑧̇ (𝑡) = 𝑡 + 𝑖𝑡
Therefore, we have
β 1
∫ 𝑓(𝑧) 𝑑𝑧 = ∫ 𝑓(𝑧(𝑡)) 𝑧̇ (𝑡)𝑑𝑡 = ∫ (𝑡 + 𝑖𝑡)𝑑𝑡
γ∗ α 0
1
𝑡2 𝑡2 1 1
=[ +𝑖 ] = +𝑖
2 2 0 2 2
(𝒂) Along the path γ = γ1 ∪ γ2
 γ1 can be represented by 𝑧 = 𝑡, (0 ≤ 𝑡 ≤ 1)
Therefore,
𝑧̇ (𝑡) = 1, 𝑓(𝑧(𝑡)) = 𝑡, 𝑓(𝑧(𝑡))𝑧̇ (𝑡) = 𝑡,
 106 Chapter 4 Complex Integration

1
1
∫ 𝑓(𝑧) 𝑑𝑧 = ∫ 𝑡𝑑𝑡 =
γ1 0 2

 γ2 can be represented by 𝑧 = 1 + 𝑖𝑡, (0 ≤ 𝑡 ≤ 1)

𝑧̇ (𝑡) = 𝑖, 𝑓(𝑧(𝑡)) = 1, 𝑓(𝑧(𝑡))𝑧̇ (𝑡) = 𝑖,
1
∫ 𝑓(𝑧) 𝑑𝑧 = ∫ 𝑖𝑑𝑡 = 𝑖
γ2 0

Therefore
1
∫ 𝑓(𝑧) 𝑑𝑧 = ∫ 𝑓(𝑧) 𝑑𝑧 + ∫ 𝑓(𝑧) 𝑑𝑧 = +𝑖
γ γ1 γ2 2
Note that this result differs from the result in (𝒂).

4.5 Cauchy’s integral theorem

In this section and the next one, we shall learn an elegant, powerful, and
a useful theorem and formula for physicists and engineers. Both the
theorem and the formula are formulated by Augustin Louis Cauchy, so
they are named Cauchy’s integral theorem and Cauchy’s integral
formula. To begin with we define the following concept.

Definition 4.1 A domain D is said to be 𝑠𝑖𝑚𝑝𝑙𝑦 𝑐𝑜𝑛𝑛𝑒𝑐𝑡𝑒𝑑 if every

simple closed contour γ lying entirely in 𝐷 can be shrunk to a point
without leaving 𝐷. A domain that is not simply connected is called
a 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑐𝑜𝑛𝑛𝑒𝑐𝑡𝑒𝑑.

𝑦 𝑦

𝑥 𝑥
simply connected domain multiply connected domain
Fig. 4.3
 Chapter 4 Complex Integration 107

In other words, in a simply connected domain every simple closed curve

lying entirely within it encloses only points of the domain D. On the
other side, a multiply connected domain has holes in it. See Fig. 4.3.
We call a domain with one hole 𝑑𝑜𝑢𝑏𝑙𝑦 𝑐𝑜𝑛𝑛𝑒𝑐𝑡𝑒𝑑, a domain with two
holes 𝑡𝑟𝑖𝑝𝑙𝑦 𝑐𝑜𝑛𝑛𝑒𝑐𝑡𝑒𝑑, and so on.

4.5.1 Cauchy’s integral theorem for simply connected domains

Theorem 4.1 (Cauchy’s integral theorem). If 𝑓(𝑧) is analytic function of

𝑧 in a simply connected domain 𝐷 and if 𝑓 ′ (𝑧) is continuous at each
point within and on a closed contour γ in 𝐷, then

∮ 𝑓(𝑧) 𝑑𝑧 = 0 (1)
γ

(i.e., the integral of the function round a closed contour is zero).

Proof. Writing 𝑧 = 𝑥 + 𝑖𝑦, and

𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦),
the line integral of 𝑓(𝑧) is given by

∮ 𝑓(𝑧) 𝑑𝑧 = ∮ (𝑢 + 𝑖𝑣)(𝑑𝑥 + 𝑖𝑑𝑦)

γ γ

= ∮ (𝑢𝑑𝑥 − 𝑣𝑑𝑦) + 𝑖 ∮ (𝑣𝑑𝑥 + 𝑢𝑑𝑦)

γ γ
′ (𝑧)
Since 𝑓(𝑧) is analytic, then 𝑓 exists in 𝐷. Moreover, 𝑓 ′ (𝑧) is
assumed to be continuous, so 𝑢 and 𝑣 have continuous partial
derivatives in 𝐷. Apply Green’s theorem for the integral (1),
𝜕𝑣 𝜕𝑢
∮ (𝑢𝑑𝑥 − 𝑣𝑑𝑦) = ∬ (− − ) 𝑑𝑥𝑑𝑦 , (2)
γ 𝑅 𝜕𝑥 𝜕𝑦

𝜕𝑢 𝜕𝑣
∮ (𝑣𝑑𝑥 + 𝑢𝑑𝑦) = ∬ ( − ) 𝑑𝑥𝑑𝑦 , (3)
γ 𝑅 𝜕𝑥 𝜕𝑦
 108 Chapter 4 Complex Integration

where 𝑅 is the region bounded by γ. From the Cauchy-Riemann

equations we have
𝜕𝑢 𝜕𝑣 𝜕𝑢 𝜕𝑣
= , =−
𝜕𝑥 𝜕𝑦 𝜕𝑦 𝜕𝑥
Hence both of the integrals in (2) and (3) will be identically zero, and
this completes the proof. ∎

Remark 1 Since the interior of a simply closed contour is a simply

connected domain, then Theorem 4.1 can be stated in
slightly more practical manner:

If 𝑓(𝑧) is analytic at all points within and on a simple closed contour

γ, then

∮ 𝑓(𝑧) 𝑑𝑧 = 0 (4)
γ

Example 1 For any closed path γ,

∮ sin 𝑧 𝑑𝑧 = ∮ cos 𝑧 𝑑𝑧 = ∮ 𝑒 𝑧 𝑑𝑧 = ∮ 𝑧 3 𝑑𝑧 = 0
γ γ γ γ

𝑑𝑧
Example 2 Evaluate ∮γ . where  is the unit circle |𝑧| = 1, counter
𝑧

clockwise.

Solution Using the parametric form of the unit circle in the form
𝑧 = 𝑒 𝑖𝑡 , 0 ≤ 𝑡 ≤ 2𝜋
Thus
𝑧̇ (𝑡) = 𝑖𝑒 𝑖𝑡 , 𝑓(𝑧(𝑡)) = 𝑒 −𝑖𝑡 , 𝑓(𝑧(𝑡))𝑧̇ (𝑡) = 𝑒 −𝑖𝑡 (𝑖𝑒 𝑖𝑡 ) = 𝑖
 Chapter 4 Complex Integration 109

2𝜋
𝑑𝑧
∮ = ∫ 𝑖𝑑𝑡 = 2𝜋𝑖
γ 𝑧 0
1
(𝑧 is not analytic in the given domain, so Cauchy’s integral theorem

cannot be applied).

Example 3 Evaluate
2𝑧 − 8
∮ 𝑑𝑧,
γ 𝑧 2 − 2𝑧
where γ is the unit circle |𝑧| = 1, counter clockwise.

Solution
2𝑧 − 8 2𝑧 − 8 4 2
∮ 2
𝑑𝑧 = ∮ 𝑑𝑧 = ∮ ( − ) 𝑑𝑧
γ 𝑧 − 2𝑧 γ 𝑧(𝑧 − 2) γ 𝑧 (𝑧 − 2)

4 2
=∮ 𝑑𝑧 − ∮ 𝑑𝑧 (5)
γ 𝑧 γ (𝑧 − 2)

The first integral in the right hand side of (5) is computed in Example
2
2. The second integral equals zero since the function (𝑧−2)
is analytic

throughout the unit circle |𝑧| = 1. Then we have

2𝑧 − 8
∮ 𝑑𝑧 = 4 × 2𝜋𝑖 − 0 = 8𝜋𝑖
γ 𝑧 2 − 2𝑧

4.5.2 Cauchy’s integral theorem for multiply connected domains

Cauchy’s integral theorem can be extended to be applied for multiply

connected domains. We first explain this for a doubly connected
domain 𝐷 with outer boundary contour γ1 and inner γ2 as shown in Fig.
4.4.
 110 Chapter 4 Complex Integration

Theorem 4.2 Suppose that γ1 and γ2 are simply closed curves with a
positive orientation. If 𝑓(𝑧) is analytic function of 𝑧 on γ1 and γ2
and at each point interior to γ1 but exterior to γ2 , then

∮ 𝑓(𝑧) 𝑑𝑧 = ∮ 𝑓(𝑧) 𝑑𝑧
γ1 γ2

Proof. As show in Fig. 4.5, we cut 𝐷 by two cuts γ̃1 and γ̃2 into two
simply connected domains 𝐷1 and 𝐷2 in which 𝑓(𝑧) is analytic. By
Cauchy’s theorem the integral over the entire boundary of 𝐷1 is zero,
and so is that over the boundary of 𝐷2 , and thus their sum. In this sum
the integrals over the cuts γ̃1 and γ̃2 cancel because we integrate over
them in both directions. Thus we have

γ2 γ2 γ෤1
𝐷2
γ1 γ෤2 γ1
𝐷 𝐷1

Fig. 4.4 Doubly connected domain

∫ 𝑓(𝑧) 𝑑𝑧 + ∫ 𝑓(𝑧) 𝑑𝑧 − ∫ 𝑓(𝑧) 𝑑𝑧 − ∫ 𝑓(𝑧) 𝑑𝑧 = 0

γ1 ⏟γ̃1 ̃2
γ γ2
𝑐𝑎𝑛𝑐𝑒𝑙

Therefore

∫ 𝑓(𝑧) 𝑑𝑧 − ∫ 𝑓(𝑧) 𝑑𝑧 = 0,
γ1 γ2

which is the required proof. ∎

For domains of higher connectivity, the idea remains the same. Thus,
for example we have for a triply connected domain (see Fig. 4.5)
 Chapter 4 Complex Integration 111

γ෤2 γ2 γ෤1
γ2 𝐷2

γ3 𝐷
γ3 𝐷1 γ1
γ1

Fig. 4.5 Triply connected domain

∫ 𝑓(𝑧) 𝑑𝑧 = ∫ 𝑓(𝑧) 𝑑𝑧 + ∫ 𝑓(𝑧) 𝑑𝑧

γ1 γ2 γ3

4.6 Cauchy’s integral formula

The most important consequence of Cauchy’s integral theorem is

Cauchy’s integral formula. This formula is useful for evaluating
integrals. Equally important is its key in proving that analytic functions
have derivatives of all orders.

Theorem 4.3 (Cauchy’s integral formula) Let 𝑓(𝑧) be analytic function

of 𝑧 in a simply connected domain 𝐷. Then for any interior point 𝑧0
of a closed contour γ in 𝐷,
1 𝑓(𝑧)
𝑓(𝑧0 ) = ∮ 𝑑𝑧 , (1)
2𝜋𝑖 γ 𝑧 − 𝑧0
where the integral is taken counter clockwise.

Proof. Let γ0 be a circle about the point 𝑧0 ,

|𝑧 − 𝑧0 | = ρ,
whose radius ρ is small enough that γ0 is interior to γ (Fig. 4.6). The
function 𝑓(𝑧)⁄(𝑧 − 𝑧0 ) is analytic at all points within and on γ except
the point 𝑧0 . Hence according to Cauchy’s integral theorem for doubly
connected domain, we have
 112 Chapter 4 Complex Integration

𝑓(𝑧) 𝑓(𝑧)
∮ 𝑑𝑧 = ∮ 𝑑𝑧,
γ 𝑧 − 𝑧0 γ0 𝑧 − 𝑧0

where both integrals are taken counter clockwise. Since the integrals
around γ and γ0 are equal, we can write
𝑓(𝑧) 𝑓(𝑧) − 𝑓(𝑧0 ) + 𝑓(𝑧0 )
∮ 𝑑𝑧 = ∮ 𝑑𝑧
γ 𝑧 − 𝑧0 γ 𝑧 − 𝑧0

𝑑𝑧 𝑓(𝑧) − 𝑓(𝑧0 )
= 𝑓(𝑧0 ) ∮ +∮ 𝑑𝑧 (2)
γ0 𝑧 − 𝑧0 γ0 𝑧 − 𝑧0

ρ
𝐷 z0 
γ0
γ1

Fig. 4.6 Proof of Cauchy’s integral formula

From Fig. 4.6, we have

𝑧 − 𝑧0 = ρ𝑒 𝑖θ , 𝑑𝑧 = 𝑖ρ𝑒 𝑖θ 𝑑θ,
so that
𝑑𝑧 𝑖ρ𝑒 𝑖θ 𝑑θ
∮ =∮ = 𝑖 ∮ 𝑑θ = 2𝜋𝑖 (3)
γ0 𝑧 − 𝑧0 γ0 ρ𝑒 𝑖θ γ0

Since 𝑓(𝑧) is analytic, then it is continuous at the point 𝑧0 . Hence,

∀ε > 0 ∃δ > 0 𝑠. 𝑡.
|𝑓(𝑧) − 𝑓(𝑧0 )| < ε for all z in the disk |𝑧 − 𝑧0 | < δ
Choosing the radius  smaller than or equal to δ. Then
𝑓(𝑧) − 𝑓(𝑧0 ) 𝑓(𝑧) − 𝑓(𝑧0 )
|∮ 𝑑𝑧| ≤ ∮ | | |𝑑𝑧|
γ0 𝑧 − 𝑧0 γ0 𝑧 − 𝑧0
ε
< (2𝜋δ) = 2𝜋ε
δ
 Chapter 4 Complex Integration 113

Since ε can be chosen arbitrary small, it follows that the last integral in
equation (2) has the value zero and we have from equations (2) and
(3)
𝑓(𝑧)
∮ 𝑑𝑧 = 2𝜋𝑖𝑓(𝑧0 ),
γ 𝑧 − 𝑧0
and the theorem is proved. ∎

Remark The Cauchy’s integral formula can be used to evaluate

contour integrals. In this case a more practical
restatement of this formula is:

If 𝑓(𝑧) is analytic at all points within and on a simply closed contour

γ, and 𝑧0 is any point interior to γ, then
𝑓(𝑧)
∮ 𝑑𝑧 = 2𝜋𝑖𝑓(𝑧0 ) (4)
γ 𝑧 − 𝑧0

6𝑒 𝑧
Example 1 Evaluate ∮γ 𝑑𝑧 where γ is the contour
2𝑧−1

(𝒊) the circle |𝑧| = 1 (𝒊𝒊) the annulus 1 ≤ 𝑧 ≤ 2.

Solution
6𝑒 𝑧 𝑒𝑧
∮ 𝑑𝑧 = 3 ∮ 𝑑𝑧
γ 2𝑧 − 1 γ 𝑧 − 1⁄2

(𝒊) Since the function 𝑒 𝑧 is entire,


𝟏⁄𝟐
so that the integral of 𝑒 𝑧 over the
circle |𝑧| = 1 enclosing 𝑧0 = 1⁄2
Fig. 4.7 The annulus 1 ≤ 𝑧 ≤ 2
(See Fig. 4.7) is given by Cauchy’s
integral formula
 114 Chapter 4 Complex Integration

6𝑒 𝑧 𝑒𝑧
∮ 𝑑𝑧 = 3 ∮ 𝑑𝑧
γ 2𝑧 − 1 γ 𝑧 − 1⁄2

= 3(2𝜋𝑖𝑒 (1⁄2) ) = 6𝜋𝑖𝑒 (1⁄2)

(𝒊𝒊) The point 𝑧0 = 1⁄2 lies outside the contour 1 ≤ 𝑧 ≤ 2, so that, by
Cauchy’s integral theorem,
6𝑒 𝑧 𝑒𝑧
∮ 𝑑𝑧 = 3 ∮ 𝑑𝑧 = 0
γ 2𝑧 − 1 γ 𝑧 − 1⁄2

tan 𝑧 
Example 2 Evaluate ∮γ 𝑑𝑧, γ is
𝑧−𝑖
𝒊
the boundary of the triangle with
vertices −1,1,2𝑖.  
Fig. 4.8 the triangle with vertices −1,1,2𝑖.

Solution Here we have the point 𝑧0 = 𝑖, lies interior to the triangle with
vertices −1,1,2𝑖 (See Fig. 4.8). Thus the direct application of Cauchy’s
integral theorem gives
tan 𝑧
∮ 𝑑𝑧 = 2𝜋𝑖tan 𝑧|𝑧=𝑖 = 2𝜋𝑖 tan 𝑖 = −2𝜋 tanh 1
γ 𝑧−𝑖

𝑧2
Example 3 Evaluate ∮γ 𝑑𝑧 , where γ is the boundary of the
𝑧 2 +2𝑧+2

circle |𝑧| = 2.

Solution The given integral can be written as

𝑧2 𝑧2
∮ 2 𝑑𝑧 = ∮ 2
𝑑𝑧
γ 𝑧 + 2𝑧 + 2 γ (𝑧 + 2𝑧 + 1) + 1
𝑧2
=∮ 𝑑𝑧
γ (𝑧 + 1 + 𝑖)(𝑧 + 1 − 𝑖)
 Chapter 4 Complex Integration 115

We can express the integral around γ as

the sum of the integral around −1 + 𝑖
−1+ i
γ
and −1 − 𝑖 where the contours are small γ1
circles around each point as shown in
Fig. 4.10. Call these contours γ1 around −1 − i
`
−1 + 𝑖 and γ2 around −1 − 𝑖. By ` γ2

Cauchy’ integral theorem for multiply

Fig. 4.9
connected theorem we get

𝑧2 𝑧2 𝑧2
∮ 𝑑𝑧 = ∮ 𝑑𝑧 + ∮ 𝑑𝑧
γ 𝑧 2 + 2𝑧 + 2 2
γ1 𝑧 + 2𝑧 + 2
2
γ2 𝑧 + 2𝑧 + 2

This equation is expressed as

𝑧2
∮ 2 𝑑𝑧
γ 𝑧 + 2𝑧 + 2

𝑧 2 ⁄(𝑧 + 1 + 𝑖) 𝑧 2 ⁄(𝑧 + 1 − 𝑖)
=∮ 𝑑𝑧 + ∮ 𝑑𝑧
γ1 (𝑧 + 1 − 𝑖) γ2 (𝑧 + 1 + 𝑖)

= 2𝜋𝑖[𝑧 2 ⁄(𝑧 + 1 + 𝑖)]𝑧=−1+𝑖 + 2𝜋𝑖[𝑧 2 ⁄(𝑧 + 1 − 𝑖)]𝑧=−1−𝑖

(−1 + 𝑖)2 (−1 − 𝑖)2
= 2𝜋𝑖 [ ] + 2𝜋𝑖 [ ]
(−1 + 𝑖 + 1 + 𝑖) (−1 − 𝑖 + 1 − 𝑖)
−2𝑖 2𝑖
= 2𝜋𝑖 ( ) + 2𝜋𝑖 ( ) = −4𝜋𝑖
2𝑖 −2𝑖

4.7 The fundamental theorem for complex integral

In this section we use Cauchy’s integral theorem to establish the

existence of an indefinite integral 𝐹(𝑧) of a given analytic function
𝑓(𝑧), that is, 𝐹 ′ (𝑧) = 𝑓(𝑧).
 116 Chapter 4 Complex Integration

Theorem 4.4 If 𝑓(𝑧) is analytic in a simply connected domain 𝐷, then

there exists an indefinite integral 𝐹(𝑧) of 𝑓(𝑧) in 𝐷. However, this
indefinite integral 𝐹(𝑧) is analytic in 𝐷, and for all paths in 𝐷 joining
any two points 𝑧0 and 𝑧1 in 𝐷, the integral of 𝑓(𝑧) from 𝑧0 to 𝑧1 is
given by the formula
𝑧1
∫ 𝑓(𝑧)𝑑𝑧 = 𝐹(𝑧1 ) − 𝐹(𝑧0 ) (1)
𝑧0

Proof.

 𝑧 + ∆𝑧
γ2
 𝒛𝟏 𝒛

𝒛𝟎 
γ1  𝒛𝟏

Fig. 4.10 Fig. 4.11

If γ1 and γ2 are two contours connecting the points 𝑧0 and 𝑧1 and lying
entirely within 𝐷, then γ1 and γ2 together form a closed curve. Thus
from Cauchy’s integral theorem we have

∫ 𝑓(𝑧)𝑑𝑧 + ∫ 𝑓(𝑧)𝑑𝑧 = 0
γ1 γ2

1 f ( z ) d z   f ( z ) d z  0 ,
2

that is, the integral from the fixed point 𝑧0 to 𝑧

𝑧
𝐹(𝑧) = ∫ 𝑓()𝑑
𝑧0

has the same value for all such paths (see Fig. 4.10).
 Chapter 4 Complex Integration 117

We shall now see that this 𝐹(𝑧) is analytic in 𝐷 and 𝐹 ′ (𝑧) = 𝑓(𝑧). The
idea of doing this is as follows (see Fig. 4.11).
𝑧+∆𝑧 𝑧 𝑧+∆𝑧
𝐹(𝑧 + ∆𝑧) − 𝐹(𝑧) = ∫ 𝑓()𝑑 − ∫ 𝑓()𝑑 = ∫ 𝑓()𝑑
𝑧0 𝑧0 𝑧

Since
𝑓(𝑧) 𝑧+∆𝑧 1 𝑧+∆𝑧
𝑓(𝑧) = ∫ 𝑑 = ∫ 𝑓(z)𝑑,
∆𝑧 𝑧 ∆𝑧 𝑧
Then
𝐹(𝑧 + ∆𝑧) − 𝐹(𝑧) 1 𝑧+∆𝑧 1 𝑧+∆𝑧
− 𝑓(𝑧) = ∫ 𝑓()𝑑 − ∫ 𝑓(z)𝑑
∆𝑧 ∆𝑧 𝑧 ∆𝑧 𝑧
1 𝑧+∆𝑧
= ∫ [𝑓() − 𝑓(𝑧)]𝑑
∆𝑧 𝑧
Since 𝑓(𝑧) is analytic, it is continuous. Thus ∀ε > 0 ∃δ > 0 such that
|𝑓() − 𝑓(𝑧)| < ε iff | − 𝑧| < δ
Using Fatou’s lemma
𝐹(𝑧 + ∆𝑧) − 𝐹(𝑧) 1 𝑧+∆𝑧
| − 𝑓(𝑧)| = | ∫ [𝑓() − 𝑓(𝑧)]𝑑|
∆𝑧 ∆𝑧 𝑧
|𝑓() − 𝑓(𝑧)| 𝑧+∆𝑧 ε
≤ ∫ 𝑑 ≤ (∆𝑧) = ε
∆𝑧 𝑧 ∆𝑧
By the definition of limit and derivative, this proves that
𝐹(𝑧 + ∆𝑧) − 𝐹(𝑧)
𝐹 ′ (𝑧) = lim = 𝑓(𝑧)
∆𝑧→0 ∆𝑧
Since 𝑧 is any point in 𝐷, this implies that 𝐹(𝑧) is analytic in 𝐷 and is
an indefinite integral of 𝑓(𝑧) in 𝐷. Namely,

𝐹(𝑧) = ∫ 𝑓(𝑧)𝑑𝑧,

and the proof completes.

 118 Chapter 4 Complex Integration

Example 1 Evaluate
2+2𝑖 3+3𝜋𝑖
(𝒊) ∫ 𝑧 2 𝑑𝑧 (𝒊𝒊) ∫ 𝑒 𝑧⁄3 𝑑𝑧
0 3−𝑖𝜋
𝜋𝑖 2+2𝑖
1
(𝒊𝒊𝒊) ∫ cos 𝑧 𝑑𝑧 (𝒊𝒗) ∫ 𝑑𝑧
−𝜋𝑖 2−2𝑖 𝑧

Solution
2+2𝑖 2+2𝑖
2
𝑧3 8 16 16 16
(𝒊) ∫ 𝑧 𝑑𝑧 = | = (1 + 𝑖)3 = (−1 + 𝑖) = − +𝑖
0 3 0 3 3 3 3
3+3𝜋𝑖
3+3𝜋𝑖
𝑒 𝑧⁄3
(𝒊𝒊) ∫ 𝑒 𝑧⁄3
𝑑𝑧 = | = 3(𝑒 (3+3𝜋𝑖)⁄3 − 𝑒 (3−𝑖𝜋)⁄3 )
3−𝑖𝜋 1⁄3 3−𝑖𝜋

= 3(𝑒 (1+𝜋𝑖) − 𝑒 (1−𝑖𝜋⁄3) ) = 3(𝑒𝑒 𝜋𝑖 − 𝑒𝑒 −𝑖𝜋⁄3 )

= 3𝑒(𝑒 𝜋𝑖 − 𝑒 −𝑖𝜋⁄3 )
= 3𝑒(cos 𝜋 + 𝑖 sin 𝜋 − cos(𝜋⁄3) + 𝑖 sin(𝜋⁄3))
1 √3 9 3√3
= 3𝑒 (−1 − −𝑖 )=− 𝑒+𝑖 𝑒
2 2 2 2
𝜋𝑖
(𝒊𝒊𝒊) ∫ cos 𝑧 𝑑𝑧 = sin 𝑧|𝜋𝑖
−𝑖𝜋
−𝜋𝑖

= 2 sin 𝜋𝑖 = 2𝑖 sinh 𝜋
2+2𝑖
1
(𝒊𝒗) ∫ 𝑑𝑧 = ln 𝑧|2+2𝑖
2−2𝑖 𝑦
2−2𝑖 𝑧
 2 + 2𝑖
= ln(2 + 2𝑖) − ln(2 − 2𝑖) √8
𝟐
𝝅⁄𝟒
𝑖𝜋⁄4 −𝑖𝜋⁄4 𝑥
= ln(√8𝑒 ) − ln(√8𝑒 ) 𝟐

𝜋 𝜋 𝜋  2 − 2𝑖
= (ln √8 + 𝑖 ) − (ln √8 − 𝑖 ) = 𝑖
4 4 2 Fig. 4.12 𝑧 = 2 ± 2𝑖
 Chapter 4 Complex Integration 119

4.8 Integral of elementary functions

In most applications, such an 𝐹(𝑧) can be found from differentiation

formulas as shown in the following table

𝑓(𝑧) 𝐹(𝑧)
𝑛+1
𝑧𝑛 𝑧
, 𝑛 ≠ −1
𝑛+1
1
ln 𝑧
𝑧
𝑒𝑧 𝑒𝑧
𝑎𝑧
𝑎𝑧
ln 𝑎
sin 𝑧 − cos 𝑧
cos 𝑧 sin 𝑧
tan 𝑧 − ln cos 𝑧 = ln sec 𝑧
cot 𝑧 ln sin 𝑧
sec 𝑧 ln(sec 𝑧 + tan 𝑧)
csc 𝑧 ln(csc 𝑧 − cot 𝑧)
sec 2 𝑧 tan 𝑧
csc 2 𝑧 − cot 𝑧
sec 𝑧 tan 𝑧 sec 𝑧
csc 𝑧 cot 𝑧 − csc 𝑧
sinh 𝑧 cosh 𝑧
cosh 𝑧 sinh 𝑧
tanh 𝑧 ln cosh 𝑧
coth 𝑧 ln sinh 𝑧
sech 𝑧 tan−1(sinh 𝑧)

csch z −coth−1 (cosh 𝑧)

sech2 𝑧 tanh 𝑧
 120 Chapter 4 Complex Integration

csch2 𝑧 − coth 𝑧
sech 𝑧 tanh 𝑧 − sech 𝑧
csch 𝑧 coth 𝑧 − csch 𝑧
1
ln (𝑧 + √𝑧 2 ± 1)
√𝑧 2 ± 1
1
tan−1 𝑧
1 + 𝑧2
1
sin−1 𝑧
√1 − 𝑧2
1
𝐬𝐞𝐜 −𝟏 𝒛
𝑧√𝑧 2 − 1

Example 1 Evaluate the following integrals

1
(𝒊) ∫ cot(4𝑧 − 5) 𝑑𝑧 (𝒊𝒊) ∫ 𝑧 cosh 𝑧 2 𝑑𝑧
−1

(𝒊𝒊𝒊) ∫ sin5 4𝑧 cos 4𝑧 𝑑𝑧 (𝒊𝒗) ∫ 𝑧𝑒 3𝑧 𝑑𝑧

𝑧+1
(𝒗) ∫ 𝑑𝑧 (𝒗𝒊) ∫ 𝑧 2 sin 4𝑧 𝑑𝑧
𝑧 2 + 2𝑧 − 15

Solution
1
(𝒊) ∫ cot(4𝑧 − 5) 𝑑𝑧 = ln sin(4𝑧 − 5) + 𝑐
4
1
1 1 1
(𝒊𝒊) ∫ 𝑧 cosh 𝑧 2 𝑑𝑧 = ∫ 2𝑧 cosh 𝑧 2 𝑑𝑧 = sinh 𝑧 2 |1−1 = 0
−1 2 −1 2
1
(𝒊𝒊𝒊) ∫ sin5 4𝑧 cos 4𝑧 𝑑𝑧 = ∫ sin5 4𝑧 (4 cos 4𝑧)𝑑𝑧
4
1
= sin6 4𝑧 + 𝑐
24
(𝒊𝒗) To evaluate the integral ∫ 𝑧𝑒 3𝑧 𝑑𝑧, we use the integration by parts
 Chapter 4 Complex Integration 121

∫ 𝑧𝑒 3𝑧 𝑑𝑧 = (𝑧)(𝑒 3𝑧 ⁄3) − (1)(𝑒 3𝑧 ⁄9) + 𝑐

𝑧 + 𝑒 3𝑧
1 1
= 𝑧𝑒 3𝑧 − 𝑒 3𝑧 + 𝑐 1 − 𝑒 3𝑧 ⁄3
3 9
0 𝑒 3𝑧 ⁄9
𝑧+1 1 2𝑧 + 2
(𝒗) ∫ 𝑑𝑧 = ∫ 2 𝑑𝑧
𝑧2 + 2𝑧 − 15 2 𝑧 + 2𝑧 − 15
1
= ln(𝑧 2 + 2𝑧 − 15) + 𝑐
2

(𝒗𝒊) Finally, on integrating by parts, we obtain

2𝜋
∫ 𝑧 2 sin 4𝑧 𝑑𝑧 =
0

= (𝑧 2 )(− cos 4𝑧⁄4) − (2𝑧)(− sin 4𝑧⁄16) + (2)(cos 4𝑧⁄64)

= (𝑧 2 )(− cos 4𝑧⁄4) − (2𝑧)(− sin 4𝑧⁄16) + (2)(cos 4𝑧⁄64)|2𝜋
0
2𝜋
1 1 1
= − 𝑧 2 cos 4𝑧 + 𝑧 sin 4𝑧 + cos 4𝑧| 𝑧2
+
sin 4𝑧
4 8 32 0
2𝑧 − cos 4𝑧⁄4
1 1 1 −
= (− (4𝜋 2 ) + 0 + ) − (0 + 0 + ) − sin 4𝑧⁄16
4 32 32 2 +
= − 𝜋2 0 cos 4𝑧⁄64

4.9 Derivatives of analytic functions

In this section Cauchy’s integral formula is used to show the basic fact
that complex analytic functions have derivatives of all orders. This is
very surprising because it differs from the situation in real calculus.
Indeed, if a real function is once differentiable, nothing follows about
the existence of second or higher derivatives. Thus, in this respect,
complex analytic functions behave much more simply than real
functions that are once differentiable.
 122 Chapter 4 Complex Integration

Theorem 4.5 Let 𝑓(𝑧) be analytic in a simply connected domain 𝐷,

and let γ be a simple closed contour lying entirely within 𝐷. If 𝑧0 is
any point interior to γ, then for 𝑛 = 1,2, ⋯
𝑛! 𝑓(𝑧)
𝑓 (𝑛) (𝑧0 ) = ∮ 𝑑𝑧 (1)
2𝜋𝑖 γ (𝑧 − 𝑧0 )𝑛+1

Remark 1 Some call the identity (1) Cauchy’s differentiation

formula.

Remark 2 It is useful to observe that Cauchy’s integral formula for

derivatives (1) is obtained formally by differentiating the
Cauchy’s integral formula under the integral sign with
respect to 𝑧0 . Thus
1 𝑓(𝑧)
𝑓 ′ (𝑧0 ) = ∮ 𝑑𝑧 , (2)
2𝜋𝑖 γ (𝑧 − 𝑧0 )2

1 𝑓(𝑧)
𝑓 ′′ (𝑧0 ) = ∮ 𝑑𝑧 , (3)
2𝜋𝑖 γ (𝑧 − 𝑧0 )3
⋮
(𝑛 − 1)! 𝑓(𝑧)
𝑓 (𝑛−1) (𝑧0 ) = ∮ 𝑛
𝑑𝑧 , (4)
2𝜋𝑖 γ (𝑧 − 𝑧0 )

and so on.
Proof. since
1 𝑓(𝑧)
𝑓(𝑧0 ) = ∮ 𝑑𝑧,
2𝜋𝑖 γ 𝑧 − 𝑧0
then
𝑓(𝑧0 + ∆𝑧) − 𝑓(𝑧0 ) 1 𝑓(𝑧) 𝑓(𝑧)
= [∮ 𝑑𝑧 − ∮ 𝑑𝑧]
∆𝑧 2𝜋𝑖∆𝑧 γ 𝑧 − 𝑧0 − ∆𝑧 γ 𝑧 − 𝑧0
 Chapter 4 Complex Integration 123

1 (𝑧 − 𝑧0 ) − (𝑧 − 𝑧0 − ∆𝑧)
= [∮ 𝑓(𝑧) 𝑑𝑧]
2𝜋𝑖∆𝑧 γ (𝑧 − 𝑧0 − ∆𝑧)(𝑧 − 𝑧0 )

1 𝑓(𝑧)
= [∮ 𝑑𝑧]
2𝜋𝑖 γ (𝑧 − 𝑧0 − ∆𝑧)(𝑧 − 𝑧0 )
Taking the limit as ∆𝑧 → 0, we get
𝑓(𝑧0 + ∆𝑧) − 𝑓(𝑧0 )
𝑓 ′ (𝑧0 ) = lim
∆𝑧→0 ∆𝑧
1 𝑓(𝑧)
= lim ∮ 𝑑𝑧
2𝜋𝑖 ∆𝑧→0 γ (𝑧 − 𝑧0 − ∆𝑧)(𝑧 − 𝑧0 )

1 𝑓(𝑧)
= [∮ 𝑑𝑧],
2𝜋𝑖 γ (𝑧 − 𝑧0 )2
which is the required formula (2). Similarly, we can prove the other
formulas.

Remark 3 Practically, we use the general formula (4) in evaluating

the line integral:
𝑓(𝑧) 2𝜋𝑖 (𝑛)
∮ 𝑛+1
𝑑𝑧 = 𝑓 (𝑧0 ), 𝑛 = 1,2, ⋯
γ (𝑧 − 𝑧0 ) 𝑛!

sin 𝑧
Example 1 Evaluate ∮γ (𝑧−𝜋𝑖)2
𝑑𝑧 where γ is any contour enclosing

the point 𝜋𝑖.

Solution
sin 𝑧 𝑑
∮ 𝑑𝑧 = 2𝜋𝑖 [ sin 𝑧] = 2𝜋𝑖 cos 𝜋𝑖 = 2𝜋𝑖 cosh 𝜋
γ (𝑧 − 𝜋𝑖)2 𝑑𝑧 𝑧=𝜋𝑖
 124 Chapter 4 Complex Integration

sinh α𝑧
Example 2 Evaluate ∮γ 𝑑𝑧 around the circle γ: |𝑧| = 2.
𝑧4

Solution
sinh α𝑧 2𝜋𝑖 𝑑 3
∮ 𝑑𝑧 = [ sinh α𝑧]
γ 𝑧4 3! 𝑑𝑧 3 𝑧=0

2𝜋𝑖 3 𝜋α3
= [α cosh α𝑧]𝑧=0 = 𝑖
3! 3

𝑒𝑧
Example 3 Evaluate ∮γ (𝑧−1)2 (𝑧 2 +4)
𝑑𝑧 where γ is any contour for

which 1 lies inside and ±2𝑖 lie outside (counter clockwise).

Solution
𝑒𝑧 𝑒 𝑧 ⁄(𝑧 2 + 4)
∮ 2 2
𝑑𝑧 = ∮ 𝑑𝑧
γ (𝑧 − 1) (𝑧 + 4) γ (𝑧 − 1)2
𝑑 𝑒𝑧 (𝑧 2 + 4)𝑒 𝑧 − 𝑒 𝑧 (2𝑧)
= 2𝜋𝑖 [ ] = 2𝜋𝑖 [ ]
𝑑𝑧 (𝑧 2 + 4) 𝑧=1 (𝑧 2 + 4)2
𝑧=1

(1 + 4)𝑒 − 2𝑒 6𝑒𝜋
= 2𝜋𝑖 ( )= 𝑖
25 25

cot 𝑧
Example 4 Evaluate ∮γ 𝜋 2
𝑑𝑧, where γ is the boundary of the
(𝑧− )
2

triangle with vertices ±𝑖 and 2 (counter clockwise).

Solution
cot 𝑧 𝑑
∮ 2 𝑑𝑧 = 2𝜋𝑖 [ cot 𝑧] = 2𝜋𝑖[−csc 2 𝑧]𝑧=𝜋⁄2 = −2𝜋𝑖
γ 𝜋 𝑑𝑧 𝑧=𝜋⁄2
(𝑧 − 2)
 Chapter 4 Complex Integration 125

Exercise 4.1

In problems 1 − 5, evaluate

𝟏. ∫ (3𝑧 2 − 2𝑧)𝑑𝑧 , 𝛾: 𝑧(𝑡) = 𝑡 + 𝑖𝑡 2 , 0 ≤ 𝑡 ≤ 1

γ

𝟐. ∫ (2𝑧 − 𝑧)𝑑𝑧 , 𝛾: 𝑥 = 𝑡, 𝑦 = 𝑡 2 + 2, 0 ≤ 𝑡 ≤ 1
γ

1+𝑧
𝟑. ∫γ 𝑑𝑧, where γ is the right half of the circle |𝑧| = 1 from 𝑧 =
𝑧

−𝑖 to 𝑧 = 𝑖.

𝟒. ∫γ sin2 𝑧 𝑑𝑧, where γ is semicircle |𝑧| = 𝜋 from 𝑧 = −𝑖 to 𝑧 = 𝜋𝑖

in the right half-plane.

1
𝟓. ∫γ 𝑑𝑧, where γ is one quarter of the circle |𝑧| = 4 from 𝑧 = 4𝑖 to
𝑧3

𝑧 = 4.
In problems 6 − 14, Evaluate the given integrals along the indicated
closed contours
4
𝟔. ∮ 𝑑𝑧 , |𝑧| = 5
γ (𝑧 − 3𝑖)

𝑒𝑧
𝟕. ∮ 𝑑𝑧 , |𝑧| = 4
γ (𝑧 − 𝜋𝑖)

cos 𝑧
𝟖. ∮ 𝑑𝑧 , |𝑧| = 2
γ (3𝑧 − 𝜋)

𝑧2
𝟗. ∮ 𝑑𝑧 , (𝒂) |𝑧 − 𝑖| = 2, (𝒃) |𝑧 + 2𝑖| = 1
γ 𝑧2 + 4

sin 𝑧
𝟏𝟎. ∮ 𝑑𝑧 , |𝑧 − 2𝑖| = 2
γ 𝑧2+ 𝜋2
 126 Chapter 4 Complex Integration

2
𝑒𝑧
𝟏𝟏. ∮ 3
𝑑𝑧 , |𝑧 − 𝑖| = 1
γ (𝑧 − 𝑖)

cos 2𝑧
𝟏𝟐. ∮ 𝑑𝑧 , |𝑧| = 1
γ 𝑧5

𝑒 −𝑧 sin 𝑧
𝟏𝟑. ∮ 𝑑𝑧 , |𝑧 − 1| = 3
γ 𝑧3

𝑒 2𝑖𝑧 𝑧4
𝟏𝟒. ∮ ( − ) 𝑑𝑧 , |𝑧| = 6
γ 𝑧4 (𝑧 − 𝑖)3
1
In problems 15 − 18, integrate counterclockwise around each of
𝑧 4 −1

the circles
𝟏𝟓. |𝑧 − 𝑖| = 1 𝟏𝟔. |𝑧 − 1| = 1
𝟏𝟕. |𝑧 + 1| = 1 𝟏𝟖. |𝑧 + 3| = 1
In problems 19 − 21, integrate 𝑓(𝑧) around the contour γ.
tan 𝑧
𝟏𝟗. 𝑓(𝑧) = , γ is any ellipse with foci ±𝑖.
𝑧2
sin 2𝜋𝑧
𝟐𝟎. 𝑓(𝑧) = , γ: |𝑧| = 2
(𝑧 − 𝑖)3
𝑒 𝑧 tan 𝑧
𝟐𝟏. 𝑓(𝑧) = , γ: |𝑧 − 𝑖| = 1
𝑧+𝑖
In problems 22 − 27, integrate the following functions
counterclockwise around the circle |𝑧| = 2
2 2
𝑒𝑧 𝑒𝑧
𝟐𝟐. 𝟐𝟑.
(𝑧 − 𝑖)2 (𝑧 − 𝑖)2
𝑧3 sin 𝜋𝑧
𝟐𝟒. 𝟐𝟓.
(𝑧 + 1)3 𝑧3
sinh 4𝑧
𝟐𝟔. tan 𝑧 𝟐𝟕.
𝑧4
In problems 28 − 31, evaluate the following integrals
 Chapter 4 Complex Integration 127

𝑖 𝑖
𝑧
𝟐𝟖. ∫ 𝑒 cos 𝑧 𝑑𝑧 𝟐𝟗. ∫ 𝑧 sin 𝑧 𝑑𝑧
𝜋 0
1+𝑖 𝜋𝑖
𝟑𝟎. ∫ 𝑧𝑒 𝑧 𝑑𝑧 𝟑𝟏. ∫ 𝑧 2 𝑒 𝑧 𝑑𝑧
𝑖 0

In problems 32 − 34, integrate

𝑧
𝟑𝟐. Counterclockwise around |𝑧| = 4.
|𝑧|
𝟑𝟑. 5 cos 𝑖𝑧 , Along the real axis from − π to π.
𝟑𝟒. cosh 𝑧 , Along the imaginary axis from 0 to 𝑖.
2
In problems 35 − 36, evaluate ∮γ 𝑧 𝑑𝑧 from 𝑧 = 0 to 𝑧 = −4 + 2𝑖

along the curve γ given by:

𝟑𝟓. the parametric equation 𝑧 = 𝑡 2 + 𝑖𝑡.
𝟑𝟔. the vertical line from 𝑧 = 0 to 𝑧 = 2𝑖 and then the horizontal line
from 𝑧 = 2𝑖 to 𝑧 = 4 + 2𝑖.

𝟑𝟕. Show that ∮γ 𝑒 −2𝑧 𝑑𝑧 is independent of the path γ joining the

points 1 − 𝜋𝑖 and 2 + 3𝜋𝑖 and determine its value.

In problems 38 − 40, verify Cauchy’s integral theorem for the given
functions and γ is the square with vertices 1 ± 𝑖 and −1 ± 𝑖.
𝟑𝟖. 3𝑧 2 + 𝑖𝑧 − 4 𝟑𝟗. 5 sin 2𝑧 𝟒𝟎. 25 cos 3𝑧
5. The residue theorem

5.1 Partial fractions

If 𝑓(𝑧) is rational, we can determine its residues using the concept

of partial fractions. In this section we shall consider the following two
cases.
Case 1 𝒇(𝒛) has only simple poles

If 𝑓(𝑧) has a simple pole at 𝑧 = 𝑧0, then we will have the fraction
𝐴
,
(𝑧 − 𝑧0 )
and the residue 𝐴 at this pole is calculated by
𝐴 = lim [(𝑧 − 𝑧0 )𝑓(𝑧)] (1)
𝑧→𝑧0

or, by the formula

𝐴 = [(𝑧 − 𝑧0 )𝑓(𝑧)]𝑧=𝑧0 (2)

Example 1 Consider the function

24(𝑧 − 1)
𝑓(𝑧) =
𝑧(𝑧 − 2)(𝑧 + 4)
𝑓(𝑧) has the representation
𝐴1 𝐴2 𝐴3
𝑓(𝑧) = + +
𝑧 (𝑧 − 2) (𝑧 + 4)
The residue at the first simple pole at 𝑧 = 0 is
24(𝑧 − 1)
𝐴1 = lim [𝑧 ]
𝑧→0 𝑧(𝑧 − 2)(𝑧 + 4)
24(𝑧 − 1) 24(−1)
= lim [ ]= = 3,
𝑧→0 (𝑧 − 2)(𝑧 + 4) (−2)(4)
and the residue at the second simple pole at 𝑧 = 2 is
 Chapter 5 The Residue Theorem 129

24(𝑧 − 1)
𝐴2 = lim [(𝑧 − 2) ]
𝑧→2 𝑧(𝑧 − 2)(𝑧 + 4)
24(𝑧 − 1) 24(1)
= lim [ ]= =2
𝑧→2 𝑧(𝑧 + 4) (2)(6)
The residue at the third pole at 𝑧 = −4 is
24(𝑧 − 1)
𝐴3 = lim [(𝑧 + 4) ]
𝑧→−4 𝑧(𝑧 − 2)(𝑧 + 4)
24(𝑧 − 1) 24(−5)
= lim [ ]= = −5
𝑧→−4 𝑧(𝑧 − 2) (−4)(−6)
Therefore, we get
24(𝑧 − 1) 3 2 5
= + −
𝑧(𝑧 − 2)(𝑧 + 4) 𝑧 (𝑧 − 2) (𝑧 + 4)

Example 2 The partial fraction decomposition of the function

156(𝑧 − 1)
𝑓(𝑧) =
(𝑧 − 2)(𝑧 2 + 9)
is
156(𝑧 − 1) 156(𝑧 − 1)
𝑓(𝑧) = =
(𝑧 − 2)(𝑧 2 + 9) (𝑧 − 2)(𝑧 + 3𝑖)(𝑧 − 3𝑖)
𝐴1 𝐴2 𝐴3
= + +
(𝑧 − 2) (𝑧 + 3𝑖) (𝑧 − 3𝑖)
The residues are calculated using (2), hence
156(𝑧 − 1) 156(1)
𝐴1 = [(𝑧 − 2) 2
] = = 12
(𝑧 − 2)(𝑧 + 9) 13
𝑧=2

156(𝑧 − 1)
𝐴2 = [(𝑧 + 3𝑖) ]
(𝑧 − 2)(𝑧 + 3𝑖)(𝑧 − 3𝑖)
𝑧=−3𝑖

156(𝑧 − 1) 156(−3𝑖 − 1)
=[ ] =[ ]
(𝑧 − 2)(𝑧 − 3𝑖) (−3𝑖 − 2)(−3𝑖 − 3𝑖)
𝑧=−3𝑖
 130 Chapter 5 The Residue Theorem

156(1 + 3𝑖) (1 + 3𝑖) (−3 − 2𝑖)

= −[ ] = −26 ∙
(2 + 3𝑖)(6𝑖) (−3 + 2𝑖) (−3 − 2𝑖)
(3 − 11𝑖)
= −26 = (−6 + 22𝑖)
(9 + 4)
156(𝑧 − 1)
𝐴3 = [(𝑧 − 3𝑖) ]
(𝑧 − 2)(𝑧 + 3𝑖)(𝑧 − 3𝑖)
𝑧=3𝑖

156(𝑧 − 1) 156(3𝑖 − 1)
=[ ] =[ ]
(𝑧 − 2)(𝑧 + 3𝑖) (3𝑖 − 2)(3𝑖 + 3𝑖)
𝑧=3𝑖

156(1 − 3𝑖) (1 − 3𝑖) (3 − 2𝑖)

=[ ] = 26 ∙
(2 − 3𝑖)(6𝑖) (3 + 2𝑖) (3 − 2𝑖)
(−3 − 11𝑖)
= 26 = (−6 − 22𝑖)
(9 + 4)
Therefore, we obtain
156(𝑧 − 1) 12 (6 − 22𝑖) (6 + 22𝑖)
2
= − −
(𝑧 − 2)(𝑧 + 9) (𝑧 − 2) (𝑧 + 3𝑖) (𝑧 − 3𝑖)

Remark The residue 𝐴2 (= −6 + 22𝑖) is the conjugate of the

residue 𝐴3 (= −6 − 22𝑖).

Case 2 𝒇(𝒛) has a pole of order 𝒎

If 𝑓(𝑧) has a pole of order 𝑚 at 𝑧 = 𝑧0 then we will have a sum of 𝑚

fractions
𝐴𝑚 𝐴𝑚−1 𝐴2 𝐴1
+ + ⋯ + +
(𝑧 − 𝑧0 )𝑚 (𝑧 − 𝑧0 )𝑚−1 (𝑧 − 𝑧0 )2 (𝑧 − 𝑧0 )
The residues 𝐴𝑛 (𝑛 = 1,2, ⋯ , 𝑚) at this pole is given by
1 𝑑 𝑚−𝑛
𝐴𝑛 = lim { 𝑚−𝑛 [(𝑧 − 𝑧0 )𝑚 𝑓(𝑧)]} (3)
(𝑚 − 𝑛)! 𝑧→𝑧0 𝑑𝑧
 Chapter 5 The Residue Theorem 131

Example 3 Consider the following function

𝑧 3 − 4𝑧 2 + 4
𝑓(𝑧) =
𝑧 2 (𝑧 − 1)(𝑧 − 2)
Here we have a pole of order 𝑚 = 2 at 𝑧 = 0 and two simple poles at
𝑧 = 1 and 𝑧 = 2, so 𝑓(𝑧) has the decomposition
𝐴2 𝐴1 𝐴3 𝐴4
𝑓(𝑧) = 2
+ + + ,
𝑧 𝑧 (𝑧 − 1) (𝑧 − 2)
where the residues are calculated as follows
1 1 𝑧 3 − 4𝑧 2 + 4 4
𝐴2 = lim{𝑧 2 𝑓(𝑧)} = lim = = 2,
0! 𝑧→0 0! 𝑧→0 (𝑧 − 1)(𝑧 − 2) (−1)(−2)
1 𝑑 𝑑 𝑧 3 − 4𝑧 2 + 4
𝐴1 = lim { [𝑧 2 𝑓(𝑧)]} = lim { [ ]}
1! 𝑧→0 𝑑𝑧 𝑧→0 𝑑𝑧 (𝑧 − 1)(𝑧 − 2)

(𝑧 − 1)(𝑧 − 2)(3𝑧 2 − 8𝑧) − (𝑧 3 − 4𝑧 2 + 4)[(𝑧 − 1) + (𝑧 − 2)]

= lim { }
𝑧→0 (𝑧 − 1)2 (𝑧 − 2)2
0 − (0 + 4)[(−1) + (−2)] 12
= = =3
(−1)2 (−2)2 4
𝑧 3 − 4𝑧 2 + 4 1−4+4
𝐴3 = [ 2 ] = = −1
𝑧 (𝑧 − 2) 𝑧=1 (−1)

𝑧 3 − 4𝑧 2 + 4 8 − 16 + 4
𝐴4 = [ 2 ] = = −1
𝑧 (𝑧 − 1) 𝑧=2 (4)(1)

Hence,
𝑧 3 − 4𝑧 2 + 4 2 3 1 1
2
= 2+ − −
𝑧 (𝑧 − 1)(𝑧 − 2) 𝑧 𝑧 (𝑧 − 1) (𝑧 − 2)

Example 4 The partial fraction decomposition of the function

27 𝐴 𝐵3 𝐵2 𝐵1
𝑓(𝑧) = 3
= + 3
+ 2
+ ,
(𝑧 − 1)(𝑧 + 2) (𝑧 − 1) (𝑧 + 2) (𝑧 + 2) (𝑧 + 2)
where
 132 Chapter 5 The Residue Theorem

27
𝐴=[ ] = 1,
(𝑧 + 2)3 𝑧=1
1 27 27
𝐵3 = lim { }= = −9
0! 𝑧→−2 (𝑧 − 1) (−2 − 1)
1 𝑑 27 1 27
𝐵2 = lim { }= lim {− } = −3
1! 𝑧→−2 𝑑𝑧 (𝑧 − 1) 1! 𝑧→−2 (𝑧 − 1)2
1 𝑑2 27 1 27
𝐵1 = lim { 2 [ ]} = lim {[2 ]} = −1
2! 𝑧→−2 𝑑𝑧 (𝑧 − 1) 2! 𝑧→−2 (𝑧 − 1)3
Hence,
27 1 9 3 1
= − − −
(𝑧 − 1)(𝑧 + 2)3 (𝑧 − 1) (𝑧 + 2)3 (𝑧 + 2)2 (𝑧 + 2)

5.2 Residues

If 𝑓(𝑧) is analytic function of 𝑧 in a simply connected domain 𝐷, then

∮γ 𝑓(𝑧) 𝑑𝑧 = 0,

where γ is a closed contour in 𝐷. But if 𝑓(𝑧) has a singularity at a point

𝑧 = 𝑧0 inside γ and is otherwise analytic on γ and inside γ, then a
positive number  exists such that the function 𝑓(𝑧) is analytic at each
point 𝑧 for which 0 < |𝑧 − 𝑧0 | < ρ. In that domain 𝑓(𝑧) is represented
by the Laurent series
∞

𝑓(𝑧) = ∑ 𝑎𝑛 (𝑧 − 𝑧0 )𝑛
𝑛=−∞
∞
𝑎−1 𝑎−2
= ∑ 𝑎𝑛 (𝑧 − 𝑧0 )𝑛 + + + ⋯,
(𝑧 − 𝑧0 ) (𝑧 − 𝑧0 )2
𝑛=0

where the coefficient 𝑎−1 is given by

1
𝑎−1 = ∮ 𝑓(𝑧) 𝑑𝑧 (1)
2𝜋𝑖 γ
 Chapter 5 The Residue Theorem 133

The coefficient 𝑎−1, the coefficient of (𝑧 − 𝑧0 )−1, is called the 𝑟𝑒𝑠𝑖𝑑𝑢𝑒

of 𝑓(𝑧) at the isolated singular point 𝑧 = 𝑧0 and is denoted by
𝑎−1 = Res 𝑓(𝑧) (2)
𝑧=𝑧0

If 𝑓(𝑡) has a pole of order 𝑚 at 𝑧 = 𝑧0, the residue is given by

1 𝑑𝑚−1
𝑎−1 = lim { [(𝑧 − 𝑧0 )𝑚 𝑓(𝑧)]} (3)
(𝑚 − 1)! 𝑧→𝑧0 𝑑𝑧 𝑚−1

5.3 The residue theorem

Theorem 5.1 Let 𝑓(𝑧) be a single-valued analytic function within and

on a simple closed curve γ except for a finite number of singular points
𝑧1 , 𝑧2 , ⋯ , 𝑧𝑘 , inside γ. If 𝑅1 , 𝑅2 , ⋯ , 𝑅𝑘 denote the residues of 𝑓(𝑧) at
those points, then
𝑘

∮ 𝑓(𝑧) 𝑑𝑧 = 2𝜋𝑖 ∑ 𝑅𝑛
γ 𝑛=1

The integral being taken counter clockwise around the path γ.

Proof. We enclose each of the singular points 𝑧0 in a circle γ𝑛 with

radius small enough that those 𝑛 circles and γ are all separated. Then
𝑓(𝑧) is analytic in the multiply connected domain 𝐷 bounded by  and
γ1 , γ2 , ⋯ , γ𝑘 and on the entire boundary of 𝐷. From Cauchy’s integral
theorem we have

∮ 𝑓(𝑧) 𝑑𝑧 = ∮ 𝑓(𝑧) 𝑑𝑧 + ∮ 𝑓(𝑧) 𝑑𝑧 + ⋯ + ∮ 𝑓(𝑧) 𝑑𝑧

γ γ1 γ2 γ𝑘
 134 Chapter 5 The Residue Theorem

Each of these integrals is equal to 2𝜋𝑖 × 𝑟𝑒𝑠𝑖𝑑𝑢𝑒 at a singular point

enclosed by the corresponding contour. Therefore, we obtain the
required assertion. ∎

Remark From the above theorem we see that the problem of

evaluating the complex integral of 𝑓(𝑧) is transformed to
a simpler problem of finding the residues of singularity
of 𝑓(𝑧). To find these residues we have to expand this
function with Laurent series, or to use the partial fraction
technique. Besides an alternate method is available if the
above two methods give some complications. This
method is illustrated as follows.
Suppose that 𝑓(𝑧) has the fractional form
𝑝(𝑧)
𝑓(𝑧) =
𝑞(𝑧)
where 𝑝 and 𝑞 are both analytic functions at 𝑧0 and 𝑝(𝑧0 ) ≠ 0. Then,
In addition to the condition 𝑝(𝑧0 ) ≠ 0, consider the two conditions of
the function 𝑞(𝑧):
 (𝒊) If 𝑞(𝑧0 ) = 0 and 𝑞 ′ (𝑧0 ) ≠ 0, then 𝑓(𝑧) has a simple pole
at 𝑧0 and the residue at this pole has the value
𝑝(𝑧0 )
𝑎−1 = Res 𝑓(𝑧) = (1)
𝑧=𝑧0 𝑞 ′ (𝑧0 )
 (𝒊𝒊) If 𝑞(𝑧0 ) = 0, 𝑞 ′ (𝑧0 ) = 0 and 𝑞 ′′ (𝑧0 ) ≠ 0 then 𝑓(𝑧) has a
second order pole at 𝑧0 and the residue at this pole has the value
𝑝′ (𝑧0 ) 2 𝑝(𝑧0 )𝑞 ′′′ (𝑧0 )
𝑎−1 = Res 𝑓(𝑧) = 2 − (2)
𝑧=𝑧0 𝑞 ′′ (𝑧0 ) 3 [𝑞 ′′ (𝑧0 )]2
 Chapter 5 The Residue Theorem 135

To prove (1) we represent the functions 𝑝(𝑧) and 𝑞(𝑧) by Taylor’s

series in |𝑧 − 𝑧0 | < ρ, (ρ is a small number),
𝑝(𝑧) 𝑝0 + 𝑝0′ 𝑢 + 𝑝0′′ 𝑢2 ⁄2! + ⋯
= ,
𝑞(𝑧) 𝑞0 + 𝑞0′ 𝑢 + 𝑞0′′ 𝑢2 ⁄2! + ⋯
where, we use the substitution 𝑢 = (𝑧 − 𝑧0 ), 𝑝(𝑧0 ) = 𝑝0 , ⋯, for
simplicity. Since 𝑞(𝑧0 ) = 𝑞0 = 0, then we have
𝑝(𝑢) 𝑝0 + 𝑝0′ 𝑢 + 𝑝0′′ 𝑢2 ⁄2! + ⋯
=
𝑞(𝑢) 𝑞0′ 𝑢 + 𝑞0′′ 𝑢2 ⁄2! + ⋯
The residue at the simple pole 𝑧0 is
𝑝(𝑧) 𝑝(𝑢)
𝑎−1 = Res 𝑓(𝑧) = lim [(𝑧 − 𝑧0 ) ] = lim [𝑢 ]
𝑧=𝑧0 𝑧→𝑧0 𝑞(𝑧) 𝑢→0 𝑞(𝑢)
𝑝0 + 𝑝0′ 𝑢 + 𝑝0′′ 𝑢2 ⁄2! + ⋯ 𝑝0
= lim [ ′ ′′ ]= ′
𝑢→0 𝑞0 + 𝑞0 𝑢⁄2! + ⋯ 𝑞0
Similarly to prove (2), we use the quotient function, obtained from
Taylor’s series (𝑞(𝑧0 ) = 0, 𝑞 ′ (𝑧0 ) = 0)
𝑝(𝑢) 𝑝0 + 𝑝0′ 𝑢 + 𝑝0′′ 𝑢2 ⁄2! + ⋯
= ,
𝑞(𝑢) 𝑞0′′ 𝑢2 ⁄2! + 𝑞0′′′ 𝑢3 ⁄3! + ⋯
thus
𝑑 𝑝(𝑧) 𝑑 2 𝑝(𝑢)
𝑎−1 = Res 𝑓(𝑧) = lim [(𝑧 − 𝑧0 )2 ] = lim [𝑢 ]
𝑧=𝑧0 𝑧→𝑧0 𝑑 𝑞(𝑧) 𝑢→0 𝑑𝑢 𝑞(𝑢)
𝑑 2 𝑝0 + 𝑝0′ 𝑢 + 𝑝0′′ 𝑢2 ⁄2! + ⋯
= lim [𝑢 ′′ 2 ]
𝑢→0 𝑑𝑢 𝑞0 𝑢 ⁄2! + 𝑞0′′′ 𝑢3 ⁄3! + ⋯
𝑑 𝑝0 + 𝑝0′ 𝑢 + 𝑝0′′ 𝑢2 ⁄2! + ⋯
= lim [ ′′ ]
𝑢→0 𝑑𝑢 𝑞0 ⁄2! + 𝑞0′′′ 𝑢 ⁄3! + ⋯
(𝑞0′′ ⁄2! + 𝑞0′′′ 𝑢⁄3! + ⋯ )(𝑝0′ + 𝑝0′′ 𝑢 + ⋯ )
= lim [
𝑢→0 (𝑞0′′ ⁄2! + 𝑞0′′′ 𝑢 ⁄3! + ⋯ )2
(𝑝0 + 𝑝0′ 𝑢 + ⋯ )(𝑞0′′′ ⁄3! + ⋯ )
− ]
(𝑞0′′ ⁄2! + 𝑞0′′′ 𝑢⁄3! + ⋯ )2
 136 Chapter 5 The Residue Theorem

(𝑝0′ 𝑞0′′ ⁄2!) − 𝑝0 𝑞0′′′ ⁄3! 2𝑝0′ 4 𝑝0 𝑞0′′′

= = ′′ −
(𝑞0′′ ⁄2!)2 𝑞0 3! (𝑞0′′ )2
2𝑝0′ 2 𝑝0 𝑞0′′′
= ′′ − ,
𝑞0 3 (𝑞0′′ )2

which is the required proof in (2). ∎

Example 1 Find the residues of the function 𝑓(𝑧) = 𝑒 𝑧 csc 2 𝑧.

Solution

𝑧 2
𝑒𝑧 𝑝(𝑧)
𝑓(𝑧) = 𝑒 csc 𝑧 = 2
=
sin 𝑧 𝑞(𝑧)
𝑞(𝑧) = sin2 𝑧 = 0 𝑎𝑡 𝑧 = 𝑚𝜋, 𝑚 = 0, ±1, ±2, ⋯ and at these points
𝑝(𝑧) ≠ 0, so
𝑞(𝑧) = sin2 𝑧 , 𝑞(𝑚𝜋) = 0
𝑞 ′ (𝑧) = 2 sin 𝑧 cos 𝑧 = sin 2𝑧 , 𝑞 ′ (𝑚𝜋) = 0
𝑞 ′′ (𝑧) = 2 cos 2𝑧 , 𝑞 ′′ (𝑚𝜋) = 2
𝑞 ′′′ (𝑧) = −4 sin 2𝑧, 𝑞 ′′′ (𝑚𝜋) = 0
since 𝑞 ′′ (𝑚𝜋) ≠ 0, then 𝑓(𝑧) has poles of second order at 𝑧 = 𝑚𝜋,
𝑚 = 0, ±1, ±2, ⋯. The residues at these poles are found from (2) as
𝑝′ (𝑧0 ) 2 𝑝(𝑧0 )𝑞 ′′′ (𝑧0 ) 𝑒 𝑚𝜋 2 𝑒 𝑧 (0)
𝑎−1 = 2 − = 2 − = 𝑒 𝑚𝜋
𝑞 ′′ (𝑧0 ) 3 [𝑞 ′′ (𝑧0 )]2 2 3 [2]2
The important residue theorem has various applications in connection
with complex and real integrals. We first consider some complex
integrals.

24(𝑧−1)
Example 2 Evaluate the integral ∮γ 𝑑𝑧, where γ is the circle
𝑧(𝑧−2)(𝑧+4)

γ: |𝑧| = 3.
 Chapter 5 The Residue Theorem 137

24(𝑧−1)
Solution The given function 𝑧(𝑧−2)(𝑧+4)

has three simple poles at 𝑧 = 0, 𝑧 = 2

and 𝑧 = −4. From these poles only the × × ×
−𝟒 𝟎 𝟐
first two poles 𝑧 = 0 and 𝑧 = 2 lie
inside the circle γ: |𝑧| = 3 as shown in
Fig. 5.1
Fig. 6.1.
The residues at the two poles are
24(𝑧 − 1)
𝐴1 = Res 𝑓(𝑧) = | = 3,
𝑧=0 (𝑧 − 2)(𝑧 + 4)
𝑧=0

24(𝑧 − 1) 24(1)
𝐴2 = Res 𝑓(𝑧) = | = =2
𝑧=2 𝑧(𝑧 + 4) 𝑧=2 2(6)
So that by the direct application of the residue theorem
24(𝑧 − 1)
∮ 𝑑𝑧 = 2𝜋𝑖 ∑ Residues = 2𝜋𝑖(3 + 2) = 10𝜋𝑖
γ 𝑧(𝑧 − 2)(𝑧 + 4)

100𝑧(𝑧−2)
Example 3 Evaluate the integral ∮γ (𝑧+1)2 (𝑧 2 +4)
𝑑𝑧, where γ is the
5
circle γ: |𝑧| = 2.
 𝟐𝐢

Solution The given function 

−𝟏
100𝑧(𝑧 − 2) 100𝑧(𝑧 − 2)
= ,  − 𝟐𝐢
(𝑧 + 1) (𝑧 + 4) (𝑧 + 1)2 (𝑧 + 2𝑖)(𝑧 − 2𝑖)
2 2

has two simple poles at 𝑧 = ±2𝑖 and a pole Fig. 5.2

of order 2 at 𝑧 = −1. All of these poles are
5
inside the contour γ: |𝑧| = 2 as shown in Fig. 6.2. The residues at these

poles are:
 138 Chapter 5 The Residue Theorem

1 𝑑 100𝑧(𝑧 − 2)
𝐴1 = lim { [ ]}
1! 𝑧→−1 𝑑𝑧 (𝑧 2 + 4)
(𝑧 2 + 4)(2𝑧 − 2) − (𝑧 2 − 2𝑧)(2𝑧)
= lim {100 [ ]}
𝑧→−1 (𝑧 2 + 4)2
(1 + 4)(−2 − 2) − (1 + 2)(−2) −20 + 6
= 100 [ 2
] = 100 [ ]
(1 + 4) 25
−14
= 100 [ ] = −56
25
100𝑧(𝑧 − 2) 100(−2𝑖)(−2𝑖 − 2)
𝐴2 = | =
(𝑧 + 1)2 (𝑧 − 2𝑖) (−2𝑖 + 1)2 (−2𝑖 − 2𝑖)
𝑧=−2𝑖

400(−𝑖)(−𝑖 − 1) 100(−1 + 𝑖)
= =
(−2𝑖 + 1)2 (−4𝑖) (−4 + 1 − 4𝑖)(−𝑖)
100(−1 + 𝑖) (−4 − 3𝑖) 100(7 − 𝑖)
= ∙ = = (28 − 4𝑖)
(−4 + 3𝑖) (−4 − 3𝑖) (16 + 9)
100𝑧(𝑧 − 2) 100(2𝑖)(2𝑖 − 2)
𝐴3 = | =
(𝑧 + 1)2 (𝑧 + 2𝑖) (2𝑖 + 1)2 (2𝑖 + 2𝑖)
𝑧=2𝑖

400(𝑖)(𝑖 − 1) 100(−1 − 𝑖)
= 2
=
(2𝑖 + 1) (4𝑖) (−4 + 1 + 4𝑖)(𝑖)
100(−1 − 𝑖) (−4 + 3𝑖) 100(7 + 𝑖)
= ∙ = = (28 + 4𝑖)
(−4 − 3𝑖) (−4 + 3𝑖) (16 + 9)
Therefore, we obtain
24(𝑧 − 1)
∮ 𝑑𝑧 = 2𝜋𝑖 ∑ Residues
γ 𝑧(𝑧 − 2)(𝑧 + 4)
= 2𝜋𝑖(−56 + 28 − 4𝑖 + 28 + 4𝑖) = 0

𝑒𝑧
Example 4 Evaluate the integral ∮γ 𝑑𝑧 where γ is the circle
sin2 𝑧

γ: |𝑧| = 1.
 Chapter 5 The Residue Theorem 139

𝑒𝑧
Solution From Example 1, the function has poles of order 2 at
sin2 𝑧

𝑧 = 𝑚𝜋, 𝑚 = 0, ±1, ±2, ⋯. The only pole that lies in γ is at 𝑧 = 0.

The residue at this pole is 𝑎−1 = [𝑒 𝑚𝜋 ]𝑚=0 = 1, therefore
𝑒𝑧
∮ 𝑑𝑧 = 2𝜋𝑖(1) = 2𝜋𝑖
γ sin2 𝑧

Example 5 Evaluate the integral ∮γ 𝑒 4⁄𝑧 𝑑𝑧 where γ is the circle

γ: |𝑧| = 1.

Solution The integrand function 𝑒 4⁄𝑧 has an essential singularity at

𝑧 = 0 and neither of the above methods is applicable to find the residue
of 𝑓 at that point. However, the Laurent series of 𝑓 at 𝑧 = 0 is
∞ ∞
4⁄𝑧
(4⁄𝑧)𝑛 4𝑛
𝑒 =∑ =∑
(𝑛!) (𝑛!)𝑧 𝑛
𝑛=0 𝑛=0

4 42 43 44
=1+ + + + +⋯
𝑧 2! 𝑧 2 3! 𝑧 3 4! 𝑧 4
From this expansion we see that the residue of the integrand 𝑒 4⁄𝑧 at 𝑧 =
0 is 4. Hence,

∮ 𝑒 4⁄𝑧 𝑑𝑧 = 2𝜋𝑖(4) = 8𝜋𝑖

γ

Exercise 5.1

In problems 1 − 6, find the residue at each pole of the given functions

1
𝟏. 𝑓(𝑧) = 𝟐. 𝑓(𝑧) = sec 𝑧
(𝑧 2 − 2𝑧 + 2)
1 𝑧
𝟑. 𝑓(𝑧) = 𝟒. 𝑓(𝑧) =
𝑧 sin 𝑧 (𝑧 2 + 16)
 140 Chapter 5 The Residue Theorem

cos 𝑧 𝑒𝑧
𝟓. 𝑓(𝑧) = 𝟔. 𝑓(𝑧) =
𝑧 (𝑧 − 𝜋)3
2 𝑒𝑧 − 1
In problems 7 − 10, use Cauchy’s integral theorem to evaluate the
given integrals along the indicated contours.
1
𝟕. ∮ 𝑑𝑧,
γ (𝑧 − 2)(𝑧 + 2)
1 3
(𝒂) |𝑧| = (𝒃) |𝑧| = (𝒄) |𝑧| = 3
2 2
(𝑧 + 1)
𝟖. ∮ 𝑑𝑧,
γ 𝑧 2 (𝑧
− 2𝑖)
(𝒂) |𝑧| = 1 (𝒃) |𝑧 − 2𝑖| = 1 (𝒄) |𝑧 − 2𝑖| = 4

𝟗. ∮ 𝑧 3 𝑒 −1⁄𝑧 𝑑𝑧,
γ

(𝒂) |𝑧| = 5 (𝒃) |𝑧 + 𝑖| = 2 (𝒄) |𝑧 − 3| = 1

1
𝟏𝟎. ∮ 𝑑𝑧
γ 𝑧 sin 𝑧
(𝒂) |𝑧 − 2𝑖| = 1 (𝒃) |𝑧 − 2𝑖| = 3 (𝒄) |𝑧| = 5

In problems 11 − 25, use Cauchy’s integral theorem to evaluate the

given integrals along the indicated contours.
1
𝟏𝟏. ∮ 𝑑𝑧 , γ: |𝑧 − 3𝑖| = 3
γ 𝑧 2 + 4𝑧 + 13

1 3
𝟏𝟐. ∮ 𝑑𝑧 , γ: |𝑧 − 2| =
γ 𝑧 3 (𝑧 − 1)4 2

1
𝟏𝟑. ∮ 𝑑𝑧 , γ: |𝑧| = 2
γ 𝑧4 − 1
 Chapter 5 The Residue Theorem 141

1
𝟏𝟒. ∮ 𝑑𝑧 , γ is the ellipse 16𝑥 2 + 𝑦 2 = 4
γ (𝑧 2 + 1)(𝑧 + 1)

𝑒 𝑖𝑧 + sin 𝑧
𝟏𝟓. ∮ 𝑑𝑧 , γ: |𝑧 − 3| = 1
γ (𝑧 − 𝜋)4

cos 𝜋𝑧 1
𝟏𝟔. ∮ 𝑑𝑧 , γ: |𝑧| =
γ 𝑧2 2

tan 𝜋𝑧
𝟏𝟕. ∮ 𝑑𝑧 , γ: |𝑧 − 1| = 2
γ 𝑧

𝑧𝑒 𝑧
𝟏𝟖. ∮ 2 𝑑𝑧 , γ: |𝑧| = 2
γ 𝑧 −1

𝑒𝑧
𝟏𝟗. ∮ 𝑑𝑧 , γ: |𝑧| = 3
γ 𝑧 3 + 2𝑧 2
2𝑧−1
𝟐𝟎. ∮γ 𝑑𝑧, γ is the rectangle defined by:
𝑧 2 (𝑧 3 +1)
1
𝑥 = −2, 𝑥 = 1, 𝑦 = , 𝑦 = 1
2
𝟐𝟏. ∮γ cos 𝜋𝑧 𝑑𝑧, γ is the rectangle defined by:
1
𝑥 = , 𝑥 = 𝜋, 𝑦 = −1, 𝑦 = 1
2
cos 𝑧
𝟐𝟐. ∮ 𝑑𝑧 , γ: |𝑧 − 1| = 1
γ (𝑧 − 1)2 (𝑧 2 + 9)

𝟐𝟑. ∮ csc 𝑧 𝑑𝑧 , γ: |𝑧 − 2𝑖| = 1

γ

2𝑧 − 1 1
𝟐𝟒. ∮ 𝑑𝑧 , γ: |𝑧| =
γ 𝑧 2 (𝑧 3+ 1) 2

𝑒 2⁄𝑧 𝑧4
𝟐𝟓. ∮ ( 4 − ) 𝑑𝑧 , γ: |𝑧| = 6
γ 𝑧 (𝑧 − 𝑖)3
 142 Chapter 5 The Residue Theorem

5.4 Definite integrals

The residue theorem has many applications in the evaluation of certain

types of definite integrals. In the present section we consider the
following four classes or types of such integrals.
2𝜋
 𝐓𝐲𝐩𝐞 𝟏: ∫0 𝐹(sin θ , cos θ)𝑑θ,
∞
 𝐓𝐲𝐩𝐞 𝟐: ∫−∞ 𝐹(𝑥)𝑑𝑥,
∞ cos 𝑚𝑥
 𝐓𝐲𝐩𝐞 𝟑: ∫−∞ 𝐹(𝑥) { 𝑑𝑥,
sin 𝑚𝑥
 𝐓𝐲𝐩𝐞 𝟒: Miscellaneous Integrals Involving Particular Contours

𝟐𝝅
𝟓. 𝟒. 𝟏 𝐓𝐲𝐩𝐞 𝟏: ∫ 𝑭(𝐬𝐢𝐧 𝛉 , 𝐜𝐨𝐬 𝛉)𝒅𝛉
𝟎

Here 𝐹(sin θ , cos θ) is a rational function of sin θ and cos θ which is

finite on the interval of integration. To evaluate this integral we take as
a contour the unit circle γ: |𝑧| = 1, the integration being taken counter
clockwise around this circle. Setting on this contour
1 1 𝑧2 − 1
sin θ = (𝑧 − ) =
2𝑖 𝑧 2𝑖𝑧
1 1 𝑧2 + 1
𝑧 = 𝑒 𝑖θ  cos θ = (𝑧 + ) = (1)
2 𝑧 2𝑧
𝑑𝑧
{ 𝑑θ = 𝑖𝑧
Thus we obtain the following integral of a rational function of 𝑧, say
𝑓(𝑧),
𝑑𝑧
∮ 𝑓(𝑧)
γ 𝑖𝑧
This contour integral can be evaluated by the residue theorem if we can
find the poles of 𝑓(𝑧).
 Chapter 5 The Residue Theorem 143

2𝜋 𝑑θ
Example 1 Evaluate the integral ∫0 .
5+4 sin θ

Solution Integrating on the unit circle γ: |𝑧| = 1, and substituting

from (1), we get
𝑧 2 − 1 5𝑖𝑧 + 2(𝑧 2 − 1) 2𝑧 2 + 5𝑖𝑧 − 2
5 + 4 sin θ = 5 + 4 = =
2𝑖𝑧 𝑖𝑧 𝑖𝑧
Hence we have

2𝜋
𝑑𝑧
𝑑θ 𝑖𝑧 1 dz
∫ =∮ 2 = ∮ 2
0 5 + 4 sin θ γ 2𝑧 + 5𝑖𝑧 − 2 2 γ 𝑧 + (5𝑖 ⁄2)𝑧 − 1
𝑖𝑧
The roots of 𝑧 2 + (5𝑖 ⁄2)𝑧 − 1 are given by

−(5𝑖 ⁄2) ± √(5𝑖 ⁄2)2 − 4(1)(−1) −(5𝑖 ⁄2) ± √− 9⁄4

𝑧= =
2 2
−(5𝑖 ⁄2) ± (3𝑖 ⁄2) −5𝑖 ± 3𝑖 1
= = = − 𝑖, −2𝑖
2 4 2
Thus
2𝜋
𝑑θ 1 dz
∫ = ∮
0 5 + 4 sin θ 2 γ (𝑧 + 1⁄2 𝑖)(𝑧 + 2𝑖) 𝟏
− 𝒊
1 𝟐
We have only one simple pole − 2 𝑖, inside the unit

circle as shown in Fig.6.3. The residue at this pole  − 𝟐𝐢

Fig. 6.3
is
1 1 1 1 1 1 1
𝐴= | = = =
2 (𝑧 + 2𝑖) 𝑧=−1𝑖 2 (− 1 𝑖 + 2𝑖) 2 (3 𝑖) 3𝑖
2 2 2
Therefore
2𝜋
𝑑θ 1 2
∫ = 2𝜋𝑖 ( ) = 𝜋
0 5 + 4 sin θ 3𝑖 3
 144 Chapter 5 The Residue Theorem

2𝜋 cos θ
Example 2 Evaluate the integral ∫0 𝑑θ
13−12 cos 2θ

Solution Let γ: 𝑧 = 𝑒 𝑖θ , then we have

1 1 𝑧2 + 1
cos θ = (𝑧 + ) =
2 𝑧 2𝑧
1 1 𝑧4 + 1
cos 2θ = (𝑧 2 + 2 ) = ,
2 𝑧 2𝑧 2
𝑑𝑧
{ 𝑑θ =
𝑖𝑧
and so
cos θ 𝑧 2 + 1⁄2𝑧
=
13 − 12 cos 2θ 13 − 12(𝑧 4 + 1⁄2𝑧 2 )
𝑧(𝑧 2 + 1) 𝑧(𝑧 2 + 1)
= =
(26𝑧 2 − 12(𝑧 4 + 1)) (26𝑧 2 − 12𝑧 4 − 12)
Now we write the given integral as
2𝜋
cos θ 𝑧(𝑧 2 + 1) 𝑑𝑧
∫ 𝑑θ = ∮ 2 4
0 13 − 12 cos 2θ γ (26𝑧 − 12𝑧 − 12) 𝑖𝑧
𝑖(𝑧 2 + 1)
=∮ 4 2
𝑑𝑧
γ (12𝑧 − 26𝑧 + 12)
The roots of the denominator are

2
26 ± √(26)2 − 4(12)(12) 26 ± 10
𝑧 = =
2(12) 2(12)
from which we get by simplification
2 3
𝑧 2 = , or 𝑧 2 =
3 2
Hence the integrand has two simple poles at 𝑧 = ±√2⁄3 inside γ. To
obtain the residues at these poles, let
 Chapter 5 The Residue Theorem 145

𝑝(𝑧) 𝑖(𝑧 2 + 1)
= ,
𝑞(𝑧) (12𝑧 4 − 26𝑧 2 + 12)
then
𝑞 ′ (𝑧) = 48𝑧 3 − 52𝑧,
and
2 5
𝑝 (±√2⁄3) = 𝑖 ( + 1) = 𝑖,
3 3
2
𝑞 ′ (±√2⁄3) = (±√2⁄3) (48 − 52) =
3
= (±√2⁄3) (32 − 52) = ∓20√2⁄3

The residues at 𝑧 = ±√2⁄3 are given by

5
𝑝(𝑧0 )
𝐴1 = ′ = 3𝑖 =−
𝑖
𝑞 (𝑧0 ) −20√2⁄3 12√2⁄3
5
𝑝(𝑧0 )
𝐴2 = ′ = 3𝑖 =
𝑖
𝑞 (𝑧0 ) 20√2⁄3 12√2⁄3

Therefore
2𝜋
cos θ
∫ 𝑑θ = 2𝜋𝑖 ∑ Residues = 2𝜋𝑖(𝐴1 + 𝐴2 ) = 0
0 13 − 12 cos 2θ
∞
𝟓. 𝟒. 𝟐 𝐓𝐲𝐩𝐞 𝟐: ∫ 𝑭(𝒙)𝒅𝒙
−∞

Here 𝐹(𝑥) is a real rational function that satisfies the following

conditions:
 (𝒂) 𝐹(𝑥) is a regular in the upper half plane
 (𝒃) The degree of its denominator is higher than the degree of
the numerator by at least two units
 (𝒄) 𝐹(𝑥) has no poles on the real axis.
 146 Chapter 5 The Residue Theorem

These conditions guarantee the

∞ 𝑦
convergence of the integral ∫−∞ 𝐹(𝑥)𝑑𝑥.

Instead of the given integral, we consider
−𝑅 𝑅 𝑥
the corresponding contour integral
Fig. 5.4
∮ 𝐹(𝑧)𝑑𝑧
γ

around a path γ shown in Fig. 6.4. Since 𝐹(𝑥) is rational, 𝐹(𝑧) has
finitely many poles in the upper half-plane, and if we choose 𝑅 large
enough, then γ enclosed all these poles. By the residue theorem we then
obtain
𝑅
∮ 𝐹(𝑧)𝑑𝑧 = ∫ 𝐹(𝑧)𝑑𝑧 + ∫ 𝐹(𝑧)𝑑𝑧 = 2𝜋𝑖 ∑ Res 𝐹(𝑧)
γ  −𝑅

From the assumptions (𝒂) − (𝒄), we can prove that

∫ 𝐹(𝑧)𝑑𝑧 = 0 as 𝑅 → ∞

Hence, as 𝑅 → ∞,
∞
∫ 𝐹(𝑥)𝑑𝑥 = 2𝜋𝑖 ∑ Res 𝐹(𝑧), (1)
−∞

where we sum over the residues of 𝐹(𝑧) corresponding to the poles of

𝐹(𝑧) in the upper half-plane.

Remark For 𝑒𝑣𝑒𝑛 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠 the following identity is satisfied

∞
1 ∞
∫ 𝐹(𝑥)𝑑𝑥 = ∫ 𝐹(𝑥)𝑑𝑥 (2)
0 2 −∞

∞ 𝑑𝑥
Example 1 Evaluate the integral ∫0 1+𝑥 4
.
 Chapter 5 The Residue Theorem 147

1
Solution Since the function 1+𝑥 4 is even function, then we have
∞
𝑑𝑥 1 ∞ 𝑑𝑥
∫ 4
= ∫
0 1+𝑥 2 −∞ 1 + 𝑥 4
1
The function 𝐹(𝑧) = 1+𝑧 4 has four simple poles at the points

𝑧1 = 𝑒 𝑖𝜋⁄4 , 𝑧2 = 𝑒 𝑖3𝜋⁄4 , 𝑧3 = 𝑒 𝑖5𝜋⁄4 , 𝑧4 = 𝑒 𝑖7𝜋⁄4 ,

around a path γ in Fig. 6.5. Let
𝑝(𝑧) 1 𝑦
𝐹(𝑧) = =
𝑞(𝑧) 1 + 𝑧 4 
× ×
then 𝑥
−𝑅 𝑅
′ (𝑧) 3 × ×
𝑞 = 4𝑧 ,
and Fig. 5.5

𝑝(𝑧1 ) 1 1
𝐴1 = Res 𝐹(𝑧) = ′
= 𝑖𝜋 ⁄4 3
= 𝑒 −𝑖3𝜋⁄4 ,
𝑧→𝑧1 𝑞 (𝑧1 ) 4(𝑒 ) 4
𝑝(𝑧2 ) 1 1
𝐴2 = Res 𝐹(𝑧) = ′
= 𝑖3𝜋 ⁄4 3
= 𝑒 −𝑖9𝜋⁄4
𝑧→𝑧2 𝑞 (𝑧2 ) 4(𝑒 ) 4
So that
𝑅
𝑑𝑧 𝑑𝑧 𝑑𝑧
∮ 4
= ∫ 4
+ ∫ 4
γ (1 + 𝑧 )  (1 + 𝑧 ) −𝑅 (1 + 𝑧 )

𝑑𝑧
Taking the limit as 𝑅 → ∞, the integral ∫ (1+𝑧 4 )
vanishes and
∞
𝑑𝑥 1 ∞ 𝑑𝑥 1
∫ 4
= ∫ 4
= 𝜋𝑖(𝑒 −𝑖3𝜋⁄4 + 𝑒 −𝑖𝜋⁄4 )
0 1 + 𝑥 2 −∞ 1 + 𝑥 4
1
= 𝜋𝑖(cos 3 𝜋⁄4 − 𝑖 sin 3 𝜋⁄4 + cos 𝜋⁄4 − 𝑖 sin 𝜋⁄4)
4
2 1 1
= 𝜋𝑖 (−𝑖 ) = 𝜋
4 √2 2√2
 148 Chapter 5 The Residue Theorem

Example 2 Show that

∞
𝑥 2 𝑑𝑥 7𝜋
∫ 2 2 2
=
−∞ (𝑥 + 1) (𝑥 + 2𝑥 + 2) 50

Solution The function

𝑧2 𝑧2
=
(𝑧 2 + 1)2 (𝑧 2 + 2𝑧 + 2) (𝑧 2 + 1)2 (𝑧 2 + 2𝑧 + 1 − 1 + 2)
𝑧2 𝑧2
= =
(𝑧 2 + 1)2 [(𝑧 + 1)2 + 1] (𝑧 + 𝑖)2 (𝑧 − 𝑖)2 (𝑧 + 1 + 𝑖)(𝑧 + 1 − 𝑖)
has a second order pole at 𝑧 = 𝑖 and a simple pole at 𝑧 = −1 + 𝑖 in the
upper half-plane. The residue at 𝑧 = 𝑖 is
1 𝑑 𝑧2
𝐴1 = Res 𝐹(𝑧) = lim { [ ]}
𝑧→𝑖 1! 𝑧→𝑖 𝑑𝑧 (𝑧 + 𝑖)2 (𝑧2 + 2𝑧 + 2)
To find the derivative
𝑑 𝑧2
[ ],
𝑑𝑧 (𝑧 + 𝑖)2 (𝑧2 + 2𝑧 + 2)
we assume that
𝑧2
𝑔(𝑧) =
(𝑧 + 𝑖)2 (𝑧 2 + 2𝑧 + 2)
So that we want to find 𝑔′ (𝑧). To do that we take the logarithm of both
sides to obtain
ln 𝑔(𝑧) = 2 ln 𝑧 − 2 ln(𝑧 + 𝑖) − ln(𝑧 2 + 2𝑧 + 2)
𝑔′ (𝑧) 2 2 2𝑧 + 2
 = − − 2 ,
𝑔(𝑧) 𝑧 (𝑧 + 𝑖) (𝑧 + 2𝑧 + 2)
2 2 2𝑧 + 2
𝑔′ (𝑧) = 𝑔(𝑧) ( − − 2 )
𝑧 (𝑧 + 𝑖) (𝑧 + 2𝑧 + 2)
Hence,
2 2 2𝑖 + 2
𝑔′ (𝑖) = 𝑔(𝑖) ( − − )
𝑖 (𝑖 + 𝑖) (−1 + 2𝑖 + 2)
 Chapter 5 The Residue Theorem 149

2 + 2𝑖 (1 − 2𝑖)
= 𝑔(𝑖) (−2𝑖 + 𝑖 − )
(1 + 2𝑖) (1 − 2𝑖)
6 − 2𝑖 −5𝑖 − 6 + 2𝑖 −6 − 3𝑖
= 𝑔(𝑖) (−𝑖 − ) = 𝑔(𝑖) ( ) = 𝑔(𝑖) ( )
5 5 5
But
−1 −1
𝑔(𝑖) = =
(2𝑖)2 (−1 + 2𝑖 + 2) (−4)(1 + 2𝑖)
1 1 (1 − 2𝑖) (1 − 2𝑖)
= =
4 (1 + 2𝑖) (1 − 2𝑖) 20
Therefore,
(1 − 2𝑖) −6 − 3𝑖 1
𝐴1 = Res 𝐹(𝑧) = 𝑔′ (𝑖) = ( )= (−12 + 9𝑖)
𝑧→𝑖 20 5 100
The residue at 𝑧 = −1 + 𝑖 is
𝑧2
𝐴2 = Res 𝐹(𝑧) = lim { 2 }
𝑧→−1+𝑖 𝑧→𝑖 (𝑧 + 1)2 (𝑧 + 1 + 𝑖)

(−1 + 𝑖)2 (−1 + 𝑖)2

= =
((−1 + 𝑖)2 + 1)2 (−1 + 𝑖 + 1 + 𝑖) (−2𝑖 + 1)2 (2𝑖)
−2𝑖 −1 (−3 + 4𝑖) (3 − 4𝑖)
= = =
(−4 + 1 − 4𝑖)(2𝑖) (−3 − 4𝑖) (−3 + 4𝑖) 25
Therefore,
𝑅
𝑥 2 𝑑𝑥 𝑥 2 𝑑𝑥
∫ 2 2 2
+∫ 2 2 2
−𝑅 (𝑥 + 1) (𝑥 + 2𝑥 + 2)  (𝑥 + 1) (𝑥 + 2𝑥 + 2)

= 2𝜋𝑖 ∑ Res 𝐹(𝑧)

As 𝑅 → ∞ the second integral will approach zero, then

∞
𝑥 2 𝑑𝑥 −12 + 9𝑖 3 − 4𝑖
∫ 2 2 2
= 2𝜋𝑖 ( + )
−∞ (𝑥 + 1) (𝑥 + 2𝑥 + 2) 100 25
−7𝑖 7𝜋
= 2𝜋𝑖 ( )= ,
100 50
which is the required assertion.
 150 Chapter 5 The Residue Theorem

∞
𝐜𝐨𝐬 𝒎𝒙
𝟓. 𝟒. 𝟑 𝐓𝐲𝐩𝐞 𝟑: ∫ 𝑭(𝒙) { 𝒅𝒙
−∞
𝐬𝐢𝐧 𝒎𝒙
These types of integrals are evaluated in a way similar to that used for
𝑇𝑦𝑝𝑒 2 integrals of the last section. The function 𝐹(𝑥) satisfies the
same conditions as in 𝑇𝑦𝑝𝑒 2. In fact, we may consider the
corresponding integral

∮ 𝐹(𝑧)𝑒 𝑖𝑚𝑧 𝑑𝑧 (1)

γ
𝑦
over the contour  as in section 6.4.2 (see 
Fig. 6.6). Then taking the real and 𝑥
−𝑅 𝑅
imaginary parts of (1), we get Fig. 5.6

∮ 𝐹(𝑧) cos 𝑚𝑧 𝑑𝑧 = Re (∮ 𝐹(𝑧)𝑒 𝑖𝑚𝑧 𝑑𝑧) (2)

γ γ

∮ 𝐹(𝑧) sin 𝑚𝑧 𝑑𝑧 = Im (∮ 𝐹(𝑧)𝑒 𝑖𝑚𝑧 𝑑𝑧) (3)

γ γ

Example 1 Prove that

∞
cos 𝑚𝑥 𝜋 −𝑚𝑎
∫ 𝑑𝑥 = 𝑒
0 (𝑥 2 + 𝑎2 ) 2𝑎

Solution
∞
cos 𝑚𝑥 1 ∞ cos 𝑚𝑥
∫ 2 2
𝑑𝑥 = ∫ 𝑑𝑥
0 (𝑥 + 𝑎 ) 2 −∞ (𝑥 2 + 𝑎2 )
Consider the contour integral
𝑒 𝑖𝑚𝑧 𝑒 𝑖𝑚𝑧
∮ 𝑑𝑧 = ∮ 𝑑𝑧 (3)
γ (𝑧 2 + 𝑎2 ) γ (𝑧 + 𝑎𝑖)(𝑧 − 𝑎𝑖)
 Chapter 5 The Residue Theorem 151

The integrand has two simple poles at 𝑦

±𝑎𝑖. The residue at the pole 𝑎𝑖 which 

× 𝒂𝒊
lies inside γ is (see Fig. 6.7).
−𝑅 𝑅 𝑥
𝑒𝑖𝑚𝑧 𝑒𝑖𝑚(𝑎𝑖) 𝑒−𝑚𝑎 Fig. 5.7
𝐴= | = =
(𝑧 + 𝑎𝑖) 𝑧=𝑎𝑖 (𝑎𝑖 + 𝑎𝑖) 2𝑎𝑖
Hence,
𝑒 𝑖𝑚𝑧 𝑒 −𝑚𝑎 𝜋 −𝑚𝑎
∮ 𝑑𝑧 = 2𝜋𝑖 = 𝑒
γ (𝑧 2 + 𝑎2 ) 2𝑎𝑖 𝑎
or
𝑅
𝑒 𝑖𝑚𝑥 𝑒 𝑖𝑚𝑧 𝜋
∫ 2 2
𝑑𝑥 + ∫ 2 2
𝑑𝑧 = 𝑒 −𝑚𝑎
−𝑅 (𝑥 + 𝑎 )  (𝑧 + 𝑎 ) 𝑎
As 𝑅 → ∞, the second integral will approach zero, thus we get
∞
𝑒 𝑖𝑚𝑥 𝜋
∫ 2 2
𝑑𝑥 = 𝑒 −𝑚𝑎
−∞ (𝑥 + 𝑎 ) 𝑎
Thus from Euler’s rule
∞
cos 𝑚𝑥 + 𝑖 sin 𝑚𝑥 𝜋
∫ 2 2
𝑑𝑥 = 𝑒 −𝑚𝑎
−∞ (𝑥 + 𝑎 ) 𝑎
Equating the real and imaginary parts we obtain
∞
cos 𝑚𝑥 𝜋 −𝑚𝑎
∫ 2 2
𝑑𝑥 = 𝑒 ,
−∞ (𝑥 + 𝑎 ) 𝑎
and finally we have the required assertion
∞
cos 𝑚𝑥 𝜋 −𝑚𝑎
∫ 2 2
𝑑𝑥 = 𝑒
0 (𝑥 + 𝑎 ) 2𝑎

Example 2 Evaluate the integral

∞
𝑥 sin 𝜋𝑥
∫ 2
𝑑𝑥
−∞ (𝑥 + 2𝑥 + 5)
 152 Chapter 5 The Residue Theorem

Solution Consider the integral

𝑧𝑒 𝑖𝜋𝑧 𝑧𝑒 𝑖𝜋𝑧
∮ 2
𝑑𝑧 = ∮ 𝑑𝑧
γ (𝑧 + 2𝑧 + 5) γ (𝑧 + 1 + 2𝑖)(𝑧 + 1 − 2𝑖)
𝑦
The integrand has a simple pole at

 − 1 + 2𝑖
𝑧 = −1 + 2𝑖 inside the contour γ
−𝑅 𝑅 𝑥
shown in Fig. 6.8, at which the  − 1 − 2𝑖
Fig. 5.8
residue is
𝑧𝑒𝑖𝜋𝑧 (−1 + 2𝑖)𝑒𝑖𝜋(−1+2𝑖)
𝐴= Res 𝐹(𝑧) = | =
𝑧→−1+2𝑖 (𝑧 + 1 + 2𝑖) 𝑧=−1+2𝑖 (−1 + 2𝑖 + 1 + 2𝑖)

(−1 + 2𝑖)𝑒 −𝑖𝜋 𝑒 −2𝜋 (−1 + 2𝑖)𝑒 −2𝜋 (cos 𝜋 − 𝑖 sin 𝜋)

= =
4𝑖 4𝑖
(−1 + 2𝑖)𝑒 −2𝜋 (−1) (1 − 2𝑖)𝑒 −2𝜋
= =
4𝑖 4𝑖
Hence we have
𝑧𝑒 𝑖𝜋𝑧 (1 − 2𝑖)𝑒 −2𝜋 𝜋
∮ 2
𝑑𝑧 = 2𝜋𝑖 ( ) = ((1 − 2𝑖)𝑒 −2𝜋 )
γ (𝑧 + 2𝑧 + 5) 4𝑖 2
𝜋 −2𝜋
= 𝑒 − 𝑖𝜋𝑒 −2𝜋 ,
2
which implies
∞
𝑥𝑒 𝑖𝜋𝑥 𝜋
∫ 2
𝑑𝑥 = 𝑒 −2𝜋 − 𝑖𝜋𝑒 −2𝜋 ,
−∞ (𝑥 + 2𝑥 + 5) 2
or,
∞
𝑥(cos 𝜋𝑥 + 𝑖 sin 𝜋𝑥) 𝜋 −2𝜋
∫ 2 + 2𝑥 + 5)
𝑑𝑥 = 𝑒 − 𝑖𝜋𝑒 −2𝜋
−∞ (𝑥 2
Equating the real and imaginary parts of both sides, we get
∞
𝑥 cos 𝜋𝑥 𝜋
∫ 𝑑𝑥 = 𝑒 −2𝜋
−∞ (𝑥 2
+ 2𝑥 + 5) 2
∞
𝑥 sin 𝜋𝑥
∫ 2 + 2𝑥 + 5)
𝑑𝑥 = −𝜋𝑒 −2𝜋
−∞ (𝑥
 Chapter 5 The Residue Theorem 153

𝟓. 𝟒. 𝟒 𝑻𝒚𝒑𝒆 𝟒: Miscellaneous integrals involving particular contours

In this section we consider some important integrals in which the
integrand does not satisfies the conditions of the integrals of 𝑇𝑦𝑝𝑒 2
and 𝑇𝑦𝑝𝑒 3. However, we apply the residue theorem after choosing a
suitable contour for each integral.

Example 1 Prove that

∞ ∞
1 𝜋
∫ cos 𝑥 2 𝑑𝑥 = ∫ sin 𝑥 2 𝑑𝑥 = √ 𝑦
0 0 2 2
𝑧 = 𝑟𝑒 𝑖π⁄4 𝑧 = 𝑅𝑒 𝑖θ 𝑖θ
𝑖𝑧 2 𝑑𝑧 = 𝑒 𝑖π⁄4 𝑑𝑟 𝑑𝑧 = 𝑖𝑅𝑒 𝑑θ
by integrating the function 𝑒
𝝅⁄𝟒
𝑧=𝑥 𝑅 𝑥
around the contour γ shown in the 𝑑𝑧 = 𝑑𝑥

indented Fig. 5.9. Fig. 5.9

Solution Note that the integrand does not satisfy the conditions of
𝑻𝒚𝒑𝒆 𝟏 integrals. From the residue theorem we have

2
∮ 𝑒 𝑖𝑧 𝑑𝑧 = 0,
γ

since there is no poles inside γ. The contour γ is broken into three curves
as shown in the figure, so we can divide the contour integral into three
integrals as
𝑅 𝜋⁄4
2 2 𝑖θ )2
∮ 𝑒 𝑖𝑧 𝑑𝑧 = ∫ 𝑒 𝑖𝑥 𝑑𝑥 + ∫ 𝑒 𝑖(𝑅𝑒 𝑑(𝑖𝑅𝑒 𝑖θ )
γ 0 0
0
𝑖π⁄4 )2
+ ∫ 𝑒 𝑖(𝑟𝑒 𝑑(𝑟𝑒 𝑖π⁄4 ) = 0, (1)
𝑅

which is written as (𝐼) + (𝐼𝐼) + (𝐼𝐼𝐼) = 0.

Beginning with the second integral
 154 Chapter 5 The Residue Theorem

𝜋⁄4 2 𝜋⁄4
𝑖(𝑅𝑒 𝑖θ ) 𝑖θ 2 (cos 2θ+𝑖 sin 2θ)
(𝐼𝐼) = ∫ 𝑒 𝑑(𝑖𝑅𝑒 ) = ∫ 𝑒 𝑖𝑅 (𝑖𝑅𝑒 𝑖θ )𝑑θ
0 0
𝜋⁄4
2 cos 2θ 2 sin 2θ
=∫ 𝑒 𝑖𝑅 𝑒 −𝑅 (𝑖𝑅𝑒 𝑖θ )𝑑θ,
0

which implies
𝜋⁄4
2 cos 2θ 2 sin 2θ
|𝐼𝐼| ≤ ∫ |𝑒 𝑖𝑅 𝑒 −𝑅 (𝑖𝑅𝑒 𝑖θ )|𝑑θ
0
𝜋⁄4
2 cos 2θ 2 sin 2θ
≤∫ |𝑒 𝑖𝑅 ||𝑒 −𝑅 ||(𝑖𝑅𝑒 𝑖θ )|𝑑θ
0
𝜋⁄4
2 sin 2θ
≤∫ 𝑅𝑒 −𝑅 𝑑θ
0

Using the substitution y y= 2 

𝑑φ 𝑦 𝑦 = 2θ⁄𝜋
2θ = φ  𝑑θ = ,
2
we get 𝑦 = sin θ

𝑅 𝜋⁄2 −𝑅2 sin φ

|𝐼𝐼| ≤ ∫ 𝑒 𝑑φ 𝜋⁄2 θ
2 0 Fig. 6.10
But, from Fig. 6.10,
sin φ ≥ 2φ⁄𝜋,
so we obtain
2 𝜋⁄2
𝑅 𝜋⁄2 −𝑅2 sin φ 𝑅 𝜋⁄2 −2𝑅2 φ⁄𝜋 𝑅 𝑒 −2𝑅 φ⁄𝜋
|𝐼𝐼| ≤ ∫ 𝑒 𝑑φ = ∫ 𝑒 𝑑φ = [ ]
2 0 2 0 2 −2𝑅 2 ⁄𝜋 0
𝜋 −2𝑅2 φ⁄𝜋 𝜋⁄2 𝜋 2
=− [𝑒 ]0 = (1 − 𝑒 −𝑅 ), (2)
4𝑅 4𝑅
which shows that as 𝑅 → ∞ (𝐼𝐼) → 0. The third integral of (1) is
0 0
𝑖π⁄4 )2 2 𝑒 𝑖π⁄2
(𝐼𝐼𝐼) = ∫ 𝑒 𝑖(𝑟𝑒 𝑑(𝑟𝑒 𝑖π⁄4 ) = ∫ 𝑒 𝑖𝑟 𝑒 𝑖π⁄4 𝑑𝑟
𝑅 𝑅
 Chapter 5 The Residue Theorem 155

0
2 (cosπ⁄2+𝑖 sinπ⁄2)
= ∫ 𝑒 𝑖𝑟 𝑒 𝑖π⁄4 𝑑𝑟
𝑅
𝑅
2
= −𝑒 𝑖π⁄4 ∫ 𝑒 −𝑟 𝑑𝑟
0

As 𝑅 → ∞, this integral will be a Gamma function with the substitution

1
𝑟 2 = 𝑡  𝑟 = √𝑡  𝑑𝑟 = 𝑑𝑡
2 √𝑡
∞
2 1 ∞ −1⁄2 −𝑡 1 1 1
 ∫ 𝑒 −𝑟 𝑑𝑟 = ∫ 𝑡 𝑒 𝑑𝑡 =  ( ) = √π
0 2 0 2 2 2
so that
∞
2 √π
(𝐼𝐼𝐼) = −𝑒 𝑖π⁄4 ∫ 𝑒 −𝑟 𝑑𝑟 = −𝑒 𝑖π⁄4 (3)
0 2
Taking the limit as 𝑅 → ∞ and substitute from (2) and (3) in (1), we
get
∞ ∞
2
∫ 𝑒 𝑖𝑥 𝑑𝑥 = ∫ (cos 𝑥 2 + 𝑖 sin 𝑥 2 )𝑑𝑥
0 0

√π 1 √π √π
= (cos(π⁄4) + 𝑖 sin(π⁄4)) = ( + 𝑖 )
2 2 √2 √2
Equating the real and imaginary parts we obtain the required assertion.

Example 2 Prove that

𝒚
∞
sin 𝑥 𝜋
∫ 𝑑𝑥 = , 𝒛 = 𝑹𝒆𝒊𝛉
0 𝑥 2 𝒅𝒛 = 𝒊𝑹𝒆𝒊𝛉 𝒅𝛉
𝒅𝒛 = 𝒊𝛒𝒆𝒊𝛉 𝒅𝛉
by integrating the function 𝑒 𝑖𝑧 ⁄𝑧 𝒛 = 𝛒𝒆𝒊𝛉

around the contour γ shown in 𝒛=𝒙 𝒛=𝒙

−𝑹 𝒅𝒛 = 𝒅𝒙
−𝛒 𝛒 𝒅𝒛 = 𝒅𝒙𝑹 𝒙

Fig. 5.11. Fig. 5.11

 156 Chapter 5 The Residue Theorem

Solution The function 𝑒 𝑖𝑧 ⁄𝑧 does not have poles inside the given
contour, then
𝑒 𝑖𝑧
∮ 𝑑𝑧 = 0
γ 𝑧
This equal to
−ρ 𝑖𝑥 0 𝑖(ρ𝑒 ) 𝑖θ 𝑅 𝑖𝑥 𝜋 𝑖(𝑅𝑒 ) 𝑖θ
𝑒 𝑒 𝑖θ
𝑒 𝑒
∫ 𝑑𝑥 + ∫ 𝑖θ )
𝑑(ρ𝑒 ) + ∫ 𝑑𝑥 + ∫ 𝑖θ
𝑑(𝑅𝑒 𝑖θ ) = 0
−𝑅 𝑥 𝜋 (ρ𝑒 ρ 𝑥 0 (𝑅𝑒 )
(4)
Therefore, we have
−ρ 𝑖𝑥 0 𝑖(ρ𝑒 ) 𝑖θ 𝑅 𝑖𝑥
𝑒 𝑒 𝑖θ
𝑒
∫ 𝑑𝑥 + ∫ 𝑖θ
(𝑖ρ𝑒 )𝑑θ + ∫ 𝑑𝑥
−𝑅 𝑥 𝜋 (ρ𝑒 ) ρ 𝑥
𝜋 𝑖θ
𝑒 𝑖(𝑅𝑒 )
+∫ 𝑖θ )
(𝑖𝑅𝑒 𝑖θ )𝑑(θ) = 0, (5)
0 (𝑅𝑒
which is written as (𝐼) + (𝐼𝐼) + (𝐼𝐼𝐼) + (𝐼𝑉) = 0. Let us compute
every term separately. In the first integral we use the substitution
𝑥 = −𝑢  𝑑𝑥 = −𝑑𝑢
and the new limits of integral will be
𝑥 = −𝑅  𝑢 = 𝑅, 𝑥 = −ρ  𝑢 = ρ
Hence we can write the first term of (5) as
−ρ 𝑖𝑥 ρ −𝑖𝑢 𝑅 −𝑖𝑥
𝑒 𝑒 𝑒
(𝐼) = ∫ 𝑑𝑥 = ∫ (−𝑑𝑢) = − ∫ 𝑑𝑥 ( 6)
−𝑅 𝑥 𝑅 −𝑢 ρ 𝑥

For (𝐼𝐼) we have we take the limit as ρ → 0,

0 0
𝑖θ
lim(𝐼𝐼) = 𝑖 ∫ 𝑒𝑖(ρ𝑒 ) 𝑑θ = 𝑖 ∫ 𝑑θ = −𝑖𝜋, ( 7)
ρ→0 𝜋 𝜋

We now turn to the fourth integral of (5). The magnitude of this

integral is
 Chapter 5 The Residue Theorem 157

𝜋 𝑖(𝑅𝑒𝑖θ ) 𝜋
𝑒 𝑖θ
(𝐼𝑉) = ∫ 𝑖𝑅𝑒 𝑑θ = 𝑖 ∫ 𝑒𝑖(𝑅𝑒 ) 𝑑θ
𝑖θ
0 (𝑅𝑒𝑖θ ) 0

𝜋 𝜋

𝑖θ
|𝐼𝑉| ≤ |𝑖 ∫ 𝑒𝑖(𝑅𝑒 ) 𝑑θ| ≤ ∫ |𝑒𝑖𝑅(cos θ+𝑖 sin θ) |𝑑θ ≤
0 0

𝜋⁄2 𝜋⁄2
≤ 2∫ 𝑒 −𝑅 sin θ
𝑑θ ≤ 2 ∫ 𝑒−2𝑅θ⁄𝜋 𝑑θ
0 0

𝜋⁄2
𝑒−2𝑅θ⁄𝜋 𝜋
≤2 | = (1 − 𝑒−𝑅 )
− 2𝑅⁄𝜋 𝑅
0

Thus we have
(𝐼𝑉) → 0 𝑎𝑠 𝑅 → ∞ (8)
Collecting (5) − (8), we obtain, as ρ → 0 and 𝑅 → ∞,
∞ −𝑖𝑥 ∞ 𝑖𝑥
𝑒 𝑒
−∫ 𝑑𝑥 − 𝑖𝜋 + ∫ 𝑑𝑥 + 0 = 0
0 𝑥 0 𝑥
Hence,
∞ 𝑖𝑥
𝑒 − 𝑒 −𝑖𝑥
∫ 𝑑𝑥 = 𝜋
0 𝑖𝑥
or,
∞
sin 𝑥 𝜋
∫ 𝑑𝑥 =
0 𝑥 2

Example 3 Prove that

𝒚
∞
−𝑥 2 √𝜋 −𝑎2
∫ 𝑒 cos 2𝑥 𝑑𝑥 = 𝑒 , 𝒛 = 𝒙 + 𝒊𝒂
0 2 𝒅𝒛 = 𝒅𝒙

by integrating the function 𝒛 = 𝒊𝒚 𝒛 = 𝑹 + 𝒊𝒚

𝒅𝒛 = 𝒊𝒅𝒚 𝒅𝒛 = 𝒊𝒅𝒚
2
𝑒 −𝑧 around the contour γ
𝒛=𝒙 𝑹 𝒙
𝒅𝒛 = 𝒅𝒙
shown in Fig. 5.12.
Fig. 5.12
 158 Chapter 5 The Residue Theorem

Solution There is no poles of the integrand inside the contour γ, so

2
∮ 𝑒−𝑧 𝑑𝑧 = 0
γ

As shown in Fig. 5.12 we have the equations of the curves of the contour
γ, therefore the given integral will be decomposed along the different
paths of this contour to give the equation
𝑅 𝑎 0
−𝑥 2 −(𝑅+𝑖𝑦)2 2
∫ 𝑒 𝑑𝑥 + ∫ 𝑒 𝑑(𝑅 + 𝑖𝑦) + ∫ 𝑒 −(𝑥+𝑖𝑎) 𝑑(𝑥 + 𝑖𝑎)
0 0 R
0
−(𝑖𝑦)2
+∫ 𝑒 𝑑(𝑖𝑦) = (𝐼) + (𝐼𝐼) + (𝐼𝐼𝐼) + (𝐼𝑉) = 0 (9)
𝑎

As 𝑅 → ∞, the first integral will be, with the substitution

1
𝑥 2 = 𝑡  𝑥 = √𝑡  𝑑𝑥 = 𝑑𝑡
2√𝑡
∞
−𝑥 2
1 ∞ −1⁄2 −𝑡 1 1 1
 (𝐼) = ∫ 𝑒 𝑑𝑥 = ∫ 𝑡 𝑒 𝑑𝑡 =  ( ) = √π, (10)
0 2 0 2 2 2
and the second integral
𝑎
2
(𝐼𝐼) = ∫ 𝑒 −(𝑅+𝑖𝑦) 𝑖𝑑𝑦 → 0 as 𝑅 → ∞ (11)
0

The third integral of (9) is

𝑅 𝑅
−(𝑥+𝑖𝑎)2 2 −𝑎2 +2𝑖𝑎𝑥)
(𝐼𝐼𝐼) = − ∫ 𝑒 𝑑𝑥 = − ∫ 𝑒 −(𝑥 𝑑𝑥
0 0
𝑅
𝑎2 2
= −𝑒 ∫ 𝑒 −𝑥 𝑒 −2𝑖𝑎𝑥 𝑑𝑥 (12)
0
𝑎
2
(𝐼𝑉) = −𝑖 ∫ 𝑒 𝑦 𝑑𝑦 (13)
0

Substitute from (10) − (13) in (9), as 𝑅 → ∞, we get

 Chapter 5 The Residue Theorem 159

∞ 𝑎
√π 𝑎2 −𝑥 2 (cos 2
+0−𝑒 ∫ 𝑒 2𝑎𝑥 + 𝑖 sin 2𝑎𝑥)𝑑𝑥 − 𝑖 ∫ 𝑒 𝑦 𝑑𝑦 = 0
2 0 0

Equating the real and imaginary parts yields

∞
√π 𝑎2 2
− 𝑒 ∫ 𝑒 −𝑥 cos 2𝑎𝑥 𝑑𝑥 = 0,
2 0

or
∞
2 √π −𝑎2
∫ 𝑒 −𝑥 cos 2𝑎𝑥 𝑑𝑥 = 𝑒
0 2

Exercise 5.2

In problems 1 − 6, evaluate the following closed integrals along the

unit circle γ.
2
𝑒 −𝑧 cosh 𝑧
𝟏. ∮ 𝑑𝑧 𝟐. ∮ 𝑑𝑧
γ sin 4𝑧 γ (𝑧 2 − 𝑖3𝑧)

𝑒𝑧
𝟑. ∮ 𝑑𝑧 𝟒. ∮ 𝑒 𝑧 cot 𝑧 𝑑𝑧
γ cos 𝜋𝑧 γ

sinh 𝑧
𝟓. ∮ cosh 𝑧 𝑑𝑧 𝟔. ∮ 𝑑𝑧
γ γ (4𝑧 2 + 1)
In problems 7 − 10, evaluate the following trigonometric integrals
2𝜋 2𝜋
1 1
𝟕. ∫ 𝑑θ 𝟖. ∫ 𝑑θ
0 1 + 0.5 sin θ 0 1 + 3 cos2 θ
2𝜋 2𝜋
cos 2 θ sin2 θ
𝟗. ∫ 𝑑θ 𝟏𝟎. ∫ 𝑑θ
0 3 − sin θ 0 5 + 4 cos θ
In problems 11 − 18, Evaluate the following improper integrals
∞ ∞
𝑑𝑥 𝑑𝑥
𝟏𝟏. ∫ 2 2
𝟏𝟐. ∫ 2 2 2
−∞ (𝑥 + 4) −∞ (𝑥 + 1) (𝑥 + 9)
∞ ∞ (𝑥 2
𝑑𝑥 + 1)
𝟏𝟑. ∫ 6
𝟏𝟒. ∫ 𝑑𝑥
−∞ (𝑥 + 4) 0 (𝑥 4 + 1)
 160 Chapter 5 The Residue Theorem

∞ ∞
𝑥 sin 𝑥 cos 𝑥
𝟏𝟓. ∫ 2
𝑑𝑥 𝟏𝟔. ∫ 𝑑𝑥
−∞ (𝑥 + 4) 0 (𝑥 2 + 4)2
∞ ∞
cos 2𝑥 cos 𝑥
𝟏𝟕. ∫ 4
𝑑𝑥 𝟏𝟖. ∫ 2 2
𝑑𝑥
−∞ (𝑥 + 1) −∞ (𝑥 + 1)(𝑥 + 9)

In problems 19 − 20, use the contour shown in Fig. 6.13 to evaluate

the following integral. 𝒚
∞
𝑑𝑥
𝟏𝟗. ∫ 6 2𝜋𝑖
−∞ (𝑥 + 1)
∞
𝑑𝑥
𝟐𝟎. ∫
−∞ (𝑥 8+ 1) −𝑅 𝑅 𝒙
Fig. 5.13
6. Multiple Integrals

In the last chapter we have finished our discussion of derivatives of

functions of more than one variable and we need to move on to integrals
of functions of two or three variables. So that present chapter is devoted
to extend the idea of a definite integral to double and triple integrals of
functions of two or three variables. Such integrals are called
𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙𝑠.

6.1 Double integrals

Let 𝑓(𝑥, 𝑦) be a continuous function, defined on a domain . We

suppose that this domain is bounded, which means that  can be
enclosed in a circle of large radius or in a rectangular region. We define
the double integral of the function 𝑓(𝑥, 𝑦) over the domain  as
follows:

(𝒙 ഥ𝒊 )
ഥ𝒊 , 𝒚
𝑦

𝒚𝒋

∆𝑦𝑗 
𝒚𝒋−𝟏 𝑥


∆𝑥𝑖

𝑥
Fig. 6.1 Division of the domain into 𝒎𝒏 sub-rectangles

Divide the domain  into sub-rectangles. This is done by drawing lines

parallel to the co-ordinate axes getting 𝑚𝑛 subrectangles which cover
 162 Chapter 6 Multiple Integrals

, as in Fig. 6.1. Next we choose a general point (𝑥̅𝑖 , 𝑦̅𝑖 ) and form the
double Riemann sum
𝑚 𝑛

∑ ∑ 𝑓 (𝑥̅𝑖 , 𝑦̅𝑖 )∆𝑥𝑖 ∆𝑦𝑗

𝑗=1 𝑖=1

Then the double integral of the function 𝑓(𝑥, 𝑦) over the domain  is
𝑚 𝑛

lim ∑ ∑ 𝑓 (𝑥̅𝑖 , 𝑦̅𝑖 )∆𝑥𝑖 ∆𝑦𝑗 ,

∬ 𝑓(𝑥, 𝑦)𝑑𝑥𝑑𝑦 = 𝑛→∞
 𝑚→∞ 𝑗=1 𝑖=1

if the limit exists.

6.1.1 Interpretation of double integrals

Just as a single integration of positive functions can be interpreted as

areas, double integrals of positive functions can be interpreted as
volumes in the following way. Suppose that 𝑓(𝑥, 𝑦) ≥ 0 and 𝑓 is
defined on . The graph of 𝑓 is a surface with equation 𝑧 = 𝑓(𝑥, 𝑦).
Let 𝑆 be the solid that lies above  and under the graph of 𝑓. Then after
the partition of the region  as above we define the volume of the solid
𝑆 (see Fig. 6.2) as the limit of
∆𝑉 = 𝑓(𝑥̅𝑖 , 𝑦̅𝑖 )∆𝐴𝑖𝑗 = 𝑓(𝑥̅𝑖 , 𝑦̅𝑖 )∆𝑥𝑖 ∆𝑦𝑗 ,
thus the volume above the region  and under the surface 𝑧 = 𝑓(𝑥, 𝑦)
is

𝑉 = ∬ 𝑓(𝑥, 𝑦)𝑑𝐴


6.1.2 Basic properties of double integrals

Notice that all the typical properties of the double integral hold. For
example, constants can be pulled out and the double integral of the sum
of two functions is the sum of the double integrals of each function.
 Chapter 6 Multiple Integrals 163

Thus, for any two functions 𝑓 and 𝑔 defined and continuous in a region
,

 ∬ (𝑓 + 𝑔)𝑑𝐴 = ∬ 𝑓𝑑𝐴 + ∬ 𝑔𝑑𝐴

 ∬ 𝑐𝑓𝑑𝐴 = 𝑐 ∬ 𝑓𝑑𝐴 (𝑐 ≡ 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡)

 ∬ 𝑓𝑑𝐴 = ∬ 𝑓𝑑𝐴 + ∬ 𝑓𝑑𝐴 ( = 1 ∪ 2 )

1 2

ഥ𝒊 )
ഥ𝒊 , 𝒚
𝒇(𝒙

𝒙
∆𝐴𝑖𝑗 (𝒙 ഥ𝒊 )
ഥ𝒊 , 𝒚

Fig. 6.2 Geometric interpretation of double integrals

6.1.3 Evaluation of the double integrals

There are two types of the domain  of the function 𝑓(𝑥, 𝑦).  is said
to be of 𝑡𝑦𝑝𝑒 𝐼 if it lies between the graphs of two continuous functions
of 𝑥, that is,
 164 Chapter 6 Multiple Integrals

 = {(𝑥, 𝑦)| 𝑎 ≤ 𝑥 ≤ 𝑏, 𝑔1 (𝑥) ≤ 𝑦 ≤ 𝑔2 (𝑥)},

where 𝑔1 (𝑥) and 𝑔2 (𝑥) are continuous functions on [𝑎, 𝑏]. Some
examples of 𝑡𝑦𝑝𝑒 𝐼 regions are shown in Fig. 6.3.
We also consider regions of 𝑡𝑦𝑝𝑒 𝐼𝐼 (Fig. 6.4), which can be expressed
as
 = {(𝑥, 𝑦)| 𝑐 ≤ 𝑦 ≤ 𝑑, ℎ1 (𝑦) ≤ 𝑥 ≤ ℎ2 (𝑦)},

𝒚 𝒚
𝒅
𝒚 = 𝒈𝟐 (𝒙)
𝒙 = 𝒉𝟏 (𝒚)

𝒕𝒚𝒑𝒆 𝑰
𝒕𝒚𝒑𝒆 𝑰𝑰
𝒙 = 𝒉𝟐 (𝒚)

𝒚 = 𝒈𝟏 (𝒙) 𝒄
𝒙 𝒙
a b
Fig. 6.3 Fig. 6.4

According to the shape of the region , we can express the double

integral in one of the following forms:
 𝑻𝒚𝒑𝒆 𝑰: For this region we choose a vertical strip to determine
the integration limits. We note that the strip starts from the
vertical line 𝑥 = 𝑎 and moves in the positive direction of the
𝑥 − 𝑎𝑥𝑖𝑠 up to the line 𝑥 = 𝑏. We assume that the strip meets
the boundary of the region  in two points at which we express
𝑦 as a function of 𝑥. So we express 𝑑𝐴 = 𝑑𝑦𝑑𝑥, and evaluate

the double integral ∬ 𝑓(𝑥, 𝑦)𝑑𝐴 as:

 Chapter 6 Multiple Integrals 165

𝑥=𝑏 𝑦=𝑔2 (𝑥)  

∬ 𝑓(𝑥, 𝑦)𝑑𝐴 = ∫ ∫ 𝑓(𝑥, 𝑦) 𝑑𝑦 𝑑𝑥 (1)
 𝑥=𝑎 𝑦=𝑔1 (𝑥)

To evaluate the integral in (1), we work from the inside integral as

indicated by the dotted rectangle. So we integrate with respect to 𝑦
when the limits are values of functions of 𝑥. The result is then integrated
with respect to 𝑥 between the limits 𝑥 = 𝑎 and 𝑥 = 𝑏.

 𝑻𝒚𝒑𝒆 𝑰𝑰: For this region we choose a horizontal strip starts from
the vertical line 𝑦 = 𝑐 and moves in the positive direction of the
𝑦 − 𝑎𝑥𝑖𝑠 up to the line 𝑦 = 𝑑. The strip meets the boundary of
the region  in two points at which we express x as a function
of 𝑦. So we express 𝑑𝐴 = 𝑑𝑥𝑑𝑦, and evaluate the double

integral ∬ 𝑓(𝑥, 𝑦)𝑑𝐴 as:

𝑦=𝑑 𝑥=ℎ2 (𝑦)

∬ 𝑓(𝑥, 𝑦)𝑑𝐴 = ∫ ∫ 𝑓(𝑥, 𝑦) 𝑑𝑥 𝑑𝑦 (2)
 𝑦=𝑐 𝑥=ℎ1 (𝑦)

To find the value of integral in (2), we first integrate with respect to 𝑥

when the limits ranges between ℎ1 (𝑦) and ℎ2 (𝑦). The result is then
integrated with respect to 𝑦 between the limits 𝑦 = 𝑐 and 𝑦 = 𝑑.

𝑥=1 𝑦=2
Example 1 Evaluate the integral ∫𝑥=0 ∫𝑦=1 (𝑥 2 + 𝑦) 𝑑𝑦𝑑𝑥.

Solution
𝑥=1 𝑦=2 𝑥=1 𝑦=2
∫ ∫ (3𝑥 2 + 2𝑦) 𝑑𝑦𝑑𝑥 = ∫ [∫ (3𝑥 2 + 2𝑦) 𝑑𝑦] 𝑑𝑥
𝑥=0 𝑦=1 𝑥=0 𝑦=1
 166 Chapter 6 Multiple Integrals

𝑥=1
=∫ [3𝑥 2 𝑦 + 𝑦 2 ]12 𝑑𝑥
𝑥=0
𝑥=1
=∫ [(3𝑥 2 (2) + 4) − (3𝑥 2 (1) + 1)]𝑑𝑥
𝑥=0
𝑥=1 𝑥=1
=∫ [(6𝑥 2 + 4) − (3𝑥 2 + 1)]𝑑𝑥 = ∫ (3𝑥 2 + 3)𝑑𝑥
𝑥=0 𝑥=0

= [𝑥 3 + 3𝑥]10 =4

Example 2 (𝒂) Sketch the region  bounded by

𝑦 = 𝑥2 & 𝑦 = 2𝑥
b) Find the volume of the solid
bounded by the region  and the 𝒚 (𝟐, 𝟒)
4
plane 𝑧 = 𝑥 + 2𝑦.
3

Solution From Fig. 6.5 we see that 2 𝒚 = 𝟐𝒙

 is as a type I region and the

1
volume is given by 
2 2𝑥 0
𝒙
𝑉 = ∫ [∫ (𝑥 + 2𝑦) 𝑑𝑦] 𝑑𝑥 0 1 2
0 𝑥2 Fig. 6.5  as a type I region
2
= ∫ [𝑥𝑦 + 𝑦 2 ]2𝑥
𝑥 2 𝑑𝑥
0
2
= ∫ [(𝑥(2𝑥) + (2𝑥)2 ) − (𝑥(𝑥 2 ) + (𝑥 2 )2 )]𝑑𝑥
0
2 2
𝑥4 𝑥5
=∫ [6𝑥 2 −𝑥 −𝑥 3 4 ]𝑑𝑥 3
= [2𝑥 − − ]
0 4 5 0
32 28
= (16 − 4 − )= .
5 5
 Chapter 6 Multiple Integrals 167

4 𝒚
Another Solution. From Fig. 6.6 (𝟐, 𝟒)
we see that  can also be written as 3
a type II region.
2 𝒚 = 𝟐𝒙
𝒚 = 𝒙𝟐
4 𝑦 1
𝑉 = ∫0 [∫𝑦√⁄2(𝑥 + 2𝑦) 𝑑𝑥] 𝑑𝑦 

4√ 𝑦 0 𝒙
𝑥2 0 1 2
= ∫ [ + 2𝑦𝑥] 𝑑𝑦
0 2 𝑦⁄2 Fig. 6.6  as a type II region

4 (𝑦⁄2)2
𝑦
= ∫ [( + 2𝑦(√𝑦)) − ( + 2𝑦(𝑦⁄2))] 𝑑𝑦
0 2 2
4 4
𝑦 3⁄2
𝑦2 2
𝑦2 2 5⁄2 9 𝑦 3
= ∫ [ + 2𝑦 − − 𝑦 ] 𝑑𝑦 = [ + 𝑦 − ]
0 2 8 4 5⁄2 8 3 0
4 3 4 28
= (4 + (32) − (64)) = 4 + (32) − 24 =
5 8 5 5

Example 3 Sketch the region  in the 𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒 bounded by

: y = 2√𝑥, 𝑦 = 0, 𝑥 = 3,

then evaluate the double integral ∬ 𝑥𝑦 𝑑𝑥𝑑𝑦.

2
Solution (3, √12)

The region D is shown in Fig. 6.7 and

3 2√𝑥
1
∬ 𝑥𝑦 𝑑𝑥𝑑𝑦 = ∫ ∫ 𝑥𝑦𝑑𝑦𝑑𝑥
 0 0
3 2 2√𝑥 3  𝑥=3
𝑦 4𝑥
= ∫ 𝑥[ ] 𝑑𝑥 = ∫ 𝑥 ( ) 𝑑𝑥
0 2 0 0 2 0
3 3 3 0 2 4
2
𝑥
= 2 ∫ 𝑥 𝑑𝑥 = 2 [ ] = 18. 𝐅𝐢𝐠. 𝟔. 𝟕
0 3 0
 168 Chapter 6 Multiple Integrals

6.1.4 Applications of double integrals

The double integrals are used to compute areas, volumes, mass and
centroids of regions.
 The area of a Region 

The area 𝐴 of a region  in the 𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒 is given from (1), or (2) by

putting 𝑓(𝑥, 𝑦) = 1, that is,

𝐴 = ∬ 𝑑𝑥𝑑𝑦


 The volume of surfaces

The volume 𝑉 under the surface 𝑧 = 𝑓(𝑥, 𝑦) and above a region  is
given by

𝑉 = ∬ 𝑓(𝑥, 𝑦)𝑑𝑥𝑑𝑦


 The mass 𝑴 in a region 

Let ρ(𝑥, 𝑦) be the density of a lamina (flat sheet) 𝑀 in the 𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒,

then the mass in  is the double integral

𝑀 = ∬ ρ(𝑥, 𝑦)𝑑𝑥𝑑𝑦


 The centroids of the mass in 

The centre of gravity (𝑥̅ , 𝑦̅) of the mass 𝑀 with a density ρ(𝑥, 𝑦) is

∬ 𝑥ρ(𝑥, 𝑦)𝑑𝑥𝑑𝑦 ∬ 𝑦ρ(𝑥, 𝑦)𝑑𝑥𝑑𝑦

𝑥̅ = , 𝑦̅ =
𝑀 𝑀
 The moment of inertia

The moment of inertia 𝐼𝑥 𝑎𝑛𝑑 𝐼𝑦 of the mass in  about the 𝑥 −

𝑎𝑛𝑑 𝑦 − 𝑎𝑥𝑒𝑠 are given, respectively, by
 Chapter 6 Multiple Integrals 169

𝐼𝑥 = ∬ 𝑦 2 ρ(𝑥, 𝑦)𝑑𝑥𝑑𝑦,


𝐼𝑦 = ∬ 𝑥 2 ρ(𝑥, 𝑦)𝑑𝑥𝑑𝑦,


and the polar moment of inertia 𝐼0 about the origin of the mass in  is

𝐼0 = ∬ (𝑥 2 + 𝑦 2 )ρ(𝑥, 𝑦)𝑑𝑥𝑑𝑦


Example 4 (𝒂) Find the area of the region bounded by

𝑦 2 = 𝑥 𝑎𝑛𝑑 𝑥 − 𝑦 = 2
(𝒃) Find the centroid of this area if the density is constant.

Solution (𝒂) From Fig. 6.8, the given region is of type II, so that the
area is
(𝟒, 𝟐)
2 𝑦+2
𝐴=∫ ∫ 𝑑𝑥𝑑𝑦
−1 𝑦 2

2
𝑦+2
= ∫ [𝑥]𝑦 2 𝑑𝑦 (𝟏, −𝟏)
−1
2 Fig. 6.8
= ∫ (𝑦 + 2 − 𝑦 2 )𝑑𝑦
−1
2
𝑦2 𝑦3 8 3 1 9
= [ + 2𝑦 − ] = (2 + 4 − ) − (− + ) = .
2 3 −1 3 2 3 2
(b) Since the density is constant, then

∬ 𝑥𝑑𝑥𝑑𝑦 ∬ 𝑦𝑑𝑥𝑑𝑦
𝑥̅ = , 𝑦̅ = 
𝐴 𝐴
But the area is computed, so we shall evaluate the numerator of each of
the above two equations.
 170 Chapter 6 Multiple Integrals

2 𝑦+2
∬ 𝑥𝑑𝑥𝑑𝑦 = ∫ ∫ 𝑥𝑑𝑥𝑑𝑦
 −1 𝑦 2

2 𝑦+2
2 (𝑦
𝑥2 + 2)2 𝑦 4
=∫ [ ] 𝑑𝑦 = ∫ ( − ) 𝑑𝑦
−1 2 𝑦 2 −1 2 2
2
1 (𝑦 + 2)3 𝑦 5 36
= [ − ] =
2 3 5 −1 5
Thus,

∬ 𝑥𝑑𝑥𝑑𝑦 36⁄5 72 8
𝑥̅ = = = =
𝐴 9⁄2 45 5
Also
2 𝑦+2
∬ 𝑦𝑑𝑥𝑑𝑦 = ∫ ∫ 𝑦𝑑𝑥𝑑𝑦
 −1 𝑦 2
2 2
𝑦+2
= ∫ [𝑥𝑦]𝑦 2 𝑑𝑦 = ∫ ((𝑦 + 2)𝑦 − (𝑦 2 )𝑦)𝑑𝑦
−1 −1
2
= ∫ (𝑦 2 + 2𝑦 − 𝑦 3 )𝑑𝑦
−1
2
𝑦3 𝑦4 3 9
= [ + 𝑦2 − ] = 3 − =
3 4 −1 4 4
Thus,

∬ 𝑦𝑑𝑥𝑑𝑦 9⁄4 1
𝑦̅ = = =
𝐴 9⁄2 2
8 1
So that the centroid of the given region is (5 , ).
2

Another solution If we want to deal with the above region of

integration as a type I region, then we must divide the total domain 
into two subdomains 1 and 2 where  ≡ 1 ∪ 2 as shown in Fig.
 Chapter 6 Multiple Integrals 171

6.9. In this case the above integral will be divided by the same way as
the region . For instant, take the integral of the area as a guide for this
division

𝐴 = ∬ 𝑑𝑦𝑑𝑥 = ∬ 𝑑𝑦𝑑𝑥 + ∬ 𝑑𝑦𝑑𝑥

 1 2

1 √𝑥 4 √𝑥
=∫ ∫ 𝑑𝑦𝑑𝑥 + ∫ ∫ 𝑑𝑦𝑑𝑥 (𝟒, 𝟐)
0 −√𝑥 1 𝑥−2
1 4
= ∫ [𝑦]√−𝑥√𝑥 𝑑𝑥 𝑥
+ ∫ [𝑦]√𝑥−2 𝑑𝑥 2
0 1 1
1 4
= ∫ 2√𝑥𝑑𝑥 + ∫ (√𝑥 − 𝑥 + 2)𝑑𝑥 (𝟏, −𝟏)
0 1
Fig. 6.9 Region of Ex. 4
3⁄2 1 4
𝑥 𝑥 3⁄2 𝑥 2
= [2 ] +[ − + 2𝑥]
3⁄2 0 3⁄2 2 1
4 8 − 1 16 − 1 15 9
= + − + (8 − 2) = 6 − +6=
3 3⁄2 2 2 2

Exercise 6.1

In problems 1 − 10, evaluate each of the following double integrals

2 4 1 5
𝑥
𝟏. ∫ ∫ (𝑥 + 2𝑦)𝑑𝑥𝑑𝑦 𝟐. ∫ ∫ 𝑑𝑦𝑑𝑥
1 2 0 2 𝑦2
1 𝑥 4 2𝑦
𝟑. ∫ ∫ √𝑥 + 𝑦𝑑𝑦𝑑𝑥 𝟒. ∫ ∫ (2𝑥 + 3𝑦)𝑑𝑥𝑑𝑦
0 0 0 𝑦

4 √𝑥 1 1−𝑥 2
𝟓. ∫ 𝑑𝑥 ∫ (2𝑦 − 5𝑥𝑦)𝑑𝑦 𝟔. ∫ ∫ (𝑥 + 𝑦)𝑑𝑦𝑑𝑥
1 0 0 0

1 |𝑥| 1 √1−𝑥 2
𝟕. ∫ ∫ (𝑥 2 𝑦 + 𝑥𝑦 2 )𝑑𝑦𝑑𝑥
𝟖. ∫ ∫ 𝑥𝑑𝑦𝑑𝑥
−1 0 −1 −√1−𝑥2
1 𝑦2 2 1+𝑥 2
𝟗. ∫ ∫ (𝑥 2 + 𝑦)𝑑𝑥𝑑𝑦 𝟏𝟎. ∫ ∫ 𝑥𝑦𝑑𝑦𝑑𝑥
−2 0 −1 −𝑥 2
 172 Chapter 6 Multiple Integrals

In problems 11 − 15, sketch the regions  and then evaluate each of

the indicated double integrals

𝟏𝟏. ∬ 𝑥𝑦𝑑𝐴 ,  = {(𝑥, 𝑦)| 0 ≤ 𝑥 ≤ 1, 𝑥 2 ≤ 𝑦 ≤ √𝑥}



𝜋 𝜋
𝟏𝟐. ∬ (3𝑥 + 𝑦)𝑑𝐴 ,  = {(𝑥, 𝑦)| ≤ 𝑥 ≤ , sin 𝑥 ≤ 𝑦 ≤ cos 𝑥}
 6 4

𝟏𝟑. ∬ (𝑥 − 2𝑦)𝑑𝐴 ,  = {(𝑥, 𝑦)| 1 ≤ 𝑥 ≤ 3, 1 + 𝑥 ≤ 𝑦 ≤ 2𝑥}



𝟏𝟒. ∬ 𝑥𝑦 2 𝑑𝐴 , 𝑤ℎ𝑒𝑟𝑒  is a parabolic segment bounded by



𝑦 2 = 𝑥, 𝑥 = 1.
𝑥
𝟏𝟓. ∬ 𝑑𝐴 , 𝑤ℎ𝑒𝑟𝑒  is a hyperbolic segment bounded by
 √𝑦
𝑥𝑦 = 4, 𝑥 + 𝑦 = 5.
In problems 16 − 18, find the mass and centre of gravity of the lamina
that occupies the given region  and has the given density function ρ.
𝟏𝟔.  = {(𝑥, 𝑦)| − 1 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1}, ρ(𝑥, 𝑦) = 𝑥 2
𝟏𝟕.  = {(𝑥, 𝑦)|0 ≤ 𝑥 ≤ 𝜋, 0 ≤ 𝑦 ≤ sin 𝑥}, ρ(𝑥, 𝑦) = 𝑦
𝟏𝟖.  is bounded by the parabola y = 9 − 𝑥 2 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑥 − 𝑎𝑥𝑖𝑠,
ρ(𝑥, 𝑦) = 𝑦.
𝟏𝟗. (𝒂) Sketch the region bounded by 𝑦 2 = 2𝑥 𝑎𝑛𝑑 𝑦 = 𝑥,
(𝒃) Find the area of the given region
(𝒄) Find the polar moment of inertia of the region assuming that
the density is constant.
𝟐𝟎. Sketch the region  in the 𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒 bounded by
𝑦 = 𝑥 2 , 𝑥 = 2 and 𝑦 = 1,

then evaluate the double integral ∬ (𝑥 2 + 𝑦 2 )𝑑𝐴.

 Chapter 6 Multiple Integrals 173

6.1.5 Interchange of the order of integration

In evaluation of the double integral sometimes we have to change the
order of integration. This is done because of the non-possibility to
evaluate the integral, or to facilitate the computations as will be shown
later. This idea is illustrated by the following examples.

Example 1. Evaluate the following double integral

∞ ∞
2 ⁄36
∫ ∫ 𝑒 −𝑥 𝑑𝑥𝑑𝑦
𝑦=0 𝑥=3𝑦

𝑦
Solution We note that the
integrand function cannot be
integrated, so we have to
interchange the order of 

integration by the aid of 𝑥

Fig. 6.10 Area of Ex. 1
sketching the region of
integration. So that we first draw the region:
: 𝑥 = 3𝑦, 𝑥 = ∞, 𝑦 = 0, 𝑦=∞
Using the sketch shown in Fig. 6.10, we get
∞ ∞ ∞ 𝑦=𝑥⁄3
−𝑥 2 ⁄36 2 ⁄36
∫ [∫ 𝑒 𝑑𝑥] 𝑑𝑦 = ∫ [∫ 𝑒 −𝑥 𝑑𝑦] 𝑑𝑥
𝑦=0 𝑥=3𝑦 𝑥=0 𝑦=0
∞ ∞
2 ⁄36 𝑥⁄3 2 ⁄36
= ∫ [𝑒 −𝑥 𝑦]0 𝑑𝑥 = ∫ 𝑒 −𝑥 (𝑥⁄3)𝑑𝑥
𝑥=0 𝑥=0
∞
2 ⁄36 2 ⁄36 ∞
= −6 ∫ 𝑒 −𝑥 (− 𝑥⁄18)𝑑𝑥 = −6[𝑒 −𝑥 ]0
𝑥=0

= −6(0 − 1) = 6
 174 Chapter 6 Multiple Integrals

Example 2. Evaluate
4 2
sin 𝑥
∫ ∫ ( ) 𝑑𝑥𝑑𝑦
0 √𝑦
𝑥2

Solution As noted in the above 𝑦

(2,4)
example, the integrand function can
not be integrated, so we have to
interchange the order of integration
using Fig. 6.11:

: 𝑦 = 0, 𝑦 = 4,
𝑥 = √𝑦 𝑦 = 𝑥 2 , 𝑥=2 𝑥=2 𝑥
Fig. 6.11 Area of Ex. 2
Thus, interchanging the order of
integration we get
4 2 2 𝑥 2
sin 𝑥 sin 𝑥
∫ ∫ ( 2 ) 𝑑𝑥𝑑𝑦 = ∫ ∫ ( 2 ) 𝑑𝑦𝑑𝑥
0 √𝑦 𝑥 0 0 𝑥
𝑥 2
2 2
sin 𝑥 sin 𝑥
= ∫ [( 2 ) 𝑦] 𝑑𝑥 = ∫ ( 2 ) ∙ 𝑥 2 𝑑𝑥
0 𝑥 0 0 𝑥
2
= ∫ sin 𝑥 𝑑𝑥 = [− cos 𝑥]20 = 1 − cos 2
0

Example 3. Evaluate the given integral by reversing the order of

integration
1 1
∫ ∫ 𝑥 3 sin 𝑦 3 𝑑𝑦𝑑𝑥
0 𝑥2
 Chapter 6 Multiple Integrals 175

Solution First we draw the region of integration. The region is bounded

by the parabola 𝑦 = 𝑥 2 and
𝑦
the straight lines 𝑥 = 0, 𝑥 =
𝒚=𝟏
1, and 𝑦 = 1, (see Fig. 6.12). (1,1)

Therefore, the given integral

𝒚 = 𝑥2 
will be computed as 𝑥
𝑥=1
Fig. 6.12 Area of Ex. 3
1 1 1 √𝑦
∫ ∫ 𝑥 3 sin 𝑦 3 𝑑𝑦𝑑𝑥 = ∫ ∫ 𝑥 3 sin 𝑦 3 𝑑𝑥𝑑𝑦
0 𝑥2 0 0
𝑦
1
𝑥4 √
3
1
3
𝑦2
= ∫ sin 𝑦 [ ] 𝑑𝑦 = ∫ sin 𝑦 ( ) 𝑑𝑦
0 4 0 0 4
1
1 1
= ∫ sin 𝑦 3 (3𝑦 2 ) 𝑑𝑦 = [− cos 𝑦 3 ]10
4×3 0 12
1
= (1 − cos 1)
12

6.1.6. Change of variables in double integrals

We often apply a change of variables when we integrate a function of

one variable independent variable, and this method is also very
important for evaluating double integrals. For functions of single
variable, when we change the variable we have to find the new
integration limits corresponding to the increment of the new variable.
In this section we study the parallel process for functions of several
variables. Consider a change of variables given by a transformation
from the 𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒 to the 𝑢𝑣 − 𝑝𝑙𝑎𝑛𝑒:
𝑢 = 𝑢(𝑥, 𝑦), 𝑣 = 𝑣(𝑥, 𝑦)
 176 Chapter 6 Multiple Integrals

Theorem 6.1 Given the transformation:

𝑥 = 𝑥(𝑢, 𝑣), 𝑦 = 𝑦(𝑢, 𝑣),
which maps a region  in the 𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒 onto a region 𝑆 in the 𝑢𝑣 −
𝑝𝑙𝑎𝑛𝑒. Suppose that 𝑓 is continuous on , then

∬ 𝑓(𝑥, 𝑦)𝑑𝑥𝑑𝑦 = ∬ 𝑓(𝑥(𝑢, 𝑣), 𝑦(𝑢, 𝑣))|𝐽|𝑑𝑢𝑑𝑣,

 S

where
𝜕𝑥 𝜕𝑥
𝜕(𝑥, 𝑦)
𝐽= = |𝜕𝑢 𝜕𝑣|
𝜕(𝑢, 𝑣) 𝜕𝑦 𝜕𝑦
𝜕𝑢 𝜕𝑣

Example 1. Using the transformation 𝑥 + 𝑦 = 𝑢, 𝑥 − 𝑦 = 𝑣, evaluate

the double integral ∬ (𝑥 2 + 𝑦 2 )𝑑𝑥𝑑𝑦, where  is the region bounded

by
: 𝑥 − 𝑦 = 0, 𝑥 + 𝑦 = 0, 𝑥 − 𝑦 = 2, 𝑥 + 𝑦 = 2

Solution From the given transformation, we have

1 1
𝑥= (𝑢 + 𝑣), 𝑦 = (𝑢 − 𝑣)
2 2
So that the Jacobian is
1 1
𝜕(𝑥, 𝑦) 1
𝐽= = |2 2 | = −
𝜕(𝑢, 𝑣) 1 1 2
−
2 2
Also we have (see Fig. 6.13)
𝑥 − 𝑦 = 0  𝑣 = 0, 𝑥−𝑦 =2  𝑣 =2
𝑥 + 𝑦 = 0  𝑢 = 0, 𝑥+𝑦 =2  𝑢 =2
 Chapter 6 Multiple Integrals 177

𝑣
𝑦

S

𝑥 𝑢

Fig. 6.13 The transformation of Example 1

So that
1 1 1
𝑥 2 + 𝑦 2 = (𝑢 + 𝑣)2 + (𝑢 − 𝑣)2 = (𝑢2 + 𝑣 2 )
4 4 2
Therefore, we have
1 2 2 2 1
∬ (𝑥 2 + 𝑦 2 )𝑑𝑥𝑑𝑦 = ∫ ∫ (𝑢 + 𝑣 2 ) |− | 𝑑𝑢𝑑𝑣
 2 0 0 2
2
1 2 𝑢3 1 2 8
= ∫ [ + 𝑢𝑣 ] 𝑑𝑣 = ∫ ( + 2𝑣 2 ) 𝑑𝑣
2
4 0 3 0
4 0 3
2
1 8 2 1 16 16 8
= [ 𝑣 + 𝑣3] = ( + ) =
4 3 3 0 4 3 3 3

6.1.7 Double Integrals in Polar Coordinate

As a special case of the change of variables is the transformation to the

polar coordinate. The connection between the polar co−ordinates 𝑟, θ
and the Cartesian coordinates 𝑥, 𝑦 is given by the relations
𝑥 = 𝑟 cos θ , 𝑦 = 𝑟 sin θ (𝑟 ≥ 0, 0 ≤ θ < 2𝜋),
if the pole coincides with the origin. From this transformation we get
the Jacobian
 178 Chapter 6 Multiple Integrals

𝜕𝑥 𝜕𝑥
𝜕(𝑥, 𝑦) 𝜕θ| = |cos θ −𝑟 sin θ
𝐽= = | 𝜕𝑟 |
𝜕(𝑟, θ) 𝜕𝑦 𝜕𝑦 sin θ 𝑟 cos θ
𝜕𝑟 𝜕θ
= 𝑟(cos2 θ + sin2 θ) = 𝑟
Therefore,

∬ 𝑓(𝑥, 𝑦)𝑑𝑥𝑑𝑦 = ∬ 𝑓(𝑥(𝑟, θ), 𝑦(𝑟, θ))𝑟𝑑𝑟𝑑θ

 S

Remark We often use the polar co-ordinates if

 The area of integration is a circle
 The integrand contains the expression 𝑥 2 + 𝑦 2

Example 1 Using polar co-ordinates, evaluate the double integral

2 +𝑦 2 )
∬ 𝑒 −(𝑥 𝑑𝑥𝑑𝑦 where  is the annulus bounded by 𝑥 2 + 𝑦 2 = 1
and 𝑥 2 + 𝑦 2 = 4.

Solution In the polar coordinates 𝑥 2 + 𝑦 2 = 𝑟 2 , so we have

𝑥2 + 𝑦2 = 1  𝑟=1 𝑦
2
𝑥 +𝑦 =4 2
 𝑟=2
Hence, the annulus is bounded by the two 

circles 𝑟 = 1 and 𝑟 = 2 in the polar 𝑥

coordinates (see Fig. 6.14). Thus we have

Fig. 6.14 Domain of Ex 1
θ=2𝜋 𝑟=2
−(𝑥 2 +𝑦 2 ) −𝑟 2
∬ 𝑒 𝑑𝑥𝑑𝑦 = ∫ ∫ 𝑒 𝑟𝑑𝑟𝑑θ
 θ=0 𝑟=1

1 θ=2𝜋 𝑟=2 −𝑟 2
= ∫ ∫ (𝑒 (−2𝑟)𝑑𝑟)𝑑θ
−2 θ=0 𝑟=1
 Chapter 6 Multiple Integrals 179

1 θ=2𝜋 −𝑟 2 2 1 θ=2𝜋 −4
= ∫ [𝑒 ]1 𝑑θ = ∫ (𝑒 − 𝑒 −1 ) 𝑑θ
−2 θ=0 −2 θ=0
(𝑒 −1 − 𝑒 −4 )
= (2𝜋) = 𝜋(𝑒 −1 − 𝑒 −4 )
2

Example 2 Using the polar coordinates, evaluate

𝑎⁄√2 √𝑎2 −𝑥 2
∫ ∫ ln(𝑥 2 + 𝑦 2 ) 𝑑𝑦𝑑𝑥
0 𝑥

Solution Changing into polar coordinates:

𝑥 = 𝑟 cos θ , 𝑦 = 𝑟 sin θ, 𝑥2 + 𝑦2 = 𝑟 2, 𝑑𝑥𝑑𝑦 = 𝑟𝑑𝑟𝑑θ,

: 𝑥 = 0, 𝑥 = 𝑎⁄√2 , 𝑦 = 𝑥, 𝑦 = √𝑎2 − 𝑥 2
From Fig. 6.15, we get

𝒚 𝒚=𝒙
𝒚= √𝒂 𝟐 − 𝒙𝟐
  (𝑎⁄√2 , 𝑎⁄√2)

𝐅𝐢𝐠. 𝟔. 𝟏𝟓

𝑎⁄√2 √𝑎2 −𝑥 2 𝜋⁄2 𝑎

∫ ∫ ln(𝑥 2 + 𝑦 2 ) 𝑑𝑦𝑑𝑥 = ∫ ∫ ln(𝑟 2 ) 𝑟𝑑𝑟𝑑θ
0 𝑥 𝜋⁄4 0
𝜋⁄2 𝑎
𝜋 𝑎
= 2∫ ∫ ln 𝑟 𝑟𝑑𝑟𝑑θ = 2 ( ) ∫ 𝑟 ln 𝑟 𝑑𝑟
𝜋⁄4 0 4 0
𝑎 𝑎
𝜋 𝑟2 𝑟2 1
= ([( ) ∙ ln 𝑟] − ∫ ( ) ( ) 𝑑𝑟)
2 2 0 0 2 𝑟

𝜋𝑎2 1 1 𝜋𝑎2 𝜋𝑎2

= (( ) ln 𝑎 − ) = (2 ln 𝑎 − 1) = (ln 𝑎2 − 1)
2 2 4 8 8
 180 Chapter 6 Multiple Integrals

Example 3 Prove that

𝑥2 𝑦2 2π
∬ √(1 − 2
− 2 ) 𝑑𝑥𝑑𝑦 = 𝑎𝑏,
 𝑎 𝑏 3

𝑥2 𝑦2
where  is the ellipse 𝑎2
+ 𝑏2 = 1.

Solution On integrating over ellipses, it is very useful to use the so-

called 𝑒𝑙𝑙𝑖𝑝𝑡𝑖𝑐 𝑝𝑜𝑙𝑎𝑟 𝑐𝑜 − 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 which is, to somewhat, a
modification of the normal co-ordinates.
𝑥 𝑦
= 𝑟 cos θ, = 𝑟 sin θ
𝑎 𝑏
Thus, we have
𝑥2 𝑦2
+ = 𝑟 2,
𝑎2 𝑏 2
and the Jacobian is
𝜕𝑥 𝜕𝑥
𝜕(𝑥, 𝑦) 𝑎 cos θ −𝑎𝑟 sin θ
𝐽= = | 𝜕𝑟 𝜕θ| = | |
𝜕(𝑟, θ) 𝜕𝑦 𝜕𝑦 𝑏 sin θ 𝑎𝑟 cos θ
𝜕𝑟 𝜕θ
= 𝑎𝑏𝑟(cos 2 θ + sin2 θ) = 𝑎𝑏𝑟,
which means that the element of area in the 𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒 will be changed
in the 𝑟 − 𝑝𝑙𝑎𝑛𝑒 into
𝑑𝑥𝑑𝑦 = |𝐽|𝑑𝑟𝑑θ = 𝑎𝑏𝑟𝑑𝑟𝑑θ
From the transformation
𝑥2 𝑦2
+ = 𝑟 2,
𝑎2 𝑏 2
𝑥2
it is evident that the elliptic polar coordinates transform the ellipse +
𝑎2
𝑦2
= 1, into a unit circle centred at the origin. So that
𝑏2
 Chapter 6 Multiple Integrals 181

2𝜋 1
𝑥2 𝑦2
∬ √(1 − 2 − 2 ) 𝑑𝑥𝑑𝑦 = ∫ ∫ √(1 − 𝑟 2 )𝑎𝑏𝑟𝑑𝑟𝑑θ
 𝑎 𝑏 0 0

1 2𝜋 1
= ∫ ∫ √(1 − 𝑟 2 )𝑎𝑏(−2𝑟)𝑑𝑟𝑑θ
−2 0 0
3⁄2 1
𝑎𝑏 2𝜋 ((1 − 𝑟 2 ))
= ∫ [( )] 𝑑θ
−2 0 3⁄2
0

𝑎𝑏 2 2𝜋𝑎𝑏
= ∙ − (2𝜋) =
−2 3 3

Example 4. By changing to polar coordinates, evaluate the given

integral
1
∬ 𝑑𝐴,
√𝑥 2 + 𝑦 2

where  is the region bounded by the cardioid 𝑟 = 1 + sin θ.

Solution Changing into polar

coordinates, we have 𝑦
𝑥 = 𝑟 cos θ 𝒓 = 𝟏 + 𝐬𝐢𝐧 𝛉
𝑦 = 𝑟 sin θ 
𝑥2 + 𝑦2 = 𝑟2
D
𝑑𝑥𝑑𝑦 = 𝑟𝑑𝑟𝑑θ, 𝑥

so we get Fig. 6.16

2𝜋 1+sin θ
1 1
∬ 𝑑𝐴 = ∫ ∫ ( ) 𝑟𝑑𝑟𝑑θ
 √𝑥 2 + 𝑦 2 0 0 𝑟
2𝜋
= ∫ (1 + sin θ)𝑑θ = [(θ − cos θ)]2𝜋
0
0
= [(2𝜋 − cos 2𝜋) − (0 − 1)] = 2𝜋
 182 Chapter 6 Multiple Integrals

Example 5 Find the volume of the solid that lies under the paraboloid
𝑧 = 𝑥 2 + 𝑦 2 , above the 𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒, and inside the cylinder 𝑥 2 + 𝑦 2 =
2𝑥.

𝑦
Solution The volume of the solid is given by
D
𝑉 = ∬ 𝑧𝑑𝐴 = ∬ (𝑥 2 + 𝑦 2 )𝑑𝑥𝑑𝑦, 
(1,0) 𝑥
 

where  is the disk

𝑥 2 + 𝑦 2 = 2𝑥  (𝑥 − 1)2 + 𝑦 2 = 1, Fig. 6.17

which is a circle of radius 1 and its centre is the point

𝑃(1,0) as shown in Fig. 6.17. In polar coordinates, this disk has the
equation
𝑥 2 + 𝑦 2 = 2𝑥  𝑟 2 = 2𝑟 cos θ  𝑟 = 2 cos θ
Thus, with the aid of Fig. 6.17, we have
𝜋 π
 = {(𝑟, θ)| − ≤ θ ≤ , 0 ≤ 𝑟 ≤ 2 cos θ}
2 2
Therefore,
𝜋⁄2 2 cos θ 𝜋⁄2 2 cos θ
𝑟4
𝑉=∫ ∫ (𝑟 2 )(𝑟𝑑𝑟𝑑θ) =∫ [ ] 𝑑θ
−𝜋⁄2 0 −𝜋⁄2 4 0

𝜋⁄2 𝜋⁄2
4
= 4∫ cos θ 𝑑θ = 8 ∫ cos4 θ 𝑑θ
−𝜋⁄2 0

3 ∙ 1 𝜋 3𝜋
=8 =
4∙2 2 2
Here, we have used the well-known law
 Chapter 6 Multiple Integrals 183

For a positive integer 𝑛 > 1,

𝜋⁄2 𝜋⁄2
𝑛
∫ cos θ 𝑑θ = ∫ sin𝑛 θ 𝑑θ
0 0

(𝑛 − 1)(𝑛 − 3)(𝑛 − 5) ⋯ 1, 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑

= 𝑓(𝑥) = {𝜋
𝑛(𝑛 − 2)(𝑛 − 4) ⋯ , 𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛
2

Example 6 Evaluate the double integral ∬ 𝑥𝑑𝐴 where  is the

region in the first quadrant that lies between the circles 𝑥 2 + 𝑦 2 = 4
and 𝑥 2 + 𝑦 2 = 2𝑥.
𝑦
𝒙𝟐 + 𝒚𝟐 = 𝟒
Solution From Fig. 6.18, we see that the

domain  is bounded in polar coordinates
by 𝒙𝟐 + 𝒚𝟐 = 𝟐𝒙
π 𝑥
 = {(𝑟, θ)|0 ≤ θ ≤ , 2 cos θ ≤ 𝑟 ≤ 2}
2
Fig. 6.18
So that the required integral is
𝜋⁄2 2
∬ 𝑥𝑑𝐴 = ∫ ∫ (𝑟 cos θ)(𝑟𝑑𝑟𝑑θ)
 0 2 cos θ

𝜋⁄2 2 𝜋⁄2
𝑟3 8 8 cos3 θ
=∫ [ ] cos θ 𝑑θ = ∫ ( − ) cos θ 𝑑θ
0 3 2 cos θ 0 3 3

8 𝜋⁄2 8 ⁄ 8 3 ∙ 1𝜋
= ∫ (cos θ − cos 4 θ)𝑑θ = ([sin θ]𝜋0 2 ) − ∙
3 0 3 3 4∙22
8 𝜋
= −
3 2
 184 Chapter 6 Multiple Integrals

Exercise 6.2

In problems 1 − 7, evaluate the given integral

1 3 1 𝑒
𝑥2
1
𝟏. ∫ ∫ 𝑒 𝑑𝑥𝑑𝑦 𝟐. ∫ ∫ 𝑑𝑦𝑑𝑥
0 3𝑦 0 𝑒 𝑥 ln 𝑦
4 2 ∞ ∞
𝑒 −𝑥
𝟑. ∫ ∫ ( 2 ) 𝑑𝑥𝑑𝑦 𝟒. ∫ ∫ cos(𝑥 2 ) 𝑑𝑥𝑑𝑦
0 √𝑦 𝑥 𝑦=0 𝑥=2𝑦

𝑥=2 √16−𝑥 2 𝑥=4 √16−𝑥 2

−(𝑥 2 +𝑦 2 ) 2 +𝑦 2 )
𝟓. ∫ ∫ 𝑒 𝑑𝑦𝑑𝑥 + ∫ ∫ 𝑒 −(𝑥 𝑑𝑦𝑑𝑥
𝑥=0 𝑦=√4−𝑥2 𝑥=2 𝑦=0

𝟔. ∬ sin(𝑥 2 ) 𝑑𝑥𝑑𝑦 , where  is the region bounded by the straight



lines 𝑥 = 0, 𝑥 = 1, 𝑦 = 0, 𝑦 = 𝑥.

𝟕. ∬ 𝑒 𝑥⁄𝑦 𝑑𝑦𝑑𝑥 , where  is the region bounded by the parabola


2
𝑦 = 𝑥, and the straight lines 𝑥 = 0, 𝑦 = 1.
𝟖. Find the mass and centre of mass of a lamina that occupies the
region D bounded by the parabola 𝑥 = 1 − 𝑦 2 and the coordinate axes
in the first quadrant if the density function is ρ(𝑥, 𝑦) = 𝑦.
9. A lamina with constant density occupies the region upper the curve
𝑦 = sin 𝑥 from 𝑥 = 0 to 𝑥 = 𝜋. Find the moments of inertia about the
𝑥 − and 𝑦 − 𝑎𝑥𝑖𝑠.
In problems 10 − 18, By changing to polar coordinates, evaluate the
given integral

𝟏𝟎. ∬ 𝑥𝑑𝐴 where  is the disk with centre the origin and radius 5.

𝟏𝟏. ∬ 𝑦𝑑𝐴 where  is the region in the first quadrant bounded by

the circle 𝑥 2 + 𝑦 2 = 9 and the lines 𝑦 = 𝑥 and 𝑦 = 0.
 Chapter 6 Multiple Integrals 185

𝟏𝟐. ∬ 𝑥𝑦𝑑𝐴 where  is the region in the first quadrant that lies
between the circles 𝑥 2 + 𝑦 2 = 4 and 𝑥 2 + 𝑦 2 = 25.

𝟏𝟑. ∬ sin(𝑥 2 + 𝑦 2 ) 𝑑𝐴 where  is the annular region

1 ≤ 𝑥 2 + 𝑦 2 ≤ 16.
1 √1−𝑥 2
2 +𝑦 2 )
𝟏𝟒. ∫ ∫ 𝑒 (𝑥 𝑑𝑦𝑑𝑥
0 0

𝑏 √𝑏 2 −𝑥 2
𝟏𝟓. ∫ ∫ (𝑥 2 + 𝑦 2 )3⁄2 𝑑𝑦𝑑𝑥
−𝑏 0

2 √4 −𝑥 2
𝟏𝟔. ∫ ∫ 𝑥 2 𝑦 2 𝑑𝑦𝑑𝑥
0 −√4 −𝑥 2

2 √2𝑥 −𝑥 2
𝟏𝟕. ∫ ∫ √𝑥 2 + 𝑦 2 𝑑𝑦𝑑𝑥
0 0
0 𝑥
𝟏𝟖. ∫ ∫ cos(𝑥 2 + 𝑦 2 ) 𝑑𝑦𝑑𝑥
−2 −√8 −𝑥 2

Describe the region of integration and evaluate the integrals in problems

19 − 22
1 2𝑥 2 𝑥
𝟏𝟗. ∫ ∫ (2 + 𝑥 2 + 𝑦 2 )𝑑𝑦𝑑𝑥 𝟐𝟎. ∫ ∫ 𝑒 𝑦 cosh 𝑥 𝑑𝑦𝑑𝑥
0 𝑥 1 −𝑥
1 𝑦 2 +1 𝜋⁄2 𝑦
2
sin 𝑦
𝟐𝟏. ∫ ∫ 𝑥 𝑦𝑑𝑥𝑑𝑦 𝟐𝟐. ∫ ∫ 𝑑𝑥𝑑𝑦
0 𝑦 0 0 𝑦
In problems 23 − 24, evaluate the following integrals by making an
appropriate change of variables.

𝟐𝟑. ∬ 𝑥𝑦𝑑𝐴 where  is the region bounded by the lines

2𝑥 − 𝑦 = 1, 2𝑥 − 𝑦 = 3, 3𝑥 + 𝑦 = 1, 3𝑥 + 𝑦 = −2
𝑥+2𝑦
𝟐𝟒. ∬ cos(𝑥−𝑦)
𝑑𝐴 where  is the parallelogram bounded by the

lines 𝑦 = 𝑥, 𝑦 = 𝑥 − 1, 𝑥 + 2𝑦 = 0, 𝑥 + 2𝑦 = 2.
 186 Chapter 6 Multiple Integrals

𝑥2
𝟐𝟓. ∬ 𝑑𝐴 over the region  bounded by the lines
𝑦

𝑥 = 0, 𝑥 = 2, 𝑦 = 1, 𝑦=2
Check your answer by integrating in the reverse order.

6.2 Triple Integrals

In this section we are going to define the integral of a function of

three independent variables, the so-called 𝑡𝑟𝑖𝑝𝑙𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙. Like double
integrals, triple integrals are widely applied to various physical and
geometrical problems.
By the same procedure used for double integrals, we may define the
triple integral of a continuous function 𝑓(𝑥, 𝑦, 𝑧) over a general
bounded region 𝑅 in three dimensional space. This integral is denoted
by

∭ 𝑓(𝑥, 𝑦, 𝑧) 𝑑𝑉 (1)
𝑅

We restrict our attention to simple types of regions. A solid region 𝑅 is

said to be of 𝑡𝑦𝑝𝑒 1 if it lies between the graphs of two continuous
functions of 𝑥 and 𝑦, that is,
𝑅 = {(𝑥, 𝑦, 𝑧)| (𝑥, 𝑦) ∈ , α1 (𝑥, 𝑦) ≤ 𝑧 ≤ α2 (𝑥, 𝑦)}, (2)
where  is the projection of 𝑅 onto the 𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒. In this case
equation (1) is written as
𝑧=α2 (𝑥,𝑦)
∭ 𝑓(𝑥, 𝑦, 𝑧) 𝑑𝑉 = ∬ ∫ 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑧𝑑𝐴 (3)
𝑅  𝑧=α1 (𝑥,𝑦)

According to the type of the area , 𝑡𝑦𝑝𝑒 𝐼 or 𝑡𝑦𝑝𝑒 𝐼𝐼 regions, the

integral in (1) is expressed in the form
𝑥=𝑏 𝑦=𝑔2 (𝑥) 𝑧=𝛼2 (𝑥,𝑦)
∭ 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑉 = ∫ ∫ ∫ 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑧𝑑𝑦𝑑𝑥 (4)
𝑅 𝑥=𝑎 𝑦=𝑔1 (𝑥) 𝑧=𝛼1 (𝑥,𝑦)
 Chapter 6 Multiple Integrals 187

or
𝑦=𝑑 𝑥=ℎ2 (𝑥) 𝑧=𝛼2 (𝑥,𝑦)
∭ 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑉 = ∫ ∫ ∫ 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑧𝑑𝑦𝑑𝑥 (5)
𝑅 𝑦=𝑐 𝑥=ℎ1 (𝑥) 𝑧=𝛼1 (𝑥,𝑦)

However, in both cases to evaluate this integral, we follow the

rules as before. Start with the innermost integral and work

outwards.




𝑦=𝑑 𝑥=ℎ2 (𝑥) 𝑧=𝛼2 (𝑥,𝑦)
∫ ∫ ∫ 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑧 𝑑𝑦 𝑑𝑥
𝑦=𝑐 𝑥=ℎ1 (𝑥) 𝑧=𝛼1 (𝑥,𝑦)

A solid region 𝑅 is of 𝑡𝑦𝑝𝑒 2 if it is of the form

𝑅 = {(𝑥, 𝑦, 𝑧)| (𝑦, 𝑧) ∈ , β1 (𝑦, 𝑧) ≤ 𝑥 ≤ β2 (𝑦, 𝑧)}, (6)
where  is the projection of 𝑅 onto the 𝑦𝑧 − 𝑝𝑙𝑎𝑛𝑒. In this case
Equation (1) will be in the form
𝑥=β2 (𝑦,𝑧)
∭ 𝑓(𝑥, 𝑦, 𝑧) 𝑑𝑉 = ∬ [∫ 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑥] 𝑑𝐴 (7)
𝑅  𝑥=β1 (𝑦,𝑧)

Finally, a 𝑡𝑦𝑝𝑒 3 region is of the form

𝑅 = {(𝑥, 𝑦, 𝑧)| (𝑥, 𝑧) ∈ , γ1 (𝑥, 𝑧) ≤ 𝑦 ≤ γ2 (𝑥, 𝑧)}, (8)
where  is the projection of 𝑅 onto the 𝑥𝑧 − 𝑝𝑙𝑎𝑛𝑒. For this type of
regions, we have
𝑦=γ2 (𝑥,𝑧)
∭ 𝑓(𝑥, 𝑦, 𝑧) 𝑑𝑉 = ∬ [∫ 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑦] 𝑑𝐴 (9)
𝑅  𝑦=γ1 (𝑥,𝑧)
 188 Chapter 6 Multiple Integrals

In each of Equations (7) and (9) there may be two possible expressions
for the integral depending on whether  is of a 𝑡𝑦𝑝𝑒 𝐼 𝑜𝑟 𝐼𝐼 plane
region.

3 1 2
Example 1 Evaluate ∫1 ∫−1 ∫0 (𝑥 + 2𝑦 − 𝑧)𝑑𝑥𝑑𝑦𝑑𝑧.

Solution
3 1 2 3 1 2
𝑥2
∫ ∫ ∫ (𝑥 + 2𝑦 − 𝑧)𝑑𝑥𝑑𝑦𝑑𝑧 = ∫ ∫ [ + 2𝑥𝑦 − 𝑥𝑧] 𝑑𝑦𝑑𝑧
1 −1 0 1 −1 2 0
3 1 3
= ∫ ∫ (2 + 4𝑦 − 2𝑧)𝑑𝑦𝑑𝑧 = ∫ [2𝑦 + 2𝑦 2 − 2𝑦𝑧]1−1 𝑑𝑧
1 −1 1
3 3
= ∫ [(2 + 2 − 2𝑧) − (−2 + 2 + 2𝑧)]𝑑𝑧 = ∫ (4 − 4𝑧)𝑑𝑧
1 1

= [4𝑧 − 2𝑧 2 ]13 = (12 − 18) − (4 − 2) = − 8

Example 2 Evaluate ∭𝑅 𝑧 𝑑𝑉 where 𝑅 is the solid tetrahedron

bounded by the four planes 𝑥 = 0, 𝑦 = 0, 𝑧 = 0 and 𝑥 + 𝑦 + 𝑧 = 1.

𝑧 𝑦

(0,0,1)
(0,1)
𝑥+𝑦=1


𝑥+𝑦+𝑧 =1

(0,1,0) 𝑦 (1,0) 𝑥

(1,0,0)
𝑥 (𝒂) (𝒃)
Fig. 6.19
 Chapter 6 Multiple Integrals 189

Solution. When setting up a triple integrals it is wise to draw two

diagrams: one of the solid region 𝑅 and one of its projection  on the
𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒 (see Fig. 6.19). The lower boundary of the tetrahedron is
the plane 𝑧 = 0 and the upper boundary is the plane 𝑧 = 1 − 𝑥 − 𝑦.
Notice that the planes 𝑥 + 𝑦 + 𝑧 = 1 and 𝑧 = 0 intersect in the line 𝑥 +
𝑦 = 1 in the 𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒. So the projection of 𝑅 is the triangular region
shown in Fig. 6.19(b) and we have
1 1−𝑥 1−𝑥−𝑦
∭ 𝑧 𝑑𝑉 = ∫ ∫ ∫ 𝑧𝑑𝑧𝑑𝑦𝑑𝑥
𝑅 0 0 0
1 1−𝑥 2 1−𝑥−𝑦 1 1−𝑥 (1
𝑧 − 𝑥 − 𝑦)2
=∫ ∫ [ ] 𝑑𝑦𝑑𝑥 = ∫ ∫ 𝑑𝑦𝑑𝑥
0 0 2 0 0 0 2
3 1−𝑥
1 1 (1 − 𝑥 − 𝑦) 1 1
= ∫ [ ] 𝑑𝑥 = ∫ (1 − 𝑥)3 𝑑𝑥
2 0 −3 0
6 0
1
1 (1 − 𝑥)4 1
= [ ] =
6 −4 0
24
6.2.1 Applications of triple integrals

The triple integrals are used to compute, volumes, masses, centroids of

regions.
 The volume of a region 𝑹
The volume 𝑉 of a region 𝑅 is given by putting 𝑓(𝑥, 𝑦, 𝑧) = 1 in (1),
that is,

𝑉𝑜𝑙𝑢𝑚𝑒 = ∭ 𝑑𝑉
𝑅
 The mass 𝑴 in a region 𝑹
Let ρ(𝑥, 𝑦, 𝑧) be the density of a solid body, its mass 𝑀 is

𝑀 = ∭ ρ(𝑥, 𝑦, 𝑧)𝑑𝑉
𝑅
 190 Chapter 6 Multiple Integrals

 The centroids of the mass in 𝑹

The centre of gravity (𝑥̅ , 𝑦̅, 𝑧̅) of the mass 𝑀 with a density ρ(𝑥, 𝑦, 𝑧)
is

∭𝑅 𝑥ρ(𝑥, 𝑦, 𝑧)𝑑𝑉
𝑥̅ = ,
𝑀
∭𝑅 𝑦ρ(𝑥, 𝑦, 𝑧)𝑑𝑉
𝑦̅ = ,
𝑀
∭𝑅 𝑧ρ(𝑥, 𝑦, 𝑧)𝑑𝑉
𝑧̅ =
𝑀
 The moment of inertia of 𝑹
The moment of inertia of 𝑅 about the three co-ordinate axes are:

𝐼𝑥 = ∭ (𝑦 2 + 𝑧 2 )ρ(𝑥, 𝑦, 𝑧)𝑑𝑉 ,
𝑅

𝐼𝑦 = ∭ (𝑥 2 + 𝑧 2 )ρ(𝑥, 𝑦, 𝑧)𝑑𝑉 ,
𝑅

𝐼𝑧 = ∭ (𝑥 2 + 𝑦 2 )ρ(𝑥, 𝑦, 𝑧)𝑑𝑉
𝑅

Example 1 Find the volume of the region bounded by the parabolic

cylinder 𝑦 2 = 4𝑥 and the planes 𝑧 = 2𝑥, 𝑥 = 4, 𝑧 = 0.
𝑦

 (4,2)
Solution The region 𝑅 is of 
𝑥
𝑡𝑦𝑝𝑒 1, so 0 2 𝒙=𝟒
4
 (4, −2)
4 2√ 𝑥 2𝑥
𝑉=∫ ∫ ∫ 𝑑𝑧𝑑𝑦𝑑𝑥 =
0 −2√𝑥 0
Fig. 6.20
 Chapter 6 Multiple Integrals 191

4 2√𝑥 4 2√𝑥 4
=∫ ∫ [𝑧]2𝑥
0 𝑑𝑦𝑑𝑥 = ∫ ∫ 2𝑥𝑑𝑦𝑑𝑥 = ∫ [2𝑥𝑦]2−2
√𝑥
√𝑥
𝑑𝑥
0 −2√𝑥 0 −2√𝑥 0
4
4
𝑥 5⁄2 16 512
= ∫ 8𝑥 √𝑥𝑑𝑥 = 8 [ ] = × 32 =
0 5⁄2 0 5 5

Example 2 Find the volume of the solid bounded by the paraboloid

𝑧 = 𝑥 2 + 𝑦 2 and the plane 𝑧 = 2𝑦 + 8.

Solution
𝑧=𝑦+2
𝑉=∬ ∫ 𝑧𝑑𝑧𝑑𝐴 = ∬ (2𝑦 + 8 − (𝑥 2 + 𝑦 2 ))𝑑𝐴
 𝑧=𝑥 2 +𝑦2 

where  is the projection of the solid into the 𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒 which is the
circle:
𝑧
𝑥 2 + 𝑦 2 = 2𝑦 + 8, 𝟐 𝟐
𝒛=𝒙 +𝒚
or
𝑥 2 + 𝑦 2 − 2𝑦 + 1 − 1 − 8 = 0,
which is the circle centered at the point 𝑦

(0,1) and of radius 3 (see Fig. (6.21)

𝑥
 2
𝑥 + (𝑦 − 1) = 9 2
𝑦
Using the polar coordinates
𝑥 = 𝑟 cos θ, (𝑦 − 1) = 𝑟 sin θ,
𝑥 2 + (𝑦 − 1)2 = 𝑟 2 ,   (0,1)
𝜕𝑥 𝜕𝑥
𝑥
𝜕(𝑥, 𝑦)
𝐽= = | 𝜕𝑟 𝜕θ|
𝜕(𝑟, θ) 𝜕𝑦 𝜕𝑦
𝜕𝑟 𝜕θ
Fig. 6.21
 192 Chapter 6 Multiple Integrals

cos θ −𝑟 sin θ
=| | = 𝑟,
sin θ 𝑟 cos θ
𝑑𝑥𝑑𝑦 = |𝐽|𝑑𝑟𝑑θ = 𝑟𝑑𝑟𝑑θ
Then, the volume is

𝑉 = − ∬ (𝑥 2 + 𝑦 2 − 2𝑦 − 8)𝑑𝐴 = − ∬ [𝑥 2 + (𝑦 − 1)2 − 9]𝑑𝐴

 
2𝜋 3 3
1 2𝜋 (𝑟 2 − 9)2
= −∫ ∫ (𝑟 2 − 9)𝑟𝑑𝑟𝑑θ = − ∫ [ ] 𝑑θ
0 0 2 0 2 0

1 2𝜋
= − ∫ ((9 − 9)2 − (0 − 9)2 )𝑑θ
4 0
81 2𝜋 81𝜋
= ∫ 𝑑θ =
4 0 2

Example 3 Find the centre of mass of the solid bounded by the

parabolic cylinder 𝑥 = 𝑦 2 and the planes 𝑥 = 𝑧, 𝑧 = 0 and 𝑥 = 1 if
ρ ≡ constant.

Solution Because of the symmetry of 𝑅 about the 𝑥𝑧 − 𝑝𝑙𝑎𝑛𝑒, 𝑦̅ = 0

and
𝑦
∭𝑅 𝑥ρ𝑑𝑉 ∭ 𝑧ρ𝑑𝑉 (1,1)
𝑥̅ = , 𝑧̅ = 𝑅
𝑀 𝑀
The solid 𝑅 is a 𝑡𝑦𝑝𝑒 1 region, then its  𝒙=𝟏 𝑥
mass is
 (1, −1)
1 1 𝑥
𝑀 = ∫ ∫ ∫ ρ𝑑𝑧𝑑𝑥𝑑𝑦 Fig. 6.22
−1 𝑦 2 0
 Chapter 6 Multiple Integrals 193

1 1 1 1
𝑥2
= ρ ∫ ∫ 𝑥𝑑𝑥𝑑𝑦 = ρ ∫ [ ] 𝑑𝑦
−1 𝑦 2 −1 2 𝑦 2

ρ 1
= ∫ (1 − 𝑦 4 )𝑑𝑦
2 −1
1
𝑦5 1 4
= ρ [𝑦 − ] = ρ (1 − ) = ρ,
5 0 5 5
1 1 𝑥
∭ 𝑥ρ(𝑥, 𝑦, 𝑧)𝑑𝑉 = ∫ ∫ ∫ 𝑥ρ𝑑𝑧𝑑𝑥𝑑𝑦 ,
𝑅 −1 𝑦 2 0

1 1 1 1
2
𝑥3
= ρ ∫ ∫ 𝑥 𝑑𝑥𝑑𝑦 = ρ ∫ [ ] 𝑑𝑦
−1 𝑦 2 −1 3 𝑦 2

ρ 1
= ∫ (1 − 𝑦 6 )𝑑𝑦
3 −1
1
2ρ 𝑦7 2ρ 1 4
= [𝑦 − ] = (1 − ) = ρ,
3 7 0 3 7 7
1 1 𝑥
∭ 𝑧ρ(𝑥, 𝑦, 𝑧)𝑑𝑉 = ∫ ∫ ∫ 𝑧ρ𝑑𝑧𝑑𝑥𝑑𝑦 ,
𝑅 −1 𝑦 2 0
1 1
𝑥2 ρ 1 31
= ρ∫ ∫ 𝑑𝑥𝑑𝑦 = ∫ [𝑥 ]𝑦 2 𝑑𝑦
−1 𝑦 2 2 6 −1
ρ 1
= ∫ (1 − 𝑦 6 )𝑑𝑦
6 −1
1
ρ 𝑦7 ρ 1 2
= [𝑦 − ] = (1 − ) = ρ
3 7 0 3 7 7
Therefore, the centre of mass is
4 2
ρ ρ 5 5
(𝑥̅ , 𝑦̅, 𝑧̅) = ( , 0, 7 ) = ( , 0,
7 )
4 4 7 14
ρ ρ
5 5
 194 Chapter 6 Multiple Integrals

Example 4 Find the volume bounded above by the paraboloid

𝑥2 𝑦2
𝑧 =1− − ,
9 4
and below by the plane 𝑧 = 0.

Solution From Fig. 6.23, we have

𝑥2 𝑦2
1− −
9 4 𝑥2 𝑦2
𝑉=∬ ∫ 𝑑𝑧𝑑𝐴 = ∬ (1 − − ) 𝑑𝐴
 0  9 4
where  is the ellipse
𝑧
𝑥2 𝑦2
: + = 1,
9 4
𝒙𝟐 𝒚 𝟐
𝒛=𝟏− −
thus using the elliptic polar co- 𝒂𝟐 𝒃𝟐

ordinates: 𝒛=𝟎
𝑦
𝑥 𝑦
= 𝑟 cos θ, = 𝑟 sin θ
3 2
𝑥2 𝑦2 𝑥
+ = 𝑟 2,
9 4 𝑦
𝑑𝑥𝑑𝑦 = 3 × 2𝑟𝑑𝑟𝑑θ
 𝒙𝟐 𝒚𝟐
The elliptic polar coordinates 𝒛 =𝟏− −
𝟗 𝟒 𝑥
transform the ellipse into a unit
circle centred at the origin.
Fig. 6.23
Therefore, the volume is
2𝜋 1
𝑥2 𝑦2
𝑉 = ∬ (1 − − ) 𝑑𝐴 = ∫ ∫ (1 − 𝑟 2 )(6𝑟𝑑𝑟𝑑θ)
 9 4 0 0
2𝜋 1 2𝜋 1
2 )(−2𝑟𝑑𝑟)
(1 − 𝑟 2 )2
= −3 ∫ ∫ (1 − 𝑟 𝑑θ = −3 ∫ [ ] 𝑑θ
0 0 0 2 0
1
=3× × 2𝜋 = 3𝜋
2
 Chapter 6 Multiple Integrals 195

Example 5 Find the volume of the elliptic paraboloid 8𝑧 = 𝑥 2 + 4𝑦 2

intercepted by 𝑧 = 2.

Solution The sketch of the

given solid and its projection in
the 𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒 is shown in Fig
(6.24) a and b.
𝑧=2
𝑉=∬ ∫ 𝑑𝑧𝑑𝐴
 𝑧=(𝑥 2 +4𝑦2 )/8

(𝑥 2 + 4𝑦 2 )
= ∬ (2 − ) 𝑑𝐴
 8
Fig. 6.24 (a)
where  is the intersection
𝑦
region of 8𝑧 = 𝑥 2 + 4𝑦 2 and
𝑧 = 2, i.e.,
𝒙𝟐 𝒚𝟐
: 8 × 2 = 𝑥 2 + 4𝑦 2 , (− 4,0) 
+
𝟏𝟔 𝟒
=𝟏
(4,0) 𝑥
or
𝑥2 𝑦2
: + =1 Fig. 6.24 (b)
16 4
Change into elliptic polar coordinates
𝑥 𝑦 𝑥2 𝑦2
= 𝑟 cos θ, = 𝑟 sin θ, + = 𝑟 2 , 𝑑𝑥𝑑𝑦 = 4 × 2𝑟𝑑𝑟𝑑θ
4 2 16 4
So the volume is
(𝑥 2 + 4𝑦 2 ) 𝑥2 𝑦2
𝑉 = ∬ (2 − ) 𝑑𝐴 = 2 ∬ (1 − − ) 𝑑𝐴
 8  16 4
2𝜋 1 2𝜋 1
2 )(8𝑟𝑑𝑟𝑑θ)
= 2∫ ∫ (1 − 𝑟 = − 8∫ ∫ (1 − 𝑟 2 )(−2𝑟𝑑𝑟𝑑θ)
0 0 0 0
2𝜋 1
(1 − 𝑟 2 )2 1
= − 8∫ [ ] 𝑑θ = 8 × × 2𝜋 = 8𝜋
0 2 0
2
 196 Chapter 6 Multiple Integrals

6.2.2 Triple integrals in spherical co-ordinates

Let the position of the moving 𝑃 in space be determined by the
following 𝑠𝑝ℎ𝑒𝑟𝑖𝑐𝑎𝑙 𝑐𝑜 − 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 (see Fig. 6.25):

 𝑃(𝑟, θ, φ)
𝑟 cos θ
𝑟
θ
φ 𝑦

𝒓 𝐬𝐢𝐧 𝛉 𝐜𝐨𝐬 𝛗
Q
𝑥

Fig. 6.25
(𝒂) The distance 𝑟 from the origin to 𝑃

(𝒃) The angle θ between 0𝑃 and the direction of the positive half-axis 𝑧.
(𝒄) The angle φ between 0𝑄 and the positive direction of the 𝑥 − 𝑎𝑥𝑖𝑠.

From Fig. 6.25, the relation between the Cartesian and the spherical co-
ordinates is given by
𝑥 = 𝑟 sin θ cos φ,
𝑦 = 𝑟 sin θ sin φ,
𝑧 = 𝑟 cos θ,
𝑥2 + 𝑦2 + 𝑧2 = 𝑟 2,
with

(𝑟 ≥ 0, 0 ≤ θ < 𝜋, 0 ≤ φ < 2𝜋 )
 Chapter 6 Multiple Integrals 197

The Jacobian of the transformation of Cartesian co-ordinates into the

spherical co-ordinates is
𝜕𝑥 𝜕𝑥 𝜕𝑥
| 𝜕𝑟 𝜕θ 𝜕φ|
𝜕(𝑥, 𝑦, 𝑧) 𝜕𝑦 𝜕𝑦 𝜕𝑦
𝐽= =
𝜕(𝑟, θ, φ) 𝜕𝑟 𝜕θ 𝜕φ
| 𝜕𝑧 𝜕𝑧 𝜕𝑧 |
𝜕𝑟 𝜕θ 𝜕φ
sin θ cos φ 𝑟 cos θ cos φ −𝑟 sin θ sin φ
= | sin θ sin φ 𝑟 cos θ sin φ 𝑟 sin θ cos φ | = 𝑟 2 sin θ
cos θ −𝑟 sin θ 0
Thus
𝑑𝑥𝑑𝑦𝑑𝑧 = |𝐽|𝑑𝑟𝑑θdφ = 𝑟 2 sin θ 𝑑𝑟𝑑θdφ

Remark We often use the 𝑠𝑝ℎ𝑒𝑟𝑖𝑐𝑎𝑙 𝑐𝑜 − 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 if

 The volume of integration is a sphereThe integrand contains
t
h
Example
e 1 Evaluate 𝑧

∭ √𝑥 2 + 𝑦 2 + 𝑧 2 𝑑𝑉 ,
e𝑅 𝒙𝟐 + 𝒚 𝟐 + 𝒛𝟐 = 𝟏
where 𝑅x is the solid hemisphere
with centre
p the origin, radius 1, 𝐳=𝟎
𝑦
that liesr above the 𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒.
𝑥
e Fig. 6.26
s
Solution From Fig. 6.26, and the spherical co-ordinates, we have
s 2𝜋 𝜋⁄2 1
∭i √𝑥 2 + 𝑦 2 + 𝑧 2 𝑑𝑉 = ∫ ∫ ∫ 𝑟 ∙ (𝑟 2 sin θ 𝑑𝑟𝑑θdφ)
𝑅 0 0 0
o
n

𝑥2+𝑦2+𝑧2
 198 Chapter 6 Multiple Integrals

2𝜋 𝜋⁄2 1
𝑟4 1 2𝜋 ⁄
= ∫ ∫ [ ] sin θ 𝑑θdφ = ∫ [− cos θ]𝜋0 2 dφ
0 0 4 0 4 0
2𝜋
1 𝜋
= ∫ dφ =
4 0 2

Example 2 Use spherical coordinates to find the volume of the solid that lies
above the cone 𝑧 = √𝑥 2 + 𝑦 2 and below the sphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 4𝑧.

Solution The equation of the sphere may be written in the form

𝑥 2 + 𝑦 2 + 𝑧 2 − 4𝑧 + 4 − 4 = 0,
or
𝑥 2 + 𝑦 2 + (𝑧 − 2)2 = 4,
which is the sphere passes through the origin and has centre (0,0,2).
Change into spherical coordinates we get
𝑥 = 𝑟 sin θ cos φ, 𝑧
𝑦 = 𝑟 sin θ sin φ,
𝒙𝟐 + 𝒚𝟐 + (𝒛 − 𝟐)𝟐 = 𝟒
𝑧 = 𝑟 cos θ,
 (𝟎, 𝟎, 𝟐)
𝑥2 + 𝑦2 + 𝑧2 = 𝑟 2,
𝑑𝑥𝑑𝑦𝑑𝑧 = 𝑟 2 sin θ 𝑑𝑟𝑑θdφ 𝑦
The equation of the sphere in spherical
𝑥 Fig. 6.27
coordinates is
𝑥 2 + 𝑦 2 + 𝑧 2 = 4𝑧  𝑟 2 = 4𝑟 cos θ  𝑟 = 4 cos θ,
and the equation of the cone is
𝑧 = √𝑥 2 + 𝑦 2  𝑟 cos θ = √(𝑟 sin θ cos φ)2 + (𝑟 sin θ sin φ)2

 𝑟 cos θ = √𝑟 2 sin2 θ (cos 2 φ + sin2 φ) = 𝑟 sin θ

or
 Chapter 6 Multiple Integrals 199

𝜋
cos θ = sin θ  tan θ = 1  θ=
4
Therefore, the description of the solid in spherical coordinates is
𝜋
𝑅 = ((𝑟, θ, φ)|0 ≤ 𝑟 ≤ 4 cos θ , 0 ≤ θ < , 0 ≤ φ < 2𝜋 ),
4
and its volume is
2𝜋 𝜋⁄4 4 cos θ
𝑉 = ∭ 𝑑𝑉 = ∫ ∫ ∫ 𝑟 2 sin θ 𝑑𝑟𝑑θdφ
𝑅 0 0 0

2𝜋 𝜋⁄4 3 4 cos θ
𝑟
=∫ ∫ [ ] sin θ 𝑑θdφ
0 0 3 0

1 2𝜋 𝜋⁄4
= ∫ ∫ (64 cos 3 θ) sin θ 𝑑θdφ
3 0 0
𝜋⁄4
64 2𝜋 cos 4 θ
= ∫ [− ] dφ
3 0 4 0

16 2𝜋 1
= ∫ (1 − ) dφ = 8𝜋
3 0 4

6.2.3 Triple integrals in cylindrical co-ordinates

Let the position of the moving 𝑃 in space be determined by the
following 𝑐𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙 𝑐𝑜 − 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 (see Fig. 6.28):

 𝑃(𝜌, 𝜑, 𝑧)

φ 𝑦
𝜌

𝝆 𝐜𝐨𝐬 𝛗 𝑄
𝑥
Fig. 6.28
 200 Chapter 6 Multiple Integrals

(𝑎) The length  of 0𝑄, where 𝑄 is the projection of 𝑃 on the 𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒,

(𝑏) The angle φ between 0𝑄 and the positive direction of the 𝑥 − 𝑎𝑥𝑖𝑠.
(𝑐) The height 𝑧 of the point 𝑃,
From Fig. 6.28, the relation between the Cartesian and the
𝑐𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙 𝑐𝑜 − 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 is given by
𝑥 = ρ cos φ
𝑦 = ρ sin φ
𝑧=𝑧
𝑥 2 + 𝑦 2 = ρ2 ,
with

(0 ≤ ρ < ∞, 0 ≤ φ < 2𝜋 )
The Jacobian of the transformation of Cartesian co-ordinates into the
cylindrical co-ordinates is
𝜕𝑥 𝜕𝑥 𝜕𝑥
|𝜕ρ 𝜕φ 𝜕𝑧 |
𝜕(𝑥, 𝑦, 𝑧) 𝜕𝑦 𝜕𝑦 𝜕𝑦
𝐽= =
𝜕(ρ, φ, 𝑧) 𝜕ρ 𝜕φ 𝜕𝑧
| 𝜕𝑧 𝜕𝑧 𝜕𝑧 |
𝜕ρ 𝜕φ 𝜕𝑧
cos φ − ρ sin φ 0
= | sin φ ρ cos φ 0| = ρ(cos 2 φ + sin2 φ) = ρ
0 0 1
Thus
𝑑𝑥𝑑𝑦𝑑𝑧 = |𝐽|𝑑ρ𝑑φd𝑧 = ρ𝑑ρ𝑑φd𝑧

Example 1 Evaluate the integral

∭ (𝑥 2 + 𝑦 2 + 𝑧 2 )𝑑𝑉 ,
𝑅
 Chapter 6 Multiple Integrals 201

where 𝑅 is the longitudinal wedge cut from a cylinder 𝐶 of height ℎ and

radius 𝑎 such that
0 ≤ ρ < 𝑎, 0 ≤ φ ≤ 𝜋, 0 ≤ z ≤ ℎ

Solution Changing into cylindrical coordinates, we get

𝑥 = ρ cos φ , 𝑦 = ρ sin φ , 𝑧 = 𝑧, 𝑥 2 + 𝑦 2 = ρ2 ,
𝑑𝑥𝑑𝑦𝑑𝑧 = ρ𝑑ρ𝑑φd𝑧
So that the given integral is computed as
ℎ 𝜋 𝑎
∭ (𝑥 2 + 𝑦 2 + 𝑧 2 )𝑑𝑉 = ∫ ∫ ∫ (ρ2 + 𝑧 2 )ρ𝑑ρ𝑑φd𝑧
𝑅 0 0 0
ℎ 𝜋ℎ 𝜋 𝑎
ρ4 ρ2 2 𝑎4 𝑎2 2
= ∫ ∫ [( + 𝑧 )] 𝑑φd𝑧 = ∫ ∫ ( + 𝑧 ) 𝑑φd𝑧
0 0 4 2 0 0 0 4 2
ℎ ℎ
𝑎4 𝑎2 2 𝑎4 𝑎2 𝑧 3
= ∫ ( + 𝑧 ) 𝜋d𝑧 = 𝜋 [ 𝑧 + ]
0 4 2 4 2 3 0

𝑎4 𝑎 2 ℎ3 𝜋𝑎2 ℎ
= 𝜋( ℎ + )= (3𝑎2 + 2ℎ2 )
4 2 3 12

Exercise 6.3

In problems 1 − 5, evaluate the integrals

𝜋 2 √4−𝑧 2
𝟏. ∫ ∫ ∫ 𝑧 sin 𝑦 𝑑𝑥𝑑𝑧𝑑𝑦
0 0 0

3 √9−𝑥 2 𝑥
𝟐. ∫ ∫ ∫ 𝑦𝑧𝑑𝑦𝑑𝑧𝑑𝑥
0 0 0
1 𝑥 𝑦
𝟑. ∫ ∫ ∫ 𝑥𝑦𝑧𝑑𝑧𝑑𝑦𝑑𝑥
0 0 0
2𝜋 𝜋 1
𝟒. ∫ ∫ ∫ 𝑟 3 sin θ 𝑑𝑟𝑑θdφ
0 0 0
 202 Chapter 6 Multiple Integrals

4 𝜋⁄2 2
𝟓. ∫ ∫ ∫ (ρ2 + 𝑧 2 )ρ𝑑ρ𝑑φd𝑧
0 0 0

In problems 6 − 7, find the mass and centre of mass of the given solid
𝑅 with the given density function ρ.
𝟔. 𝑅 is bounded by the parabolic cylinder 𝑧 = 1 − 𝑦 2 and the planes
𝑥 + 𝑧 = 1, 𝑥 = 0, 𝑧 = 0, ρ(𝑥, 𝑦, 𝑧) = 4.
𝟕. 𝑅 is the cube given by 0 ≤ 𝑥 ≤ 𝑎, 0 ≤ 𝑦 ≤ 𝑎, 0 ≤ 𝑧 ≤ 𝑎 and
ρ(𝑥, 𝑦, 𝑧) = 𝑥 2 + 𝑦 2 + 𝑧 2 .
𝟖. Find the volume of the region bounded by 𝑧 = 𝑥 2 + 𝑦 2 and 𝑧 = 2𝑥.
𝟗. Find the volume of a sphere of radius 𝑎 with its centre at the origin.
𝟏𝟎. Find the volume of the solid that lies between the spheres
𝑟 = 1, 𝑟 = 3 and above the cone θ = 𝜋⁄4.
𝟏𝟏. Find the volume of a cylinder of radius 𝑎 and height ℎ.
In problems 12 − 13, evaluate the given iterated integral by changing
to spherical coordinates.
3 √9−𝑥2 √9−𝑥2 −𝑦2
𝟏𝟐. ∫ ∫ ∫ 𝑧√𝑥 2 + 𝑦 2 + 𝑧 2 𝑑𝑧𝑑𝑦𝑑𝑥
−3 −√9−𝑥 2 0

3 √9−𝑦 2 √18−𝑥2 −𝑦 2
𝟏𝟑. ∫ ∫ ∫ (𝑥 2 + 𝑦 2 + 𝑧 2 )𝑑𝑧𝑑𝑥𝑑𝑦
0 0 √𝑥 2 +𝑦2

In problems 14 − 15, evaluate the given iterated integral by changing

to cylindrical coordinates.

𝟏𝟒. ∭𝑅 𝑧𝑑𝑉 , where 𝑅 satisfies

𝑅: 1 ≤ 𝑧 ≤ 2, 𝑥2 + 𝑦2 ≤ 1

𝟏𝟓. ∭𝑅 (𝑥 2 + 𝑦 2 )𝑑𝑉 , where 𝑅 is described by

𝑅: 𝑥 ≥ 0, 𝑦 ≥ 0, 0 ≤ 𝑧 ≤ 1, 𝑥 2 + 𝑦 2 ≤ 2
6.3 Line Integrals

In this chapter we define an integral that is similar to the single integral

𝑏
∫ 𝑓(𝑥) 𝑑𝑥 ,
𝑎

except that instead of integrating over an interval [𝑎, 𝑏], we integrate

over a curve 𝐶. Such integrals are called 𝑙𝑖𝑛𝑒 𝑜𝑟 𝑐𝑢𝑟𝑣𝑒𝑑 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙𝑠 and
is denoted by

∫ 𝑓 𝑑𝑙
𝐶

The curve 𝐶 is called the 𝑝𝑎𝑡ℎ 𝑜𝑓 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛. Indeed, the line

integral can be considered as a fairly simple generalization of the
fundamental theorem of calculus.
We call 𝐶 a 𝑠𝑚𝑜𝑜𝑡ℎ 𝑐𝑢𝑟𝑣𝑒 if its parametric representation
𝑟(𝑡) = 𝑥(𝑡)𝒊 + 𝑦(𝑡)𝒋 + 𝑧(𝑡)𝒌,
is continuous and possess the continuous derivative 𝑟 ′ (𝑡) i.e., the curve
has a unique tangent at each of its points. A continuous curve is a
𝑝𝑖𝑒𝑐𝑒𝑤𝑖𝑠𝑒 𝑠𝑚𝑜𝑜𝑡ℎ curve if it is a union of a finite number of smooth
curves.

Remark 1 In this course, every path of integration of a line integral

is assumed to be piecewise smooth.

6.3.1 Line integrals of scalar functions in the plane

Let 𝐶 be a smooth curve represented parametrically by

𝑥 = 𝑥(𝑡), 𝑦 = 𝑦(𝑡), α ≤ 𝑡 ≤ β,
 204 Chapter 6 Line Integrals

and 𝑓(𝑥, 𝑦) be a function defined along the curve. Then the line
integral of 𝑓(𝑥, 𝑦) over the contour 𝐶, connected between the two
points 𝐴(𝑎1 , 𝑏1 ) and 𝐵(𝑎2 , 𝑏2 ), is defined by

𝑦
 𝑩(𝒂𝟐 , 𝒃𝟐 )
𝑑𝑙
𝑑𝑦

𝑑𝑥

 𝑨(𝒂𝟏 , 𝒃𝟏 )

𝑥
Fig. 6.29. Contour of integration

𝐵
∫ 𝑓(𝑥, 𝑦) 𝑑𝑙 = ∫ 𝑓(𝑥, 𝑦) √(𝑑𝑥)2 + (𝑑𝑦)2
𝐶 𝐴
𝑎2
= ∫ 𝑓(𝑥, 𝑦(𝑥)) √1 + (𝑑𝑦⁄𝑑𝑥 )2 𝑑𝑥
𝑎1
𝑏2
= ∫ 𝑓(𝑥(𝑦), 𝑦) √(𝑑𝑥⁄𝑑𝑦)2 + 1 𝑑𝑦
𝑏1
𝐵
= ∫ 𝑓(𝑥(𝑡), 𝑦(𝑡)) √(𝑑𝑥⁄𝑑𝑡)2 + (𝑑𝑦⁄𝑑𝑡)2 𝑑𝑡
𝐴

A line integral over a closed contour 𝐶 is often denoted by the symbol

∮𝐶 𝑓 𝑑𝑙 .

Remark 2 If 𝐶 is a piecewise smooth curve such that 𝐶 is a union of

the finite smooth curves 𝐶1 , 𝐶2 , ⋯ , 𝐶𝑛 , then we can
define the integral of 𝑓 along 𝐶 as the sum of the integrals
of 𝑓 along each of the smooth pieces of 𝐶:

∫ 𝑓(𝑥, 𝑦) 𝑑𝑙 = ∫ 𝑓(𝑥, 𝑦) 𝑑𝑙 + ⋯ + ∫ 𝑓(𝑥, 𝑦) 𝑑𝑙

𝐶 𝐶1 𝐶𝑛
 Chapter 6 Line Integrals 205

Example 1 Evaluate ∫𝐶 2𝑥 𝑑𝑙 where 𝐶 consists of the arc of the

parabola 𝑦 = 𝑥 2 from (0,0) to (1,1) followed by the vertical line
segment from (1,1) to (1,2).

Solution The curve 𝐶 = 𝐶1 ∪ 𝐶2 is shown in Fig. 6.30. We compute

the integral along each contour separately.

𝑦
 Along the path 𝑪𝟏 . On 𝐶1 we choose
 (1,2)
𝑥 as the parameter so that

𝒙=𝟏
𝐶2
𝐶1 : 𝑦 = 𝑥 2  𝑦 ′ = 2𝑥, 0 ≤ 𝑥 ≤ 1,
 (1,1)
1
∫ 2𝑥 𝑑𝑙 = ∫ 2𝑥 √1 + (𝑑𝑦⁄𝑑𝑥)2 𝑑𝑥 𝐶1
𝐶1 0
1 𝑥
= ∫ 2𝑥 √1 + 4𝑥 2 𝑑𝑥
0
Fig. 6.30
1
1
= ∫ 4 × 2𝑥 √1 + 4𝑥 2 𝑑𝑥
4 0
1
1 (1 + 4𝑥 2 )3⁄2 1 5√5 − 1
= [( )] = (53⁄2 − 1) = (1)
4 3⁄2 0
6 6
 Along the path 𝑪𝟐 . On this path we have
𝐶2 : 𝑥 = 1  𝑑𝑥⁄𝑑𝑦 = 0, 1 ≤ 𝑦 ≤ 2,
and
2
∫ 2𝑥 𝑑𝑙 = ∫ 2(1) √(𝑑𝑥⁄𝑑𝑦)2 + 1 𝑑𝑦
𝐶2 1
1
= ∫ 2𝑑𝑦 = 2 (2)
0
Thus from (1) and (2)
5√5 − 1
∫ 2𝑥 𝑑𝑙 = ∫ 2𝑥 𝑑𝑙 + ∫ 2𝑥 𝑑𝑙 = +2
𝐶 𝐶1 𝐶2 6
 206 Chapter 6 Line Integrals

Example 2 Evaluate the line integral ∫𝐶 𝑥𝑦 2 𝑑𝑙 , where 𝐶 is the right

half of the circle 𝑥 2 + 𝑦 2 = 9.

𝑦
Solution From Fig. 6.31 we have the

𝒕 = 𝝅 ⁄𝟐
parametric form: 𝒙𝟐 + 𝒚𝟐 = 𝟗
𝑥 = 3 cos 𝑡 , 𝑦 = 3 sin 𝑡,
𝜋 𝜋 𝑥

𝒕 = − 𝝅 ⁄𝟐
where − 2 ≤ 𝑡 ≤ 2 , then

𝑑𝑙 = √(𝑑𝑥⁄𝑑𝑡)2 + (𝑑𝑦⁄𝑑𝑡)2 𝑑𝑡

= √(−3 sin 𝑡)2 + (3 cos 𝑡)2 𝑑𝑡 = 3𝑑𝑡, Fig. 6.31

so that
𝜋⁄2
2 (3 cos 𝑡)(3 sin 𝑡)2 (3𝑑𝑡)
∫ 𝑥𝑦 𝑑𝑙 = ∫
𝐶 −𝜋⁄2
𝜋⁄2
⁄
= 81 ∫ sin2 𝑡 cos 𝑡 𝑑𝑡 = 27[sin3 𝑡]𝜋−𝜋2⁄2 = 54
−𝜋⁄2

Example 3 Evaluate the line integrals, where 𝐶 is the given contour

(𝒊) ∫ 𝑥 𝑑𝑙 , 𝐶: 𝑥 = 𝑡 3 , 𝑦 = 𝑡, 0≤𝑡≤1
𝐶

(𝒊𝒊) ∫𝐶 (𝑥 − 2𝑦 2 ) 𝑑𝑦, 𝐶 is the arc of the parabola 𝑦 = 𝑥 2 from (−2,4)

to (1,1).

Solution
𝑑𝑥 𝑑𝑦
(𝒊) Here we have 𝑥 = 𝑡 3 , 𝑦 = 𝑡  = 3𝑡 2 , =1
𝑑𝑡 𝑑𝑡
Substitute in the parametric formula of the line integral we get
 Chapter 6 Line Integrals 207

1
∫ 𝑥 𝑑𝑙 = ∫ (𝑡 3 )√(𝑑𝑥⁄𝑑𝑡)2 + (𝑑𝑦⁄𝑑𝑡)2 𝑑𝑡
𝐶 0
1
= ∫ 𝑡 3 √(3𝑡 2 )2 + (1)2 𝑑𝑡
0
1
1 1 1 (9𝑡 4 + 1)3⁄2
= ∫ √9𝑡 4 + 1(36𝑡 3 )𝑑𝑡 = [ ]
36 0 36 3⁄2 0

1 10√10 − 1
= [(10)3⁄2 − 1] =
54 54
(𝒊𝒊) Here we have
𝑦 = 𝑥 2  𝑑𝑦 = 2𝑥𝑑𝑥,
so that
1
∫ (𝑥 − 2𝑦 2 ) 𝑑𝑦 = ∫ (𝑥 − 2(𝑥 2 )2 ) (2𝑥𝑑𝑥)
𝐶 −2
1 1
2𝑥 3 4𝑥 6
=∫ (2𝑥 2 − 4𝑥 5)
𝑑𝑥 = [ − ]
−2 3 6 −2
2 4 2(−8) 4(64)
=( − )−( − )
3 6 3 6
16 128 144
=( + )= = 48
3 3 3

6.3.2 Line integrals of scalar functions in space

We now suppose that 𝐶 is a smooth curve given by parametric equations

𝐶: 𝑥 = 𝑥(𝑡), 𝑦 = 𝑦(𝑡), 𝑧 = 𝑧(𝑡), α≤𝑡≤β
If 𝑓 is a function of three variables that is continuous on some region
containing 𝐶, then we define the line integral of 𝑓 along 𝐶

𝛽
𝑑𝑥 2 𝑑𝑦 2 𝑑𝑧 2
√
∫ 𝑓(𝑥, 𝑦, 𝑧) 𝑑𝑙 = ∫ 𝑓(𝑥(𝑡), 𝑦(𝑡), 𝑧(𝑡)) ( ) + ( ) + ( ) 𝑑𝑡
𝐶 𝛼 𝑑𝑡 𝑑𝑡 𝑑𝑡
 208 Chapter 6 Line Integrals

Example 1 Evaluate the line integrals

π
∫ 𝑥𝑦𝑧 𝑑𝑙 , 𝐶: 𝑥 = 2𝑡, 𝑦 = 3 sin 𝑡 , 𝑧 = 3 cos 𝑡 , 0 ≤ 𝑡 ≤
𝐶 2

Solution The direct substitution in the parametric formula of the line

integral needs to evaluate
𝑑𝑥 𝑑𝑦 𝑑𝑧
= 2, = 3 cos 𝑡, = −3 sin 𝑡
𝑑𝑡 𝑑𝑡 𝑑𝑡
Hence we have

𝜋⁄2 2 2 2
𝑑𝑥 𝑑𝑦 𝑑𝑧
∫ 𝑥𝑦𝑧 𝑑𝑙 = ∫ 𝑥(𝑡)𝑦(𝑡)𝑧(𝑡)√( ) +( ) + ( ) 𝑑𝑡
𝐶 0 𝑑𝑡 𝑑𝑡 𝑑𝑡
𝜋⁄2
=∫ (2𝑡)(3 sin 𝑡)(3 cos 𝑡)√(2)2 + (3 cos 𝑡)2 + (−3 sin 𝑡)2 𝑑𝑡
0
𝜋⁄2
= 18 ∫ 𝑡 sin 𝑡 cos 𝑡 √4 + 9 cos2 𝑡 + 9 sin2 𝑡 𝑑𝑡
0
18√13 𝜋⁄2 𝜋⁄2
= ∫ 𝑡 sin 2𝑡 𝑑𝑡 = 9√13 ∫ 𝑡 sin 2𝑡 𝑑𝑡
2 0 0

cos 2𝑡 sin 2𝑡 𝜋⁄2

= 9√13 [(𝑡) (− ) − (1) (− )]
2 4 0

𝜋 −1 9𝜋√13
= 9√13 [( ) (− ) − (1)(0)] =
2 2 4
6.3.3 Line integrals of vector fields
Suppose that
𝐹⃗ = 𝑃(𝑥, 𝑦, 𝑧)𝑖 + 𝑄(𝑥, 𝑦, 𝑧)𝑗 + 𝑅(𝑥, 𝑦, 𝑧)𝑘,
is a vector field that is defined on a region containing the curve 𝐶,
represented by
𝑟(𝑡) = 𝑥(𝑡)i + 𝑦(𝑡)j + 𝑧(𝑡)k,
 Chapter 6 Line Integrals 209

from 𝑡 = α to 𝑡 = β. Then the line integral of 𝐹⃗ along 𝐶 is

∫ 𝐹⃗ (𝑟) ∙ 𝑑𝑟 = ∫ 𝑃 𝑑𝑥 + 𝑄 𝑑𝑦 + 𝑅 𝑑𝑧
𝐶 𝐶
β
= ∫ 𝐹⃗ (𝑟(𝑡)) ∙ 𝑟̇ (𝑡) 𝑑𝑡
α
β
𝑑𝑥 𝑑𝑦 𝑑𝑧
= ∫ (𝑃 +𝑄 + 𝑅 ) 𝑑𝑡
α 𝑑𝑡 𝑑𝑡 𝑑𝑡

Remark 3 (The physical interpretation of line integral). In physics

the line integral represents the work done by the force
field 𝐹 on moving a particle moves along the curve
𝐶 in the direction of the tangent to this curve.

Example 1 Find the work done by the force field 𝐹⃗ = −𝑦 2 𝑖 + 𝑥𝑦𝑗 in

moving a particle along the semicircle 𝑟(𝑡) = cos 𝑡 i + sin 𝑡 j, 0 ≤ 𝑡 ≤ 𝜋.

Solution
𝑟(𝑡) = cos 𝑡 i + sin 𝑡 j
𝑟̇ (𝑡) = − sin 𝑡 i + cos 𝑡 j
𝐹⃗ (𝑟(𝑡)) = − sin2 𝑡 i + sin 𝑡 cos 𝑡 j
𝐹⃗ (𝑟(𝑡)) ∙ 𝑟̇ (𝑡) = sin3 𝑡 + sin 𝑡 cos2 𝑡 }
Therefore, the work done is
π
∫ 𝐹⃗ (𝑟) ∙ 𝑑𝑟 = ∫ 𝐹⃗ (𝑟(𝑡)) ∙ 𝑟̇ (𝑡) 𝑑𝑡
𝐶 0
π
= ∫ (sin3 𝑡 + sin 𝑡 cos2 𝑡) 𝑑𝑡
0
π
= ∫ (sin3 𝑡 + sin 𝑡 (1 − sin2 𝑡)) 𝑑𝑡
0
π
= ∫ sin 𝑡 𝑑𝑡 = [− cos 𝑡]𝜋0 = 2
0
 210 Chapter 6 Line Integrals

Example 2 Evaluate the line integral ∫𝐶 𝐹⃗ (𝑟) ∙ 𝑑𝑟, where 𝐶 is given

by the vector function 𝑟(𝑡).
(𝒊) 𝐹⃗ (𝑥, 𝑦) = 𝑥 2 𝑦𝑖 − 𝑥𝑦𝑗, 𝑟(𝑡) = 𝑡 3 i + 𝑡 4 j, 0 ≤ 𝑡 ≤ 1
(𝒊𝒊) 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑥𝑖 − 𝑧𝑗 + 𝑦k, 𝑟(𝑡) = 2𝑡i + 3𝑡j − 𝑡 2 𝑘, −1 ≤ 𝑡 ≤ 1
(𝒊𝒊𝒊) 𝐹⃗ (𝑥, 𝑦, 𝑧) = sin 𝑥 𝑖 + cos 𝑦 𝑗 + 𝑥𝑧k, 𝑟(𝑡) = 𝑡 3 i − 𝑡 2 j + t𝑘, 0 ≤ 𝑡 ≤ 1

Solution
1
(𝒊) ∫ 𝐹⃗ (𝑟) ∙ 𝑑𝑟 = ∫ 𝐹⃗ (𝑟(𝑡)) ∙ 𝑟̇ (𝑡) 𝑑𝑡
𝐶 0

𝑟(𝑡) = 𝑡 3 i + 𝑡 4 j
𝑟̇ (𝑡) = 3𝑡 2 i + 4𝑡 3 j
𝐹⃗ (𝑟(𝑡)) = ((𝑡 3 )2 )(𝑡 4 )𝑖 − (𝑡 3 )(𝑡 4 )𝑗 = 𝑡10 𝑖 − 𝑡 7 𝑗

𝐹⃗ (𝑟(𝑡)) ∙ 𝑟̇ (𝑡) = (𝑡10 )(3𝑡 2 ) + (−𝑡 7 )(4𝑡 3 ) = 3𝑡12 − 4𝑡10

Then
1
∫ 𝐹⃗ (𝑟) ∙ 𝑑𝑟 = ∫ (3𝑡12 − 4𝑡10 ) 𝑑𝑡
𝐶 0

𝑟(𝑡) = 2𝑡i + 3𝑡j − 𝑡 2 𝑘

𝑟̇ (𝑡) = 2i + 3j − 2t𝑘
𝐹⃗ (𝑟(𝑡)) = 2𝑡𝑖 + 𝑡 2 𝑗 + 3𝑡k

𝐹⃗ (𝑟(𝑡)) ∙ 𝑟̇ (𝑡) = 4𝑡 + 3𝑡 2 − 6𝑡 2
1 1
𝑡3
∫ 𝐹⃗ (𝑟) ∙ 𝑑𝑟 = ∫ (4𝑡 − 3𝑡 2 ) 𝑑𝑡 = 2 [−3 ]
𝐶 −1 3 0
3 4 19
= −( − )=−
13 11 143
 Chapter 6 Line Integrals 211

1
(𝒊𝒊) ∫ 𝐹⃗ (𝑟) ∙ 𝑑𝑟 = ∫ (2𝑡𝑖 + 𝑡 2 𝑗 + 3𝑡k) ∙ (2i + 3j − 2𝑡k) 𝑑𝑡
𝐶 −1
1 1
= ∫ (4𝑡 + 3𝑡 2 − 6𝑡 2 ) 𝑑𝑡 = ∫ (4𝑡 − 3𝑡 2 ) 𝑑𝑡
−1 −1
1
𝑡3
= 0 + 2 [−3 ] = −2
3 0

(𝒊𝒊𝒊) ∫ 𝐹⃗ (𝑟) ∙ 𝑑𝑟
𝐶
1
= ∫ (sin 𝑡 3 𝑖 + cos(−𝑡 2 ) 𝑗 + (𝑡 3 )(𝑡)k) ∙ (3𝑡 2 i − 2𝑡j + 𝑘) 𝑑𝑡
0
1
= ∫ (3𝑡 2 sin 𝑡 3 − 2𝑡 cos 𝑡 2 + 𝑡 4 ) 𝑑𝑡
0
1
3
𝑡5 2
1
= [− cos 𝑡 − sin 𝑡 + ] = (− cos 1 − sin 1 + + 1)
5 0 5
6
= − cos 1 − sin 1
5

Example 3 Find the work done by the force field

𝐹⃗ (𝑥, 𝑦) = 𝑥𝑖 + (𝑦 + 2)𝑗,
in moving an object along an arch of the cycloid
𝑟(𝑡) = (𝑡 − sin 𝑡)i + (1 − cos 𝑡)j, 0 ≤ 𝑡 ≤ 2𝜋

Solution From the physical interpretation of the line integral of vector

fields, we define the work done by
𝑟(𝑡) = (𝑡 − sin 𝑡)i + (1 − cos 𝑡)j
𝑟̇ (𝑡) = (1 − cos 𝑡)i + sin 𝑡 j
𝐹⃗ (𝑟(𝑡)) = (𝑡 − sin 𝑡)𝑖 + ((1 − cos 𝑡) + 2)𝑗

𝐹⃗ (𝑟(𝑡)) ∙ 𝑟̇ (𝑡) = (𝑡 − sin 𝑡)(1 − cos 𝑡) + sin 𝑡 (3 − cos 𝑡)

 212 Chapter 6 Line Integrals

Hence,
2𝜋
∫ 𝐹⃗ (𝑟) ∙ 𝑑𝑟 = ∫ [(𝑡 − sin 𝑡)(1 − cos 𝑡) + sin 𝑡 (3 − cos 𝑡)] 𝑑𝑡
𝐶 0
2𝜋
(𝑡 − sin 𝑡)2 (3 − cos 𝑡)2
=[ + ]
2 2 0

(2𝜋 − 0)2 (3 − 1)2 (3 − 1)2

=( + ) − (0 + ) = 2𝜋 2
2 2 2

6.3.4 Independence of path

Definition 6.1 The line integral is said to be

𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑎𝑡ℎ in the region  provided that given any
two points 𝐴 and 𝐵 of , the integral has the same value along every
piecewise smooth curve or path in  from 𝐴 to 𝐵.

In this case, we may write

𝐵
∫ 𝐹⃗ (𝑟) ∙ 𝑑𝑟 = ∫ 𝐹⃗ (𝑟) ∙ 𝑑𝑟,
𝐶 𝐴

because the value of the integral depends only on the points 𝐴 and 𝐵
and not on the particular choice of the path 𝐶 joining them.
𝑏
The formula ∫𝑎 𝑓 ′ (𝑥) 𝑑𝑥 = 𝑓(𝑏) − 𝑓(𝑎), expresses one version of the
fundamental theorem of calculus. To generalize this theorem to higher
dimensions, we have to generalize the integral itself. This section
enables us to give an easy extension of the fundamental theorem in
which the derivative of 𝑓 is replaced by the gradient of 𝑓. The formula
is
𝑏
∫ ∇𝑓 ∙ 𝑑𝑟 = 𝑓(𝑏) − 𝑓(𝑎)
𝑎
 Chapter 6 Line Integrals 213

6.3.5 The fundamental theorem for line integrals

Definition 6.2 The vector field F defined on a region  is called

conservative if there exists a scalar function  defined on  such that
𝐹⃗ = ∇φ,
at each point of . In this case,  is called a 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 for
the vector field 𝐹⃗ .

Now we give two important theorems concerning the line integral.

Theorem 6.2 The vector field

𝐹⃗ = 𝑃(𝑥, 𝑦, 𝑧)𝑖 + 𝑄(𝑥, 𝑦, 𝑧)𝑗 + 𝑅(𝑥, 𝑦, 𝑧)𝑘,
is 𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 on a region  and hence has a potential function
(𝑥, 𝑦, 𝑧) if and only if
∇ × 𝐹⃗ = 0

That is to say, for conservative vector field 𝐹⃗ defined in a region ,

independence of the path in the line integral ∫𝐶 𝐹⃗ (𝑟) ∙ 𝑑𝑟 means that

∫ 𝐹⃗ (𝑟) ∙ 𝑑𝑟 = ∫ 𝐹⃗ (𝑟) ∙ 𝑑𝑟,

𝐶1 𝐶2

where 𝐶1 and 𝐶2 are any two piecewise smooth curves in  having

initial point 𝐴 and terminate point 𝐵. An alternative formulation of the
independence property is that

∫ 𝐹⃗ ∙ 𝑑𝑟 = 0,
𝐶

for every piecewise smooth closed curve 𝐶 lying in . The equivalence

of the two properties follows from the observation that the path 𝐶1
followed by 𝐶1 in reverse direction is a closed path.
 214 Chapter 6 Line Integrals

Theorem 6.3 Let

𝐹⃗ = 𝑃(𝑥, 𝑦, 𝑧)𝑖 + 𝑄(𝑥, 𝑦, 𝑧)𝑗 + 𝑅(𝑥, 𝑦, 𝑧)𝑘,
be a 𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 on a region  with a potential function
φ(𝑥, 𝑦, 𝑧), then the line integral of 𝐹⃗ along every piecewise smooth
curve 𝐶 in  , having initial point 𝐴 and terminate point 𝐵 , is
independent of the path and
𝐵 𝐵
∫ 𝐹⃗ ∙ 𝑑𝑟 = ∫ ∇φ ∙ 𝑑𝑟 = φ(𝐵) − φ(𝐴)
𝐴 𝐴

Example 1 Find φ if
∇φ(𝑥, 𝑦) = (𝑦 2 + 2𝑥𝑦)𝑖 + (2𝑥𝑦 + 𝑥 2 )𝑗

Solution Here we have from the equality of two vectors

∂φ
= 𝑦 2 + 2𝑥𝑦 (1)
∂𝑥
∂φ
= 2𝑥𝑦 + 𝑥 2 (2)
∂𝑦
Now, integrating both sides of (1) with respect to 𝑥, we get
 φ(𝑥, 𝑦) = 𝑥𝑦 2 + 𝑥 2 𝑦 + 𝑐(𝑦) (3)
The differentiation of (3) with respect to y gives
∂φ
= 2𝑥𝑦 + 𝑥 2 + 𝑐 ′ (𝑦) (4)
∂𝑦
Equations (2) and (4) imply
𝑐 ′ (𝑦) = 0,
that is
𝑐(𝑦) = 𝑐,
where 𝑐 is the numerical constant. Substitute into (3) we obtain
φ(𝑥, 𝑦) = 𝑥𝑦 2 + 𝑥 2 𝑦 + 𝑐
 Chapter 6 Line Integrals 215

Example 2 Prove that

𝐹⃗ = (𝑦 + 𝑧)𝑖 + (𝑥 + 𝑧)𝑗 + (𝑥 + 𝑦)𝑘,
is conservative (irrotational), then find its potential function φ. Hence
evaluate

∫ (𝑦 + 𝑧) 𝑑𝑥 + (𝑥 + 𝑧) 𝑑𝑦 + (𝑥 + 𝑦) 𝑑𝑧,
𝐶

where 𝑟(𝑡) = 𝑡i + 𝑡j + 𝑡𝑘 between the two points 𝐴(0,0,0) and

𝐵(1,1,1).

Solution First we evaluate the curl of the vector field 𝐹⃗ , i.e.,

𝑖 𝑗 𝑘
𝜕 𝜕 𝜕
∇ × 𝐹⃗ = | |
𝜕𝑥 𝜕𝑦 𝜕𝑧
(𝑦 + 𝑧) (𝑥 + 𝑧) (𝑥 + 𝑦)
= (1 − 1)𝑖 − (1 − 1)𝑗 + (1 − 1)𝑘 = 0
Therefore, 𝐹⃗ is conservative and this implies that
𝐹⃗ = ∇φ,
or
𝜕φ 𝜕φ 𝜕φ
(𝑦 + 𝑧)𝑖 + (𝑥 + 𝑧)𝑗 + (𝑥 + 𝑦)𝑘 = 𝑖+ 𝑗+ 𝑘,
𝜕𝑥 𝜕𝑦 𝜕𝑧
from which we obtain
∂φ
= (𝑦 + 𝑧)  φ = 𝑥𝑦 + 𝑥𝑧 + 𝑐1 ,
∂𝑥
∂φ
= (𝑥 + 𝑧)  φ = 𝑥𝑦 + 𝑦𝑧 + 𝑐2 ,
∂y
∂φ
= (𝑥 + 𝑦)  φ = 𝑥𝑧 + 𝑦𝑧 + 𝑐3
∂z
 216 Chapter 6 Line Integrals

Therefore, with the sum of the last equations and take the repeated terms
one time
φ = 𝑥𝑦 + 𝑥𝑧 + 𝑦𝑧 + 𝑐
From the fundamental theorem of the line integral we have
𝐵(1,1,1)
∫ 𝐹⃗ (𝑟) ∙ 𝑑𝑟 = ∫ 𝐹⃗ (𝑟) ∙ 𝑑𝑟
𝐶 𝐴(0,0,0)
𝐵(1,1,1)
=∫ ∇φ ∙ 𝑑𝑟 = φ(1,1,1) − φ(0,0,0)
𝐴(0,0,0)

= (1 + 1 + 1 + 𝑐) − (0 + 0 + 0 + 𝑐) = 3

Example 3 Show that the vector field

𝐹⃗ = 𝑦𝑖 + (𝑥 + 𝑧)𝑗 + 𝑦𝑘,
is conservative (irrotational), then find its potential function φ. Hence
(8,3,−1)
evaluate ∫(2,1,4) 𝑦 𝑑𝑥 + (𝑥 + 𝑧) 𝑑𝑦 + 𝑦 𝑑𝑧, where 𝐶 is the line

segment joining the two points.

Solution
𝑖 𝑗 𝑘
𝜕 𝜕 𝜕
∇ × 𝐹⃗ = | |
𝜕𝑥 𝜕𝑦 𝜕𝑧
𝑦 (𝑥 + 𝑧) 𝑦
= (1 − 1)𝑖 − (0 − 0)𝑗 + (1 − 1)𝑘 = 0
Therefore, 𝐹⃗ is conservative, that is, we can write 𝐹⃗ as
𝐹⃗ = ∇φ,
∂φ
= 𝑦  φ = 𝑥𝑦 + 𝑐1 ,
∂𝑥
∂φ
= (𝑥 + 𝑧)  φ = 𝑥𝑦 + 𝑦𝑧 + 𝑐2 ,
∂y
 Chapter 6 Line Integrals 217

∂φ
=𝑦  φ = 𝑦𝑧 + 𝑐3
∂z
Thus we have
φ = 𝑥𝑦 + 𝑦𝑧 + 𝑐
From the fundamental theorem of the line integral we have
(8,3,−1) (8,3,−1)
∫ 𝐹⃗ (𝑟) ∙ 𝑑𝑟 = ∫ 𝐹⃗ (𝑟) ∙ 𝑑𝑟 = ∫ ∇φ ∙ 𝑑𝑟
𝐶 (2,1,4) (2,1,4)

= φ(8,3, −1) − φ(2,1,4)

= (24 − 3 + 𝑐) − (2 + 4 + 𝑐) = 15

Remark 4 In 2-dimensional spaces, the assertion of theorem 6.2 and

theorem 6.3 will be: The vector field
𝐹⃗ = 𝑃(𝑥, 𝑦, 𝑧)𝑖 + 𝑄(𝑥, 𝑦, 𝑧)𝑗,
is conservative on a region  and hence has a potential
function φ(𝑥, 𝑦, 𝑧) if and only if
𝜕𝑃 𝜕𝑄
=
𝜕𝑦 𝜕𝑥
and the line integral along every piecewise smooth curve
𝐶 in  is independent of the path with
𝐵 𝐵
∫ 𝐹⃗ ∙ 𝑑𝑟 = ∫ ∇φ ∙ 𝑑𝑟 = φ(𝐵) − φ(𝐴)
𝐴 𝐴

Exercise 6.4

In problems 1 − 5, evaluate the line integrals, where 𝐶 is the given

contour.

𝟏. ∫𝐶 𝑥𝑦 𝑑𝑙 , 𝐶 is the line segment joining (−1,1) to (2,3).

 218 Chapter 6 Line Integrals

𝟐. ∫𝐶 𝑥 √𝑦 𝑑𝑥 + 2𝑦√𝑥, consists of the arc of the circle 𝑥 2 + 𝑦 2 = 1

from (1,0) to (0,1) and the line segment from (0,1) to (4,3).

𝟑. ∫ 𝑥 2 𝑧 𝑑𝑙 , 𝐶: 𝑥 = sin 2𝑡 , 𝑦 = 3𝑡, 𝑧 = cos 2𝑡 , 0 ≤ 𝑡 ≤ 4.

𝐶

𝟒. ∫𝐶 𝑥𝑦 2 𝑑𝑙 , where 𝐶 is the circumference of the circle 𝑥 2 + 𝑦 2 = 9.

(1,3)
𝟓. ∫(0,0) (𝑥𝑦 + 𝑦 3 ) 𝑑𝑥, along the path 𝑦 = 3𝑥 2 .

In problems 6 − 8, evaluate the line integral ∫𝐶 𝐹⃗ (𝑟) ∙ 𝑑𝑟, where 𝐶 is

given by the vector function r (t ) .

𝟔. 𝐹⃗ (𝑥, 𝑦) = 𝑒 𝑥 𝑖 + 𝑥𝑦𝑗, 𝑟(𝑡) = 𝑡 2 i + 𝑡 3 j, 0 ≤ 𝑡 ≤ 1.
𝟕. 𝐹⃗ (𝑥, 𝑦, 𝑧) = (𝑦 + 𝑧)𝑖 − 𝑥 2 𝑗 − 4𝑦 2 𝑘, 𝑟(𝑡) = 𝑡i + 𝑡 2 j − 𝑡 4 k,
0 ≤ 𝑡 ≤ 1.
𝟖. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑥 2 𝑖 + 𝑥𝑦𝑗 + 𝑧 2 𝑘, 𝑟(𝑡) = sin 𝑡 i + cos 𝑡 j + 𝑡 2 k,
0 ≤ 𝑡 ≤ 𝜋⁄2.
𝟗. Find the work done by the force field 𝐹⃗ (𝑥, 𝑦) = 𝑥 sin 𝑦 𝑖 + 𝑦𝑗 on a
particle that moves along the parabola 𝑦 = 𝑥 2 from (−1,1) to (2,4).
𝟏𝟎. Find the work done by the force field
𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑥𝑧𝑖 + 𝑦𝑥𝑗 + 𝑧𝑦𝑘,
on a particle that moves along the curve
𝑟(𝑡) = 𝑡 2 i − 𝑡 3 j + 𝑡 4 k, 0≤𝑡≤1
𝟏𝟏. Prove that 𝐹⃗ (𝑥, 𝑦, 𝑧) = (2𝑥𝑧 + sin 𝑦)𝑖 + 𝑥 cos 𝑦 𝑗 + 𝑥 2 𝑘 is
conservative (irrotational), then find its potential function φ. Hence
evaluate

∫ (2𝑥𝑧 + sin 𝑦) 𝑑𝑥 + 𝑥 cos 𝑦 𝑑𝑦 + 𝑥 2 𝑑𝑧,

𝐶

where 𝑟(𝑡) = cos 𝑡 i + sin 𝑡 j + 𝑡k, 0 ≤ 𝑡 ≤ 2𝜋.

 Chapter 6 Line Integrals 219

In problems 12 − 13, show that the line integrals are independent of

path and evaluate the integrals.

𝟏𝟐. ∫𝐶 2𝑥 sin 𝑦 𝑑𝑥 + (𝑥 2 cos 𝑦 − 3𝑦 2 ) 𝑑𝑦, 𝐶 is any path from (−1,0)

to (5,1).

𝟏𝟑. ∫𝐶 (2𝑦 2 − 12𝑥 3 𝑦 3 ) 𝑑𝑥 + (4𝑥𝑦 − 9𝑥 4 𝑦 2 ) 𝑑𝑦, 𝐶 is any path from

(−1,0) to (5,1).
𝟏𝟒. Evaluate the line integral
(𝑎,𝑎)
∫ (𝑥 2 + 𝑦) 𝑑𝑥 + (𝑦 2 + 𝑥) 𝑑𝑦,
(0,0)

along any two different curves and show that the integral is path
independent.

6.3.6 Green’s Theorem

Green’s theorem gives the relation between a line integral around a

simple closed curve 𝐶 and a double integral over the plane region 
bounded by 𝐶. Green’s theorem was named after the British scientist
George Green and is a special case of the more general Stokes’ theorem.
Before stating Green’s theorem, we need the following definition.

Definition 6.3 We say the curve 𝐶 is simple if it does not intersect itself.

An ellipse is simple while a figure eight is not simple (see Fig. 6.32).

Simple
Not Simple

Fig. 6.32 Simple and not simple curves

 220 Chapter 6 Line Integrals

Definition 6.4 A simple closed curve 𝐶, given by the vector function

𝑟(𝑡) = 𝑥(𝑡)i + 𝑦(𝑡)j, α ≤ 𝑡 ≤ β , has a 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑜𝑟𝑖𝑒𝑛𝑡𝑎𝑡𝑖𝑜𝑛 if the
region  is always on the left as the point 𝑟(𝑡)traverses 𝐶.

Here, in Fig. 6.33 is a sketch of simple and closed curve 𝐶. First, notice
that because the curve is simple and closed there are no holes in the
region . Also notice that a direction has been put on the curves.

Positively oriented Negatively oriented

Fig. 6.33 Orientation of

curves

Theorem 6.4 (Green Theorem) Let 𝐶 be a positive oriented, piecewise-

smooth, simple closed curve in the plane and let  be the region
bounded by 𝐶. If 𝑃 and 𝑄 have continuous partial derivatives on an
open region that contains , then
𝜕𝑄 𝜕𝑃
∮ 𝑃 𝑑𝑥 + 𝑄 𝑑𝑦 = ∬ ( − ) 𝑑𝑥 𝑑𝑦
𝐶  𝜕𝑥 𝜕𝑦

𝑦
Proof. For simplicity we shall
𝐵(𝑎2 , 𝑏2 )
prove the theorem for simple 

closed curve 𝐶 as shown in Fig. 

6.34 We assume that the closed 
𝑨(𝒂𝟏 , 𝒃𝟏 )
curve passes through two points
𝐴(𝑎1 , 𝑏1 ) and 𝐵(𝑎2 , 𝑏2 ), which
Fig. 6.34 𝑥
 Chapter 6 Line Integrals 221

divides the curve 𝐶 into two parts; the lower 𝑦 = 𝑢(𝑥) and the upper
𝑦 = 𝑣(𝑥). Since

∮ 𝑃(𝑥, 𝑦) 𝑑𝑥 + 𝑄(𝑥, 𝑦) 𝑑𝑦 = ∮ 𝑃(𝑥, 𝑦) 𝑑𝑥 + ∮ 𝑄(𝑥, 𝑦) 𝑑𝑦,

𝐶 𝐶 𝐶

we can work with each term on right separately. We have

𝐵 𝐴
∮ 𝑃(𝑥, 𝑦) 𝑑𝑥 = ∫ 𝑃(𝑥, 𝑢(𝑥))𝑑𝑥 + ∫ 𝑃(𝑥, 𝑣(𝑥)) 𝑑𝑥
𝐶 𝐴 𝐵

So that we get
𝐵
∮ 𝑃(𝑥, 𝑦) 𝑑𝑥 = ∫ [𝑃(𝑥, 𝑢(𝑥)) − 𝑃(𝑥, 𝑣(𝑥))]𝑑𝑥
𝐶 𝐴
𝐵 𝑣(𝑥)
𝜕𝑃(𝑥, 𝑦)
= − ∫ [∫ 𝑑𝑦] 𝑑𝑥
𝐴 𝑢(𝑥) 𝜕𝑦

𝜕𝑃(𝑥, 𝑦)
= −∬ 𝑑𝑦 𝑑𝑥
 𝜕𝑦
Similarly, we can prove that
𝜕𝑄(𝑥, 𝑦)
∮ 𝑄(𝑥, 𝑦) 𝑑𝑦 = ∬ 𝑑𝑦 𝑑𝑥
𝐶  𝜕𝑥
and this completes the proof. ∎

Remark 5 The above equation can be put in the vector form using
the nabla operators as follows:

∮ 𝐹⃗ ∙ 𝑑𝑟 = ∬ (∇ × 𝐹⃗ ) ∙ 𝑘 𝑑𝑥 𝑑𝑦,
𝐶 

where
𝐹⃗ = 𝑃(𝑥, 𝑦)𝑖 + 𝑄(𝑥, 𝑦) 𝑗,
𝑑𝑟 = 𝑑𝑥𝑖 + 𝑑𝑦𝑗
 222 Chapter 6 Line Integrals

Example 1 Use Green’s theorem to evaluate

𝑦
∮ 𝑥𝑦 𝑑𝑥 + 𝑥 2 𝑦 3 𝑑𝑦, (𝟏, 𝟐)
𝐶

where 𝐶 is the triangle with vertices

(0,0), (1,0), (1,2).


Solution We first sketch the closed curve
𝐶 as shown in Fig.6.35. From this figure (𝟎, 𝟎) (𝟏, 𝟎) 𝑥
the closed region  is defined by Fig. 6.35

: 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 2𝑥
We can identify 𝑃 and 𝑄 from the line integral. Here they are
𝜕𝑃
𝑃 = 𝑥𝑦  = 𝑥,
𝜕𝑦
𝜕𝑄
𝑄 = 𝑥2𝑦3  = 2𝑥𝑦 3
𝜕𝑥
From Green's theorem, we get
𝜕𝑄 𝜕𝑃
∮ 𝑃 𝑑𝑥 + 𝑄 𝑑𝑦 = ∬ ( − ) 𝑑𝑦 𝑑𝑥
𝐶  𝜕𝑥 𝜕𝑦
1 2𝑥
 ∮ 𝑥𝑦 𝑑𝑥 + 𝑥 2 𝑦 3 𝑑𝑦 = ∫ ∫ (2𝑥𝑦 3 − 𝑥) 𝑑𝑦 𝑑𝑥
𝐶 0 0
1 2𝑥
2𝑥𝑦 4
=∫ [ − 𝑥𝑦] 𝑑𝑥
0 4 0
1
= ∫ (8𝑥 5 − 2𝑥 2 ) 𝑑𝑥
0
1
8𝑥 6 2𝑥 3 2
=[ − ] =
6 3 0 3
 Chapter 6 Line Integrals 223

Example 2 Verify Green’s theorem for the line integral

∮ −𝑦 𝑑𝑥 + 𝑥 𝑑𝑦,
𝐶

where 𝐶 is the closed contour bounded by 𝑦 = 𝑥 2 and 𝑥 = 𝑦 2 .

Solution The closed contour 𝐶 is

𝑦  (𝟏, 𝟏)
sketched in Fig. 6.36. From the
given integral, we have

𝜕𝑃
𝑃 = −𝑦  = −1, (5)
𝜕𝑦
(𝟎, 𝟎) 𝑥
𝜕𝑄
𝑄=𝑥  =1 (6) Fig. 6.36
𝜕𝑥
Therefore, Green's theorem tells us that

∮ −𝑦 𝑑𝑥 + 𝑥 𝑑𝑦 = ∬ 2 𝑑𝑥 𝑑𝑦 (7)
𝐶 

 L.H.S. of (7)

∮ −𝑦 𝑑𝑥 + 𝑥 𝑑𝑦
𝐶=𝐶1 +𝐶2

 Along the path 𝐶1

𝐶1 : 𝑦 = 𝑥 2  𝑑𝑦 = 2𝑥𝑑𝑥

∫ (−𝑦 𝑑𝑥 + 𝑥 𝑑𝑦) =
𝐶1

1 1 1
𝑥3 1
=∫ (−𝑥 2 𝑑𝑥 + 𝑥 (2𝑥𝑑𝑥)) = ∫ 𝑥 𝑑𝑥 = [ ] = 2 (8)
0 0 3 0 3

 Along the path 𝐶2

𝐶2 : 𝑥 = 𝑦 2  𝑑𝑥 = 2𝑦𝑑𝑦
 224 Chapter 6 Line Integrals

0
∫ (−𝑦 𝑑𝑥 + 𝑥 𝑑𝑦) = ∫ (−𝑦 (2𝑦𝑑𝑦) + 𝑦 2 𝑑𝑦)
𝐶2 1
0 0
2
𝑦3 1
= ∫ −𝑦 𝑑𝑦 = − [ ] = (9)
1 3 1 3
Adding (8) and (9) we get
2
∮ −𝑦 𝑑𝑥 + 𝑥 𝑑𝑦 = (10)
𝐶 3
 R.H.S. of (7)
1 √𝑦 1
∬ 2 𝑑𝑥 𝑑𝑦 = ∫ ∫ 2𝑑𝑥𝑑𝑦 = 2 ∫ (√𝑦 − 𝑦 2 )𝑑𝑦
 0 𝑦2 0
3 1
𝑦 3⁄2 𝑦 2
= 2[ − ] = (11)
3⁄2 3 0 3
From (10) and (11), Green’s theorem is verified.

Remark 6 The above example showed that if

𝜕𝑄 𝜕𝑃
− 𝜕𝑦 = 1, then the
𝜕𝑥

line integral gives the area of the enclosed region. So that

the following three forms
1
∫ 𝑥 𝑑𝑦 , ∫ −𝑦 𝑑𝑦 , ∫ (−𝑦 𝑑𝑥 + 𝑥 𝑑𝑦),
𝐶 𝐶 2 𝐶
can be used to calculate the area of the region  whose
boundary is the positively oriented smooth curve 𝐶.

Example 3 Use Green’s theorem to 𝑦

evaluate the line integral  (𝝅⁄𝟐 , 𝝅⁄𝟐)

∮ 𝑦 sin 𝑥 𝑑𝑥 + 𝑥𝑒 𝑦 𝑑𝑦,
𝐶

where 𝐶 is the closed curve in Fig. 6.37. 𝑥

Fig. 6.37
 Chapter 6 Line Integrals 225

Solution
𝜕𝑃
𝑃 = 𝑦 sin 𝑥  = sin 𝑥,
𝜕𝑦
𝜕𝑄
𝑄 = 𝑥𝑒 𝑦  = 𝑒𝑦
𝜕𝑥
From Green's theorem, we get
𝜕𝑄 𝜕𝑃
∮ 𝑃 𝑑𝑥 + 𝑄 𝑑𝑦 = ∬ ( − ) 𝑑𝑦 𝑑𝑥
𝐶  𝜕𝑥 𝜕𝑦
𝜋⁄2 𝑥
= ∫0 ∫0 (𝑒 𝑦 − sin 𝑥) 𝑑𝑦 𝑑𝑥
𝜋⁄2 𝜋⁄2
=∫ [𝑒 𝑦 − 𝑦 sin 𝑥]0𝑥 𝑑𝑥 = ∫ (𝑒 𝑥 − 1 − 𝑥 sin 𝑥) 𝑑𝑥
0 0
𝜋⁄2
= [𝑒 𝑥 − 𝑥 − [(𝑥)(− cos 𝑥) − (1)(− sin 𝑥)]]0
⁄
= [𝑒 𝑥 − 𝑥 + 𝑥 cos 𝑥 − sin 𝑥]𝜋0 2
𝜋 𝜋
= 𝑒 𝜋⁄2 − 1 − + 0 − 0 − 1 = 𝑒 𝜋⁄2 − − 2
2 2

Example 4 Evaluate ∮𝐶 𝑦 3 𝑑𝑥 − 𝑥 3 𝑑𝑦, where 𝐶 is the positively

oriented circle of radius 2 centered at the origin.

Solution Here we have

𝜕𝑃
𝑃 = 𝑦3  = 3𝑦 2 ,
𝜕𝑦
𝜕𝑄
𝑄 = −𝑥 3  = −3𝑥 2
𝜕𝑥
Now using Green's theorem on the line integrals gives

∮ 𝑦 3 𝑑𝑥 − 𝑥 3 𝑑𝑦 = ∬ (−3𝑥 2 − 3𝑦 2 ) 𝑑𝑦 𝑑𝑥,
𝐶 
 226 Chapter 6 Line Integrals

where  is a disk of radius 2 centred at the origin. So, change into polar
coordinates:
𝑥 = 𝑟 cos θ , 𝑦 = 𝑟 sin θ, 𝑥2 + 𝑦2 = 𝑟 2, 𝑑𝑥𝑑𝑦 = 𝑟𝑑𝑟𝑑θ
Thus we have

∮ 𝑦 3 𝑑𝑥 − 𝑥 3 𝑑𝑦 = −3 ∬ (𝑥 2 + 𝑦 2 ) 𝑑𝑦 𝑑𝑥
𝐶 
2𝜋 2
= −3 ∫ ∫ (𝑟 2 )(𝑟𝑑𝑟𝑑θ)
0 0
2
𝑟4
= −3(2𝜋) [ ] = −24𝜋
4 0

Exercise 6.5

In problems 1 − 10, use Green’s theorem to evaluate the line integral

along the given positively-oriented curve.

𝟏. ∮𝐶 (𝑦 + 𝑒 √𝑥 ) 𝑑𝑥 + (2𝑥 + cos 𝑦 2 ) 𝑑𝑦, where the contour 𝐶 is the

boundary of the region enclosed by the parabolas 𝑦 = 𝑥 2 and 𝑥 = 𝑦 2 .

𝟐. ∮𝐶 𝑦 𝑑𝑥 − 𝑥 2 𝑑𝑦, where 𝐶 is the square with corners (±1, ±1),

traced counter clockwise.

𝟑. ∮𝐶 𝑥𝑦 𝑑𝑥 + 2𝑥 2 𝑑𝑦, where 𝐶 consists of the line segment from

(−2,0) to (2,0) and the top half of the circle 𝑥 2 + 𝑦 2 = 4.

𝟒. ∫𝐶 𝐹⃗ (𝑟) ∙ 𝑑𝑟, 𝐹⃗ = (𝑦 2 − 𝑥 2 𝑦)𝑖 + 𝑥𝑦 2 𝑗, and 𝐶 consists of the

circle 𝑥 2 + 𝑦 2 = 4 from (√2, √2) to (0,0) and from (0,0) to (2,0).

𝟓. ∮𝐶 𝑦 𝑑𝑥 − 𝑥 2 𝑑𝑦, where 𝐶 is the closed path

𝑥 = cos 𝑡 , 𝑦 = sin 𝑡 , 0 ≤ 𝑡 ≤ 2𝜋
 Chapter 6 Line Integrals 227

𝟔. ∮𝐶 𝑦 2 𝑑𝑥 − 𝑥 2 𝑑𝑦, where the contour 𝐶 is the boundary of the unit

square.

𝟕. ∮𝐶 𝑒 𝑥 𝑑𝑦 + 𝑒 𝑦 𝑑𝑥, where the contour 𝐶 is the boundary of the unit

square.

𝟖. ∮𝐶 sin 𝜋𝑦 𝑑𝑥, where 𝐶 is the boundary of the region : 𝑦 = 0 and

𝑦 = 𝑥 for 0 ≤ 𝑥 ≤ 1.

𝟗. ∮𝐶 (𝑥 2 + 𝑦 2 ) (𝑑𝑥 + 𝑑𝑦), where 𝐶 is the boundary of the region :

𝑦 = 0 and 𝑦 = 1 − 𝑥 2 .

𝟏𝟎. ∮𝐶 𝑥 2 𝑑𝑦 − 2𝑦 𝑑𝑥, where 𝐶 is the boundary of the region

: 𝑦 = 0, 𝑦 = 1 − 𝑥2.
In problems 11 − 13, evaluate the following line integrals by whether
method seems simplest.

𝟏𝟏. ∮𝐶 𝑒 𝑥 cos 𝑦 𝑑𝑥 + 𝑒 𝑥 sin 𝑦 𝑑𝑦, where 𝐶 is the triangle with vertices

(0,0), (1,0), (1, 𝜋⁄2).

𝟏𝟐. ∮𝐶 𝑒 𝑥 cos 𝑦 𝑑𝑥 + 𝑒 𝑥 sin 𝑦 𝑑𝑦, where 𝐶 is the square with vertices

(0,0), (1,0), (1,1), (0,1).

𝟏𝟑. ∮𝐶 (𝑥 2 − 𝑦 2 ) 𝑑𝑥 + (𝑥 2 + 𝑦 2 ) 𝑑𝑦, where 𝐶 is the circle of radius 1

centred at (0,0) and traced clockwise.
𝟏𝟒. Use Green’s theorem to find the area of a disk of radius 𝑎.
𝟏𝟓. Verify Green’s theorem for the line integral

∮ 𝑥 2 𝑦 𝑑𝑥 + 𝑥𝑦 3 𝑑𝑦,
𝐶

where 𝐶 is the square with vertices (0,0), (1,0), (1,1), (0,1).

 228 Chapter 6 Line Integrals

𝟏𝟔. Show that if  is a simple region bounded by a piecewise smooth

curve 𝐶, then the area of  is given by
1
Area of  = ∫ (−𝑦 𝑑𝑥 + 𝑥 𝑑𝑦)
2 𝐶
In problems 17 − 19, evaluate the area of the regions bounded by the
following curves.
𝟏𝟕. 𝐶: 𝑟(𝑡) = (𝑡 3 − 𝑡)i + (𝑡 2 − 𝑡)j, 0≤𝑡≤1
𝟏𝟖. 𝐶: 𝑟(𝑡) = sin 𝜋𝑡 i + (𝑡 3 − 𝑡)j, 0≤𝑡≤1
𝟏𝟗. 𝐶: 𝑟(𝑡) = sin3 𝑡 i + cos3 𝑡 j, 0 ≤ 𝑡 ≤ 2𝜋
20. Let  be a triangle with vertices (𝑥1 , 𝑦1 ), (𝑥2 , 𝑦2 ) 𝑎𝑛𝑑 (𝑥3 , 𝑦3 ).
Find three different formulas for finding the area of a triangle given the
coordinates of its vertices. Apply each result to the triangle with vertices
(1,2), (7,1) and (5,4).

6.4. Surface Integrals

In mathematics, a surface integral is a definite integral taken over some

surface that may be a curved set in space. Defining an integral over a
surface 𝑆 leads to a different geometric situation having, however, a
close analogy with the line integral. Given a surface, one can integrate
over it scalar functions and vector fields.
6.4.1 Surface integrals of scalar functions
To begin, let 𝑆 be a smooth surface determined by an equation
𝑔(𝑥, 𝑦, 𝑧) = 0, (𝑥, 𝑦) ∈ 1 where 1 is the projection of 𝑆 on the
𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒 which is a bounded closed domain. Let the normal vector
𝑛 to 𝑆 have the angles α, β, γ with the positive directions of the co-
ordinate axes. From Fig. 6.38 we get
𝑑𝑥 𝑑𝑦 = 𝑑𝑆 cos γ
 Chapter 6 Line Integrals 229

𝑘
𝑧
𝑛

γ
γ
𝑑𝑥𝑑𝑦
𝑺

𝟏

𝑥 Fig. 6.38

𝑛
The direction cosines of the unit normal vector 𝑛̂ = |𝑛| is obtained using
the notation of the differential operator nabla; namely
𝑛 = ∇𝑔 = 𝑔𝑥 𝑖 + 𝑔𝑦 𝑗 + 𝑔𝑧 𝑘,
which implies
∇𝑔 𝑔𝑥 𝑖 + 𝑔𝑦 𝑗 + 𝑔𝑧 𝑘
𝑛̂ = =
|∇𝑔| √𝑔𝑥2 + 𝑔𝑦2 + 𝑔𝑧2
Now we can evaluate cos γ from this relation by
𝑔𝑧
cos γ = 𝑛̂ ∙ 𝑘 =
√𝑔𝑥2 + 𝑔𝑦2 + 𝑔𝑧2
Suppose that 𝑓(𝑥, 𝑦, 𝑧) is a bounded function defined on the surface 𝑆.
Then we define the 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 of 𝑓 over 𝑆 by
√𝑔𝑥2 + 𝑔𝑦2 + 𝑔𝑧2
∬ 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑆 = ∬ 𝑓(𝑥, 𝑦, 𝑧(𝑥, 𝑦)) 𝑑𝑥 𝑑𝑦
𝑆 1 𝑔𝑧
Similarly,
 230 Chapter 6 Line Integrals

√𝑔𝑥2 + 𝑔𝑦2 + 𝑔𝑧2

∬ 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑆 = ∬ 𝑓(𝑥, 𝑦(𝑥, 𝑧), 𝑧) 𝑑𝑥 𝑑𝑧,
𝑆 2 𝑔𝑦
and
√𝑔𝑥2 + 𝑔𝑦2 + 𝑔𝑧2
∬ 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑆 = ∬ 𝑓(𝑥(𝑦, 𝑧), 𝑦, 𝑧) 𝑑𝑦 𝑑𝑧,
𝑆 3 𝑔𝑥
where 1 , 2 and 3 are the projections of 𝑆 on 𝑥𝑦−, 𝑥𝑧 − and 𝑦𝑧 −
𝑝𝑙𝑎𝑛𝑒, respectively.

Remark Note that if 𝑓(𝑥, 𝑦, 𝑧) = 1, then the above surface

integral is equal to the surface integral of 𝑆.

Example 1 Evaluate ∬𝑆 3(𝑥 + 𝑦 + 𝑧)𝑑𝑆 where the surface 𝑆 is given

by 𝑆: 𝑧 = 𝑥 + 𝑦, 0 ≤ 𝑦 ≤ 𝑥, 0 ≤ 𝑥 ≤ 1.

Solution Here we have 𝑓(𝑥, 𝑦, 𝑧) = 3(𝑥 + 𝑦 + 𝑧) and 𝑔(𝑥, 𝑦, 𝑧) =

𝑥 + 𝑦 − 𝑧 so that
√𝑔𝑥2 + 𝑔𝑦2 + 𝑔𝑧2
∬ 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑆 = ∬ 𝑓(𝑥, 𝑦, 𝑧(𝑥, 𝑦)) 𝑑𝑥 𝑑𝑦
𝑆 1 𝑔𝑧

√1 + 1 + 1
 ∬ 3(𝑥 + 𝑦 + 𝑧)𝑑𝑆 = ∬ 3(𝑥 + 𝑦 + (𝑥 + 𝑦)) 𝑑𝑥 𝑑𝑦
𝑆 1 −1
1 𝑥 1 𝑥
𝑦2
= ∫ ∫ − 6√3(𝑥 + 𝑦) 𝑑𝑦 𝑑𝑥 = −6√3 ∫ [(𝑥𝑦 + )] 𝑑𝑥
0 0 0 2 0
1 1
𝑥2 2
3 𝑥3
= −6√3 ∫ (𝑥 + ) 𝑑𝑥 = −6√3 × [ ] = −3√3
0 2 2 3 0

Example 2 Evaluate ∬𝑆 𝑦𝑑𝑆, where 𝑆 is the surface

𝑧 = 𝑥 + 𝑦2 , 0 ≤ 𝑥 ≤ 1, 0≤𝑦≤2
 Chapter 6 Line Integrals 231

Solution Since,
𝑔(𝑥, 𝑦, 𝑧) = 𝑥 + 𝑦 2 − z  ∇𝑔 = 𝑖 + 2𝑦𝑗 − 𝑘,
then,
√1 + 4𝑦2 + 1
∬ 𝑦𝑑𝑆 = ∬ 𝑦 𝑑𝑥 𝑑𝑦
𝑆 1 −1
2
2 1
1 (2 + 4𝑦2 )3⁄2
= − ∫ ∫ 𝑦 √2 + 4𝑦2 𝑑𝑥 𝑑𝑦 = − ([ ] ) ∙ [𝑥]10
0 0 8 3⁄2 0

2 1
=− ((18)3⁄2 − (2)3⁄2 ) = ((18)3⁄2 − (2)3⁄2 )
24 12
1
= ((2 × 9)3⁄2 − (2)3⁄2 )
12
1
= ((2)3⁄2 (9)3⁄2 − (2)3⁄2 )
12
2√2 13√2
= ((9)3⁄2 − 1) =
12 3

Example 3 Find the surface area of the part of the plane 𝑧 = 8𝑥 + 4𝑦

that lies inside the cylinder 𝑥 2 + 𝑦 2 = 4.

Solution Here, the surface 𝑆 is given by 𝑔(𝑥, 𝑦, 𝑧) = 𝑧 − 8𝑥 − 4𝑦, so

that
√64 + 16 + 1
𝑆 = ∬ 𝑑𝑆 = ∬ 𝑑𝑥 𝑑𝑦 = 9 ∬ 𝑑𝑥 𝑑𝑦,
𝑆 1 1 1

where 1 is the circle of radius 2 and centred at the origin, thus

𝑆 = 9(4𝜋) = 36𝜋
 232 Chapter 6 Line Integrals

Example 4 Find the surface area of the part of the paraboloid

𝑧 = 2 − 𝑥2 − 𝑦2, 𝑧 ≥ 0
𝑧
𝒛=𝟐

Solution The surface 𝑆 (Fig. 6.39) is given by 𝑺

2 2
𝑔(𝑥, 𝑦, 𝑧) = 𝑧 + 𝑥 + 𝑦 − 2 = 0, 1
𝑦
𝑥
so Fig. 6.39
√𝑔𝑥2 + 𝑔𝑦2
+ 𝑔𝑧2
𝑆 = ∬ 𝑑𝑆 = ∬ 𝑑𝑥 𝑑𝑦
𝑆 1 𝑔𝑧
√4𝑥 2 + 4𝑦 2 + 1
=∬ 𝑑𝑥 𝑑𝑦,
1 1
where 1 is the circle 𝑥 2 + 𝑦 2 = 2. Now this is double integral over a
circle, so we change into polar coordinates which implies
1 2𝜋 √2
𝑆 = ∫ ∫ √4𝑟 2 + 1(8𝑟 𝑑𝑟)𝑑θ
8 0 0
√2
1 (4𝑟 2 + 1)3⁄2 𝜋 13𝜋
= ∙ 2𝜋 [ ] = (27 − 1) =
8 3⁄2 0
6 3

Example 5 Find the surface area of the cylinder 𝑥 2 + 𝑧 2 = 4 inside the

cylinder 𝑥 2 + 𝑦 2 = 4.

Solution Here, we have

𝑔(𝑥, 𝑦, 𝑧) = 𝑥 2 + 𝑧 2 − 4 = 0,
which implies
𝑔𝑥 = 2𝑥, 𝑔𝑦 = 0, 𝑔𝑧 = 2𝑧
Hence the required surface area S is computed by
 Chapter 6 Line Integrals 233

√𝑔𝑥2 + 𝑔𝑦2 + 𝑔𝑧2

𝑆 = ∬ 𝑑𝑆 = ∬ 𝑑𝑦 𝑑𝑥,
𝑆 1 𝑔𝑧
where 1 is the circle 𝑥 2 + 𝑦 2 = 4 the 𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒. Hence, we get
√4𝑥 2 + 4𝑧 2 4
𝑆=∬ 𝑑𝑦 𝑑𝑥 = ∬ 𝑑𝑦 𝑑𝑥
1 2𝑧 1 2√4 − 𝑥
2

2 √4−𝑥 2
1 1
= 2∬ 𝑑𝑦 𝑑𝑥 = 8 ∫ ∫ 𝑑𝑦 𝑑𝑥
1 √4 − 𝑥 2 0 0 √4 − 𝑥 2
2 2
2 2 1
= 8∫ [𝑦]√4−𝑥
0 𝑑𝑥 = 8 ∫ (√4 − 𝑥 2 ) 𝑑𝑥
√4 − 𝑥 2 √4 − 𝑥 2
0 0
2
= 8 ∫ 𝑑𝑥 = 16
0

Exercise 6.6

In problems 1 − 5, evaluate each of the surface integrals:

𝟏. ∬𝑆 (𝑥 2 𝑧 + 𝑦 2 𝑧)𝑑𝑆, where 𝑆 is the hemisphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 4,

𝑧 ≥ 0.

𝟐. ∬𝑆 (𝑥 2 𝑧 + 𝑦 2 𝑧)𝑑𝑆, where 𝑆 is the part of the cylinder 𝑥 2 + 𝑦 2 = 9

between the planes 𝑧 = 0 and 𝑧 = 2.

𝟑. ∬𝑆 𝑥𝑑𝑆, where 𝑆 is the surface 𝑦 = 𝑥 2 + 4𝑧, 0 ≤ 𝑥 ≤ 2,

0 ≤ 𝑥 ≤ 2.

𝟒. ∬𝑆 𝑧𝑑𝑆, where 𝑆 the part of the paraboloid 𝑧 = 𝑥 2 + 𝑦 2 that lies

inside the cylinder 𝑥 2 + 𝑦 2 = 1.

𝟓. ∬𝑆 (𝑥𝑦 + 𝑧)𝑑𝑆, where 𝑆 is that part of the plane 𝑥 + 𝑦 + 𝑧 = 2,

in the first octant.
In problems 6 − 10, find the surface area of the given surfaces
 234 Chapter 6 Line Integrals

𝟔. The part of the plane 𝑥 + 2𝑦 + 𝑧 = 4 that lies inside the cylinder

𝑥 2 + 𝑦 2 = 4.
𝟕. The part of the hyperbolic paraboloid 𝑧 = 𝑦 2 − 𝑥 2 that lies between
the cylinder 𝑥 2 + 𝑦 2 = 1 and 𝑥 2 + 𝑦 2 = 4.
𝟖. The part of the sphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑎2 that lies inside the
cylinder 𝑥 2 + 𝑦 2 = 𝑎𝑥.
𝟗. The cylinder 𝑥 2 + 𝑧 2 = 4 and inside the cylinder 𝑥 2 + 𝑦 2 = 4.
𝟏𝟎. The sphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 9 that lies inside the cylinder
𝑥 2 + 𝑦 2 = 3𝑦.

6.4.2 Surface integrals of vector fields

To define a surface integral of vectors, we take a piecewise smooth

surface S, given by the equation 𝑔(𝑥, 𝑦, 𝑧) = 0, so that it has a normal
vector 𝑛 = ∇𝑔 at every point of the surface 𝑆. We define the
𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑜𝑓 𝑡ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 𝑓𝑖𝑒𝑙𝑑 𝐹⃗ over 𝑆 by
𝐹⃗ ∙ 𝑛
∬ 𝐹⃗ ∙ 𝑑𝑆 = ∬ 𝑑𝑥𝑑𝑦,
𝑆 1 𝑛. 𝑘

where 1 is the projection of 𝑆 on the 𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒.

In a similar way we define the surface integral in terms of the double
integrals in the 𝑥𝑧 − and 𝑦𝑧 − 𝑝𝑙𝑎𝑛𝑒, respectively as follows:
𝐹⃗ ∙ 𝑛
∬ 𝐹⃗ ∙ 𝑑𝑆 = ∬ 𝑑𝑥𝑑𝑧,
𝑆 2 𝑛. 𝑗

𝐹⃗ ∙ 𝑛
∬ 𝐹⃗ ∙ 𝑑𝑆 = ∬ 𝑑𝑦𝑑𝑧,
𝑆 3 𝑛. 𝑖

where 2 and 3 are the projections of 𝑆 on 𝑥𝑧 − and 𝑦𝑧 − 𝑝𝑙𝑎𝑛𝑒,

respectively.
 Chapter 6 Line Integrals 235

Example 1 Evaluate ∬𝑆 𝐹⃗ ∙ 𝑑𝑆 where 𝐹⃗ is the vector field 𝐹⃗ = 𝑖 + 2𝑗 + 7𝑘

and the surface 𝑆 is represented by 𝑆: 𝑧 = 4(𝑥 2 + 𝑦 2 ), 𝑧 ≤ 4.

Solution The surface 𝑆 is the paraboloid given in Fig. 6.40 (a) with its
projection on the 𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒 is the unit circle 1 centred at the origin
shown in Fig. 6.40 (b). Therefore,

𝑧
𝑦

𝒛=𝟒
𝒙𝟐 + 𝒚𝟐 = 𝟏

𝟏 𝑥
𝒛 = 𝟒(𝒙𝟐 + 𝒚𝟐 )

𝑥
(𝒂) (𝒃)
Fig. 6.40

𝑔(𝑥, 𝑦, 𝑧) = 4(𝑥 2 + 𝑦 2 ) − 𝑧 = 0,
𝐹⃗ = 𝑖 + 2𝑗 + 7𝑘,
𝑛 = ∇𝑔 = 8𝑥𝑖 + 8𝑦𝑗 − 𝑘,
𝐹⃗ ∙ 𝑛 = 8𝑥 + 16𝑦 − 7,
𝑛. 𝑘 = −1
This gives
𝐹⃗ ∙ 𝑛 8𝑥 + 16𝑦 − 7
∬ 𝐹⃗ ∙ 𝑑𝑆 = ∬ 𝑑𝑥𝑑𝑦 = ∬ 𝑑𝑥𝑑𝑦
𝑆 1 𝑛. 𝑘 1 −1
Change into polar co-ordinates:
 236 Chapter 6 Line Integrals

𝑥 = 𝑟 cos θ , 𝑦 = 𝑟 sin θ, 𝑥2 + 𝑦2 = 𝑟 2, 𝑑𝑥𝑑𝑦 = 𝑟𝑑𝑟𝑑θ,

Hence,
2𝜋 1
∬ 𝐹⃗ ∙ 𝑑𝑆 = − ∫ ∫ (8𝑟 cos θ + 16𝑟 sin θ − 7)𝑟𝑑𝑟𝑑θ
𝑆 0 0
2𝜋 1
𝑟3 𝑟3 𝑟2
= −∫ [8 cos θ + 16 sin θ − 7 ] 𝑑θ
0 3 3 2 0
2𝜋
8 16 7
= −∫ ( cos θ + sin θ − ) 𝑑θ = 7𝜋
0 3 3 2

Example 2 Evaluate ∬𝑆 (𝑥𝑖 + 2𝑦𝑗 + 𝑧𝑘) ∙ 𝑑𝑆 where 𝑆 is given by

𝑆: 𝑧 = (𝑥 + 𝑦 2 ), 0 ≤ 𝑥 ≤ 3, 2 ≤ 𝑦 ≤ 5.

Solution Here, we have

𝑔(𝑥, 𝑦, 𝑧) = 𝑥 + 𝑦 2 − 𝑧 = 0,
𝐹⃗ = 𝑥𝑖 + 2𝑦𝑗 + 𝑧𝑘,
𝑛 = ∇𝑔 = 𝑖 + 2𝑦𝑗 − 𝑘,
𝐹⃗ ∙ 𝑛 = 𝑥 + 4𝑦 2 − 𝑧 = 𝑥 + 4𝑦 2 − 𝑥 − 𝑦 2 = 3𝑦 2 ,
𝑛. 𝑘 = −1
which implies
3𝑦 2
∬ (𝑥𝑖 + 2𝑦𝑗 + 𝑧𝑘) ∙ 𝑑𝑆 = ∬ ( ) 𝑑𝑥𝑑𝑦
𝑆 1 −1
5 3 5
= − ∫ ∫ 3𝑦 𝑑𝑥𝑑𝑦 = − ∫ 3𝑦 2 [𝑥]30 𝑑𝑦
2
2 0 2
3 5
𝑦
= −9 [ ] = −3(53 − 23 ) = −351
3 2
 Chapter 6 Line Integrals 237

Example 3 Evaluate the surface integral ∬𝑆 𝐹⃗ ∙ 𝑑𝑆, where 𝐹⃗ = 𝑒 𝑦 𝑖 +

𝑦𝑒 𝑥 𝑗 + 𝑥 2 𝑦𝑘, and 𝑆 is the part of the paraboloid 𝑧 = 𝑥 2 + 𝑦 2 that lies
above the square 0 ≤ 𝑥 ≤ 1, 2 ≤ 𝑦 ≤ 1.

Solution
𝐹⃗ ∙ 𝑛
∬ 𝐹⃗ ∙ 𝑑𝑆 = ∬ 𝑑𝑥𝑑𝑦,
𝑆 1 𝑛. 𝑘

𝑔(𝑥, 𝑦, 𝑧) = 𝑥 2 + 𝑦 2 − 𝑧 = 0,
𝐹⃗ = 𝑒 𝑦 𝑖 + 𝑦𝑒 𝑥 𝑗 + 𝑥 2 𝑦𝑘,
𝑛 = ∇𝑔 = 2𝑥𝑖 + 2𝑦𝑗 − 𝑘,
𝐹⃗ ∙ 𝑛 = 2𝑥𝑒 𝑦 + 2𝑦 2 𝑒 𝑥 − 𝑥 2 𝑦,
𝑛. 𝑘 = −1
Thus we have
1 1
∬ 𝐹⃗ ∙ 𝑑𝑆 = − ∫ ∫ (2𝑥𝑒 𝑦 + 2𝑦 2 𝑒 𝑥 − 𝑥 2 𝑦)𝑑𝑥𝑑𝑦
𝑆 0 0
1 1
𝑥3
2 𝑦 2 𝑥
= − ∫ [𝑥 𝑒 + 2𝑦 𝑒 − 𝑦] 𝑑𝑦
0 3 0
1
1
= − ∫ (𝑒 𝑦 + 2𝑦 2 𝑒 − 𝑦 − 2𝑦 2 ) 𝑑𝑦
0 3
1
𝑦
𝑦3 1 𝑦3 𝑦3
= − [𝑒 + 2 𝑒 − −2 ]
3 3 2 3 0
2 1 2
= − (𝑒 + 𝑒 − − − 1)
3 6 3
5 11
= −( 𝑒 − )
3 6
 238 Chapter 6 Line Integrals

Exercise 6.7

In problems 1 − 5, evaluate the surface integral ∬𝑆 𝐹⃗ ∙ 𝑑𝑆 for the

given vector field 𝐹⃗ and the given surface 𝑆.

𝟏. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘, 𝑆 is the sphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 9.
𝟐. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑦𝑗 − 𝑧𝑘, 𝑆 consists of the paraboloid 𝑦 = 𝑥 2 + 𝑧 2 ,
0 ≤ 𝑦 ≤ 1 and 𝑥 2 + 𝑧 2 ≤ 1.
𝟑. 𝐹⃗ (𝑥, 𝑦, 𝑧) = −𝑥𝑖 − 𝑦𝑗 + 𝑧 2 𝑘, 𝑆 is the part of the cone 𝑧 = √𝑥 2 + 𝑦 2
between the planes 𝑧 = 1 𝑎𝑛𝑑 𝑧 = 2.
𝟒. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑦 2 𝑖 + 𝑥 4 𝑗 + 𝑧 2 𝑘, 𝑆: 𝑧 = 𝑥 2 + 𝑦 2 , 𝑦 ≥ 0, 𝑧 = 9
𝟓. 𝐹⃗ = cosh 𝑥 𝑖 + sinh 𝑦 𝑗, 𝑆: 𝑧 = 𝑥 + 𝑦 2 , 0 ≤ 𝑦 ≤ 𝑥, 0 ≤ 𝑥 ≤ 1.

6.4.3 The Divergence theorem

In this section we study the first “big” integral theorem, which

transforms surface integral into triple integral and conversely. This
theorem which is known also Gauss’s Theorem and Ostrogradski-
Gauss theorem, has many applications in engineering and physics.

Theorem 6.5 (Gauss’s Theorem). Let 𝑅 be a closed bounded region in

space whose boundary is a piecewise smooth surface 𝑆 . Let
𝐹⃗ (𝑥, 𝑦, 𝑧) be a vector field that is continuous and has continuous first
partial derivatives in 𝑅. Then

∬ 𝐹⃗ ∙ 𝑑𝑆 = ∭ (∇ ∙ 𝐹⃗ )𝑑𝑉 (1)
𝑆 𝑅

Proof. Let us begin by defining the vector field 𝐹⃗ (𝑥, 𝑦, 𝑧) and the unit
normal to 𝑆 by
 Chapter 6 Line Integrals 239

𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝐹1 (𝑥, 𝑦, 𝑧)𝑖 + 𝐹2 (𝑥, 𝑦, 𝑧)𝑗 + 𝐹3 (𝑥, 𝑦, 𝑧)𝑘,

𝑛̂ = cos α 𝑖 + cos β 𝑗 + cos γ 𝑘
where α , β and γ are the direction cosines of 𝑛̂ with the positive
directions of 𝑥−, 𝑦 − , and 𝑧 − 𝑎𝑥𝑖𝑠 respectively. Thus we want to
prove that
𝜕𝐹1 𝜕𝐹2 𝜕𝐹3
∭ ( + + ) 𝑑𝑥𝑑𝑦𝑑𝑧
𝑅 𝜕𝑥 𝜕𝑦 𝜕𝑧

= ∬ (𝐹1 cos α + 𝐹2 cos β + 𝐹3 cos γ)𝑑𝐴 (2)



Clearly equation (2) is true if the integrals of each component on both

sides of (2) are equal. We first prove that
𝜕𝐹3
∭ 𝑑𝑥𝑑𝑦𝑑𝑧 = ∬ 𝐹3 cos γ 𝑑𝐴 , (3)
𝑅 𝜕𝑧 

for a special region 𝑅 that is bounded by two piecewise smooth surfaces

𝑆1 and 𝑆2 determined by
𝑧 = 𝑧1 (𝑥, 𝑦), 𝑧 = 𝑧2 (𝑥, 𝑦), 𝑧2 (𝑥, 𝑦) ≥ 𝑧1 (𝑥, 𝑦), (4)
and by a lateral cylindrical surface 𝑆3 with 𝑧
𝑺𝟐
elements parallel to the 𝑧 − 𝑎𝑥𝑖𝑠 as shown in
𝑺𝟑
Fig. 6.41. The union of the three surfaces 𝑆1,
𝑆2 and 𝑆3 forms the whole boundary of the 𝑺𝟏

region 𝑅. Now let us integrate the left-hand

side of (3), we obviously have 𝑦

𝑥
𝑧2 (𝑥,𝑦) Fig. 6.41
𝜕𝐹3 𝜕𝐹3
∭ 𝑑𝑥𝑑𝑦𝑑𝑧 = ∬ [∫ 𝑑𝑧] 𝑑𝑥𝑑𝑦
𝑅 𝜕𝑧  𝑧1 (𝑥,𝑦) 𝜕𝑧

= ∬ [𝐹3 (𝑥, 𝑦, 𝑧2 (𝑥, 𝑦)) − 𝐹3 (𝑥, 𝑦, 𝑧1 (𝑥, 𝑦))] 𝑑𝑥𝑑𝑦 (5)


 240 Chapter 6 Line Integrals

The first integral on the right hand side of (5) can be written in the form
of the surface integral of the function 𝐹3 (𝑥, 𝑦, 𝑧) taken over the upper
side of the surface
𝑆2 : 𝑧 = 𝑧2 (𝑥, 𝑦)
Similarly, the second integral

∬ 𝐹3 (𝑥, 𝑦, 𝑧1 (𝑥, 𝑦)) 𝑑𝑥𝑑𝑦



can be regarded as the surface integral of the function 𝐹3 (𝑥, 𝑦, 𝑧) taken

over the upper side of the surface
𝑆1 : 𝑧 = 𝑧1 (𝑥, 𝑦)
with the minus sign attached to it. Hence we obtain
𝜕𝐹3
∭ 𝑑𝑥𝑑𝑦𝑑𝑧 = ∬ 𝐹3 𝑑𝑥𝑑𝑦 + ∬ 𝐹3 𝑑𝑥𝑑𝑦 + ∬ 𝐹3 𝑑𝑥𝑑𝑦
𝑅 𝜕𝑧 𝑆1 𝑆2 𝑆3

= ∬ 𝐹3 𝑑𝑥𝑑𝑦 = ∬ 𝐹3 cos γ 𝑑𝑆 (6)

𝑆 

In a similar way, we can prove that

𝜕𝐹1
∭ 𝑑𝑥𝑑𝑦𝑑𝑧 = ∬ 𝐹1 cos α 𝑑𝑆 (7)
𝑅 𝜕𝑥 

𝜕𝐹2
∭ 𝑑𝑥𝑑𝑦𝑑𝑧 = ∬ 𝐹2 cos β 𝑑𝑆 (8)
𝑅 𝜕𝑦 

Adding together equations (6) − (8), we get (2) and the proof
completes. ∎

Example 1 Use Gauss' theorem to evaluate ∬𝑆 𝐹⃗ ∙ 𝑑𝑆, where 𝑆 is the

unit sphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 1 and 𝐹⃗ is the vector field 𝐹⃗ (𝑥, 𝑦, 𝑧) =

2𝑥𝑖 + 𝑦 2 𝑗 + 𝑧 2 𝑘.
 Chapter 6 Line Integrals 241

Solution The direct computation of this integral is quite difficult, but

can be simplified using Gauss' theorem:

∬ 𝐹⃗ ∙ 𝑑𝑆 = ∭ (∇ ∙ 𝐹⃗ )𝑑𝑉 = ∭ (2 + 2y + 2z)𝑑𝑉
𝑆 𝑅 𝑅

Change into spherical coordinates,

𝑥 = 𝑟 sin θ cos φ , 𝑦 = 𝑟 sin θ sin φ , 𝑧 = 𝑟 cos θ,
𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑟 2 , 𝑑𝑥𝑑𝑦𝑑𝑧 = 𝑟 2 sin θ 𝑑𝑟𝑑θdφ,
we get

∬ (2𝑥𝑖 + 𝑦 2 𝑗 + 𝑧 2 𝑘) ∙ 𝑑𝑆
𝑆
2𝜋 𝜋 1
=∫ ∫ ∫ (2 + 2𝑟 sin θ sin φ + 2𝑟 cos θ)(𝑟 2 sin θ 𝑑𝑟𝑑θdφ)
0 0 0
2𝜋 𝜋 1
=∫ ∫ ∫ (2𝑟 2 sin θ + 2𝑟 3 sin2 θ sin φ + 2𝑟 3 sin θ cos θ)𝑑𝑟𝑑θdφ
0 0 0
2𝜋 𝜋 1
𝑟3 𝑟4 𝑟4
=∫ ∫ (2 sin θ + 2 sin2 θ sin φ + 2 sin θ cos θ) 𝑑θdφ
0 0 3 4 4 0
2𝜋 𝜋
2 1 θ − sin 2θ⁄2 1
=∫ (− cos θ + ( ) sin φ + sin2 θ) dφ
0 3 2 2 2 0
2𝜋
4 8𝜋
=∫ dφ =
0 3 3

Example 2 Let 𝑆 be the surface of the region 𝑅 bounded by the planes

𝑧 = 0, 𝑦 = 0, 𝑦 = 2, and the paraboloid 𝑧 = 1 − 𝑥 2 (see Fig. 6.42).

Apply the divergence theorem to compute ∬𝑆 𝐹⃗ ∙ 𝑑𝑆 given that

𝐹⃗ (𝑥, 𝑦, 𝑧) = (𝑥 + cos 𝑦)𝑖 + (𝑦 + sin 𝑧)𝑗 + (𝑧 + 𝑒 𝑥 )𝑘

 242 Chapter 6 Line Integrals

Solution To evaluate the 𝑧

surface integral directly

𝒚=𝟐
would be quite lengthy. But
∇ ∙ 𝐹⃗ = 1 + 1 + 1 = 3 𝒛 = 𝟏 − 𝒙𝟐
𝑦
So the divergence theorem
𝒙=𝟏
may be applied easily and 𝑥
Fig. 6.42
yields
1 2 1−𝑥 2
∬ 𝐹⃗ ∙ 𝑑𝑆 = ∭ (∇ ∙ 𝐹⃗ )𝑑𝑉 = ∫ ∫ ∫ 3𝑑𝑧𝑑𝑦𝑑𝑥
𝑆 𝑅 −1 0 0
1 2 1
2 )𝑑𝑦𝑑𝑥
= ∫ ∫ 3(1 − 𝑥 = 6 ∫ (1 − 𝑥 2 )𝑑𝑥
−1 0 −1
3 1
𝑥 2
= 12 [𝑥 − ] = 12 × = 8
3 0 3

Example 3 Use the Divergence theorem to calculate the surface

integral ∬𝑆 𝐹⃗ ∙ 𝑑𝑆, where 𝑧

𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑥 3 𝑖 + 𝑦 3 𝑗 + 𝑧 3 𝑘,
𝒛=𝟐
and 𝑆 is the surface of the solenoid
bounded by the cylinder 𝑥 2 + 𝑦 2 = 1
𝒙𝟐 + 𝒚𝟐 = 𝟏
and the planes 𝑧 = 0 𝑎𝑛𝑑 𝑧 = 2 (see
Fig. 6.43). 𝒛=𝟎
𝑦

𝑥
Solution Since Fig. 6.43

𝑑𝑖𝑣𝐹⃗ = 3(𝑥 2 + 𝑦 2 + 𝑧 2 )
Then the divergence theorem may be applied easily and yields
 Chapter 6 Line Integrals 243

∬ 𝐹⃗ ∙ 𝑑𝑆 = ∭ (∇ ∙ 𝐹⃗ )𝑑𝑉
𝑆 𝑅
2
= ∬ ∫ 3(𝑥 2 + 𝑦 2 + 𝑧 2 )𝑑𝑧𝑑𝑦𝑑𝑥
 0
2
𝑧3 2 2
= 3 ∬ [𝑥 𝑧 + 𝑦 𝑧 + ] 𝑑𝑦𝑑𝑥
 3 0

= ∬ (6(𝑥 2 + 𝑦 2 ) + 8)𝑑𝑦𝑑𝑥


Change into polar coordinates, we get

2𝜋 1 2𝜋 1
∬ 𝐹⃗ ∙ 𝑑𝑆 = ∫ ∫ (6𝑟 2 + 8)𝑟𝑑𝑟𝑑θ = ∫ ∫ (6𝑟 3 + 8𝑟)𝑑𝑟𝑑θ
𝑆 0 0 0 0
1
𝑟4 𝑟2 11
= 2𝜋 [6 + 8 ] = 2𝜋 ( ) = 11𝜋
4 2 0 2

Example 4 Verify Gauss' theorem for the vector field

𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑥𝑧𝑖 + 𝑦𝑧𝑗 + 𝑧 2 𝑘,
on the solid bounded by the paraboloid 𝑧 = 1 − 𝑥 2 − 𝑦 2 and the 𝑥𝑦 −
𝑝𝑙𝑎𝑛𝑒.

Solution Gauss' theorem states that

∬ 𝐹⃗ ∙ 𝑑𝑆 = ∭ (∇ ∙ 𝐹⃗ )𝑑𝑉 (9)
𝑆 𝑅

We begin by calculating the right hand side of this equation. Since

𝑑𝑖𝑣𝐹⃗ = 𝑧 + 𝑧 + 2𝑧 = 4𝑧,
then
1−𝑥 2 −𝑦2
∭ (∇ ∙ 𝐹⃗ )𝑑𝑉 = ∬ ∫ 4𝑧𝑑𝑧𝑑𝑦𝑑𝑥
𝑅  0
 244 Chapter 6 Line Integrals

= 2 ∬ [1 − 𝑥 2 − 𝑦 2 ]2 𝑑𝑦𝑑𝑥


Here  is the circle 𝑥 + 𝑦 = 1, so that we have by the change into

2 2

polar coordinates
2𝜋 1
∭ 𝑑𝑖𝑣𝐹⃗ 𝑑𝑉 = 2 ∫ ∫ (1 − 𝑟 2 )2 𝑟𝑑𝑟𝑑θ
𝑅 0 0
2𝜋 1
1
= × 2 ∫ ∫ (1 − 𝑟 2 )2 (−2𝑟)𝑑𝑟𝑑θ
−2 0 0
1
(1 − 𝑟 2 )3 1 2𝜋
= −2𝜋 [ ] = −2𝜋 [0 − ] = (10)
3 0
3 3
Now we compute the left hand side of (9). The surface of the given solid
is divided into two parts; namely the surface of the solid and the surface
of the bottom plane. Thus we can write

∬ 𝐹⃗ ∙ 𝑑𝑆 = (∬ 𝐹⃗ ∙ 𝑑𝑆) + (∬ 𝐹⃗ ∙ 𝑑𝑆) (11)

𝑆 𝑆 𝑠𝑖𝑑𝑒𝑠 𝑆 𝑏𝑜𝑡𝑡𝑜𝑚

Now we evaluate each term separately. Writing

𝐹⃗ ∙ 𝑛
∬ 𝐹⃗ ∙ 𝑑𝑆 = ∬ 𝑑𝑥𝑑𝑦, 𝑧
𝑆 1 𝑛. 𝑘 𝒏𝒔𝒊𝒅𝒆𝒔

with
𝑔(𝑥, 𝑦, 𝑧) = 𝑥 2 + 𝑦 2 + 𝑧 − 1 = 0,
𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑥𝑧𝑖 + 𝑦𝑧𝑗 + 𝑧 2 𝑘,
𝑦
𝑛𝑠𝑖𝑑𝑒𝑠 = ∇𝑔 = 2𝑥𝑖 + 2𝑦𝑗 + 𝑘,
𝑛𝑠𝑖𝑑𝑒𝑠 . 𝑘 = 1 𝒏𝒃𝒐𝒕𝒕𝒐𝒎
𝑥
𝐹⃗ ∙ 𝑛𝑠𝑖𝑑𝑒𝑠 = 2𝑥 2 𝑧 + 2𝑦 2 𝑧 + 𝑧 2 ,
Fig. 6.44
= 2𝑧(𝑥 2 + 𝑦 2 ) + 𝑧 2
= 2𝑧(1 − 𝑧) + 𝑧 2
= 2𝑧 − 𝑧 2
= 𝑧(2 − 𝑧)
 Chapter 6 Line Integrals 245

This gives

(∬ 𝐹⃗ ∙ 𝑑𝑆) = ∬ 𝑧(2 − 𝑧) 𝑑𝑥𝑑𝑦

𝑆 𝑠𝑖𝑑𝑒𝑠 1

= ∬ (1 − 𝑥 2 − 𝑦 2 )[2 − (1 − 𝑥 2 − 𝑦 2 )] 𝑑𝑥𝑑𝑦
1

= ∬ (1 − (𝑥 2 + 𝑦 2 ))(1 + (𝑥 2 + 𝑦 2 )) 𝑑𝑥𝑑𝑦
1

= ∬ [1 − (𝑥 2 + 𝑦 2 )2 ] 𝑑𝑥𝑑𝑦
1
2𝜋 1
=∫ ∫ (1 − 𝑟 4 )𝑟𝑑𝑟𝑑θ
0 0
2 1
𝑟 𝑟6 4 2𝜋
= 2𝜋 [ − ] = 2𝜋 ( ) = (12)
2 6 0 12 3
On the bottom of the surface, we have
𝑔(𝑥, 𝑦, 𝑧) = 𝑧 = 0 (𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒),
𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑥𝑧𝑖 + 𝑦𝑧𝑗 + 𝑧 2 𝑘,
𝑛𝑏𝑜𝑡𝑡𝑜𝑚 = − 𝑘,
𝐹⃗ ∙ 𝑛𝑠𝑖𝑑𝑒𝑠 = − 𝑧 2 = 0,

(∬ 𝐹⃗ ∙ 𝑑𝑆) =0 (13)
𝑆 𝑏𝑜𝑡𝑡𝑜𝑚
The verification of Gauss' theorem is concluded from equations (10) −
(13).
Exercise 6.8

In problems 1 − 5, use the divergence theorem to evaluate ∬𝑆 𝐹⃗ ∙ 𝑑𝑆.

𝟏. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 3𝑦 2 𝑧 3 𝑖 + 9𝑥 2 𝑦𝑧 2 𝑗 − 4𝑥𝑦 2 𝑘, 𝑆 is the surface of the

cube with vertices (±1, ±1, ±1).
2
𝑥 𝑦2 𝑧2
𝟐. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑥𝑧𝑖 − 𝑦𝑧𝑗 + 𝑧 2 𝑘, 𝑆 is ellipsoid vertices 𝑎2 + 𝑏2 + 𝑐 2 = 1.

𝟑. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑧 cos 𝑦 𝑖 + 𝑥 sin 𝑧 𝑗 + 𝑥𝑧𝑘, 𝑆 is the surface of the

tetrahedron bounded by the planes 𝑥 = 0, 𝑦 = 0, 𝑧 = 0, 𝑥 + 𝑦 + 𝑧 = 2.
 246 Chapter 6 Line Integrals

𝟒. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑦 2 𝑖 + 𝑧 2 𝑗 + 𝑥 2 𝑧𝑘, 𝑆 is the surface of 𝑥 2 + 𝑦 2 ≤ 4,

𝑥 ≥ 0, 𝑦 ≥ 0, 𝑎𝑛𝑑 |𝑧| ≤ 1.
𝟓. 𝐹⃗ (𝑥, 𝑦, 𝑧) = sin2 𝑥 𝑖 − 𝑧(1 + sin 2𝑥)𝑘, 𝑆 is the surface of 𝑥 2 + 𝑦 2 ≤ 𝑧,
𝑧 ≤ 1⁄2.
In problems 6 − 9, verify that the divergence theorem is true for the
given vector field 𝐹⃗ on the given region 𝑅.
𝟔. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 3𝑥𝑖 + 𝑥𝑦𝑗 − 2𝑥𝑧𝑘, 𝑅 is the cube bounded by the
planes 𝑥 = 0, 𝑥 = 1, 𝑦 = 0, 𝑦 = 1, 𝑧 = 0, 𝑧 = 1.
𝟕. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑥𝑧𝑖 + 𝑦𝑧𝑗 + 3𝑧 2 𝑘, 𝑅 is the solid bounded by the
paraboloid 𝑧 = 𝑥 2 + 𝑦 2 and the plane 𝑧 = 1.
𝟖. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 3𝑦 2 𝑧 3 𝑖 + 9𝑥 2 𝑦𝑧 2 𝑗 − 4𝑥𝑦 2 𝑘, 𝑅 is the solid bounded
by 𝑥 2 + 𝑦 2 ≤ 𝑎2 , |𝑧| ≤ ℎ.
𝟗. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑦 3 𝑖 + 𝑧 3 𝑗 + 𝑥 3 𝑧𝑘, 𝑅 is the cube bounded by 𝑥 2 +
𝑦 2 + 𝑧 2 = 1.
𝟏𝟎. For 𝐹⃗ (𝑥, 𝑦, 𝑧) = −𝑧𝑦𝑖 + 𝑧𝑥𝑗 − 𝑥𝑦𝑘, evaluate

(𝒂) ∫ 𝐹⃗ (𝑟) ∙ 𝑑𝑟, 𝐶: cos 𝑡 𝑖 + sin 𝑡 𝑗 − 𝑘, 0 ≤ 𝑡 ≤ 2𝜋.

𝐶

(𝒃) the surface integral ∬ 𝐹⃗ ∙ 𝑑𝑆 , 𝑆: 𝑧 = 𝑥 2 + 𝑦 2 , 𝑧 ≤ 1.

𝑆

(𝒄) using Gauss’ theorem, evaluate ∬𝑇 𝑐𝑢𝑟𝑙𝐹⃗ ∙ 𝑑𝑆 , 𝑇: 𝑥 2 +

𝑦 2 + 𝑧 2 = 25.
6.4.4 Stokes ‘theorem

The second “big” integral theorem in this chapter is Stokes’ theorem,

which may be regarded as a higher dimensional version of Green’s
theorem. Whereas Green’s theorem relates a double integral over a
plane region  to a line integral around its plane boundary curve,
Stokes’ theorem relates a surface integral over a surface 𝑆 to a line
integral around the boundary curve of 𝑆.
 Chapter 6 Line Integrals 247

Theorem 6.5 (Stokes' Theorem). Let 𝑆 be an oriented piecewise

smooth surface that is bounded by a simple, closed, piecewise-
smooth boundary curve 𝐶 with positive orientation. Let 𝐹⃗ be a vector
field whose components have continuous partial derivatives in a
space region containing 𝑆. Then

∮ 𝐹⃗ (𝑟) ∙ 𝑑𝑟 = ∬ 𝑐𝑢𝑟𝑙𝐹⃗ ∙ 𝑑𝑆 (1)

𝐶 𝑆

Proof. First we define

𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝐹1 (𝑥, 𝑦, 𝑧)𝑖 + 𝐹2 (𝑥, 𝑦, 𝑧)𝑗 + 𝐹3 (𝑥, 𝑦, 𝑧)𝑘, (2)
the surface 𝑠 by
𝑆: 𝑔(𝑥, 𝑦, 𝑧) = 0, (3)
and we assume that the normal to the surface is
𝑛̂ = cos α 𝑖 + cos β 𝑗 + cos γ 𝑘 (4)
where α , β and γ are the direction 𝑧
𝑛̂
cosines of 𝑛̂ with the positive
directions of 𝑥−, 𝑦 −, and 𝑧 − 𝑎𝑥𝑖𝑠
respectively. Let 𝑆 be bounded by a 
𝐶
smooth curve 𝐶 and let the functions
𝐹1 , 𝐹2 , 𝐹3 and their first order partial
derivatives be continuous in the 𝑦

given domain. We shall transform
the line integral 𝑥 Fig. 6.45

∮ 𝐹⃗ (𝑟) ∙ 𝑑𝑟 = ∮ 𝐹1 𝑑𝑥 + 𝐹2 𝑑𝑦 + 𝐹3 𝑑𝑧, (5)

𝐶 𝐶

into a surface integral over the surface 𝑆. Let us begin by the integral
 248 Chapter 6 Line Integrals

∮ 𝐹1 (𝑥, 𝑦, 𝑧) 𝑑𝑥 = ∮ 𝐹1 (𝑥, 𝑦, 𝑧(𝑥, 𝑦)) 𝑑𝑥, (6)

𝐶 𝐶

where 𝑧 can be expressed as a function of 𝑥 and 𝑦 from the equation

of the surface. Applying Green's theorem to the integral (6), we get
𝜕𝐹1
∮ 𝐹1 (𝑥, 𝑦, 𝑧(𝑥, 𝑦)) 𝑑𝑥 = ∬ − 𝑑𝑦 𝑑𝑥
𝐶  𝜕𝑦
𝜕𝐹1 𝜕𝐹1 𝜕𝑧
= −∬ ( + ) 𝑑𝑦 𝑑𝑥 (7)
 𝜕𝑦 𝜕𝑧 𝜕𝑦
But from equation (3) we obtain
𝜕𝑧 𝑔𝑦 cos β
=− =−
𝜕𝑦 𝑔𝑧 cos γ
Hence
𝜕𝐹1 𝜕𝐹1 cos β
∮ 𝐹1 (𝑥, 𝑦, 𝑧(𝑥, 𝑦)) 𝑑𝑥 = − ∬ ( − ) 𝑑𝑦 𝑑𝑥
𝐶  𝜕𝑦 𝜕𝑧 cos γ
𝜕𝐹1 𝜕𝐹1 cos β
= −∬ ( − ) cos γ 𝑑𝑆
 𝜕𝑦 𝜕𝑧 cos γ
𝜕𝐹1 𝜕𝐹1
= −∬ ( cos γ − cos β) 𝑑𝑆 (8)
 𝜕𝑦 𝜕𝑧
Similarly
𝜕𝐹2 𝜕𝐹2
∮ 𝐹2 (𝑥, 𝑦, 𝑧) 𝑑𝑦 = − ∬ ( cos α − cos γ) 𝑑𝑆 (9)
𝐶 S 𝜕𝑧 𝜕𝑥
𝜕𝐹3 𝜕𝐹3
∮ 𝐹3 (𝑥, 𝑦, 𝑧) 𝑑𝑧 = − ∬ ( cos β − cos α) 𝑑𝑆 (10)
𝐶 S 𝜕𝑥 𝜕𝑦
The summation of equations (8) − (10) yields
𝜕𝐹3 𝜕𝐹2
∮ 𝐹1 𝑑𝑥 + 𝐹2 𝑑𝑦 + 𝐹3 𝑑𝑧 = ∬ [( − ) cos α +
𝐶 S 𝜕𝑦 𝜕𝑧
𝜕𝐹1 𝜕𝐹3 𝜕𝐹2 𝜕𝐹1
+( − ) cos β + ( − ) cos γ] 𝑑𝑆 (11)
𝜕𝑧 𝜕𝑥 𝜕𝑥 𝜕𝑦
 Chapter 6 Line Integrals 249

Equation (11) can be written in the form

∮ 𝐹1 𝑑𝑥 + 𝐹2 𝑑𝑦 + 𝐹3 𝑑𝑧 = ∬ 𝑐𝑢𝑟𝑙𝐹⃗ ∙ 𝑛̂𝑑𝑆 = ∬ 𝑐𝑢𝑟𝑙𝐹⃗ ∙ 𝑑𝑆

𝐶 𝑆 𝑆

which is the required proof. ∎

Example 1 Use Stokes’ theorem to evaluate ∮𝐶 𝐹⃗ (𝑟) ∙ 𝑑𝑟 where

𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑦𝑖 + 𝑧𝑗 − 𝑥𝑦𝑘, and 𝑆 is the plane 𝑧 = 4 − 𝑥 − 2𝑦 (see

Fig. 6.46).

𝑧
Solution From Stokes’ theorem (𝟎, 𝟎, 𝟒) 

∮ 𝐹⃗ (𝑟) ∙ 𝑑𝑟 = ∬ 𝑐𝑢𝑟𝑙𝐹⃗ ∙ 𝑑𝑆
𝐶 𝑆
𝑺
𝑐𝑢𝑟𝑙𝐹⃗ ∙ 𝑛
=∬ ∙ 𝑑𝑥𝑑𝑦, 
 𝑛. 𝑘 (𝟎, 𝟐, 𝟎) 𝑦
where  is the projection of  (𝟒, 𝟎, 𝟎)
𝑆: 𝑧 = 4 − 𝑥 − 2𝑦, 𝑥 𝑦
on the 𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒. (𝟎, 𝟐)

𝑖 𝑗 𝑘 
𝑥
𝜕 𝜕 𝜕 (𝟒, 𝟎)
∇ × 𝐹⃗ = | |=
𝜕𝑥 𝜕𝑦 𝜕𝑧 Fig. 6.46
𝑦 𝑧 −𝑥𝑦
= (−𝑥 − 1)𝑖 − (−𝑦 − 0)𝑗 + (0 − 1)𝑘
= −(𝑥 + 1)𝑖 + 𝑦𝑗 − 𝑘,
𝑔(𝑥, 𝑦, 𝑧) = 𝑥 + 2𝑦 + 𝑧 − 4,
𝑛 = ∇𝑔 = 𝑖 + 2𝑗 + 𝑘,
𝑐𝑢𝑟𝑙𝐹⃗ ∙ 𝑛 = −(𝑥 + 1) + 2𝑦 − 1 = −2 − 𝑥 + 2𝑦,
𝑛. 𝑘 = 1
 250 Chapter 6 Line Integrals

Thus we have

∮ 𝐹⃗ (𝑟) ∙ 𝑑𝑟 = ∮ (𝑦𝑖 + 𝑧𝑗 − 𝑥𝑦𝑘) ∙ 𝑑𝑟

𝐶 𝐶
2 4−2𝑦
=∫ ∫ (−2 − 𝑥 + 2𝑦)𝑑𝑥𝑑𝑦
0 0
2 4−2𝑦
𝑥2
= ∫ [−2𝑥 − + 2𝑥𝑦] 𝑑𝑦
0 2 0
2 (4 − 2𝑦)2
= ∫ (−2(4 − 2𝑦) − + 2(4 − 2𝑦)𝑦) 𝑑𝑦
0 2
2
= ∫ (−8 + 12𝑦 − 2(2 − 𝑦)2 − 4𝑦 2 )𝑑𝑦
0
2
(2 − 𝑦)3
2
𝑦3
= [−8𝑦 + 6𝑦 − 2 −4 ]
−3 3 0
0−8 8
= −16 + 24 − 2 −4
−3 3
= −16 + 24 − 16 = − 8

Example 2 Verify Stokes’ theorem for the vector field

𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑦𝑖 + 𝑧𝑗 + 𝑥𝑘, 𝑧

and 𝑆 is the paraboloid 𝑧 = 1 − 𝑥 2 − 𝑦 2 ,

𝑧 ≥ 0. (see Fig. 6.47). 𝑺: 𝒛 = 𝟏 − 𝒙𝟐 − 𝒚𝟐

𝑪
 𝑦
Solution The closed contour 𝐶 is 𝑦
𝑥
𝑟(𝑡) = cos 𝑡 i + sin 𝑡 j + 0𝑘

𝑟̇ (𝑡) = − sin 𝑡 i + cos 𝑡 j + 0𝑘 𝒙𝟐 + 𝒚𝟐 = 𝟏 𝑥

𝐹⃗ (𝑟(𝑡)) = sin 𝑡 𝑖 + 0𝑗 + cos 𝑡 𝑘

𝐹⃗ (𝑟(𝑡)) ∙ 𝑟̇ (𝑡) = − sin2 𝑡 + 0 + 0

Fig. 6.47
 Chapter 6 Line Integrals 251

Hence,
2𝜋 2𝜋
∮ 𝐹⃗ (𝑟) ∙ 𝑑𝑟 = ∫ 𝐹⃗ (𝑟(𝑡)) ∙ 𝑟̇ (𝑡)𝑑𝑡 = ∫ − sin2 𝑡 𝑑𝑡
𝐶 0 0
2𝜋
1 − cos 2𝑡
= −∫ 𝑑𝑡 = −𝜋 (12)
0 2
On the other hand, we now consider the integral over
𝑆: 𝑔(𝑥, 𝑦, 𝑧) = 𝑥 2 + 𝑦 2 + 𝑧 − 1,
𝑐𝑢𝑟𝑙𝐹⃗ ∙ 𝑛
∬ 𝑐𝑢𝑟𝑙𝐹⃗ ∙ 𝑑𝑆 = ∬ ∙ 𝑑𝑥𝑑𝑦,
𝑆  𝑛. 𝑘
𝑖 𝑗 𝑘
𝜕 𝜕 𝜕
𝑐𝑢𝑟𝑙𝐹⃗ = | | = (0 − 1)𝑖 − (1 − 0)𝑗 + (0 − 1)𝑘
𝜕𝑥 𝜕𝑦 𝜕𝑧
𝑦 𝑧 𝑥
= −𝑖 − 𝑗 − 𝑘,
𝑛 = ∇𝑔 = 2𝑥𝑖 + 2𝑦𝑗 + 𝑘,
𝑐𝑢𝑟𝑙𝐹⃗ ∙ 𝑛 = −2𝑥 − 2𝑦 − 1,
𝑛. 𝑘 = 1
The domain  is the circle 𝑥 2 + 𝑦 2 = 1, so the change into polar
coordinates yields
2𝜋 1
∬ 𝑐𝑢𝑟𝑙𝐹⃗ ∙ 𝑑𝑆 = ∫ ∫ (−2𝑟 cos θ − 2𝑟 sin θ − 1)𝑟𝑑𝑟𝑑θ
𝑆 0 0
2𝜋 1
=∫ ∫ (−2𝑟 2 cos θ − 2𝑟 2 sin θ − 𝑟)𝑑𝑟𝑑θ
0 0
2𝜋 1
𝑟3 𝑟3 𝑟2
=∫ [−2 cos θ − 2 sin θ − ] 𝑑θ
0 3 3 2 0
2𝜋
2 2 1
=∫ (− cos θ − sin θ − ) 𝑑θ
0 3 3 2
 252 Chapter 6 Line Integrals

2 2 1 2𝜋
= [− sin θ + cos θ − θ] = − 𝜋 (13)
3 3 2 0
From (12) and (13), Stokes’ theorem is verified.

Example 3 Use Stokes’ theorem to evaluate ∬𝑆 𝑐𝑢𝑟𝑙𝐹⃗ ∙ 𝑑𝑆 for the

vector field 𝐹⃗ (𝑥, 𝑦, 𝑧) = 2𝑦𝑖 + 𝑧𝑗 + 3𝑦𝑘 and 𝑆 is the paraboloid

𝑧 = (𝑥 2 + 𝑦 2 ), 𝑧 ≤ 𝑏 2 .
𝑧

Solution Stokes’ theorem states that 

𝒛 = (𝒙𝟐 + 𝒚𝟐 )
𝑆
∮ 𝐹⃗ (𝑟) ∙ 𝑑𝑟 = ∬ 𝑐𝑢𝑟𝑙𝐹⃗ ∙ 𝑑𝑆 𝑦
𝐶 𝑆 𝑦
𝑥
So that, we can use the left hand side
𝒙𝟐 + 𝒚𝟐 = 𝒂𝟐
𝑥
in order to calculate the required
𝐶
surface integral as follows
Fig. 6.48
∮ 𝐹⃗ (𝑟) ∙ 𝑑𝑟 = ∫ 𝐹⃗ (𝑟(𝑡)) ∙ 𝑟̇ (𝑡) 𝑑𝑡,
𝐶 C

where
𝑟(𝑡) = 𝑏 cos 𝑡 i + 𝑏 sin 𝑡 j + 𝑏 2 𝑘
𝑟̇ (𝑡) = − 𝑏 sin 𝑡 i + 𝑏 cos 𝑡 j + 0𝑘
𝐹⃗ (𝑟(𝑡)) = 2(𝑏 sin 𝑡)𝑖 + (𝑏 2 )𝑗 + 3(𝑏 sin 𝑡)𝑘

𝐹⃗ (𝑟(𝑡)) ∙ 𝑟̇ (𝑡) = −2(𝑏 sin 𝑡)2 + 𝑏 3 cos 𝑡

Therefore
2𝜋
∬ 𝑐𝑢𝑟𝑙𝐹⃗ ∙ 𝑑𝑆 = ∮ 𝐹⃗ (𝑟) ∙ 𝑑𝑟 = ∫ (−2𝑏 2 sin2 𝑡 + 𝑏 3 cos 𝑡)𝑑𝑡
𝑆 𝐶 0
𝜋⁄2
1 𝜋
= −2𝑏 2 × 4 ∫ sin2 𝑡 𝑑𝑡 = −8𝑏 2 × = −2𝜋𝑏 2
0 2 2
 Chapter 6 Line Integrals 253

Exercise 6.9

In problems 1 − 5, use Stokes’ theorem to evaluate ∬𝑆 𝑐𝑢𝑟𝑙𝐹⃗ ∙ 𝑑𝑆

𝟏. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑥𝑦𝑧𝑖 + 𝑥𝑗 + 𝑒 𝑥𝑦 cos 𝑧 𝑘 and 𝑆 is the hemisphere

𝑥 2 + 𝑦 2 + 𝑧 2 = 1, 𝑧 ≥ 0.
𝟐. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑦𝑧 3 𝑖 + sin 𝑥𝑦𝑧 𝑗 + 𝑥 3 𝑘 and 𝑆 is the part of paraboloid
𝑦 = (1 − 𝑥 2 − 𝑧 2 ) that lies to the right of the 𝑥𝑧 − 𝑝𝑙𝑎𝑛𝑒, oriented
toward the 𝑥𝑧 − 𝑝𝑙𝑎𝑛𝑒.
𝟑. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑧 3 𝑖 − 3𝑥𝑦𝑗 + 𝑥 3 𝑦 3 𝑘 and 𝑆 is the part of 𝑧 = 5 − 𝑥 2 − 𝑦 2
above the plane 𝑧 = 1.
𝟒. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑦𝑖 + 𝑧𝑗 + 𝑥𝑘 and 𝑆 is the paraboloid 𝑧 = 1 − (𝑥 2 + 𝑦 2 ),
𝑧≥0
𝟓. ⃗⃗⃗
𝐹(𝑥, 𝑦, 𝑧) = 𝑒2𝑧 𝑖 + 𝑒𝑧 sin 𝑦 𝑗 + 𝑒𝑧 cos 𝑦 𝑘 and 𝑆: 0 ≤ 𝑥 ≤ 2,
0 ≤ 𝑦 ≤ 1, 𝑧 = 𝑦2 .

In problems 6 − 10, use Stokes’ theorem to evaluate ∮𝐶 𝐹⃗ (𝑟) ∙ 𝑑𝑟

𝟔. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑥𝑧𝑖 + 2𝑥𝑦𝑗 + 3𝑥𝑦𝑘 and 𝐶 is the boundary of the part

of the plane 3𝑥 + 𝑦 + 𝑧 = 3 in the first octant.
𝟕. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 2𝑧𝑖 + 4𝑥𝑗 + 5𝑦𝑘 and 𝐶 is the curve of intersection of
the plane 𝑧 = 𝑥 + 4 and the cylinder 𝑥 2 + 𝑦 2 = 4.
𝟖. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑥 2 𝑧𝑖 + 𝑥𝑦 2 𝑗 + 𝑧 2 𝑘 and 𝐶 is the curve of intersection
of the plane 𝑥 + 𝑦 + 𝑧 = 1 and the cylinder 𝑥 2 + 𝑦 2 = 9.
𝟗. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑧 2 𝑖 + 𝑦 2 𝑗 + 𝑥𝑘 and 𝐶 is the triangle with vertices
(1,0,0), (0,1,0), and (0,0,1).
𝟏𝟎. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 2𝑦𝑖 + 𝑧𝑗 + 3𝑦𝑘 and the contour 𝐶 is the circle
𝑥 2 + 𝑦 2 + 𝑧 2 = 6𝑧, 𝑧 = 𝑥 + 3.
 254 Chapter 6 Line Integrals

In problems 11 − 15, verify that Stokes’ theorem is true for the given
vector field 𝐹⃗ and the surface 𝑆.
𝟏𝟏. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑦 3 𝑖 − 𝑥 3 𝑗 and 𝑆 is the circular disk 𝑥 2 + 𝑦 2 ≤ 1,
𝑧 = 0.
𝟏𝟐. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 3𝑦𝑖 + 4𝑧𝑗 − 6𝑥𝑘 𝑆 is the part of the paraboloid
𝑧 = 9 − 𝑥 2 − 𝑦 2 that lies above the 𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒, oriented upward.
𝟏𝟑. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 3𝑦𝑖 + 4𝑧𝑗 − 6𝑥𝑘 𝑆 is the part of the plane 𝑥 + 𝑦 + 𝑧 = 1
that lies in the first octant, oriented upward.
𝟏𝟒. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑥𝑦𝑖 + 𝑦𝑧𝑗 + 𝑥𝑧𝑘 over the surface 𝑧 = √𝑎2 − 𝑥 2 − 𝑦 2 ,
oriented upward.

𝟏𝟓. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 2𝑦𝑖 + 𝑧𝑗 − 3𝑦𝑘 over the surface 𝑆: 𝑧 = 𝑥 2 + 𝑦 2 ,

𝑧 ≤ 9.
 Chapter 6 Line Integrals 255

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