Calculus(II)

Department of Computer Science and Information Engineering

Chaoyang University of Technology
Taichung, Taiwan, Republic of China

Instructor: De-Yu Wang

E-mail: dywang@mail.cyut.edu.tw
Phone: (04)23323000 ext. 4538
Office: E738
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February 10, 2012

CALCULUS CALCULUS

Calculus

• Instructor:

1. Name: De-Yu Wang

2. Email: dywang@mail.cyut.edu.tw
3. Phone: (04)23323000 ext. 4538
4. Office: E738

• Textbook:
R. T. Smith and R. B. Minton, ”Calculus: Early Transcendental Functions,”
3e, 2006.

• Reference:

1. M. D. Weir, J. Hass and F. R. Giordano, “Thomas’ Calculus,” Eleventh

Edition, Greg Tobin, 2005.
2. J. Stewart, ”Early Transcendentals Calculus,” Five Edition, Thomson
Learning Inc., 2003.
3. R. Larson, R. Hostetler and B. H. Edwards ”Essential Calculus: Early
Transcendental Functions,” 2006.

• Grade:

1. Performance in class
2. Attendance condition
3. Weekly tests
4. Midterm
5. Final

De-Yu Wang CSIE CYUT Calculus

CONTENTS CONTENTS

Contents

1 LIMITS AND CONTINUITY 1

1.1 The Concept of Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Computation of Limits . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.3 Continuity and its Consequences . . . . . . . . . . . . . . . . . . . . . 9
1.4 Limits Involving Infinity . . . . . . . . . . . . . . . . . . . . . . . . . 13

2 DIFFERENTIATION 15
2.1 Tangent Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.2 The Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.3 Computation of Derivatives: the Power Rule . . . . . . . . . . . . . . 19
2.4 The Product and Quotient Rules . . . . . . . . . . . . . . . . . . . . 21
2.5 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.6 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . 26
2.7 Derivatives of Trigonometric Functions . . . . . . . . . . . . . . . . . 28
2.8 Derivatives of Exponential and Logarithmic Functions . . . . . . . . . 32

3 APPLICATIONS OF DIFFERENTIATION 35
3.1 The Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . 36
3.2 Linearization and Newton’s Method . . . . . . . . . . . . . . . . . . . 37
3.3 Indeterminate Forms and L’Hôpital’s Rule . . . . . . . . . . . . . . . 41
3.4 Maximum and Minimum Values . . . . . . . . . . . . . . . . . . . . . 43
3.5 Monotonic Functions and the First Derivative Test . . . . . . . . . . 45
3.6 Concavity and the Second Derivative Test . . . . . . . . . . . . . . . 47
3.7 Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

4 INTEGRATION 54
4.1 Antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
4.2 Area Under a Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
4.3 The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
4.4 Numerical Integration . . . . . . . . . . . . . . . . . . . . . . . . . . 69
4.5 The Natural Logarithm as an Integral . . . . . . . . . . . . . . . . . . 72
4.6 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . 75
4.7 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

De-Yu Wang CSIE CYUT i

CONTENTS

5 INTEGRATION TECHNIQUES 81
5.1 Integration by Substitution . . . . . . . . . . . . . . . . . . . . . . . . 82
5.2 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
5.3 Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
5.4 Integrals Involving Powers of Trigonometric Functions . . . . . . . . . 89
5.5 Trigonometric Substitution . . . . . . . . . . . . . . . . . . . . . . . . 94

6 DIFFERENTIAL EQUATIONS 97
6.1 Growth and Decay Problems . . . . . . . . . . . . . . . . . . . . . . . 98
6.2 Separable Differential Equations . . . . . . . . . . . . . . . . . . . . . 99
6.3 Euler’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

7 INFINITE SERIES 105

7.1 Sequences of Real Numbers . . . . . . . . . . . . . . . . . . . . . . . 106
7.2 Infinite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
7.3 The Integral Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
7.4 Comparison Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
7.5 Alternating Series and Absolute Convergence . . . . . . . . . . . . . . 121
7.6 The Ratio Test and The Root Test . . . . . . . . . . . . . . . . . . . 125
7.7 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
7.8 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

8 PARAMETRIC EQUATIONS AND POLAR COORDINATES 136

8.1 Plane Curves and Parametric Equations . . . . . . . . . . . . . . . . 137
8.2 Calculus and Parametric Equations . . . . . . . . . . . . . . . . . . . 138
8.3 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
8.4 Calculus and Polar Coordinates . . . . . . . . . . . . . . . . . . . . . 144

9 VECTORS 149
9.1 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150
9.2 The Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
9.3 The Cross Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156
9.4 Vector-Valued Functions . . . . . . . . . . . . . . . . . . . . . . . . . 160
9.5 The Calculus of Vector-Valued Functions . . . . . . . . . . . . . . . . 163
9.6 Arc Length and Curvature . . . . . . . . . . . . . . . . . . . . . . . . 167

10 PARTIAL DIFFERENTIATION 170

10.1 Functions of Several Variables . . . . . . . . . . . . . . . . . . . . . . 171
10.2 Limits and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . 172
10.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175
10.4 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176
10.5 The Gradient and Directional Derivatives . . . . . . . . . . . . . . . . 180
10.6 Extrema of Functions of Several Variables . . . . . . . . . . . . . . . 182

De-Yu Wang CSIE CYUT ii

CONTENTS

11 MULTIPLE INTEGRALS 186

11.1 Double Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187
11.2 Double Integrals in Polar Coordinates . . . . . . . . . . . . . . . . . . 192
11.3 Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194
11.4 Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196
11.5 Triple Integrals in Cylindrical and Spherical Coordinates . . . . . . . 199
11.6 Change of Variables: Jacobians . . . . . . . . . . . . . . . . . . . . . 204

De-Yu Wang CSIE CYUT iii

Chapter 7

INFINITE SERIES
7.1. SEQUENCES OF REAL NUMBERS

7.1 Sequences of Real Numbers

Definition 7.1.1. (a) Sequence {an }∞
n=n0 : A function whose domain is the set of
integers.

(b) Sequence {an }∞

n=n0 converges to L:

lim an = L
n→∞

if for each ε > 0, there exist N > 0 such that |an − L| < ε whenever n > N . If
the limit does not exist, then {an }∞
n=n0 diverges.

 ∞
1
Example 7.1.1. Write out the terms of the sequence {an }∞
n=1 = and then
n2 n=1
show that {an }∞
n=1 converges to 0.

Solution:
 ∞  
1 1 1 1
= , ,··· , 2,···
n2 n=1 1 4 n

Given any ε > 0, we must find N sufficiently large so that for every n > N ,

1
−0 <ε
n2
1
<ε
n2
1
n2 >
ε
r
1
n> = N.
ε

Theorem 7.1.1. Suppose that {an }∞ ∞

n=n0 and {bn }n=n0 both converge. Then

(a) lim (an + bn ) = lim an + lim bn ,

n→∞ n→∞ n→∞

(b) lim (an − bn ) = lim an − lim bn ,

n→∞ n→∞ n→∞
  
(c) lim (an bn ) = lim an lim bn ,
n→∞ n→∞ n→∞

an lim an  
(d) lim = n→∞ assuming lim bn 6= 0 .
n→∞ bn lim bn n→∞
n→∞

De-Yu Wang CSIE CYUT 106

7.1. SEQUENCES OF REAL NUMBERS

5n + 7
Example 7.1.2. Evaluate lim .
n→∞ 3n − 5

Solution:
7
5n + 7 5+ n 5 5
lim = lim 5 = the sequence converges to .
n→∞ 3n − 5 n→∞ 3 − 3 3
n

n2 + 1
Example 7.1.3. Evaluate lim .
n→∞ 2n − 3

Solution:
n2 + 1 n + n1
lim = lim =∞ the sequence diverges.
n→∞ 2n − 3 n→∞ 2 − 3
n

Theorem 7.1.2. Suppose that lim f (x) = L. Then, lim f (n) = L, also.
x→∞ n→∞
Remark. If lim f (n) = L, it need not be true that lim f (x) = L. For exapmle,
n→∞ x→∞

lim cos(2πn) = 1, lim cos(2πx) does not exist.

n→∞ x→∞

cn f (x)
6 6
2.0 1
1.5 0.5
1.0 s s s s s s s s s s s s s s s s -x
2π 4π 6π
0.5 −0.5
-n −1
5 10 15

n+1
Example 7.1.4. Evaluate lim
n→∞ en

Solution: Apply L’Hôpital’s rule for

d
x+1 (x + 1) 1 cn
lim = lim dx = lim = 0 6
0.8 s
x→∞ ex x→∞ d x x→∞ ex
e
dx
n+1 0.6
s
∴ lim n
= 0, also.
n→∞ e 0.4
s
s
0.2
s s s s s s s s s s s-s
n
5 10 15

De-Yu Wang CSIE CYUT 107

7.1. SEQUENCES OF REAL NUMBERS

Theorem 7.1.3. Squeeze theorem for sequences

If lim an = L = lim bn ,
n→∞ n→∞

and there exists an integer N such that an ≤ cn ≤ bn for all n > N , then

lim cn = L.
n→∞

Proof: Given any ε1 , ε2 > 0, we have

|an − L| < ε1 , |bn − L| < ε2 , for n > N

L − ε 1 < an < L + ε 1 , L − ε 2 < bn < L + ε 2
L − ε 1 < an < c n < b n < L + ε 2
|cn − L| < max{ε1 , ε2 }

Theorem 7.1.4. Absolute value theorem

If lim |an | = 0, then lim an = 0 also.

n→∞ n→∞

Proof: For all n,

− |an | ≤ an ≤ |an | and lim |an | = 0 = lim (−|an |)

n→∞ n→∞
∴ lim an = 0, also.
n→∞

 ∞
sin n
Example 7.1.5. Determine whether the sequence converges or diverges.
n2 n=1

Solution:

− 1 ≤ sin n ≤ 1, for all n,

−1 sin n 1
2
≤ 2 ≤ 2 , for all n ≥ 1,
n n n
−1 1 sin n
lim 2 = 0 = lim 2 ∴ lim = 0, also.
n→∞ n n→∞ n n→∞ n2

Definition 7.1.2. For any integer n ≥ 1, the factorial,

n! = 1 · 2 · 3 · · · n, and 0! = 1.

De-Yu Wang CSIE CYUT 108

7.1. SEQUENCES OF REAL NUMBERS

Definition 7.1.3. Monotonic sequence {an }∞

n=1

(a) The sequence {an }∞

n=1 is increasing if a1 ≤ a2 ≤ · · · ≤ an ≤ an+1 · · · .

(b) The sequence {an }∞

n=1 is decreasing if a1 ≥ a2 ≥ · · · ≥ an ≥ an+1 · · · .

Definition 7.1.4. Bounded sequence {an }∞

n=n0 :

If there is a number M > 0 for which |an | ≤ M , for all n.

Theorem 7.1.5. Every bounded, monotonic sequence converges.

 ∞
2n
Example 7.1.6. Prove that the sequence {an }∞
n=1 = converges.
n! n=1

Proof: We cannot apply L’Hôpital’s rule to n!

(a) {an }∞
n=1 is monotonic (decreasing).

2n+1
an+1 (n + 1)! 2
= n = ≤ 1, for n ≥ 1
an 2 n+1
n!
an+1 ≤ an , for n ≥ 1,

(b) {an }∞
n=1 is bounded.

2n 2 · 2 · 2···2 · 2 2
0< = ≤ = 2, for n ≥ 1
n! n · (n − 1) · · · 2 · 1 1
|an | ≤ 2, for n ≥ 1.

∴ {an }∞
n=1 converges.

Exercise 7.1.1. Write out the terms of the squence

2n − 1 4
(a) an = (c) an = √
n2 n+1
3 n
(b) an = (d) an = (−1)n
n! 3n − 1

Exercise 7.1.2. Find the limit of the squence

n 4
(a) an = (b) an = √
2n + 1 n+1

De-Yu Wang CSIE CYUT 109

7.2. INFINITE SERIES

Exercise 7.1.3. Determine whether the sequence converges or diverges of

 ∞
n!
(a) an = ne−n (c)
2n n=1
n2 + 3  ∞
(b) an = 3 nn + 3
n +1 (d) (−1)
n + 2 n=1

Exercise 7.1.4. Determine whether the sequence is increasng, decreasing or neither.

 ∞
en
(a)
n n=1
n+3
 ∞ (c) an =
2 n n+2
(b)
(n + 1)! n=1

Exercise 7.1.5. Show that the sequence is bounded

 ∞  ∞
3n2 − 2 6n − 1
(a) (b)
n2 + 1 n+3 n=1
n=1

7.2 Infinite Series

Definition 7.2.1. Infinite Series

∞
X n
X
ak = lim ak = lim Sn = S,
n→∞ n→∞
k=1 k=1

where Sn = a1 + a2 + · · · + an is the nth partial sum.

De-Yu Wang CSIE CYUT 110

7.2. INFINITE SERIES

Summary. Convergence tests for infinte series

Page Test Series Conditions of Comment
Convergence Divergence
∞
X
111 kth-Term ak lim ak 6= 0 Cannot be
k→∞
k=k0 used to show
convergence.
∞
X ark0
112 Geometric ark |r| < 1 |r| ≥ 1 S=
1−r
series k=k0
f is continuous and decreasing
ak = f (k) ≥ 0
∞
X Z ∞ Z ∞
114 Integral ak f (x) dx f (x) dx 0Z < Rn <
∞
1 1
k=1 converges diverges f (x) dx
n
∞
X 1
116 p-series p>1 p≤1
kp
k=1
X∞
118 Comparison ak 0 ≤ ak ≤ bk and 0 ≤ bk ≤ ak and
∞ ∞
(ak , bk > 0) k=1 X X
bk converges bk diverges
k=1 k=1
ak
lim =L>0
k→∞ bk
∞
X X∞ X∞
119 Limit Com- ak ak and bk both converge or
parison k=1 k=1 k=1 both diverge
(ak , bk > 0)
∞
X
122 Alternating (−1)k+1 ak lim ak = 0 and 0 < |S − Sn | ≤
k→∞
Series k=1 ak+1≤ ak an+1
X∞ X∞
124 Absolute ak |ak |
Conver- k=1 k=1
converges
gence
ak+1
lim =L
k→∞ ak
∞
X
125 Ratio ak (k!, bk ) L<1 L>1 L = 1 (no
k=1
p conclusion)
k
lim |ak | = L
k→∞
∞
X
126 Root Test ak (k k , bk ) L<1 L>1 L = 1 (no
k=1 conclusion)

Theorem 7.2.1. kth-term test for divergence

∞
X
(a) If ak converges, then lim ak = 0.
k→∞
k=1

∞
X
(b) If lim ak 6= 0, then ak diverges.
k→∞
k=1

Remark. This test cannot be used to show convergence.

De-Yu Wang CSIE CYUT 111

7.2. INFINITE SERIES

Proof: Assume that

∞
X
ak = lim Sn = L
n→∞
k=1
n
X n−1
X
an = ak − ak = Sn − Sn−1
k=1 k=1
lim an = lim (Sn − Sn−1 ) = lim Sn − lim Sn−1 = L − L = 0
n→∞ n→∞ n→∞ n→∞

∞
X k
Example 7.2.1. Use the kth-term test to determine if the series diverges.
k=1
k+1

k
Solution: lim ak = lim = 1 6= 0, ∴ this series diverges.
k→∞ k→∞ k + 1

∞
X 1
Example 7.2.2. Use the kth-term test to determine if the series diverges.
k=1
k

1
Solution: lim ak = lim = 0.
k→∞ k→∞ k
Remark. This does not say that the series converges.

∞
X ark0
Theorem 7.2.2. For a 6= 0, the geometric seies ark converges to if |r| < 1
k=k0
1−r
and diverges if |r| ≥ 1.
Proof:
Sn = ark0 + ark0 +1 + ark0 +2 + · · · + arn
− rSn = ark0 +1 + ark0 +2 + ark0 +3 + · · · + arn+1
(1 − r)Sn = ark0 − arn+1

ark0 − arn+1
Sn =
1−r  k0
ar k0
− ar n+1  ar
if |r| < 1;
S = lim Sn = lim = 1−r
n→∞ n→∞ 1−r 
does not exist if |r| ≥ 1

De-Yu Wang CSIE CYUT 112

7.2. INFINITE SERIES

∞  k
X 1 2
Example 7.2.3. Determine if the geometric series converges or diverges.
k=3
3 3
Find the sum of the series if it converges.

Solution:
∞  k
2 2 X 1 2
r= , |r| = < 1. ∴ converges.
3 3 k=3
3 3
1

3
ar3 3
· 23 8
S= = 2 =
1−r 1− 3 27

∞  k
X 1 3
Example 7.2.4. Determine if the geometric series − converges or di-
k=1
3 2
verges. Find the sum of the series if it converges.

Solution:
∞  k
3 3 X 1 3
r=− , |r| = > 1 ∴ − diverges.
2 2 k=1
3 2

Exercise 7.2.1. Use the kth-term test to determine if the series diverges.
∞ ∞
X 2 X 2k
(a) (c) (−1)k+1
k k=0
k+1
k=1
∞ ∞  
X 4k X 1 1
(b) (d) k
−
k+2 k=0
2 k+1
k=0

Exercise 7.2.2. Determine if the geometric series converges or diverges. Find the
sum of the series if it converges.
∞  k ∞
X 1 X 1
k
(a) 5 − (c) 4−1
2 k=0
3
k=2
∞  k ∞  k
X 1 X 1 7
(b) 3 (d) −
4 k=3
5 2
k=1

Exercise 7.2.3. Prove Theorem 7.2.2 and Theorem 7.2.1.

De-Yu Wang CSIE CYUT 113

7.3. THE INTEGRAL TEST

7.3 The Integral Test

Theorem 7.3.1. Integral Test
(a) If f (k) = ak for all k = 1, 2, · · · , f is continuous and decreasing, and f (x) ≥ 0
Z ∞ X∞
for x ≥ 1, then f (x) dx and ak either both converge or both diverge.
1 k=1
∞
X
(b) If f (k) converges. Then, the remainder Rn satisfies
k=1
∞
X Z ∞
0 ≤ Rn = ak ≤ f (x) dx.
k=n+1 n

Z n
Proof: (a) 0 ≤ a2 + a3 + · · · + an = Sn − a1 ≤ f (x) dx ≤ a1 + a2 + · · · + an−1 = Sn−1
1

y
6
y
y = f (x) 6
y = f (x)
s s a1 )
(2, a2 )
s
(1,
s s s
(5, a5 )
s s
(4, a4 )
-x
1 2 3 4 5 6
-x
1 2 3 4 5 6
Z ∞
(i) If f (x) dx converges to L
1
Z n
Sn − a1 ≤
f (x) dx
1
Z n Z ∞
lim (Sn − a1 ) ≤ lim f (x) dx = f (x) dx = L
n→∞ n→∞ 1 1
lim Sn ≤ lim (L + a1 ) = L + a1 , converges, too.
n→∞ n→∞
Z ∞
(ii) If f (x) dx diverges
1
Z n
f (x) dx ≤ Sn−1
1
Z n Z ∞
lim f (x) dx = f (x) dx = ∞ ≤ lim Sn−1
n→∞ 1 1 n→∞

lim Sn−1 > ∞, diverges, too.

n→∞

De-Yu Wang CSIE CYUT 114

7.3. THE INTEGRAL TEST

∞
X
(iii) If ak converges to L
k=1
Z n
f (x) dx ≤ Sn−1
Z n Z 1∞
lim f (x) dx = f (x) dx ≤ lim Sn−1 = L
n→∞ 1 1 n→∞
Z ∞
f (x) dx ≤ L, converges, too.
1
∞
X
(iv) If ak diverges
k=1
Z n
f (x) dx
Sn − a1 ≤
1
Z n Z ∞
lim (Sn − a1 ) = ∞ ≤ lim f (x) dx = f (x) dx
n→∞ n→∞ 1 1
Z ∞
f (x) dx ≥ ∞, diverges, too.
1
∞
X Z ∞
(b) Rn = ak ≤ f (x) dx y
k=n+1 n 6

y = f (x)

s(n + 1, an+1 )
s s s

-x
nn+1

∞
X 3
Example 7.3.1. Use the integral test to determine if the series converges
k=0
(2 + k)2
or diverges.
3
Solution: f (x) = is continuous and decreasing, and f (x) ≥ 0 for x ≥ 0
(2 + x)2
Z ∞ Z R R
3 3 −3
dx = lim dx = lim
0 (2 + x)2 R→∞ 0 (2 + x)2 R→∞ (2 + x)
0
 
1 1 3
= −3 lim − = , converges
R→∞ 2 + R 2 2
∞
X 3
∴ converges, too.
k=0
(2 + k)2

De-Yu Wang CSIE CYUT 115

 7.3. THE INTEGRAL TEST

Example 7.3.2. Estimate the error for the integral test in using the partial sum S100
∞
X 3
to approximate the sum of the series .
k=0
(2 + k)2

Solution:
Z ∞ Z R R
3 3 −3
0 ≤ R100 ≤ dx = lim dx = lim
100 (2 + x)2 R→∞ 100 (2 + x)
2 R→∞ (2 + x)
100
 
1 1 3
= −3 lim − = .
R→∞ 2 + R 102 102

Example 7.3.3.
Determine the number of terms needed to obtain an approximation to the sum of the
∞
X 3
series 2
correct to within 10−6 with error estimate for the integral test.
k=0
(2 + k)

Solution:
Z ∞ Z R R
3 3 −3
0 ≤ Rn ≤ dx = lim dx = lim
n (2 + x)2 R→∞ n (2 + x)
2 R→∞ (2 + x)
n
 
1 1 3
= −3 lim − = ≤ 10−6 ,
R→∞ 2 + R n+2 n+2
3
Rn ≤ ≤ 10−6 , ∴ n ≥ 3 × 106 − 2.
n+2

Theorem 7.3.2. p series

∞
X 1
If p-series p
converges if p > 1 and diverges if p ≤ 1.
k=1
k

1
Proof: f (x) = = x−p is continuous and decreasing, and f (x) ≥ 0 for x ≥ 1
xp

(a) For p 6= 1
Z ∞ Z R R
−p x−p+1
x dx = lim x−p dx = lim
1 R→∞ 1 R→∞ −p + 1 1
1  
= lim R−p+1 − 1
−p + 1 R→∞

 1 , if p > 1
= p−1
∞, if p < 1

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 7.3. THE INTEGRAL TEST

(b) For p = 1
Z ∞ Z R
1 1
dx = lim dx = lim ln |x||R
1
1 x R→∞ 1 x R→∞

= lim [ln |R| − 1] = ∞.

R→∞

Example 7.3.4. Determine convergence or divergence of the p-series

∞ ∞ ∞
X X X 1
k −11/10 , k −10/11 , and .
k=1 k=1 k=1
k
∞ ∞
X
−11/10
X 1
Solution: (a) k =
k=1 k=1
k 11/10
∞
11 X 1
∵ p= >1 ∴ k −11/10 converges to 11 = 10.
10 k=1 10
−1
∞ ∞
X
−10/11
X 1
(b) k =
k=1 k=1
k 10/11
∞
10 X
∵ p= <1 ∴ k −10/11 diverges.
11 k=1

∞
X 1
(c)
k=1
k
∞
X 1
∵ p=1 ∴ diverges.
k=1
k

Exercise 7.3.1. Use the integral test to determine if the series converges or diverges.
If the series converges:
(1) Estimate the error in using the partial sum S100 to approximate the sum of the
series.
(2) Determine the number of terms needed to obtain an approximation to the sum
of the series correct to within 10−6 with error estimate for the integral test.

∞
∞
X k+1
X 1
(a) (c)
2
k + 2k + 3 k=3
(1 + 2k)2
k=3
∞
X∞
2
X e1/k
(b) (d)
k ln k k=3
k2
k=2

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7.4. COMPARISON TESTS

Exercise 7.3.2. Determine convergence or divergence of the p-series

∞ ∞
X
X 4 (c) k −2/3
(a) √
5
k=1
k k=1
∞ ∞
X X 4
(b) k −9/10 (d) √
k=6
k3
k=1

Exercise 7.3.3. Prove Theorem 7.3.1 and Theorem 7.3.2

7.4 Comparison Tests

Theorem 7.4.1. Comparison test
Suppose that 0 ≤ ak ≤ bk , for all k
∞
X ∞
X
(a) If bk converges, then ak converges, too.
k=1 k=1

∞
X ∞
X
(b) If ak diverges, then bk diverges, too.
k=1 k=1

∞
X
Proof for (a): If bk = B
k=1

∞
X
0 ≤ S n = a1 + a2 + · · · + an ≤ b 1 + b 2 + · · · + b n ≤ bk = B
k=1
{Sn }∞
n=1 is bounded, (∵ 0 ≤ Sn ≤ B)
Sn = a1 + a2 + · · · + an ≤ a1 + a2 + · · · + an + an+1 = Sn+1
{Sn }∞
n=1 is increasing, (∵ Sn ≤ Sn+1 )
∴ lim Sn = S converges.
n→∞

∞
X
Proof for (b): If ak = ∞
k=1

∞
X ∞
X
bk = lim (b1 + b2 + · · · + bn ) ≥ lim (a1 + a2 + · · · + an ) = ak = ∞.
n→∞ n→∞
k=1 k=1
∞
X
∴ bk =∞ diverges, too.
k=1

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7.4. COMPARISON TESTS

∞
X 2
Example 7.4.1. Use the comparison test to determine if the series con-
k=1
3k 2+1
verges or diverges.
Solution:
2 2
< 2, for k ≥ 1
3k 2 +1 3k
∞
X 2
∵ converges (∵ p = 2),
k=1
3k 2
∞
X 2
∴ converges, too.
k=1
3k 2+1

∞
X 2
Example 7.4.2. Use the comparison test to determine if the series √
3
k=1
3 k2 − 1
converges or diverges.
Solution:
2 2
√
3
> √ 3
, for k ≥ 1
3 k2 − 1 3 k2
∞ ∞  
X 2 X 2 2
∵ √ = 2 diverges ∵ p = ,
k=1
3
3 k2 k=1 3k
3 3
∞
X 2
∴ √
3
diverges, too.
3 k 2−1
k=1

Theorem 7.4.2. Limit Comparison test

∞
ak X
Suppose that ak , bk > 0 and lim = L > 0 for k large. Then, either ak and
k→∞ bk
k=1
X∞
bk both converge or both diverge.
k=1

ak ak L
Proof: If lim = L > 0, we can make − L < ε = , for all k > N
k→∞ bk bk 2
L ak 3L
< <
2 bk 2
L 3L
bk <ak < bk , for all k > N.
2 2

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7.4. COMPARISON TESTS

∞
X
(a) If bk converges to B, then
k=1
∞ N ∞ ∞
N N
X X X 3L X X X 3L
ak = ak + ak < ak + bk < ak + B, converges,
k=1 k=1 k=N +1 k=1
2 k=N +1 k=1
2
too.
∞
X
(b) If bk diverges, then
k=1
∞ N ∞ N ∞
X X X X L X
ak = ak + ak > ak + bk , diverges, too.
k=1 k=1 k=N +1 k=1
2 k=N +1
| {z }
∞

∞
X
(c) If ak converges to A, then
k=1
∞ N ∞ N ∞ N
X X X X 2 X X 2
bk = bk + bk < bk + ak < bk + A, converges, too.
k=1 k=1 k=N +1 k=1
L k=N +1 k=1
L

∞
X
(d) If ak diverges, then
k=1
∞ N ∞ N ∞
X X X 2 X
X
bk = bk + bk > bk + ak , diverges, too.
k=1 k=1 k=N +1 k=1
3L k=N +1
| {z }
∞

∞
X 1
Example 7.4.3. Use the limit comparison test to determine if the series
k=3
k3 − 5k
converges or diverges.

1 1
Solution: Let ak = > 0, bk = 3 > 0 for k large.
k3 − 5k k

1
ak 3 1
lim = lim k − 5k = lim =1>0
k→∞ bk k→∞ 1 k→∞ 1 − 52
k
k3
∞ ∞
X 1 X 1
Since 3
is a convergent p-series (p = 3 > 1), 3
is also convergent.
k=3
k k=3
k − 5k

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7.5. ALTERNATING SERIES AND ABSOLUTE CONVERGENCE

Example 7.4.4. Use the limit comparison test to determine if the series
∞
X k 2 − 2k + 7
converges or diverges.
k=1
k 5 + 5k 4 − 3k 3 + 2k − 1

Solution: Let
k 2 − 2k + 7 1 − k2 + k72 1
ak = = , bk =
k 5 + 5k 4 − 3k 3 + 2k − 1 k 3 + 5k 2 − 3k + k2 − 1
k2
k3
1 − k2 + k72
ak k 3 + 5k 2 − 3k + k2 − 1
k2
1 − k2 + 7
k2
lim = lim = lim =1>0
k→∞ bk k→∞ 1 k→∞ 1 + 5 − 32 + 2
− 1
k k k3 k5
k3
∞ ∞
X 1 X k 2 − 2k + 7
Since 3
is a convergent p-series (p = 3 > 1), 5 + 5k 4 − 3k 3 + 2k − 1
is also
k=1
k k=1
k
convergent.

Exercise 7.4.1. Use the comparison test to determine if the series converges or
diverges.
∞ ∞
X 2k X 2k
(a) (c)
3
k +4 k=3
k4 + 5
k=1
∞ ∞
X 2 X 2
(b) (d)
3k 2+1 k=2
3k 2+k
k=1

Exercise 7.4.2. Use the limit comparison test to determine if the series converges
or diverges.
∞ ∞
X k 3 + 2k + 3 X k2 + 3
(a) (b)
k 4 + 2k 2 + 4 k=3
k5 + k2 + 4
k=1

Exercise 7.4.3. Prove Theorem 7.4.1 and Theorem 7.4.2

7.5 Alternating Series and Absolute Convergence

Definition 7.5.1. Alternating Series
Any series of the form
∞
X
(−1)k+1 ak = a1 − a2 + a3 − a4 + a5 − a6 + · · · ,
k=1

where ak > 0, for all k.

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7.5. ALTERNATING SERIES AND ABSOLUTE CONVERGENCE

Theorem 7.5.1. Alternating Series Test

Suppose that lim ak = 0 and 0 < ak+1 ≤ ak , for all k ≥ 1. Then, the alternating
k→∞
∞
X
series (−1)k+1 ak converges.
k=1

Proof: (a) The even-index partial sums S2n

{S2n }∞
n=1 is monotonic(increasing).

S2 = a1 − a2 > 0
S4 = S2 + (a3 − a4 ) ≥ S2 , ∵ a3 ≥ a4
..
.
S2n = S2n−2 + (a2n−1 − a2n−2 ) ≥ S2n−2
S2n ≥ S2n−2 ≥ · · · ≥ S4 ≥ S2 > 0.

{S2n }∞
n=1 is bounded.

0 < S2n = a1 + (−a2 + a3 ) + (−a4 + a5 ) + · · · + (−a2n−2 + a2n−1 ) −a2n

| {z } | {z } | {z } | {z }
≤0 ≤0 ≤0 ≤0

≤ a1

∴ lim S2n converges.

n→∞

(b) The odd-index partial sums S2n+1

lim S2n+1 = lim (S2n + a2n+1 ) = lim S2n + lim a2n+1 = lim S2n
n→∞ n→∞ n→∞ n→∞ n→∞

converges, too.

∞
X k2
Example 7.5.1. Determine if the alternating series (−1)k+1 converges or
k=0
k2 + 3
diverges.
∞
k2 X k2
Solution: lim ak = lim = 1 6= 0, ∴ (−1)k+1 diverges.
k→∞ k→∞ k 2 + 3 k2 + 3
k=0

∞
X 1
Example 7.5.2. Determine if the alternating series (−1)k+1 converges or
k=0
k2 +3
diverges.
1 1 1
Solution: lim ak = lim = 0 and a k+1 = < = ak ,
k→∞ k→∞ k 2 + 3 (k + 1)2 + 3 k2 + 3
∞
X 1
∴ (−1)k+1 converges.
k=0
k2 +3

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7.5. ALTERNATING SERIES AND ABSOLUTE CONVERGENCE

∞
X
Theorem 7.5.2. If the alternating series (−1)k+1 ak converges to some number
k=1
S, then the error in approximating S by the nth partial sum Sn satisfies

|S − Sn | ≤ an+1 .

Proof: (a) n = 2m (even)

S − S2m = a2m+1 + (−a2m+2 + a2m+3 ) + (−a2m+4 + a2m+5 ) + · · ·

| {z } | {z }
≤0 ≤0

≤ a2m+1 = an+1

(b) n = 2m + 1 (odd)

S − S2m+1 = −a2m+2 + (a2m+3 − a2m+4 ) + (a2m+5 − a2m+6 ) + · · ·

| {z } | {z }
≥0 ≥0

≥ −a2m+2 = −an+1
∴ |S − Sn | ≤ an+1

Example 7.5.3. Determine how many terms needed to estimate the sum of the
∞
X 4
alternating series (−1)k+1 2 correct to within 10−8 .
k=1
k

Solution:

|S − Sn | ≤ an+1 ≤ 10−8
4
2
≤ 10−8
(n + 1)
n + 1 ≥ 2 × 104
n ≥ 19999

Definition 7.5.2. Absolute and conditional convergence

∞
X ∞
X
(a) ak is absolutely convergent if |ak | converges.
k=1 k=1

∞
X ∞
X ∞
X
(b) ak is conditionally convergent if ak converges but |ak | diverges.
k=1 k=1 k=1

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7.5. ALTERNATING SERIES AND ABSOLUTE CONVERGENCE

Theorem 7.5.3. Absolute Convergence

∞
X ∞
X
If |ak | converges, then ak converges.
k=1 k=1
∞
X
Proof: If |ak | converges to L
k=1

∞
X
ak = a1 + a2 + a3 + · · ·
k=1
∞
X
≤ |a1 | + |a2 | + |a3 | + · · · = |ak | = L, converges, too.
k=1

∞
X 2
Example 7.5.4. Use the absolute convergence to determine if the series (−1)k 2
k=1
k
is absolutely convergent, conditionally convergent or divergent.
Solution:
∞ ∞ ∞
X X 2 k
X 2
|ak | = (−1) 2 = 2
, is a convergent p-series (p = 2 > 1)
k=1 k=1
k k=1
k
∞ ∞
X X 2
∴ ak = (−1)k is absolutely convergent.
k=1 k=1
k2

Example 7.5.5. Use the absolute convergence to determine if the series

∞
X 2
(−1)k √
3
is absolutely convergent, conditionally convergent or divergent.
k=1
k2
Solution:
∞ ∞ ∞  
X X 2 k
X 2 2
|ak | = (−1) √ = 2 , is a divergent p-series p=
k=1 k=1
3
k2 k=1 k
3 3
2
lim ak = lim 2 = 0
k→∞ k→∞ k 3
2 2
and ak+1 = 2 < 2 = ak ,
(k + 1) 3 k3
∞
X 2
∴ (−1)k 2 is conditionally convergent.
k=0 k3

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7.6. THE RATIO TEST AND THE ROOT TEST

Exercise 7.5.1. Determine if the alternating series is absolutely convergent, condi-

tionally convergent or divergent. If the series converges, then determine how many
terms needed to estimate the sum of the alternating series correct to within 10−6 .
∞ ∞
X 2 X 2k − 1
(a) (−1)k+1 3 (c) (−1)k+1
k k=7
k3
k=0
∞ ∞
X 3 X k2
(b) (−1)k+1 √ (d) (−1)k+1
k k=1
k+1
k=2

Exercise 7.5.2. Prove Theorem 7.5.1, Theorem 7.5.3 and Theorem 7.5.2.

7.6 The Ratio Test and The Root Test

Theorem∞
7.6.1. Ratio Test
X ak+1
Given ak with ak 6= 0 for all k, suppose that lim = L. Then,
k=1
k→∞ a k

(a) if L < 1, the series converges absolutely,

(b) if L > 1, the series diverges and
(c) if L = 1, there is no conclusion.

ak+1
Proof for (a): For L < 1, pick any number r with lim = L < r < 1.
k→∞ ak
Then, we can make
ak+1
< r, for k > N
ak
|ak+1 | < r|ak |
|ak+2 | < r|ak+1 | < r2 |ak |
..
.
|ak+m | < r|ak+m−1 | < rm |ak |

∞
X N
X ∞
X
|ak | = |ak | + |ak |
k=1 k=1 k=N +1
N ∞ N
X X
k−N −1
X |aN +1 |
< |ak | + r |aN +1 | = |ak | + < ∞, converges.
k=1 k=N +1 k=1
1−r
∞
X
∴ ak converges absolutely.
k=1

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7.6. THE RATIO TEST AND THE ROOT TEST

ak+1
Proof for (b): For L > 1, we have lim = L > 1.
k→∞ ak
Then, we can make

ak+1
> 1, for k > N
ak
|ak+1 | > |ak | ⇒ lim |ak | 6= 0
k→∞

∞
X
By the kth-term test, ak diverges.
k=1

∞
X 2
Example 7.6.1. Use the ratio test to determine if the alternating series (−1)k
k=0
k!
is absolutely convergent, conditionally convergent or divergent.
2
ak+1 (k+1)! 1
Solution: lim = lim 2 = lim =0<1
k→∞ ak k→∞
k!
k→∞ k + 1
∞
X 2
∴ (−1)k is absolutely convergent.
k=0
k!

Theorem
∞
7.6.2. Root Test
X p
Given ak , suppose that lim k |ak | = L. Then,
k→∞
k=1

(a) if L < 1, the series converges absolutely,

(b) if L > 1, the series diverges and

(c) if L = 1, there is no conclusion.

p
k
Proof for (a): For L < 1, pick any number r with lim |ak | = L < r < 1.
k→∞
Then, we can make
p
k
|ak | < r, for k > N
|ak | < rk
|ak+1 | < rk+1
..
.
|am | < rm

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7.6. THE RATIO TEST AND THE ROOT TEST

∞
X N
X ∞
X
|ak | = |ak | + |ak |
k=1 k=1 k=N +1
N ∞ N
X X
k
X rN +1
< |ak | + r = |ak | + < ∞, converges.
k=1 k=N +1 k=1
1−r
∞
X
∴ ak converges absolutely.
k=1

p
k
Proof for (b): For L > 1, we have lim |ak | = L > 1.
k→∞
Then, we can make
pk
|ak | = L > 1, for k > N
|ak | > Lk
|ak+1 | > Lk+1 ⇒ lim |ak | 6= 0
k→∞
∞
X
By the kth-term test, ak diverges.
k=1

∞
X 3
Example 7.6.2. Use the root test to determine if the series (−1)k is absolutely
k=1
kk
convergent, conditionally convergent or divergent.
Solution:
s √
k
p
k k 3 3
lim |ak | = lim k
= lim =0<1
k→∞ k→∞ k k→∞ k
∞
X 3
∴ (−1)k is absolutely convergent.
k=1
kk

Exercise 7.6.1. Use the ratio test to determine if the alternating series is absolutely
convergent, conditionally convergent or divergent.
∞ ∞
X X 10k
(a) (−1)k 2k (d) (−1)k
k=4
k!
k=0
∞ ∞
X k! X (−1)k k!
(b) (−1)k+1 (e)
4k k=1
ek
k=1
∞ ∞ 2 k
kk 3
X 2 X
(c) (f) (−1)
k! k=3
2k
k=1

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7.7. POWER SERIES

Exercise 7.6.2. Use the root test to determine if the series is absolutely convergent,
conditionally convergent or divergent.
∞ ∞
X 2 X e3k
(a) (c)
(k + 3)k k=2
k 3k
k=1
∞  k ∞  k
X 6k X 4
(b) (d)
5k + 1 k=1
3k + 2
k=1

Exercise 7.6.3. Prove Theorem 7.6.1 and Theorem 7.6.2.

7.7 Power Series

Definition 7.7.1. Power Series
∞
X
bk (x − c)k = b0 + b1 (x − c) + b2 (x − c)2 + b3 (x − c)3 + · · ·
k=0

Theorem 7.7.1. Convergence of a Power Series

∞
X
For a power series centered at c, bk (x − c)k , precisely one of the following is ture.
k=0
The series converges

(a) only for x = c, r = 0;

(b) for x ∈ (c − r, c + r), 0 < r < ∞;

(c) for all x ∈ (−∞, ∞), r = ∞.

where r is the radius of convergence of the power series.

2
Example 7.7.1. Find a power series representation of about c = 0. Also
1−x
determine the radius of convergence and interval of convergence.

Solution:
∞
X a
∵ abk = , for |b| < 1
k=0
1−b
∞
2 X
∴ = 2 · xk for |x| < 1,
1 − x k=0

the radius of convergence r = 1 and the interval of convergence x ∈ (−1, 1).

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7.7. POWER SERIES

Example 7.7.2. Determine the radius and interval of convergence for the geometric
∞
X 10k
power series (x − 1)k .
k=0
k!

Solution: From the ratio test, we have

10k+1
(x − 1)k+1
ak+1 (k + 1)! 1
lim = lim k
= 10|x − 1| lim =0<1
k→∞ ak k→∞ 10 k→∞ k + 1
(x − 1)k
k!
∞
X 10k
∴ (x − 1)k converges absolutly for all x ∈ (−∞, ∞) and r = ∞.
k=0
k!

Example 7.7.3. Determine the radius and interval of convergence for the geometric
∞
X xk
power series k
.
k=1
k 4

Solution: From the ratio test, we have

xk+1
ak+1 (k + 1)4k+1 |x| k |x|
lim = lim k
= lim = < 1.
k→∞ ak k→∞ x 4 k→∞ k + 1 4
k 4k
For x = 4,
∞ ∞ ∞
X xk X 4k X 1
k
= k
= , is a divergent p-series (p = 1).
k=1
k4 k=1
k4 k=1
k

For x = −4,
∞ ∞ ∞
X xk X (−4)k X (−1)k
k
= k
= , is a convergent alternating series.
k=1
k 4 k=1
k 4 k=1
k

∴ r = 4, x ∈ [−4, 4).

Theorem 7.7.2. If the function given by

∞
X
f (x) = bk (x − c)k = b0 + b1 (x − c) + b2 (x − c)2 + b3 (x − c)3 + · · ·
k=0

has a radius of convergence of r > 0, then

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7.7. POWER SERIES

(a) the derivative

d
f ′ (x) = (b0 + b1 (x − c) + b2 (x − c)2 + b3 (x − c)3 + · · · )
dx
= b1 + 2b2 (x − c) + 3b3 (x − c)2 + · · ·
X∞
= bk k(x − c)k−1 ,
k=1

(b) and the antiderivative

Z Z X ∞ ∞ Z
X
k
f (x) dx = bk (x − c) dx = bk (x − c)k dx
k=0 k=0
∞ k+1
X (x − c)
= bk +K
k=0
k+1

have the same radii of convergence r.

Remark. The interval of convergence may differ at the endpoints.
∞
X
Example 7.7.4. Use the power series (−1)k xk to find power series representations
k=0
1
of . Also determine the radii of convergence and intervals of convergence.
(1 + x)2
∞ ∞
X
k k
X 1 1
Solution: First, (−1) x = (−x)k = = , for | − x| = |x| < 1.
k=0 k=0
1 − (−x) 1+x
Therefore, r = |x| = 1, and x ∈ (−1, 1).
d 1 −1
=
dx 1 + x (1 + x)2
∞
1 d 1 d X
2
= − = − (−x)k
(1 + x) dx 1 + x dx k=0
∞
X ∞
X
=− k(−x)k−1 · (−1) = (−1)k+1 kxk−1
k=1 k=1

For x = −1,
∞
X ∞
X ∞
X
k+1 k−1 2k
(−1) kx = (−1) k = k, diverges (k-term)
k=1 k=1 k=1

For x = 1,
∞
X ∞
X
(−1)k+1 kxk−1 = (−1)k+1 k, diverges (k-term)
k=1 k=1

∴ r = 1, and x ∈ (−1, 1).

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7.8. TAYLOR SERIES

Exercise 7.7.1. Find a power series representation of f (x) about c = 0. Also deter-
mine the radius of convergence and interval of convergence.

3 3
(a) f (x) = (c) f (x) =
4+x x−1
2 3
(b) f (x) = (d) f (x) =
1 + x2 6−x

Exercise 7.7.2. Determine the radius and interval of convergence for the power series
∞ ∞
X X 2k
(a) (x + 2)k (c) (x − 2)k
k=0
k!
k=0
∞ ∞
X  x k X k!
(b) (−1)k (d) (−1)k+1
2 k=1
4k
k=0

∞
X
Exercise 7.7.3. Use the power series (−1)k xk to find power series representations
k=0
of f (x). Also determine the radii of convergence and intervals of convergence.

(a) f (x) = ln(1 + x2 ) (c) f (x) = 3 tan−1 x

2 3x
(b) f (x) = (d) f (x) =
(x − 1)2 (1 − x2 )2

7.8 Taylor Series

Theorem 7.8.1. (a) Taylor series expansion of f (x) about x = c is given
∞
X f (k) (c)
f (x) = (x − c)k
k=0
k!
f ′′ (c) f (k) (c)
=f (c) + f ′ (c)(x − c) + (x − c)2 + · · · + (x − c)k + · · ·
2! k!

(b) Taylor polynomial of degree n for f expanded about x = c

f (x) ≈Pn (x)

n
X f (k) (c)
= (x − c)k
k=0
k!
f ′′ (c) f (n) (c)
=f (c) + f ′ (c)(x − c) + (x − c)2 + · · · + (x − c)n
2! n!
Remark. A Taylor series expansion about x = 0 (i.e., take c = 0) is called a Maclaurin
series.

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7.8. TAYLOR SERIES

Proof:
∞
X
f (x) = bk (x − c)k
k=0
= b0 + b1 (x − c) + b2 (x − c)2 + b3 (x − c)3 + · · · f (c) = b0 ,
f ′ (x) = b1 + 2b2 (x − c) + 3b3 (x − c)2 + 4b4 (x − c)3 + · · · f ′ (c) = b1 ,
f ′′ (x) = 2b2 + 3 · 2b3 (x − c) + 4 · 3b4 (x − c)2 + · · · f ′′ (c) = 2b2 ,
f ′′′ (x) = 3 · 2b3 + 4 · 3 · 2b4 (x − c) + · · · f ′′′ (c) = 3! b3 ,
..
.
··· f (k) (c) = k! bk ,
f (k) (c)
∴ bk = , for k = 0, 1, 2, · · ·
k!

Example 7.8.1. Find the Taylor series about c = 0 and its interval of convergence
for ex
Solution: (a) Taylor series expansion
∞
X f (k) (0)
f (x) = ex = xk , f (0) = 1
k=0
k!
′ x
f (x) = e , f ′ (0) = 1
f ′′ (x) = ex , f ′′ (0) = 1
f (3) (x) = ex , f (3) (0) = 1
.. ..
. .
f (k) (x) = ex , f (k) (0) = 1
∞
X 1 k 1 1
∴ ex = x = 1 + x + x2 + x3 + · · ·
k=0
k! 2! 3!

(b) To test the convergence by ratio test,

xk+1
ak+1 (k + 1)! 1
lim = lim k
= |x| lim =0<1
k→∞ ak k→∞ x k→∞ k+1
k!
∞
X 1 k
∴ x converges for all x ∈ (−∞, ∞) and r = ∞.
k=0
k!

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7.8. TAYLOR SERIES

Example 7.8.2. For f (x) = ex , find the Taylor polynomial of degree n expanded
about x = 0 and use it to approximate the number e.
Solution: f (k) (x) = ex , for all k. So, the nth-degree Taylor polynomial is
n n
X f (k) (0) X 1 k
f (x) ≈ Pn (x) = (x − 0)k = x
k=0
k! k=0
k!
1 2 1 3 1
=1+x+ x + x + · · · + xn .
2! 3! n!
1 1 1
e = e1 = 1 + 1 + 12 + 13 + · · · + 1n ≈ 2.718281828.
2! 3! n!

Example 7.8.3. Expand f (x) = sin x in a Taylor series about x = π2 (i.e., take
c = π2 ) and find its interval of convergence.
∞

X f (k) π2  π k
Solution: (a) Taylor series expansion sin x = x−
k=0
k! 2
π 
f (x) = sin x, f =1
2
π
f ′ (x) = cos x, f′ =0
2
π
f ′′ (x) = − sin x, f ′′ = −1
2
π
f (3) (x) = − cos x, f (3) =0
2
π 
f (4) (x) = sin x, f (4) =1
2
..
.
π 
··· f (k) = (−1)k/2 for k = 0, 2, 4, · · ·
2
∞
X (−1)k/2  π k
∴ sin x = x− , for k = 0, 2, 4, · · ·
k=0
k! 2
∞
X (−1)m  π 2m k
= x− , where m =
m=0
(2m)! 2 2
1  
π 2 1  
π 4 1  π 6
=1− x− + x− − x− + ···
2 2 4! 2 6! 2
(b) To test the convergence by ratio test,
(−1)k+1
2k+2
x − π2
ak+1 (2k + 2)!
lim = lim
k→∞ ak k→∞ (−1)k
2k
x − π2
(2k)!
 π 2 1
= x− lim =0<1
2 k→∞ (2k + 2)(2k + 1)

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7.8. TAYLOR SERIES

∞
X (−1)k  π 2k
∴ x− converges absolutly for all x ∈ (−∞, ∞) and r = ∞.
k=0
(2k)! 2

2
Example 7.8.4. Find Taylor series in power of x for e2x , ex and e−2x .
Solution:
∞
x
X 1 k 1 1
e = x = 1 + x + x2 + x3 + · · · for x ∈ (−∞, ∞)
k=0
k! 2! 3!
∞ ∞
2x
X 1 k
X 2k k 22 23
e = (2x) = x = 1 + 2x + x2 + x3 + · · ·
k=0
k! k=0
k! 2! 3!
∞ ∞
2
X 1 2 k X 1 2k 1 1
ex = (x ) = x = 1 + x2 + x4 + x6 + · · ·
k=0
k! k=0
k! 2! 3!
∞ ∞
−2x
X 1 k
X (−1)k 2k k 22 23
e = (−2x) = x = 1 − 2x + x2 − x3 + · · ·
k=0
k! k=0
k! 2! 3!

Summary. Common Taylor series

Taylor Interval of
series Convergence
∞
X 1 k 1 2 1
ex = x =1+x+ x + x3 + · · · (−∞, ∞)
k! 2! 3!
k=0
∞
X (−1)k 2k+1 1 3 1 5 1
sin x = x =x− x + x − x7 + · · · (−∞, ∞)
(2k + 1)! 3! 5! 7!
k=0
∞
X (−1)k  π 2k 1 
π 2 1  π 4
= x− =1− x− + x− − ··· (−∞, ∞)
(2k)! 2 2 2 4! 2
k=0
∞
X (−1) k 1 2 1 1
cos x = x2k =1− x + x4 − x6 + · · · (−∞, ∞)
(2k)! 2! 4! 6!
k=0
∞
X (−1)k+1 1 1
ln x = (x − 1)k = (x − 1) − (x − 1)2 + (x − 1)3 − · · · (0, 2]
k 2 3
k=1
∞
−1
X (−1)k 2k+1 1 1 1
tan x = x = x − x3 + x5 − x7 + · · · [−1, 1]
(2k + 1) 3 5 7
k=0

Exercise 7.8.1. Find the Maclaurin series (i.e. Taylor series about c = 0) and its
interval of convergence.
1
(a) f (x) = ln(1 + x) (b) f (x) =
1−x

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7.8. TAYLOR SERIES

Exercise 7.8.2. Find the Taylor series about c and detremine the interval of conver-
gence.

(a) f (x) = ex−1 , c = 1 π

(c) f (x) = cos x, c = −
1 2
(b) f (x) = , c = −1 (d) f (x) = ln x, c = e
x

Exercise 7.8.3. Find Taylor series in the summary table and prove Theorem 7.8.1.

De-Yu Wang CSIE CYUT 135

Chapter 8

PARAMETRIC EQUATIONS
AND POLAR COORDINATES
8.1. PLANE CURVES AND PARAMETRIC EQUATIONS

8.1 Plane Curves and Parametric Equations

Definition 8.1.1.
Given any pair of functions x(t) and y(t) defined on the same domain D, the equations

x = x(t), y = y(t)

are called parametric equations.

The collection of all points (x(t), y(t)) is a plane curve.

Example
√ 8.1.1. Sketch the plane curve defined by the parametric equations x =
t, y = t, for 0 ≤ t ≤ 4.

Solution:
y
t = 0, (x, y) = (0, 0) 3 6 t=4
s: s
(4, 2)
t = 1, (x, y) = (1, 1) 2 s1
s *
 √ 
1 t=√2
t = 2, (x, y) = 2, 2
s
 (2, 2)
 √  -x
t = 3, (x, y) = 3, 3 t=0 1 2 3 4
 √  (0, 0)
t = 4, (x, y) = 4, 4

Example 8.1.2. Find the parametric equation for the line segment from (0, 1) to
(5, 6).

Solution: For a line segment, the parametric equations can be chosen

(
x = a + bt
, t1 ≤ t ≤ t2 .
y = c + dt

For t = t1 = 0, (x, y) = (x(0), y(0)) = (a, c) = (0, 1),

∴ a = 0, c = 1.

For t = t2 = 1, (x, y) = (x(1), y(1)) = (a + b, c + d) = (b, 1 + d) = (5, 6),

∴ b = 5, d = 5.
(
x = 5t
We now have that , 0 ≤ t ≤ 1.
y = 1 + 5t

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8.2. CALCULUS AND PARAMETRIC EQUATIONS

Example 8.1.3. Find the parametric equation for the portion of the parabola y =
x2 + 1 from (1, 2) to (2, 5).
Solution:
( (
x=t x = 3t 1 2
, 1 ≤ t ≤ 2, or , ≤t≤ .
y = t2 + 1 y = 9t2 + 1 3 3

Exercise 8.1.1. Sketch the plane curve defined by the parametric equations
( (
x = −1 + 2t x = 2 cos t
(a) ,1 ≤ t ≤ 2 (c) , 0 ≤ t ≤ 2π
y = 3t + 1 y = 3 sin t
( (
x = −1 + 2t x = −2t2
(b) ,0 ≤ t ≤ 2 (d) , −2 ≤ t ≤ 1
y = t2 − 1 y =2−t

Exercise 8.1.2. Find the parametric equation for the line segment

(a) from (0, 1) to (3, 4) (c) from (2, −5) to (5, 1)

(b) from (−3, 1) to (1, 3) (d) from (4, 1) to (2, −3)
Exercise 8.1.3. Find the parametric equation for the portion of the parabola

(a) y = x2 + 1 from (1, 2) to (3, 10) (c) y = 2 − x2 from (2, −2) to (0, 2)
2
(b) y = 2x2 − 1 from (0, −1) to (2, 7) (d) y = x + 1 from (1, 2) to (−1, 2)

8.2 Calculus and Parametric Equations

Theorem 8.2.1. Parametric form of the derivative
For the parametric equations x(t) and y(t), then
 
dy d dy
dy d2 y dt dx dx
= dt , and = , if 6= 0.
dx dx dx2 dx dt
dt dt
Proof: By the chain rule
dy
dy dy dt dx
= = dt , if 6= 0
dx dt dx dx dt
dt  
d dy
 
d2 y d dy dt dx dx
= = , if 6= 0.
dx2 dx dx dx dt
dt

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8.2. CALCULUS AND PARAMETRIC EQUATIONS

Example 8.2.1. Find the slope of the tangent lines to the curve x = 2 cos t +
π
sin 2t, y = 2 sin t + cos 2t at (a) t = 0, (b) t = and (c) the point (0, −3).
4

dy
dy 2 cos t − 2 sin 2t
Solution: = dt =
dx dx −2 sin t + 2 cos 2t
dt

(a) t = 0

dy 2 cos t − 2 sin 2t 2 cos 0 − 2 sin 0

= = = 1.
dx t=0 −2 sin t + 2 cos 2t t=0 −2 sin 0 + 2 cos 0

π
(b) t =
4
π π √
dy 2 cos t − 2 sin 2t 2 cos − 2 sin 2−2 √
4 2
dx
=
−2 sin t + 2 cos 2t
= π π = −√2 = 2 − 1.
t= π4 t= π4 −2 sin + 2 cos
4 2

(c) The point (0, −3)

( (
x = 2 cos t + sin 2t = 0 sin 2t = −2 cos t
y = 2 sin t + cos 2t = −3 cos 2t = −3 − 2 sin t

sin2 2t + cos2 2t = 1 = (−2 cos t)2 + (−2 sin t − 3)2

1 = 4 cos2 t + 9 + 12 sin t + 4 sin2 t
3π
−1 = sin t, ⇒ t= .
2

 
dy 2 cos t − 2 sin 2t 0
= , By the L’Hôpital rule
dx t= 3π −2 sin t + 2 cos 2t t= 3π 0
2 2
3π
−2 sin t − 4 cos 2t −2 sin − 4 cos 3π
= = 2 =∞
−2 cos t − 4 sin 2t t= 3π
3π
2 −2 cos − 2 sin 3π
2

does not exist.

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8.2. CALCULUS AND PARAMETRIC EQUATIONS

Theorem 8.2.2. Arc length in parametrical equations

For the curve defined parametrically by x = x(t), y = y(t), a ≤ t ≤ b, if x′ and y ′
are continuous on [a, b] and the curve does not intersect itself, then the arc length of
the curve is given by
Z bp
s= [x′ (t)]2 + [y ′ (t)]2 dt.
a

Proof: Divide the t-interval [a, b] into n subinterval of equal length, ∆t:
y
a = t0 < t1 < t2 < · · · < tn = b, 6

b−a t =s a = t0 st = b = t
where ti − ti−1 = ∆t = , n
n
for each i = 1, 2, 3, · · · , n.
-x
The arc length of the subinterval [ti−1 , ti ] is
p
si ≈ [x(ti ) − x(ti−1 )]2 + [y(ti ) − y(ti−1 )]2
s 2  2
x(ti ) − x(ti−1 ) y(ti ) − y(ti−1 )
= + ∆t
∆t ∆t

By the mean value theorem,

p
si ≈ [x′ (ci )]2 + [y ′ (di )]2 ∆t, where ci , di ∈ (ti−1 , ti )
Xn X n
p
s = lim si = lim [x′ (ci )]2 + [y ′ (di )]2 ∆t
n→∞ n→∞
i=1 i=1
Z bp
= [x′ (t)]2 + [y ′ (t)]2 dt.
a

Example 8.2.2. Find the arc length of the curve x = 2 cos t, y = 2 sin t for 0 ≤ t ≤
2π.
Solution:
Z bp y
s= [x′ (t)]2 + [y ′ (t)]2 dt 6
Z a I
2π p
= [−2 sin t]2 + [2 cos t]2 dt rt = 0- x
0
Z 2π √ Z 2π t = 2π
= 4 dt = 2 dt = 4π.
0 0

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8.2. CALCULUS AND PARAMETRIC EQUATIONS

Example 8.2.3. Find the arc length of the curve x = 2 cos t+sin 2t, y = 2 sin t+cos 2t
for 0 ≤ t ≤ 2π.

Solution:
Z bp
s= [x′ (t)]2 + [y ′ (t)]2 dt
a
Z 2π p
= [−2 sin t + 2 cos 2t]2 + [2 cos t − 2 sin 2t]2 dt
Z0 2π
√
= 8 − 8 sin t cos 2t − 8 cos t sin 2t dt
0
Z 2π p Z 2π s  
3t 2 3t 3t 3t
= 8(1 − sin 3t) dt = 8 cos2 + sin − 2 sin cos dt
0 0 2 2 2 2
s  2
Z 2π
3t 3t y
= 8 cos − sin dt
0 2 2 6

t =s 5π t s= π6
Z 2π √
3t 3t
s
= 8 cos − sin dt 6
0 2 2
Z 5π √   t=0
6 3t 3t -x
=3 8 sin − cos dt
π
6
2 2
  5π
√ 2 3t 3t 6

−3 st = 3π
= 3 8· − cos − sin = 16.
3 2 2 π
6
2

Exercise 8.2.1. Sketch the graph and find the slope of the given curves at the
indicated points
(
x = t2 − 2
(a) at (i) t = −1, (ii) t = 1, (iii) (−2, 0)
y = t3 − t
(
x = t3 − t
(b) at (i) t = −1, (ii) t = 1, (iii) (0, 4)
y = t4 − 5t2 + 4
(
x = 2 cos t
(c) at (i) t = π4 , (ii) t = π2 , (iii) (0, 3)
y = 3 sin t
(
x = 2 cos 2t
(d) at (i) t = π4 , (ii) t = π2 , (iii) (−2, 0)
y = 3 sin 2t

De-Yu Wang CSIE CYUT 141

8.3. POLAR COORDINATES

Exercise 8.2.2. Sketch the graph and find the arc length of the given curves.

( (
3
x = t − 4t x = 2 cos t
(a) , −2 ≤ t ≤ 2 (c) , 0 ≤ t ≤ 2π
y = t2 − 3 y = 3 sin t
( (
x = t3 − 4t x = 2 cos 2t
(b) , −2 ≤ t ≤ 2 (d) , 0 ≤ t ≤ 2π
y = t2 − 3t y = 3 sin 7t

8.3 Polar Coordinates

Definition 8.3.1. Polar coordinates P = (r, θ)

r = directed distance from the pole (origin) O to P

θ = directed angle, counterclockwise from polar axis to segment OP

s P = (r, θ)
"
"
"
P "
O"
"
=
r"
"
"
"
s" K θ
"
-
O Polar axis

Theorem 8.3.1. Coordinate conversion y

s (r, θ)
6

r y = r sin θ

s Kθ - x
O x = r cos θ
(
x = r cos θ
(a) , (x, y) = (r cos θ, r sin θ)
y = r sin θ

( p
r 2 = x2 + y 2 , r = ± x2 + y 2
(b) y ,
tan θ = , θ = tan−1 xy
x
θ can be any angle for which tan θ = xy , while − π2 < tan−1 y
x
< π2 , so that

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8.3. POLAR COORDINATES

p y
(r, θ) = (+ x2 + y 2 , tan−1 + 2nπ), for x > 0
x
−1 y
p
= (− x2 + y 2 , tan + π + 2nπ),
x
−1 y
p
(r, θ) = (+ x2 + y 2 , tan + π + 2nπ), for x < 0
x
−1 y
p
= (− x2 + y 2 , tan + 2nπ),
x
where n is an integer.

Example 8.3.1. Find the rectangular coordinate for the polar point (2, −π/3).
Solution:
 
−π −π
(x, y) = (r cos θ, r sin θ) = 2 cos , 2 sin
3 3
√ !
1 − 3 √
= 2· , 2· = (1, − 3).
2 2

Example

8.3.2. Find the
polar coordinate representations for the points (r, θ) =
π 7π
2, 6 and (r, θ) = 2, 6 with r = −2.
Solution:
 π  π   7π

2, = −2, + π + 2nπ = −2, + 2nπ
6 6 6
      y
7π 7π π
2, = −2, + π + 2nπ = −2, + 2nπ 6 π
6 6 6 θ=
'$ s""

6
"
"K" 2, 6
π

" " -x
U"
s" &%
" 1 2
""

" 2, 7π
" 7π
6
θ= 6

Example 8.3.3. Find all polar coordinate representations for the rectangular point
(2, −2).
Solution:
( p √
r2 = x2 + y 2 , r = ± 22 + (−2)2 = ± 8
y .
tan θ = , θ = tan−1 −2
2
= tan−1 (−1) = − π4
x √   √ 
π π
∵ x = 2 > 0 ∴ (r, θ) = 8, − + 2nπ = − 8, − + π + 2nπ .
4 4

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8.4. CALCULUS AND POLAR COORDINATES

Example 8.3.4. Find all polar coordinate representations for the rectangular point
(−2, 2).

Solution:
 p √
r2 = x2 + y 2 , r = ± (−2)2 + 22 = ± 8
.
tan θ = y , 2
θ = tan−1 −2 = tan−1 (−1) = − π4
x √   √ 
π π
∵ x = −2 < 0 ∴ (r, θ) = 8, − + π + 2nπ = − 8, − + 2nπ .
4 4

Exercise 8.3.1. Find the rectangular coordinate for the polar point

(a) (−2, −π/3) (c) (0, 3)

(b) (3, π/5) (d) (−3, 1)

Exercise 8.3.2. Find all polar coordinate representations for the rectangular points
√
(a) (−2, −1) (c) (−2, 3)
(b) (0, 3) (d) (3, −4)

8.4 Calculus and Polar Coordinates

Theorem 8.4.1. Slope in polar form
If f is a differentiable function of θ, then the slope of the tangent line to r = f (θ) at
the point (r, θ) is

dy
dy f (θ) cos θ + f ′ (θ) sin θ dx
= dθ = provided that 6= 0 at (r, θ).
dx dx −f (θ) sin θ + f ′ (θ) cos θ dθ
dθ
Proof: Using the product rule

dy d(r sin θ) d[f (θ) sin θ]

dy
= dθ = dθ = dθ
dx dx d(r cos θ) d[f (θ) cos θ]
dθ dθ dθ
f (θ) cos θ + f ′ (θ) sin θ dx
= ′
if 6= 0.
−f (θ) sin θ + f (θ) cos θ dθ

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8.4. CALCULUS AND POLAR COORDINATES

Example 8.4.1. Find the slope of the tangent line to the polar curve at the point
r = cos 2θ at θ = 0, θ = π4 and the point (−1, π2 ). y

x
−1 1

−1
Solution:

dy
dy f ′ (θ) sin θ + f (θ) cos θ −2 sin(2θ) sin θ + cos(2θ) cos θ
= dθ = ′ =
dx dx f (θ) cos θ − f (θ) sin θ −2 sin(2θ) cos θ − cos(2θ) sin θ
dθ

(a) θ = 0

dy 1
= does not exist.
dx θ=0 0

π
(b) θ =
4

dy −2 sin(2θ) sin θ + cos(2θ) cos θ

=
dx θ= π4 −2 sin(2θ) cos θ − cos(2θ) sin θ θ= π4
√ √
−2 sin( π2 ) sin π4 + cos( π2 ) cos π4 −2 · 1 · 2
2
+0· 2
2
= π π π π =
√ √ = 1.
−2 sin( 2 ) cos 4 − cos( 2 ) sin 4 −2 · 1 · 2
−0· 2
2 2

(c) The point (−1, π2 ), θ = π

2

dy −2 sin(2θ) sin θ + cos(2θ) cos θ

=
dx θ= π2 −2 sin(2θ) cos θ − cos(2θ) sin θ θ= π
π

π

2
−2 sin(π) sin 2 + cos(π) cos 2
=


−2 sin(π) cos π2 − cos(π) sin π2
−2 · 0 · 1 + (−1) · 0
= = 0.
−2 · 0 · 0 − (−1) · 1

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8.4. CALCULUS AND POLAR COORDINATES

(
n petals if n is odd
Note. Rose curves sin nθ, cos nθ :
2n petals if n is even

cos θ cos 2θ cos 3θ cos 4θ cos 5θ

y y y y y

x x x x x

sin θ sin 2θ sin 3θ sin 4θ sin 5θ

y y y y y

x x x x x

Theorem 8.4.2. Area in polar coordinates

If f is continuous and nonnegative on the interval [a, b], 0 < b − a ≤ 2π, then the
area of the region bounded by the graph of r = f (θ) between the radial lines θ = a
and θ = b is given by
Z b
1
A= [f (θ)]2 dθ.
a 2
Proof: Divide the θ-interval [a, b] into n subinterval of equal length, ∆θ:
a = θ0 < θ1 < θ2 < · · · < θn = b,
b−a
where θi − θi−1 = ∆θ = , for each i = 1, 2, 3, · · · , n. The area of the circular
n
sector of radius f (θi ) and central angle ∆θ is
y y
6 6
θ=b θ=b θ = θi





θ = θi−1


. . . Ai

A  θ=a
 θ=a

 
 



 

  r = f (θ)
  r = f (θ)

 - x

  - x

1 1
Ai ≈ r · r∆θ = [f (θi )]2 ∆θ
2 2
n n Z b
X X 1 2 1
A = lim Ai = lim [f (θi )] ∆θ = [f (θ)]2 dθ.
n→∞
i=1
n→∞
i=1
2 a 2

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8.4. CALCULUS AND POLAR COORDINATES

Example 8.4.2. Find the area of one leaf of r = cos 3θ.

π π 5π y
Solution: If r = cos 3θ = 0 then θ = , , , · · · .
6 2 6
1
Z π Z π
2 1 1 2
2
A= (cos 3θ) dθ = (1 + cos 6θ) dθ
π
6
2 4 π6
  π2
1 1 x
= θ + sin 6θ −1 1
4 6 π
6
π
= .
12
−1

Theorem 8.4.3. Arc length in polar coordinates

Let f be a function whose derivaive is continuous on [a, b]. The length of the graph
of r = f (θ) from θ = a to θ = b is
Z bp
s= [f ′ (θ)]2 + [f (θ)]2 dθ.
a

Proof: Divide the θ-interval [a, b] into n subinterval of equal length, ∆θ:
a = θ0 < θ1 < θ2 < · · · < θn = b,
b−a
where θi −θi−1 = ∆θ = , for each i = 1, 2, 3, · · · , n. The arc length from θ = θi−1
n
to θ = θi is
p
si ≈ [x(θi ) − x(θi−1 )]2 + [y(θi ) − y(θi−1 )]2
p
= [f (θi ) cos θi − f (θi−1 ) cos θi−1 ]2 + [f (θi ) sin θi − f (θi−1 ) sin θi−1 ]2
p
= f 2 (θi ) + f 2 (θi−1 ) − 2f (θi )f (θi−1 )[cos θi cos θi−1 + sin θi sin θi−1 ]
p
= f 2 (θi ) + f 2 (θi−1 ) − 2f (θi )f (θi−1 ) cos ∆θ
p
= [f (θi ) − f 2 (θi−1 )]2 + 2f (θi )f (θi−1 )[1 − cos ∆θ]
v " 
u 2 #
u ∆θ
= t[f (θi ) − f 2 (θi−1 )]2 + 2f (θi )f (θi−1 ) 2 sin
2
p
≈ [f (θi ) − f 2 (θi−1 )]2 + [f (θi )]2 (∆θ)2
p
≈ [f ′ (θi )]2 + f 2 (θi )∆θ
X n X n
p
s = lim si = lim [f ′ (θi )]2 + [f (θi )]2 ∆θ
n→∞ n→∞
i=1 i=1
Z bp
= [f (θ)]2 + [f ′ (θ)]2 dθ.
a

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8.4. CALCULUS AND POLAR COORDINATES

Example 8.4.3. Sketch the graph and find the arc length of the curve r = 2−2 cos θ.

Solution:
Z bp
s= [f (θ)]2 + [f ′ (θ)]2 dθ y
a
Z 2π p 2
= (2 − 2 cos θ)2
dθ + (2 sin θ)2 1
0
Z 2π p Z 2π r x
θ −4 −3 −2 −1 1
= 8(1 − cos θ) dθ = 4 sin2 dθ
0 0 2 −1
Z 2π 2π
θ θ −2
=4 sin dθ = −8 cos = 16.
0 2 2 θ=0

Exercise 8.4.1. Sketch the graph and find the slope of the given curves at the
indicated points

(a) r = sin 3θ at θ = π (c) r = 3 sin θ at θ = 0

3
π
(b) r = cos 2θ at θ = 0 (d) r = cos 2θ at θ = 4

Exercise 8.4.2. Sketch the graph and find the area of the indicated region.

(a) One leaf of r = sin 3θ (c) Bounded by r = 2 − 2 cos θ

(b) One leaf of r = sin 2θ (d) BOunded by r = 3 − 3 sin θ

Exercise 8.4.3. Sketch the graph and find the arc length of the given curve.

(a) r = sin 3θ (c) r = 3 − 3 cos θ

(b) r = 2 − 2 sin θ (d) r = 1 + 2 sin 2θ

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Chapter 9

VECTORS
9.1. VECTORS

9.1 Vectors
Definition 9.1.1. Vectors

(a) Notation

Vectors in the Plane: V2 = {hx, yi |x, y ∈ ℜ}

−→
P Q = hx1 − x0 , y1 − y0 i = ha1 , a2 i = a

 s Q(x1 , y1 )
*
  terminal point
−→ 
P Q


s
initial point
P (x0 , y0 )
Vectors in the Space: V3 = {hx, y, zi |x, y, z ∈ ℜ} ]
−→
P Q = hx1 − x0 , y1 − y0 , z1 − z0 i = ha1 , a2 , a3 i = a

s Q(x1 , y1 , z1 )
*

−→  terminal point
P Q


s
initial point
P (x0 , y0 , z0 )

(b) Magnitude:
q
kak = a21 + a22 for a ∈ V2
q
kak = a21 + a22 + a23 for a ∈ V3 .

(c) Addition and subtraction:

a ± b = ha1 , a2 i ± hb1 , b2 i = ha1 ± b1 , a2 ± b2 i for a ∈ V2

a ± b = ha1 , a2 , a3 i ± hb1 , b2 , b3 i
= ha1 ± b1 , a2 ± b2 , a3 ± b3 i for a ∈ V3

(d) Zero vector

0 = h0, 0i for a ∈ V2
0 = h0, 0, 0i for a ∈ V3

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9.1. VECTORS

(e) The negative of a is

−a = h−a1 , −a2 i for a ∈ V2

−a = h−a1 , −a2 , −a3 i for a ∈ V3

(f) Parallel:

b = ca, for any real number c.

Example 9.1.1. For vector a = h5, −2i and b = h2, 1i, compute k5a − 2bk .

Solution:

5a − 2b = 5 h5, −2i − 2 h2, 1i

= h25, −10i − h4, 2i = h21, −12i
p √
k5a − 2bk = k h21, −12i k = 212 + (−12)2 = 585.

Example 9.1.2. Determine whether or not the given pair of vectors is parallel: (a)
a = h2, 3i and b = h4, 5i, (b) a = h2, 3i and b = h−4, −6i.

5
Solution: (a) If b = ca, h4, 5i = h2c, 3ci , then c = 2 and c = . This is a
3
contradiction and so a and b are not parallel.

(b) If b = ca, h−4, −6i = h2c, 3ci , then c = −2. This says that b = −2a and so a
and b are parallel.

Example 9.1.3. Find the distance between the points (1, −3, 5) and (5, 2, −3).

Solution:

d = k h5 − 1, 2 − (−3), −3 − 5i k = k h4, 5, −8i k

p √
= 42 + 52 + (−8)2 = 105.

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9.1. VECTORS

Theorem 9.1.1. Properties of vector operations

Let a, b and c be vectors, and let d and e be scales.

1. a + b = b + a (commutativity)
2. a + (b + c) = (a + b) + c (associativity)
3. a + 0 = a (zero vector)
4. a + (−a) = 0 (additive inverse)
5. d(a + b) = da + db (distributive law)
6. (d + e)a = da + ea (distributive law)
7. (1)a = a (multiplication by 1) and
8. (0)a = 0 (multiplication by 0).

Proof:

1. a + b = ha1 , a2 i + hb1 , b2 i = ha1 + b1 , a2 + b2 i

= hb1 + a1 , b2 + a2 i = b + a

2. a + (b + c) = ha1 , a2 i + (hb1 , b2 i + hc1 , c2 i)

= ha1 , a2 i + hb1 + c1 , b2 + c2 i
= ha1 + b1 + c1 , a2 + b2 + c2 i
= ha1 + b1 , a2 + b2 i + hc1 , c2 i = (a + b) + c

Theorem 9.1.2. (a) Scalar multiplication:

ca = c ha1 , a2 i = hca1 , ca2 i for a ∈ V2

ca = c ha1 , a2 , a3 i = hca1 , ca2 , ca3 i for a ∈ V3

and kcak = |c| kak.

(b) For any nonzero position vector a, a unit vector having the same direction as a
is given by
1
u= a.
kak

p p
Proof: (a) kcak = (ca1 )2 + (ca2 )2 = |c| a21 + a22 = |c| kak
1 1
(b) kuk = a = kak = 1.
kak kak

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9.1. VECTORS

Definition 9.1.2. The standard basis (unit) vectors

i = h1, 0i and j = h0, 1i for a ∈ V2

i = h1, 0, 0i , j = h0, 1, 0i and k = h0, 0, 1i for a ∈ V3

For any vector a, we can write

a = ha1 , a2 i = a1 h1, 0i + a2 h0, 1i

= a1 i + a2 j for a ∈ V2
a = ha1 , a2 , a3 i = a1 h1, 0, 0i + a2 h0, 1, 0i + a3 h0, 0, 1i
= a1 i + a2 j + a3 k for a ∈ V3

Example 9.1.4. Find a vector a with the magnitude 3 and in the same direction as
the vector v = h4, 1, −2i.
Solution:
 
1 1 4 1 −2
u= v = √ h4, 1, −2i = √ , √ , √
kvk 21 21 21 21
   
4 1 −2 12 3 −6
a = 3u = 3 √ , √ , √ = √ ,√ ,√
21 21 21 21 21 21

Exercise 9.1.1. For a = h4, −1, −3i and b = h−1, 3, 1i

(a) Compute 2a − 3b.
(b) Compute k2a − 3bk
(c) Determine if the vectors a and b are parallel.
(d) Find a unit vector in the same direction as a.

Exercise 9.1.2. Repeat Exercise 9.1.1 with

(a) a = h2, 1i , b = h3, 4i (c) a = 2i − 3k, b = 6i + 2j

(b) a = 2i + j − 3k, b = 6i + 2j − k (d) a = h3, 1, 0i , b = h−3, 0, 4i

Exercise 9.1.3. Find a vector with the given magnitude and in the same direction
as the given vector v.

(a) Magnitude 6, vector v = h2, 2, 1i (c) Magnitude 3, vector v = 2i − j + 3k

(b) Magnitude 10, vector v = h3, 0, −4i(d) Magnitude 7, vector v = 3i − 5j + k

Exercise 9.1.4. Prove Theorem 9.1.1

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9.2. THE DOT PRODUCT

9.2 The Dot Product

Definition 9.2.1. The dot product of two vectors
a · b = ha1 , a2 i · hb1 , b2 i = a1 b1 + a2 b2 for a ∈ V2
a · b = ha1 , a2 , a3 i · hb1 , b2 , b3 i = a1 b1 + a2 b2 + a3 b3 for a ∈ V3
Note. The dot product of two vectors is a scalar (i.e., a number, not a vector). For
this reason, the dot product is also called the scalar product.

Example 9.2.1. For vector a = h5, −2, 1i and b = h2, 1, −3i, compute the dot
product a · b.
Solution:
a · b = h5, −2, 1i · h2, 1, −3i = 10 − 2 − 3 = 5.

Theorem 9.2.1. Let a, b and c be vectors, and let d be scale.

1. a · b = b · a (commutativity)
2. a · (b + c) = a · b + a · c (distributive law)
3. (da) · b = d(a · b) = a · (db)
4. 0 · a = 0
5. a · a = kak2 .
Proof:
1. a · b = ha1 , a2 i · hb1 , b2 i = a1 b1 + a2 b2
= b 1 a1 + b 2 a2 = b · a
5. a · a = ha1 , a2 i · ha1 , a2 i
= a21 + a22 = kak2

Theorem 9.2.2. Let θ be the angle between nonzero vectors a and b. Then
a · b = kak kbk cos θ
Proof: (a) If a and b are not parallel, then
ka − bk2 = (kak sin θ)2 + (kbk − kak cos θ)2
= kak2 + kbk2 − 2kak kbk cos θ
i
P
 PPP
PP
PP
kak kak sin θP ka − bk
PP
PP
PP
θ PP
I P
-
kak cos θ kbk − kak cos θ

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9.2. THE DOT PRODUCT

(b) Using the properties of the dot product,

ka − bk2 = (a − b) · (a − b)
=a·a−a·b−b·a+b·b
= kak2 − 2a · b + kbk2
= kak2 + kbk2 − 2kak kbk cos θ
∴ a · b = kak kbk cos θ

Definition 9.2.2. Two vector a and b are orthogonal if a · b = 0.

Example 9.2.2. For a = h4, −1, −3i and b = h−1, 3, 1i

(a) Determine if the vectors a and b are orthogonal.

(b) Compute the angle between the vectors a and b.

Solution: (a) a · b = −4 − 3 − 3 = −10 6= 0, so that a and b are not orthogonal.

(b) From Theorem 9.2.2

a·b −10
cos θ = =√ √
kak kbk 26 11
 
−1 −10
0 ≤ θ = cos √ √ ≤ π.
26 11

Exercise 9.2.1. For a = h3, −2, 1i and b = h−1, 4, 2i

(a) Compute the dot product a · b.

(b) Determine if the vectors a and b are orthogonal.

(c) Compute the angle between the vectors a and b.

Exercise 9.2.2. Repeat Exercise 9.2.1 with

(a) a = h2, 1i , b = h3, 4i (c) a = 2i − 3k, b = 6i + 2j

(b) a = h3, 1, 0i , b = h−3, 0, 4i (d) a = 2i + j − 3k, b = 6i + 2j − k

Exercise 9.2.3. Prove Theorem 9.2.1 and Theorem 9.2.2

De-Yu Wang CSIE CYUT 155

9.3. THE CROSS PRODUCT

9.3 The Cross Product

Definition 9.3.1. The determinate of

(a) 2 × 2

a1 a2
= a1 b 2 − a2 b 1 .
b1 b2

(b) 3 × 3

a1 a2 a3
b b b b b b
b 1 b 2 b 3 = a1 2 3 − a2 1 3 + a3 1 2 .
c2 c3 c1 c3 c1 c2
c1 c2 c3

Definition 9.3.2. For two vectors a = ha1 , a2 , a3 i and b = hb1 , b2 , b3 i in V3 , the cross
(vector) product of a and b is

i j k
a a a a a a
a × b = a1 a2 a3 = 2 3 i − 1 3 j + 1 2 k
b2 b3 b1 b3 b1 b2
b1 b2 b3

Example 9.3.1. For vectors a = h2, 1, 4i , b = h−1, 2, −1i, compute the cross prod-
uct a × b.

Solution:
i j k
1 4 2 4 2 1
a×b= 2 1 4 = i− j+ k = h−9, −2, 5i .
2 −1 −1 −1 −1 2
−1 2 −1

Theorem 9.3.1. Let a, b and c be vectors in V3 , and let d be scale.

1. a × b = −(b × a) (anticommutativity)
2. a × (b + c) = a × b + a × c (distributive law)
(a + b) × c = a × c + b × c
3. (da) × b = d(a × b) = a × (db)
4. a · (b × c)=(a × b) · c (scalar triple product)
=(b × c) · a = b · (c × a)
=(c × a) · b
5. a × a = 0
6. a × 0 = 0 × a = 0

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9.3. THE CROSS PRODUCT

Proof:
i j k
a a a a a a
1. a × b = a1 a2 a3 = 2 3 i − 1 3 j + 1 2 k
b2 b3 b1 b3 b1 b2
b1 b2 b3
b2 b3 b b b b
=− i + 1 3 j − 1 2 k = −(b × a)
a2 a3 a1 a3 a1 a2

a1 a2 a3 a1 a2 a3 c1 c2 c3
4. a · (b × c) = b1 b2 b3 = − c1 c2 c3 = a1 a2 a3
c1 c2 c3 b1 b2 b3 b1 b2 b3
= c · (a × b) = (a × b) · c

Theorem 9.3.2. Let θ be the angle between nonzero vectors a and b in V3 . Then
(a) a × b is orthogonal to both a and b.
(b) ka × bk = kak kbk sin θ.
(c) a and b are parallel if and only if a × b = 0.

Proof for (a):

a1 a2 a3 0 0 0
a · (a × b) = a1 a2 a3 = a1 a2 a3 = 0
b1 b2 b3 b1 b2 b3
b1 b2 b3 0 0 0
b · (a × b) = a1 a2 a3 = a1 a2 a3 = 0
b1 b2 b3 b1 b2 b3

Proof for (b):

ka × bk2 = (a × b) · (a × b)
= [a2 b3 − a3 b2 ]2 + [a1 b3 − a3 b1 ]2 + [a1 b2 − a2 b1 ]2
= (a21 + a22 + a23 )(b21 + b22 + b23 ) − (a1 b1 + a2 b2 + a3 b3 )2
= kak2 kbk2 − (a · b)2
= kak2 kbk2 − kak2 kbk2 cos2 θ
= kak2 kbk2 (1 − cos2 θ)
= kak2 kbk2 sin2 θ

Since sin θ ≥ 0, for 0 ≤ θ ≤ π, so that ka × bk = kak kbk sin θ.

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9.3. THE CROSS PRODUCT

Proof for (c): • If a and b are parallel (θ = 0 or π), then

ka × bk = kak kbk sin θ = kak kbk · 0 = 0.

• If ka × bk = 0 = kak kbk sin θ then sin θ = 0

⇒ θ = 0 or π, (a and b are parallel).

Theorem 9.3.3. The area of the parallelogram with two adjacent edges formed by
the vectors a and b.

A = ka × bk

Proof: 3

kak kak sin θ

A = kbk kak sin θ = ka × bk 
|{z} | {z } Iθ -
base altitude kbk

Example 9.3.2. Find the area of the parallelogram with two adjacent edges formed
by the vectors a = h3, −1, 1i , and b = h−4, 0, 1i .
Solution:
i j k
−1 1 3 1 3 −1
a × b = 3 −1 1 = i− j+ k = h−1, −7, −4i
0 1 −4 1 −4 0
−4 0 1
√
A = ka × bk = k h−1, −7, −4i k = 66.

Theorem 9.3.4. The distance form the point Q, to the line through the points P
and R.
−→ −→
kP Q × P Rk
d= −→ .
kP Rk
Proof:
−→ sQ
−→ −→ kP Rk 
d = kP Qk sin θ = kP Qk sin θ · −→
kP Rk −→
−→ −→ d = kP Qk sin θ
kP Q × P Rk
s Iθ -s
= −→ .
kP Rk
P R

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9.3. THE CROSS PRODUCT

Example 9.3.3. Find the distance form the point Q(1, 2, −1), to the line through
the points P (0, 1, 3) and R(5, −2, 1).
Solution:
−→ −→
P Q = h1, 1, −4i and P R = h5, −3, −2i
−→ −→
P Q × P R = h1, 1, −4i × h5, −3, −2i
i j k
= 1 1 −4 = h−14, −18, −8i
5 −3 −2
−→ −→ √
kP Q × P Rk k h−14, −18, −8i k 584
d= −→ = = √
kP Rk k h5, −3, −2i k 38

Theorem 9.3.5. The volume of the parallelepiped with three adjacent edges formed
by the vectors a, b and c.

V = |c · (a × b)|.

Proof: a×b
 
6 


V =A·h 


= ka × bk · |kck cos θ|
 c



| {z } | {z }




Area of base altitude









|c · (a × b)| h




= ka × bk · b

ka × bk θ
*







  A 
= |c · (a × b)|. -

a

Example 9.3.4. Find the volume of the parallelepiped with three adjacent edges
formed by the vectors a = h−4, −1, 0i , b = h2, 2, −1i , and c = h1, 4, 2i .
Solution:
1 4 2
c · (a × b) = −4 −1 0
2 2 −1
−1 0 −4 0 −4 −1
=1 −4 +2 = −27
2 −1 2 −1 2 2
V = |c · (a × b)| = | − 27| = 27.

De-Yu Wang CSIE CYUT 159

9.4. VECTOR-VALUED FUNCTIONS

Exercise 9.3.1. For a = h3, −2, 1i and b = h−1, 4, 2i

(a) Compute the cross product a × b

(b) Determine if the vectors a and b are parallel.

(c) Find the area of the parallelogram with two adjacent edges formed by the vectors
a and b.

Exercise 9.3.2. Repeat Exercise 9.3.1 with

(a) a = h2, 1, −5i , b = h3, 4, 1i (c) a = 2i − 3k, b = 6i + 2j

(b) a = h3, 1, 0i , b = h−3, 0, 4i (d) a = 2i + j − 3k, b = 6i + 2j − k

Exercise 9.3.3. Find the distance form the point Q, to the line through the points
P and R.
(a) Q = (1, 2, 0), P = (6, −1, 2) and R = (3, −4, 1)

(b) Q = (1, 0, −3), P = (−2, 1, 7) and R = (−3, 1, 1)

(c) Q = (3, 2, 0), P = (5, 1, 2) and R = (4, 1, −1)

(d) Q = (1, 3, 1), P = (−2, 1, 3) and R = (1, 0, −2)

Exercise 9.3.4. Find the volume of the parallelepiped with three adjacent edges
formed by the vectors a, b and c.
(a) a = h−4, −1, 0i , b = h2, 2, −1i , and c = h3, 0, 2i .

(b) a = h1, −1, 2i , b = h1, 2, 3i , and c = h0, 4, 2i .

(c) a = h−3, 2, 1i , b = h3, 2, −5i , and c = h1, 5, −2i .

(d) a = h−2, −1, 5i , b = h−3, 6, −1i , and c = h−1, 4, 2i .

Exercise 9.3.5. Prove Theorem 9.3.1, Theorem 9.3.2, Theorem 9.3.3 and Theorem
9.3.4

9.4 Vector-Valued Functions

Definition 9.4.1. A function of the form

r(t) = f (t)i + g(t)j + h(t)k = hf (t), g(t), h(t)i

is a vector-valued function, where f, g, and h are real-valued functions of the param-

eter t.

De-Yu Wang CSIE CYUT 160

9.4. VECTOR-VALUED FUNCTIONS

Example 9.4.1. Plot the values of the two-dimensional vector-valued function

r(t) = ht + 1, t2 − 2i at t = −2, t = 0 and t = 2.
y
Solution: 3 6
h−1, 2i h3, 2i
r(−2) = hf (−2), g(−2)i = h−1, 2i 2
AK r(2)3

r(0) = hf (0), g(0)i = h1, −2i r(−2)A1A 

r(2) = hf (2), g(2)i = h3, 2i 
A -x
−2 −1 A 1 2 3 4
−1 A r(0)
A
−2 AU
h1, −2i

Example 9.4.2. Sketch a graph of the curve traced out by the endpoint of the
vector-valued function r(t) = ht + 1, t2 − 2i from t = −2 to t = 2.
Solution: The endpoint of all position vector r lie on the curve
y
( 3 6
s 2 s 2i
x = f (t) = t + 1 h−1, 2i
C: , −2 ≤ t ≤ 2 h3,
y = g(t) = t2 − 2 AK r(2)3


y = t2 − 2 = (x − 1)2 − 2. r(−2)A1 

NA  
A -x
−2 −1 A 1 2 3 4
−1 U A r(0)
A
AUs

−2
h1, −2i

Note. If an object moves along the curve C traced out by the endpoint of r, then r′
and r′′ are the velocity and acceleration vectors, respectively.
Definition 9.4.2. Limit of Vector-Valued Functions
For a vector-valued function r(t) = hf (t), g(t), h(t)i, the limit of r(t) as t approaches
a is given by
D E
lim r(t) = lim hf (t), g(t), h(t)i = lim f (t), lim g(t), lim h(t)
t→a t→a t→a t→a t→a

provided all of the indicated limits exist. If any one of the limits fails to exist, then
lim r(t) does not exist.
t→a
D √ E
Example 9.4.3. Find the limit lim t2 , e2t , t2 + 2t .
t→1

Solution:
D √ E D √ E D √ E
lim t2 , e2t , t2 + 2t = lim t2 , lim e2t , lim t2 + 2t = 1, e2 , 3
t→1 t→1 t→1 t→1

De-Yu Wang CSIE CYUT 161

9.4. VECTOR-VALUED FUNCTIONS

Definition 9.4.3. Continuity of Vector-Valued Functions

The vector-valued function r(t) = hf (t), g(t), h(t)i is continuous at t = a whenever

lim r(t) = r(a)

t→a

Theorem 9.4.1. A vector-valued function r(t) = hf (t), g(t), h(t)i is continuous at

t = a if and only if all of f, g and h are continuous at t = a.

Example
 9.4.4.
 Determine all values of t at which the vector-valued function r(t) =
t+1 2
, t , 2t is continuous.
t−1
Solution:

{t|t 6= 1}

Example 9.4.5. Determine all values of t at which the vector-valued function

r(t) = he5t , ln(t + 1), cos ti is continuous.

Solution:

{t|t > −1}

Exercise 9.4.1. Sketch a graph of the curve traced out by the endpoint of the vector-
valued function
π
(a) r(t) = h3t, t2 i , t = 0, t = 1 (c) r(t) = hcos 2t, sin ti , t = 0, t = 2
π
(b) r(t) = ht2 , 2t − 1i , t = 0, t = 1 (d) r(t) = hcos t, sin 2ti , t = 0, t = 2

Exercise 9.4.2. Find the limit if it exists

 
2 2t sin t t+1
(a) lim t − 1, e , sin t (b) lim , cos t,
t→0 t→0 t t−1

Exercise 9.4.3. Determine all values of t at which the given vector-valued function
is continuous.
 
3 (b) r(t) = htan t, cos t2 , 2t + 3i
(a) r(t) = , cos t, 2t
t

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9.5. THE CALCULUS OF VECTOR-VALUED FUNCTIONS

9.5 The Calculus of Vector-Valued Functions

Definition 9.5.1. The derivative r′ (t) of the vector-valued function r(t) is defined
as

r(t + ∆t) − r(t)

r′ (t) = lim ,
∆t→0 ∆t

for any values of t for which the limit exists. When the limit exists for t = a, we say
that r is differentiable at t = a.

Theorem 9.5.1. Let r(t) = hf (t), g(t), h(t)i and suppose that the components f, g
and h are all differentiable for some value of t. Then

r′ (t) = hf ′ (t), g ′ (t), h′ (t)i

Proof:

r(t + ∆t) − r(t)

r′ (t) = lim ,
∆t→0 ∆t
hf (t + ∆t), g(t + ∆t), h(t + ∆t)i − hf (t), g(t), h(t)i
= lim
∆t→0 ∆t
hf (t + ∆t) − f (t), g(t + ∆t) − g(t), h(t + ∆t) − h(t)i
= lim
∆t→0
 ∆t 
f (t + ∆t) − f (t) g(t + ∆t) − g(t) h(t + ∆t) − h(t)
= lim , ,
∆t→0 ∆t ∆t ∆t
 
f (t + ∆t) − f (t) g(t + ∆t) − g(t) h(t + ∆t) − h(t)
= lim , lim , lim
∆t→0 ∆t ∆t→0 ∆t ∆t→0 ∆t
′ ′ ′
= hf (t), g (t), h (t)i

Example 9.5.1. Find the derivative of the vector-valued function r(t) =

h3t3 + 2t, sin(t2 + 3t), e3 t3 i and s(t) = h7tet , cos t3 , te2 i.

Solution:

r′ (t) = 9t2 + 2, (2t + 3) cos(t2 + 3t), 3e3 t2

s′ (t) = 7et + 7tet , −3t2 sin t3 , e2

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9.5. THE CALCULUS OF VECTOR-VALUED FUNCTIONS

Theorem 9.5.2. Suppose that r(t) and s(t) are differentiable vector-valued functions,
f (t) is a differentiable scalar function and c is any scalar constant. Then

d
(a) [r(t) + s(t)] = r′ (t) + s′ (t)
dt
d
(b) [cr(t)] = cr′ (t)
dt
d
(c) [f (t)r(t)] = f ′ (t)r(t) + f (t)r′ (t)
dt
d
(d) [r(t) · s(t)] = r′ (t) · s(t) + r(t) · s′ (t)
dt
d
(e) [r(t) × s(t)] = r′ (t) × s(t) + r(t) × s′ (t)
dt
Proof: Let r(t) = hf1 (t), g1 (t), h1 (t)i and s(t) = hf2 (t), g2 (t), h2 (t)i

d d
(a) [r + s] = [hf1 , g1 , h1 i + hf2 , g2 , h2 i]
dt dt
d
= [hf1 + f2 , g1 + g2 , h1 + h2 i]
dt
= hf1′ + f2′ , g1′ + g2′ , h′1 + h′2 i
= hf1′ , g1′ , h′1 i + hf2′ , g2′ , h′2 i = r′ + s′

d d
(d) [r · s] = [hf1 , g1 , h1 i · hf2 , g2 , h2 i]
dt dt
d
= [f1 f2 + g1 g2 + h1 h2 ]
dt
= f1′ f2 + f1 f2′ + g1′ g2 + g1 g2′ + h′1 h2 + h1 h′2
= [f1′ f2 + g1′ g2 + h′1 h2 ] + [f1 f2′ + g1 g2′ + h1 h′2 ]
= r′ · s + r · s ′

d
Example 9.5.2. For f (t) = 3t3 + 2t, and r(t) = h7tet , cos t3 , te2 i, find [f (t)r(t)].
dt
Solution:
d d
[f (t)r(t)] = [(3t3 + 2t) 7tet , cos t3 , te2 ]
dt dt
=(9t2 + 2) 7tet , cos t3 , te2 +
(3t3 + 2t) 7et + 7tet , −3t2 sin t3 , e2

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9.5. THE CALCULUS OF VECTOR-VALUED FUNCTIONS

Example 9.5.3. For r(t) = hln t, t2 , 3t + 1i, and s(t) = h3t , t2 − 1, te2 i, find
d
[r(t) · s(t)].
dt
Solution:
d
[r(t) · s(t)] =r′ (t) · s(t) + r(t) · s′ (t)
dt
d
= ln t, t2 , 3t + 1 · 3t , t2 − 1, te2
dt
d t 2
+ ln t, t2 , 3t + 1 · 3 , t − 1, te2
  dt
1
= , 2t, 3 · 3t , t2 − 1, te2 + ln t, t2 , 3t + 1 · 3t ln 3, 2t, e2
t
3t
= + 2t3 − 2t + e2 3t + 3t ln t ln 3 + 2t3 + e2 (3t + 1)
t

D E d
t2
Example 9.5.4. For r(t) = 5 , 3t , and s(t) = hsec t, sin ti, find [r(t) × s(t)].
dt
Solution:
d
[r(t) × s(t)] =r′ (t) × s(t) + r(t) × s′ (t)
dt
d D t2 E D 2 E d
= 5 , 3t × hsec t, sin ti + 5t , 3t × hsec t, sin ti
dt
D 2 E D 2 Edt
t t
= 5 2t ln 5, 3 × hsec t, sin ti + 5 , 3t × hsec t tan t, cos ti

Definition 9.5.2. The vector-valued function R(t) is an antiderivative of the vector-

valued function r(t) whenever

R′ (t) = r(t)

Definition 9.5.3. If R(t) is any antiderivative of r(t) = hf (t), g(t), h(t)i, then

(a) the indefinite integral of r(t) is defined to be

Z Z
r(t) dt = hf (t), g(t), h(t)i dt
Z Z Z 
= f (t) dt, g(t) dt, h(t) dt = R(t) + c.

where c is an arbitrary constant vector.

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9.5. THE CALCULUS OF VECTOR-VALUED FUNCTIONS

(b) the definite integral of r(t) is defined to be

Z b Z b
r(t) dt = hf (t), g(t), h(t)i dt
a a
Z b Z b Z b 
= f (t) dt, g(t) dt, h(t) dt
a a a
= R(t)|bt=a = R(b) − R(a)

Z
Example 9.5.5. Evaluate the indefinite integral cos t, 2t2 + 3, e4t dt.

Solution:
Z  
2 4t 2 3 e4t
cos t, 2t + 3, e dt = sin t + c1 , t + 3t + c2 , + c3
3 4
 
2 3 e4t
= sin t, t + 3t, +c
3 4

Z 4  
t2
Example 9.5.6. Evaluate the definite integral sin t, dt.
1 3

Solution:
Z 4   4
2 t3
sin t, t /3 dt = − cos t, = h− cos 4 + cos 1, 7i
1 9 t=1

Exercise 9.5.1. Find the derivative of the vector-valued function

(a) r(t) = h3t3 , sin t2 , te3t i (b) r(t) = h3t3 + 2t, sin(t2 + 3t), e3 t3 i

Exercise 9.5.2. Evaluate the indefinite or definite integral

Z Z 4
(a) 2 4t
cos t, 2t + 3, e dt (c) sin t, t2 /3 dt
1
Z   Z 2  
t2 3t 4
(b) te , 3t sin t, 2 dt (d) , et−2 , tet , dt
t +1 0 t+1

De-Yu Wang CSIE CYUT 166

9.6. ARC LENGTH AND CURVATURE

9.6 Arc Length and Curvature

Theorem 9.6.1. Arc length in the vector-valued function
The arc length of the curve traced out by the endpoint of the vector-valued function
r(t) = hf (t), g(t), h(t)i , a ≤ t ≤ b is given by
Z b
s= kr′ (t)k dt.
a

Proof: The curve described parametrically by

C : x = f (t), y = g(t), z = h(t), a ≤ t ≤ b.

The arc length (see page 140) is

Z bp Z b
s= ′ 2 ′ 2 ′ 2
[f (t)] + [g (t)] + [h (t)] dt = kr′ (t)k dt.
a a

Example 9.6.1. Find the arc length of the curve traced out by the endpoint of the
vector-valued function r = h2t, ln t, t2 i, for 1 ≤ t ≤ e.
Solution:
Z bp
s= [f ′ (t)]2 + [g ′ (t)]2 + [h′ (t)]2 dt
a z
Z e
s  2 6
1
= 22 + + (2t)2 dt
1 t
s
Z er Z e  2
1 1 
= 4 + 2 + 4t dt =2 + 2t dt y
t t :

1 1
 

Z e   e
1 t2  
= + 2t dt = ln |t| + 2 PP
t 2 1 PP
1 PP
Pq
P
= (ln e + e2 ) − (ln 1 + 1) = e2 x

Definition 9.6.1. The curvature of a curve C is

dT
κ= ,
ds
where T is the unit tangent vector.
Note. The curvature of a straight line is zero and
1
the curvature for a circle of radius a > 0 is a constant .
a

De-Yu Wang CSIE CYUT 167

9.6. ARC LENGTH AND CURVATURE
z
6
C

Ti
y




:




PP
  s

PP 1
s 
T
 PP i−1
Pq P
x

Theorem 9.6.2. The curvature of the curve given by the vector-valued function r(t)
is
kT′ (t)k kr′ (t) × r′′ (t)k
κ(t) = = .
kr′ (t)k kr′ (t)k3
r′ (t)
where T(t) = kr′ (t)k
.
ds
p
Proof: Since dt
= [f ′ (t)]2 + [g ′ (t)]2 + [h′ (t)]2 = kr′ (t)k, so that
dT
dT dt kT′ (t)k
κ(t) = = ds
= .
ds dt
kr′ (t)k

Example 9.6.2. Using

D two different
E methods in Theorem 9.6.2 to find the curvature
2 t3
of the curve r(t) = 2t, t , − 3 at the point t = 0.

kT′ (t)k
Solution with Method 1: κ(t) =
kr′ (t)k
r′ (t) = 2, 2t, −t2
√
kr′ (t)k = 4 + 4t2 + t4 = t2 + 2
r′ (t) h2, 2t, −t2 i
T(t) = ′ =
kr (t)k t2 + 2
h0, 2, −2ti · (t2 + 2) − h2, 2t, −t2 i · 2t
T′ (t) =
(t2 + 2)2
h−4t, 4 − 2t2 , −4ti
=
(t2 + 2)2
√
16t2 + 16 − 16t2 + 4t4 + 16t2 2
kT′ (t)k = 2 2
= 2
(t + 2) t +2
′
kT (t)k 2 1
κ(t) = ′ = 2 2
, κ(0) = .
kr (t)k (t + 2) 2

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9.6. ARC LENGTH AND CURVATURE

kr′ (t) × r′′ (t)k

Solution with Method 2: κ(t) =
kr′ (t)k3

r′ (t) = 2, 2t, −t2

√
kr′ (t)k = 4 + 4t2 + t4 = t2 + 2
r′′ (t) = h0, 2, −2ti
i j k
r′ (t) × r′′ (t) = 2 2t −t2 = −4t2 + 2t2 , 4t, 4 = −2t2 , 4t, 4
0 2 −2t
√
kr′ (t) × r′′ (t)k = 4t4 + 16t2 + 16 = 2(t2 + 2)
kr′ (t) × r′′ (t)k 2(t2 + 2) 2 1
κ(t) = ′ 3
= 2 3
= 2 2
, κ(0) = .
kr (t)k (t + 2) (t + 2) 2

Exercise 9.6.1. Find the arc length of the curve traced out by the of the vector-
valued function

(a) r(t) = hcos t, sin t, 2ti , 0 ≤ t ≤ 2π

√
(b) r(t) = 8t, t2 , ln(t2 ) , 1 ≤ t ≤ 2

(c) r(t) = ht2 + 1, 2t, t2 − 1i , 0 ≤ t ≤ 2

(d) r(t) = ht, t2 − 1, t3 i , 0 ≤ t ≤ 2

Exercise 9.6.2. Using two different methods to find the curvature of the curve at
the given point.

(a) r(t) = he−2t , 2t, 4i , t = 0 (c) r(t) = ht, t, t2 − 1i , t = 2

(b) r(t) = h2, sin πt, ln ti , t = 1 (d) r(t) = ht, t2 + t − 1, ti , t = 0

De-Yu Wang CSIE CYUT 169

Chapter 10

PARTIAL DIFFERENTIATION
10.1. FUNCTIONS OF SEVERAL VARIABLES

10.1 Functions of Several Variables

Definition 10.1.1. A function f is a rule that assign a unique real number

z = f (x1 , x2 , · · · , xn ),

to each n-tuples of real numbers (x1 , x2 , · · · , xn ) in D. The set D = {(x1 , x2 , · · · , xn )}

is the domain of f , and the set R = {z} is the range of f .
Note. x1 , x2 , · · · and xn are the independent variables and z is the dependent variable.

Example
√ 10.1.1. Find the domain and range of the function z = f (x, y) =
x + 3y − 2 and evaluate the value of f at the point (2, 1).

Solution: (a) The domain

D = {(x, y)|x + 3y − 2 ≥ 0}.

(b) The range

R = {z|z ≥ 0}
√ √ √
(c) f (2, 1) = x + 3y − 2 = 2+3·1−2= 3.

Example 10.1.2. Find the domain and range of the function z = f (x, y) = cos(x2 +
y 2 ) and evaluate the value of f at the point (−2, 1).

Solution: (a) The domain

D = {(x, y)|all x, y ∈ ℜ}.

(b) The range

R = {z| − 1 ≤ z ≤ 1}

(c) f (−2, 1) = cos(x2 + y 2 ) = cos[(−2)2 + 12 ] = cos(5).

Exercise 10.1.1. Find the domain and range of the function f and evaluate the
value of f at the given point.

(a) f (x, y) = 1
, (3, 1) (c) f (x, y) = sin2 x + cos2 y, (π, 0)
x+y
eyz
(b) f (x, y) = ln(2 + x + y), (−3, 7) (d) f (x, y, z) = z−x2 −y 2
, (2, −3, 1)

De-Yu Wang CSIE CYUT 171

10.2. LIMITS AND CONTINUITY

10.2 Limits and Continuity

Definition 10.2.1. Formal definition of limit
Let f be defined on an open disk centered at (a, b), except possibly at (a, b). We say
that

lim f (x, y) = L,
(x,y)→(a,b)

if for every ε > 0 there exists a δ > 0 such that

p
|f (x, y) − L| < ε whenever 0 < (x − a)2 + (y − b)2 < δ.

Remark. If f (x, y) approaches L1 as (x, y) approaches (a, b) along a path P1 and

f (x, y) approaches L2 6= L1 as (x, y) approaches (a, b) along a path P2 , then
lim f (x, y) does not exist.
(x,y)→(a,b)

y z y
6
'$
6 6
P5'$
P1
C s
f P7
δ Cs (x, y) s P
@
Rsf (x, y)
L+ε @?
R
P3 -

&% &%
4
(a, b) 6 I
@
@
P8 P6
L P2
-x -x
L−ε

ex+y−z
Example 10.2.1. Compute the limit lim .
(x,y,z)→(1,1,2) x − z

Solution:
ex+y−z e0
lim = = −1.
(x,y,z)→(1,1,2) x − z −1

y
Example 10.2.2. Compute the limit lim .
(x,y)→(1,0) x + y − 1

Solution:
y
P1 : x = 1 lim = lim 1 = 1
(1,y)→(1,0) 1 + y − 1 y→0
0
P2 : y = 0 lim = lim 0 = 0 6= 1
(x,0)→(1,0) x + 0 − 1 x→1
y
∴ lim does not exist.
(x,y)→(1,0) x + y − 1

De-Yu Wang CSIE CYUT 172

10.2. LIMITS AND CONTINUITY

xy 2
Example 10.2.3. Evaluate lim .
(x,y)→(0,0) x2 + y 4

Solution:
0
P1 : x = 0 lim = lim 0 = 0
(0,y)→(0,0) 0 + y 4 y→0
0
P2 : y=0 lim = lim 0 = 0
(x,0)→(0,0) x2 + 0 x→0

x3 x
P3 : y=x lim 2 4
= lim =0
(x,x)→(0,0) x + x x→0 1 + x2
y 2 (y 2 ) y4 1
P4 : x = y2 lim 2 2 4
= lim 4
= 6= 0
(y ,y)→(0,0) (y ) + y
2 y→0 2y 2
2
xy
∴ lim does not exist.
(x,y)→(0,0) x + y 4
2

Theorem 10.2.1. Suppose that |f (x, y) − L| ≤ g(x, y) for all (x, y) in the interior of
some circle centered at (a, b), except possibly at (a, b). If lim g(x, y) = 0, then
(x,y)→(a,b)

lim f (x, y) = L.
(x,y)→(a,b)

Proof: If lim g(x, y) = 0, given any ε > 0 there exists a δ > 0 such that
(x,y)→(a,b)
p
|g(x, y) − 0| = |g(x, y)| < ε whenever 0 < (x − a)2 + (y − b)2 < δ
|f (x, y) − L| ≤ g(x, y) ≤ |g(x, y)| < ε
p
|f (x, y) − L| < ε whenever 0 < (x − a)2 + (y − b)2 < δ
∴ lim f (x, y) = L.
(x,y)→(a,b)

xy 2
Example 10.2.4. Show that the limit lim exist.
(x,y)→(0,0) x2 + y 2

Proof:
xy 2 0
lim 2 2
= lim = 0,
(x,0)→(0,0) x + y (x,0)→(0,0) x2

xy 2 0
lim 2 2
= lim =0
(0,y)→(0,0) x + y (0,y)→(0,0) y 2

xy 2 xy 2
∵ |f (x, y) − L| = 2 −0 ≤ = |x| and lim |x| = 0
x + y2 y2 (x,y)→(0,0)

xy 2
∴ lim =0
(x,y)→(0,0) x2 + y 2

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10.2. LIMITS AND CONTINUITY

Definition 10.2.2. Suppose that f (x, y) is defined in the interior of a circle centered
at (a, b). We say that f is continuous at (a, b) if lim f (x, y) = f (a, b).
(x,y)→(a,b)

Example 10.2.5. Determine all points at which the function f (x, y) = ln(x2 +y 2 −5)
is continuous.

Solution:

{(x, y)|x2 + y 2 > 5}

Exercise 10.2.1. Compute the indicated limit

ex+y−z 4xz
(a) lim . (c) lim .
(x,y,z)→(1,0,2) y 2
+ z2
(x,y,z)→(1,1,2) x − z
cos xy exy
(b) lim (d) lim
(x,y)→(π,2) y 2 + 1 (x,y)→(−3,0) x2 + y 2

Exercise 10.2.2. Show that the indicated limit exist.

xy 2 3x3
(a) lim (c) lim
(x,y)→(0,0) x2 + y 2 (x,y,z)→(0,0,0) x2 + y 2 + z 2

x2 y x2 y 2 z 2
(b) lim (d) lim
(x,y)→(0,0) x2 + y 2 (x,y,z)→(0,0,0) x2 + y 2 + z 2

Exercise 10.2.3. Show that the indicated limit does not exist.

3x2 x2 yz
(a) lim (c) lim
(x,y)→(0,0) x2 + y 2 (x,y,z)→(0,0,0) x4 + y 4 + z 4

2x2 y xyz
(b) lim (d) lim
(x,y,z)→(0,0,0) x + y 3 + z 3
3
(x,y)→(0,0) x4 + y 2

Exercise 10.2.4. Determine all points at which the function is continuous.

p
(a) f (x, y) = x2 − y 2 − 5 (d) f (x, y) = ln(5 − x2 + y)
p
(b) f (x, y) = ln(x2 + y 2 − 5) (e) f (x, y) = 9 − x2 − y 2
(c) f (x, y) = 4xy + sin 3x2 y (f) f (x, y) = tan(x + y)

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10.3. PARTIAL DERIVATIVES

10.3 Partial Derivatives

Definition 10.3.1. The partial derivative of f (x, y) with respect to x and y are
defined as
∂f f (x + h, y) − f (x, y)
fx (x, y) = (x, y) = lim ,
∂x h→0 h
∂f f (x, y + h) − f (x, y)
fy (x, y) = (x, y) = lim ,
∂y h→0 h
provided the limits exist.
∂f
Example 10.3.1. For f (x, y) = 3x2 + x3 y + 4y 2 , compute ∂x
(x, y), ∂f
∂y
(x, y), fx (1, 0)
and fy (2, −1).
Solution:
∂f
= 6x + 3x2 y + 0 = 6x + 3x2 y
∂x
∂f
= 0 + x3 · 1 + 8y = x3 + 8y
∂y
fx (1, 0) = 6 + 0 = 6
fy (2, −1) = 8 − 8 = 0

Guidelines 10.3.1. Higher order partial derivatives

 
∂ ∂f ∂ 2f
= =fxx
∂x ∂x ∂x2
 
∂ ∂f ∂ 2f
= =fyy
∂y ∂y ∂y 2
 
∂ ∂f ∂ 2f
= =fxy
∂y ∂x ∂x∂y
 
∂ ∂f ∂ 2f
= =fyx = fxy
∂x ∂y ∂y∂x
Example 10.3.2. Find all second-order partial derivatives of f (x, y) = x2 y−y 3 +ln x.
Solution:
1
fx = 2xy +
x
1
fxx = 2y − 2
x
fxy = 2x
fy = x2 − 3y 2
fyx = 2x = fxy
fyy = −6y

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10.4. THE CHAIN RULE

Example 10.3.3. Find the partial derivative fyxy of f (x, y) = 2 sec−1 (x3 ) − 4xy 2 +
x ln 3y.

Solution:
x
fy = −8xy +
y
1
fyx = −8y +
y
1
fyxy = −8 − 2
y

Exercise 10.3.1. Evaluate fx and fy at the indicated point.

(a) f (x, y) = cos−1 xy, (1, 0) (c) f (x, y) = tan−1 xy , (1, −2)
xy
(b) f (x, y) = 6xy
, (1, 1) (d) f (x, y) = x−y
, (2, −2)
4x2 +5y 2

Exercise 10.3.2. Find all second-order partial derivatives

(a) f (x, y) = x3 − 4xy 2 + y 4 (c) f (x, y, z) = √ 2

x2 +y 2 +z 2
(b) f (x, y) = x3 sin xy 2 − 3y 2

Exercise 10.3.3. Find the partial indicated derivative

(a) fyxy ; f (x, y) = 2 sec−1 (x3 ) − 4xy 2 + x ln 3y

(b) fyxy ; f (x, y) = 2 cos(y 3 ) − 3x3 y 2 + ln 3xy

z2
(c) fxx , fyy , fyyzz ; f (x, y, z) = e2xy − y
+ xz sin y
√
(d) fxx , fyy , fwxyz ; f (w, x, y, z) = wyz − x3 sin w

10.4 The Chain Rule

Definition 10.4.1. For z = f (x, y), the increment of z is given by

∆z = f (x + ∆x, y + ∆y) − f (x, y)

where ∆x and ∆y are the increments of x and y.

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10.4. THE CHAIN RULE

Theorem 10.4.1. The differential of z is

dz = fx (x, y) dx + fy (x, y) dy.
Proof:
∆z =f (x + ∆x, y + ∆y) − f (x, y)
=f (x + ∆x, y + ∆y) − f (x, y + ∆y) + f (x, y + ∆y) − f (x, y)
f (x + ∆x, y + ∆y) − f (x, y + ∆y) f (x, y + ∆y) − f (x, y)
= ∆x + ∆y
∆x ∆y
dz = lim ∆z
∆x→0
∆y→0
f (x + ∆x, y + ∆y) − f (x, y + ∆y) f (x, y + ∆y) − f (x, y)
= lim ∆x + lim ∆y
∆x→0 ∆x ∆x→0 ∆y
∆y→0 ∆y→0

=fx (x, y)dx + fy (x, y)dy.

Example 10.4.1. Find the total differential of f (x, y) = yex + 2 sin x

Solution:
dz = fx (x, y) dx + fy (x, y) dy
= (yex + 2 cos x) dx + (ex ) dy

Theorem 10.4.2. Chain Rule

If z = f (x(t), y(t)), where x(t) and y(t) are differentiable and f (x, y) is a differentiable
function of x and y, then
s
z = f (x, y)
dz ∂f dx ∂f dy
= + . JJ
dt ∂x dt ∂y dt
∂f
J ∂f
∂x
J ∂y
J
J
x y

dx dy
dt dt
t t

Proof:
dz =fx (x, y)dx + fy (x, y)dy
dz dx dy
=fx (x, y) + fy (x, y) .
dt dt dt

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10.4. THE CHAIN RULE

Example 10.4.2. Use the chain rule to find the derivative g ′ (t) where
g(t) = f (x(t), y(t)), f (x, y) = x2 y + sin y, x(t) = t3 + 1, y(t) = e2t .
Solution:
∂f dx ∂f dy
g ′ (t) = +
∂x dt ∂y dt
= 2xy · 3t + (x2 + cos y) · 2e2t
2
 
= 2(t3 + 1)e2t · 3t2 + (t3 + 1)2 + cos(e2t ) · 2e2t

Theorem 10.4.3. Chain Rule

If z = f (x(s, t), y(s, t)), where f (x, y) is a differentiable function of x and y and
where x = x(s, t) and y = y(s, t) both have first-order partial derivatives. Then

s
∂z ∂f ∂x ∂f ∂y z = f (x, y)
= +
∂s ∂x ∂s ∂y ∂s JJ
∂z ∂f ∂x ∂f ∂y ∂f
J ∂f
= + J
∂t ∂x ∂t ∂y ∂t ∂x ∂y
J
J
xs ys

A
A
∂x
A ∂x ∂y
A ∂y
Comments:
A ∂t
A ∂t
∂s ∂s

A
A
dz =fx (x, y)dx + fy (x, y)dy
A
A
dz dx dy s t s t
=fx (x, y) + fy (x, y)
dt dt dt
∂z ∂x ∂y
=fx (x, y) + fy (x, y)
∂t ∂t ∂t
∂z ∂x ∂y
=fx (x, y) + fy (x, y) .
∂s ∂s ∂s

∂g ∂g
Example 10.4.3. Use the chain rule to find the derivative ∂u and ∂v where
g(u, v) = f (x(u, v), y(u, v)), f (x, y) = 4x2 y 3 , x(u, v) = u3 − v sin u, y(u, v) = 4u2 .
Solution:
∂g ∂f ∂x ∂f ∂y
= +
∂u ∂x ∂u ∂y ∂u
= (8xy 3 )(3u2 − v cos u) + (12x2 y 2 )(8u)
= [8(u3 − v sin u)(4u2 )3 ](3u2 − v cos u) + [12(u3 − v sin u)2 (4u2 )2 ](8u)
∂g ∂f ∂x ∂f ∂y
= +
∂v ∂x ∂v ∂y ∂v
= (8xy )(− sin u) + (12x2 y 2 )(0)
3

= − sin u[8(u3 − v sin u)(4u2 )3 ]

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10.4. THE CHAIN RULE

Theorem 10.4.4. Implicit differentiation

Suppose that the equation F (x, y, z) = 0 implicitly defines a function of z = f (x, y),
then
∂z Fx ∂z Fy
=− , =− .
∂x Fz ∂y Fz
Proof: Let w = F (x, y, z) = 0 and by the chain rule,

∂w ∂x ∂y ∂z ∂w ∂x ∂y ∂z
= Fx + Fy + Fz = Fx + Fy + Fz
∂x ∂x ∂x ∂x ∂y ∂y ∂y ∂y
∂z ∂z
0 = Fx + Fz 0 = Fy + Fz
∂x ∂y
∂z Fx ∂z Fy
=− for Fz 6= 0; =− for Fz 6= 0;
∂x Fz ∂y Fz

Example 10.4.4. For F (x, y, z) = xy 2 +z 3 +sin(xyz) = 0, use implicit differentiation

∂z ∂z
to find ∂x and ∂y .
Solution:
∂z Fx y 2 + yz cos(xyz)
=− =− 2 ;
∂x Fz 3z + xy cos(xyz)
∂z Fy 2xy + xz cos(xyz)
=− =− 2 .
∂y Fz 3z + xy cos(xyz)

Exercise 10.4.1. Find the total differential of f (x, y)

(a) f (x, y) = yex + sin x (c) f (x, y) = x2 y 2 + 3y − 4x,

√ (d) f (x, y) = x+3
(b) f (x, y) = x + y y

Exercise 10.4.2. Use the chain rule to find the indicated derivative(s)
2
(a) g ′ (t) where g(t) = f (x(t), y(t)), f (x, y) = ex y + sin y, x(t) = t3 + 1, y(t) = e2t
(b) g ′ (t) where g(t) = f (x(t), y(t)), f (x, y) = ln xy 2 +sin(xy), x(t) = 2t3 +t, y(t) =
e2t
∂g ∂g
(c) ∂u
and ∂v where g(u, v) = f (x(u, v), y(u, v)), f (x, y) = 4x2 y 3 , x(u, v) = u3 −
v sin u, y(u, v) = 4u2 .
∂g ∂g
(d) and ∂v where g(u, v) = f (x(u, v), y(u, v)), f (x, y) = xy 3 − 4x2 , x(u, v) =
∂u
u2
√
e , y(u, v) = v 2 + 1 sin u
∂z ∂z
Exercise 10.4.3. Use implicit differentiation to find ∂x
and ∂y
.
(a) F (x, y, z) = xyz − 4y 2 z 2 + cos(xy) = 0
(b) F (x, y, z) = 3y 2 z − e4x cos(3z) − 3y 2 = 0.

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10.5. THE GRADIENT AND DIRECTIONAL DERIVATIVES

10.5 The Gradient and Directional Derivatives

Definition 10.5.1. The directional derivative of f (x, y) at the point (a, b) and in the
direction of the unit vector u = hu1 , u2 i is given by
f (a + hu1 , b + hu2 ) − f (a, b)
Du f (a, b) = lim ,
h→0 h
provided the limit exists.

Theorem 10.5.1. Suppose that f is differentiable at (a, b) and u = hu1 , u2 i is any

unit vector. Then,
Du f (a, b) = fx (a, b)u1 + fy (a, b)u2 .
Proof: Let g(h) = f (a + hu1 , b + hu2 ) = f (x, y), where x = a + hu1 and y = b + hu2 .
Then g(0) = f (a, b) and so
∂g ∂f dx ∂f dy
g ′ (h) = = + = fx (x, y)u1 + fy (x, y)u2
∂h ∂x dh ∂y dh
g ′ (0) = fx (a, b)u1 + fy (a, b)u2
f (a + hu1 , b + hu2 ) − f (a, b) g(h) − g(0)
Du f (a, b) = lim = lim
h→0 h h→0 h
′
= g (0) = fx (a, b)u1 + fy (a, b)u2 .

Example
D √ 10.5.1.
E For f (x, y) = x2 y − 4y 3 , compute Du f (2, 1) for the directions (a)
u = 12 , 23 and (b) u in the direction from (2, 1) to (4, 0).
Solution:
fx (x, y) = 2xy,
fy (x, y) = x2 − 12y 2
D √ E
(a) For u = 12 , 23

Du f (a, b) = fx (2, 1)u1 + fy (2, 1)u2

√
1 3 √
=4· −8· = 2 − 4 3.
2 2
(b) For u in the direction from (2, 1) to (4, 0).

 
h4 − 2, 0 − 1i h2, −1i 2 1
u= = = √ , −√
k h4 − 2, 0 − 1i k k h2, −1i k 5 5
Du f (a, b) = fx (2, 1)u1 + fy (2, 1)u2
 
2 1 16
= 4 · √ − 8 · −√ =√ .
5 5 5

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10.5. THE GRADIENT AND DIRECTIONAL DERIVATIVES

Definition 10.5.2. The gradient of f (x, y) is the vector-valued function

 
∂f ∂f ∂f ∂f
∇f (x, y) = , = i+ j
∂x ∂y ∂x ∂y

provided both partial derivatives exist.

Example 10.5.2. Find the gradient of f (x, y) = x2 + 4xy 2 − y 5 .

Solution:

▽f (x, y) = 2x + 4y 2 , 8xy − 5y 4

Theorem 10.5.2. If f is differentiable function of x and y and u is any unit vector,

then

Du f (x, y) = ∇f (x, y) · u.

Proof:

Du f (x, y) =fx (x, y)u1 + fy (x, y)u2

 
∂f ∂f
= , · hu1 , u2 i
∂x ∂y
=∇f (x, y) · u.

Example 10.5.3. Using the gradient of f (x, y) = x2 y−4y 2 to compute the directional
derivative of f (x, y) at the point (2, −1) in the direction of the vector h1, 2i.

Solution:

∇f (x, y) = 2xy, x2 − 8y
∇f (2, −1) = h−4, 12i
 
h1, 2i 1 2
u= √ = √ , √
5 5 5
 
1 2 20
Du f (2, −1) = h−4, 12i · √ , √ =√
5 5 5

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10.6. EXTREMA OF FUNCTIONS OF SEVERAL VARIABLES

Exercise 10.5.1. Compute the directional derivative of f at the given point in the
direction of the indicated vector. D √ E
(a) f (x, y) = x2 y + 4y 2 , (2, 1), u = 21 , 23
D E
3 2 1 √1
(b) f (x, y) = x y − 4y , (2, −1), u = √
2
, 2
p
(c) f (x, y) = x2 + y 2 , (3, −4), u in the direction of h3, −2i
(d) f (x, y) = cos(2x − y), (π, 0), u in the direction from (π, 0) to (2π, π)
Exercise 10.5.2. Find the gradient of the given function at the indicated point.

(a) f (x, y) = x3 e3y − y 4 , (2, −1) (c) f (x, y) = e3y/x − x2 y 4 , (1, 2)

2
(b) f (x, y) = sin 3xy + y 2 , (π, 1) (d) f (x, y) = xexy + cos y 2 , (1, −1)
Exercise 10.5.3. Repeat Exercise 10.5.1 using Theorem 10.5.2.

10.6 Extrema of Functions of Several Variables

Definition 10.6.1. Absolute extremum (maximum or minimum)
Let f be a function with domain D. Then
(a) f (a, b) is an absolute maximum of f if
f (x, y) ≤ f (a, b) for all (x, y) in D
(b) f (a, b) is an absolute minimum of f if
f (x, y) ≥ f (a, b) for all (x, y) in D

Definition 10.6.2. Local extremum

(a) f (a, b) is a local maximum of f if there is an open disk R centered at (a, b), for
which f (a, b) ≥ f (x, y) for all (x, y) ∈ R.
(b) f (c) is a local minimum of f if there is an open disk R centered at (a, b), for
which f (a, b) ≤ f (x, y) for all (x, y) ∈ R.
z
6 absolute maximum
s9



local maximum
Ps
PP
q

s
O XX
 X XX
absolute
 XXminimum
s9
   XXz
6   Xy


local minimum
+

x

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10.6. EXTREMA OF FUNCTIONS OF SEVERAL VARIABLES

Definition 10.6.3. The point (a, b) is a critical number of the function f (x, y) if
(a, b) is in the domain of f and one of the following is true.

(a) fx (a, b) = fy (a, b) = 0

(b) fx (a, b) or fy (a, b) does not exist.

Theorem 10.6.1. If f (x, y) has a local extremum at (a, b), then (a, b) must be a
critical number of f .

Proof: Suppose that f has a local extremum at (a, b).

(a) Holding y = b, then g(x) = f (x, b) has a local extremum at x = a. By the

Fermat’s theorem (page 44) either g ′ (a) = fx (a, b) = 0 or g ′ (a) does not exist.

(b) Holding x = a, then h(y) = f (a, y) has a local extremum at y = b. It follows

that h′ (b) = fy (a, b) = 0 or h′ (b) does not exist.

fx (a, b) = g ′ (a) = 0 or does not exist
⇒ (a, b) must be a critical number of f .
fy (a, b) = h′ (b) = 0 or does not exist

x2 y3
Example 10.6.1. Find all critical numbers of f (x, y) = xe− 2 − 3 +y .

Solution:
x2 y3 x2 y3 x2 y3
fx (x, y) = e− 2 − 3 +y + x(−x)e− 2 − 3 +y = (1 − x2 )e− 2 − 3 +y = 0
⇒ x = ±1
x2 y3
fy (x, y) = x(−y 2 + 1)e− 2 − 3 +y = 0
⇒ x = 0 or y = ±1.

The critical numbers are (1, 1), (−1, 1), (1, −1) and (−1, −1).

Theorem 10.6.2. Second Derivatives Test

Suppose that f (x, y) has continuous second-order partial derivatives in some open disk
containing the point (a, b) and that fx (a, b) = fy (a, b) = 0. Define the discriminant

fxx (a, b) fxy (a, b)

D(a, b) = fxx (a, b)fyy (a, b) − [fxy (a, b)]2 = .
fyx (a, b) fyy (a, b)

(a) If D(a, b) > 0 and fxx (a, b) > 0, then f has a local minimum at (a, b).

(b) If D(a, b) > 0 and fxx (a, b) < 0, then f has a local maximum at (a, b).

(c) If D(a, b) < 0, then f has a saddle point at (a, b).

(d) If D(a, b) = 0, then no conclusion.

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 10.6. EXTREMA OF FUNCTIONS OF SEVERAL VARIABLES

Comments:

(
fxx (a, b) > 0 and fyy (a, b) > 0 a local minimum at (a, b)
D(a, b) > 0
fxx (a, b) < 0 and fyy (a, b) < 0 a local maximum at (a, b)
(
fxx (a, b) > 0 and fyy (a, b) < 0 a saddle point at (a, b)
D(a, b) < 0
fxx (a, b) < 0 and fyy (a, b) > 0 a saddle point at (a, b)

z
6
4 saddle point
3
2

1
O
1 XX1XX
2 3
XXX4 5
2 
XX z
Xy
3 
4 

x+


Example 10.6.2. Locate and classify all critical numbers for f (x, y) = 2x2 −y 3 −2xy.

Solution:

(a) The critical numbers

fx = 4x − 2y = 0, ⇒ y = 2x
2
fy = −3y − 2x = 0
= −12x2 − 2x = −2x(6x + 1) = 0
1
⇒ x = 0 or x = −
6  
1 1
f have two critical numbers(0, 0) and − ,− .
6 3

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 10.6. EXTREMA OF FUNCTIONS OF SEVERAL VARIABLES

(b) Second derivatives test:

fxx = 4, fyy = −6y, fxy = −2

D(0, 0) = fxx (0, 0)fyy (0, 0) − [fxy (0, 0)]2
= 4 · 0 − (−2)2 = −4 < 0,
∴ (0, 0) is a saddle point.
        2
1 1 1 1 1 1 1 1
D − ,− = fxx − , − fyy − , − − fxy − , −
6 3 6 3 6 3 6 3
= 4 · 2 − (−2)2 = 4 > 0
 
1 1
and fxx − , − = 4 > 0,
6 3
 
1 1
∴ − ,− is a local minimum.
6 3

Exercise 10.6.1. Locate and classify (use second derivatives test) all critical numbers

(a) f (x, y) = e−x

2 −y 2
(c) f (x, y) = x3 − 3xy + y 3
2
(b) f (x, y) = e−x (y 2 + 1) (d) f (x, y) = y 2 + x2 y + x2 − 2y

De-Yu Wang CSIE CYUT 185

Chapter 11

MULTIPLE INTEGRALS
11.1. DOUBLE INTEGRALS

11.1 Double Integrals

Definition 11.1.1. Double integrals over rectangles
For any function f (x, y) defined on a rectangle R = {(x, y)|a ≤ x ≤ b and c ≤ y ≤ d},
we define the double integral of f over R by
ZZ Xn
f (x, y) dA = lim f (ui , vi )∆Ai ,
R k∆k→0
i=1

provided the limit exists and is the same for every choice of the evaluation points
(ui , vi ) in Ri , for i = 1, 2, · · · , n. When this happens, we say that f is integrable over
R. z
6
y
z = f (x, y) 6
d

Ri
O XX c
 XX
a X XX d c
 XXXX XXzXy
   X
b   
 
XX R  -x
 XXX
+
 O a b
x ZZ ZZ
Note. Area of R: A = 1 dA and Volume V = f (x, y) dA.
R R

Theorem 11.1.1. Fubini’s Theorem (First Form)

Suppose that f is integrable over the rectangle R = {(x, y)|a ≤ x ≤ b and c ≤ y ≤ d}.
Then,
ZZ Z bZ d Z dZ b
f (x, y) dA = f (x, y) dy dx = f (x, y) dx dy
R a c c a

Proof for dA = dydx: For each fixed value of x, the area of the cross section is the
area under the curve z = f (x, y) for c ≤ y ≤ d, which is given by
Z d
z
A(x) = f (x, y) dy 6
ZcZ
V = f (x, y) dA z = f (x, y)
R
Z b
= A(x) dx
a
Z bZ d O XX c

= f (x, y) dy dx. a  A(x) XXXXX d
 XXz
a
| c {z   XXXXX Xy

} b
  
A(x) XX R 
 XXX
+

x

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11.1. DOUBLE INTEGRALS

Proof for dA = dxdy: For each fixed value of y, the area of the cross section is the
area under the curve z = f (x, y) for a ≤ x ≤ b, which is given by
Z b
A(y) = f (x, y) dx z
ZZ a 6
V = f (x, y) dA
Z
R z = f (x, y)
d
= A(y) dy
c
Z dZ b
O XX c
= f (x, y) dx dy.  X
c
| a {z
a

A(y)XXXX d XXz
XX Xy
  XXX
}
A(y) b   
 
XXX R 
 XX
+

x

ZZ
Example 11.1.1. Evaluate (6x2 + 4xy 3 ) dA, where R = {(x, y)|0 ≤ x ≤ 2, 1 ≤
R
y ≤ 4}.

Solution:
ZZ Z 4 Z 2
2 3
(6x + 4xy ) dA = (6x2 + 4xy 3 ) dx dy
R
Z1 4 0

2
= 2x3 + 2x2 y 3 x=0
dy
1
Z 4

= 16 + 8y 3 dy
1
4
= (16y + 2y 4 ) y=1
= 558.

Definition 11.1.2. Double integrals over general regions

For any function f (x, y) defined on a bounded region R ⊂ ℜ2 , we define the double
integral of f over R by
ZZ n
X
f (x, y) dA = lim f (ui , vi )∆Ai ,
R k∆k→0
i=1

provided the limit exists and is the same for every choice of the evaluation points
(ui , vi ) in Ri , for i = 1, 2, · · · , n. When this happens, we say that f is integrable over

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11.1. DOUBLE INTEGRALS

R. z
6
y
z = f (x, y) 6
d

R
O XX c
 XXX
a XXX
d
 Xz
 Xy c
b 
 R -x

+
 O a b
x
Theorem 11.1.2. Fubini’s Theorem (Stronger Form)
Suppose that f (x, y) is continuous on a region R ⊂ ℜ2 .
(a) If R is defined by R = {(x, y)|a ≤ x ≤ b and g1 (x) ≤ y ≤ g2 (x)}, with g1 and
g2 continuous on [a, b], then
ZZ Z b Z g2 (x)
f (x, y) dA = f (x, y) dy dx.
R a g1 (x)

(b) If R is defined by R = {(x, y)|c ≤ y ≤ d and h1 (y) ≤ x ≤ h2 (y)}, with h1 and

h2 continuous on [c, d], then
ZZ Z d Z h2 (y)
f (x, y) dA = f (x, y) dx dy.
R c h1 (y)

Proof for (a): For each fixed value of x, the area of the cross section is the area under
the curve z = f (x, y) for g1 (x) ≤ y ≤ g2 (x), which is given by
Z g2 (x)
A(x) = f (x, y) dy
g1 (x)
ZZ Z b Z bZ g2 (x)
V = f (x, y) dA = A(x) dx = f (x, y) dy dx.
R a a g1 (x)
| {z }
A(x)
z
6
y
r
z = f (x, y) 6 y = g2 (x)

O XX c
r

a  A(x) XXXXX d
 XXz
X
 y
b  y = g1 (x)
 R -x

+
 O a b
x

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11.1. DOUBLE INTEGRALS

Proof for (b): For each fixed value of y, the area of the cross section is the area under
the curve z = f (x, y) for h1 (y) ≤ x ≤ h2 (y), which is given by
Z h2 (y)
A(y) = f (x, y) dx
h1 (y)
ZZ Z h2 (y) Z d Z h2 (y)
V = f (x, y) dA = A(y) dy = f (x, y) dx dy.
R h1 (y) c h1 (y)
| {z }
A(y)
z
6
y
z = f (x, y) 6
d

x = h1 (y)r r x = h (y)
R
O XX c
 X
A(y)XXXXX
2
a d
 Xz
Xy c

b  R
 -x

+
 O
x

2
Example
Z Z 11.1.2. Let R be the region bounded by the graphs of x = y and x = 2−y.
Write f (x, y) dA as the iterated integrals with dA = dxdy and dA = dydx.
R

Solution: y
3 6
2
x=y =2−y x = y2
2
2
y +y−2=0 @
1 @
y = 2, −1 @
@ -x
−2 −1 1 @
2 3 4
−1 R @@
ZZ Z 1 Z 2−y −2 @
f (x, y) dA = f (x, y) dx dy @ x=2−y
R −2 y2
Z Z √ Z Z
1 x 4 2−x
= √
f (x, y) dy dx + √
f (x, y) dy dx
0 − x 1 − x

√
Example 11.1.3. Z ZLet R be the region bounded by the graphs of y = x, x = 0 and
y = 3. Evaluate (2xy 2 + 2y sin x) dA.
R

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11.1. DOUBLE INTEGRALS
y
Solution: 4 6 √
y=3 y= x
3
2 R
1
-x
−3 3 6 9 12 15
ZZ −1 Z 3Z y2
2
(2xy + 2y sin x) dA = (2xy 2 + 2y sin x) dx dy
R
Z0 3 0
y2
= (x2 y 2 − 2y cos x) x=0
dy
0
Z 3
= [(y 6 − 2y cos y 2 ) − (0 − 2y cos 0)] dy
Z0 3
(y 6 − 2y cos y 2 + 2y) dy
=
0
 7 3
y 2 2 37
= − sin y + y = − sin 9 + 9
7 y=0 7
ZZ Z 9Z 3
2
(2xy + 2y sin x) dA = √
(2xy 2 + 2y sin x) dy dx
R 0 x

Exercise 11.1.1. Evaluate the double integral

Z 2Z 1
(a) (6 − 2x − 3y) dy dx
−2 0
Z 2 Z 1
(b) (6 − 2x2 + y 2 ) dx dy
−1 0
ZZ
(c) (2xy + y 2 ) dA, where R = {(x, y)|0 ≤ x ≤ 2, 0 ≤ y ≤ 1}
R
ZZ
(d) 4xe2y dA, where R = {(x, y)|1 ≤ x ≤ 4, 0 ≤ y ≤ 1}
R

Exercise 11.1.2. Evaluate the double integral

Z 2 Z x2
(a) (2x + 1) dy dx
0 0
Z 2 Z y
(b) 3ey dx dy
0 0

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11.2. DOUBLE INTEGRALS IN POLAR COORDINATES

ZZ
(c) (2x − y) dA, where R is bound by y = −3 and y = 1 − x2
R
ZZ
2
(d) ex dA, where R is bound by y = x2 and y = 1
R

11.2 Double Integrals in Polar Coordinates

Theorem 11.2.1. Fubini’s Theorem
Suppose that f (r, θ) is continuous on a region R = {(r, θ)|α ≤ θ ≤ β and g1 (θ) ≤ r ≤
g2 (θ)}. Then
ZZ Z β Z g2 (θ)
f (r, θ) dA = f (r, θ) r dr dθ.
R α g1 (θ)

y y θ = θ2
6
θ=β 6 

 ∆r r = r2

   ∆A

   θ = θ1

 R    " "

    θ = α r = r1 ""

     "

   r = g2 (θ) ∆θ"""

  
 k "
 ""




  r = g1 (θ) - "
 -
x x

Proof:

∆θ ∆θ 1
∆A = πr22 · − πr12 · = (r22 − r12 )∆θ
2π 2π 2
1
= (r2 + r1 )(r2 − r1 )∆θ
2
= r∆r∆θ
1
where r = (r2 + r1 ), ∆r = r2 − r1 .
2
Xn n
X
V = lim Vi = lim f (ri , θi ) ∆Ai
n→∞ n→∞ | {z } |{z}
i=1 i=1 area of base
height
n
X
= lim f (ri , θi )ri ∆ri ∆θi
n→∞
i=1
Z β Zg2 (θ)
= f (r, θ) r dr dθ
α g1 (θ)

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11.2. DOUBLE INTEGRALS IN POLAR COORDINATES

Example 11.2.1. Find the area of the region bounded by the curves r = 2 − 2 sin θ.

Solution:
ZZ Z 2π Z 2−2 sin θ
A= dA = r dr dθ
R 0 0
Z 2π 2 2−2 sin θ
y
r
= dθ 1
0 2 r=0
Z x
1 2π −3 −2 −1 1 2 3
= [(2 − 2 sin θ)2 − 0] dθ
2 0 −1
Z 2π R
−2
= (2 − 4 sin θ + 2 sin2 θ) dθ
Z0 2π −3
= [2 − 4 sin θ + (1 − cos 2θ)] dθ −4
0
  2π
1
= 3θ + 4 cos θ − sin 2θ = 6π
2 0

Example 11.2.2. Evaluate the iterated integral by converting to polar coordinates

Z 2 Z √4−x2 p
√
x2 + y 2 dy dx.
−2 − 4−x2

√ √
Solution: R = {(x, y)| − 2 ≤ x ≤ 2 and − 4 − x2 ≤ y ≤ 4 − x2 } is a circle of
radius centered at the origin.

Z Z √ Z Z y
2 4−x2 p 2π 2 √ 6
√
x2 + y 2 dy dx = r2 r dr dθ
−2 − 4−x2 0 0
r = 2
2 
Z 2π R
r3
= dθ  M θ - x
0 3 r=0 2
Z 2π
8 16π
= dθ =
0 3 3

Exercise 11.2.1. Find the area of the region bounded by the curves

(a) r = 3 + 2 sin θ (c) r = 2 sin θ

(b) r = 2 − 2 cos θ (d) r = 3 cos θ

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11.3. SURFACE AREA

Exercise 11.2.2. Evaluate the iterated integral by converting to polar coordinates

Z 2 Z √4−x2 p Z 2Z 0
(a) x2 + y 2 dy dx (c) √
y dy dx
−2 0 0 − 4−x2
√ Z √
Z 2 Z 4−x2 2Z 4−x2
2 −y 2
(b) e−x
2 −y 2
dy dx (d) √
e−x dy dx
√ 0 − 4−x2
−2 − 4−x2

11.3 Surface Area

Theorem 11.3.1. Surface Area
If f and its first partial derivatives are continuous on the closed region R in the
xy-plane, then the area of the surface S given by z = f (x, y) over R is given by
ZZ q
S= [fx (x, y)]2 + [fy (x, y)]2 + 1 dA.
R
Note.
Z b
Length on x-axis: dx = b − a
a
Z b Z bp
Arc length in xy-plane: ds = 1 + [f ′ (x)]2 dx
ZaZ a

Area in xy-plane: dA = A
Z ZR ZZ q
Surface area in space: dS = [fx (x, y)]2 + [fy (x, y)]2 + 1 dA
R R

Proof: Let Ti be the portion of the tangent plane lying above Ri . To find the area of
the parallelogram ∆Ti note that its sides are given by the vectors
ai =< ∆xi , 0, fx (xi , yi )∆xi >,
bi =< 0, ∆yi , fy (xi , yi )∆yi >,
i j k
ai × bi = ∆xi 0 fx (xi , yi )∆xi
0 ∆yi fy (xi , yi )∆yi
=< −fx (xi , yi )∆xi ∆yi , −fy (xi , yi )∆xi ∆yi , ∆xi ∆yi >,
q
∆Ti = kai × bi k = [fx (xi , yi )]2 + [fy (xi , yi )]2 + 1 ∆xi ∆yi
| {z }
∆Ai
n
X n q
X
S = lim ∆Ti = lim [fx (xi , yi )]2 + [fy (xi , yi )]2 + 1 ∆Ai
k∆|→0 k∆k→0
Z Z qi=1 i=1

= [fx (x, y)]2 + [fy (x, y)]2 + 1 dA.

R

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11.3. SURFACE AREA

Example 11.3.1. Find the surface area of that portion of the surface z = y 2 + 4x
lying above the triangular region R in the xy-plane with vertices at (0, 0), (0, 2) and
(2, 2). y
z 6
6 3
hhh 4
hhhh
2 s s
3 (0, 2)
z = y 2 + 4x (2, 2)
2 (((
(((( R
1
Os
1 y=x
s2 3
1 XX1XX
2  R  XXX4XX
s s
5
3  z
Xy -x
4  (0, 0) 1 2 3
+

x
Solution:
ZZ q
S= [fx (x, y)]2 + [fy (x, y)]2 + 12 dA
R
ZZ p Z 2Z yp
= 2 2
4 + 4y + 1 dA = 4y 2 + 17 dx dy
Z 2Rp y
0
Z 2 0p
= 4y 2 + 17 x dy = 4y 2 + 17 y dy
0 x=0 0
2
1 2 3 2 1  3 3

= (4y + 17) 2 · = 33 − 17
2 2
8 3 y=0 12

Example 11.3.2. Find the surface area of that portion of the paraboloid z = 1 +
x2 + y 2 that lies below the plane z = 5.
z
6 z = 1 + x2 + y 2
8
 6  y
 =5
4 z 6

2 r = 2
R 
O  M θ -x
1 XX1XX
2 3
XXX4 5 2
2  XX
3  z
Xy
4 

x+


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11.4. TRIPLE INTEGRALS

Solution:

z = 5 = 1 + x2 + y 2 , ⇒ x2 + y 2 = 4
ZZ q
S= [fx (x, y)]2 + [fy (x, y)]2 + 1 dA
R
ZZ p
= 4x2 + 4y 2 + 1 dA
R
Z 2π Z 2√
= 4r2 + 1 rdr dθ
0 0
Z r=2
1 2π 2 2 3
= (4r + 1) 2 dθ
8 0 3 r=0
Z 2π
1 3 3
= (17 2 − 1 2 ) dθ
12 0
π 3
= (17 2 − 1)
6

Exercise 11.3.1. Evaluate or estimate the surface area.

p
(a) The portion of z = x2 + y 2 between y = x2 and y = 4.

(b) The portion of x + 3y + z = 6 in the first octant.

(c) The portion of z = 2y + 3x between y = 2x, y = 0 and x = 3.

(d) The portion of the surface z = 3x + y lying above the triangular region R in
the xy-plane with vertices at (0, 0), (2, 2) and (2, 0).

11.4 Triple Integrals

Definition 11.4.1. Triple integrals over rectangles
For any function f (x, y, z) defined on a rectangle box Q = {(x, y, z)|a ≤ x ≤ b, c ≤
y ≤ d and r ≤ z ≤ s}, we define the triple integral of f over Q by
ZZZ n
X
f (x, y, z) dV = lim f (ui , vi , wi )∆Vi ,
Q k∆k→0
i=1

provided the limit exists and is the same for every choice of the evaluation points
(ui , vi , wi ) in Qi , for i = 1, 2, · · · , n. When this happens, we say that f is integrable

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11.4. TRIPLE INTEGRALS

over Q.
z z
6 6
4 XX Q 4 XX Q
XX
XXXX  XXX

XX X X X  XXX

X 3
 X XX XX 3
XX  XX XX
X X XX  X
X X  

XX X XXX X X  
X
X X X
2X X X 
XXX 2 
X X X X  XXX 
XXX XX X X  XX
X
X
XXXXX 1XXX 1
XXXXXXX 
X X 
XXX1  O
XXX 1
XXXX 2 

3
O XX1 2
1  XX XX 3 
XX XXXXX  X
X

4  X
X XX4 5
2XX X X X 5
XX 2   XXXz
3XXXXXXXX z
Xy XX
3 XXX X
X  Xy
4  XXX 4  
XXX

+ 
x x+
 ∆Vi = ∆zi ∆yi ∆xi
ZZZ
Note. The volume of box Q: V = 1 dV.
Q

Theorem 11.4.1. Fubini’s Theorem (First Form)

Suppose that f (x, y, z) is continuous on the box Q = {(x, y, z)|a ≤ x ≤ b, c ≤ y ≤
d and r ≤ z ≤ s}. Then,
ZZZ Z sZ d Z b Z bZ d Z s
f (x, y, z) dV = f (x, y, z) dx dy dz = f (x, y, z) dz dy dx,
Q r c a a c r

or in any of the four remaining orders.

ZZZ
Example 11.4.1. Evaluate 2xey sin z dV , where Q = {(x, y, z)|1 ≤ x ≤ 2, 0 ≤
Q
y ≤ 1 and 0 ≤ z ≤ π}.

Solution:
ZZZ Z π Z 1 Z 2
y
2xe sin z dV = 2xey sin z dx dy dz
Q 0 0 1
Z π Z 1 2
2x2
= ey sin z dy dz
0 0 2 x=1
Z π Z 1
= 3ey sin z dy dz
0 0
Z π
=3 ey sin z|1y=0 dz
0
1 π
= 3(e − 1)(− cos z) z=0
= 6(e − 1).

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11.4. TRIPLE INTEGRALS

Definition 11.4.2. Triple integrals over general regions

For any function f (x, y, z) defined on a bounded solid Q, we define the triple integral
of f over Q by
ZZZ n
X
f (x, y, z) dV = lim f (ui , vi , wi )∆Vi ,
Q k∆k→0
i=1

provided the limit exists and is the same for every choice of the evaluation points
(ui , vi , wi ) in Qi , for i = 1, 2, · · · , n. When this happens, we say that f is integrable
over Q.
z z
6 6
4 XX 4
XX
XXXX

XX X X X

X 3
X X XX XX 3
XX  X
X X XX X  X
XX X

XXX XXX X X
X
X X X2 X X 2
X X X 
XXXXX X XXX
X
XXX X X
1XX 
X
1
XXXXXXXX Q Q
XXXX X 
1X
X
OXX
 XX
1 
XXX2 
3

 O XX1 2
1 XX XX 3
XX X  X
 X
XX XX 
2X X4
XX 5 2  X
X XXX4 5
XX
3XXXXXXXX XzXy 3  X X z
Xy
4  XXX 4 

x+ x+
  ∆Vi = ∆zi ∆yi ∆xi

Theorem 11.4.2. Fubini’s Theorem (Stronger Form)

Suppose that f (x, y, z) is continuous on a region Q = {(x, y, z)|(x, y) ∈
R and g1 (x, y) ≤ z ≤ g2 (x, y)}, with R is some region in the xy-plane, then
ZZZ ZZ Z g2 (x,y)
f (x, y, z) dV = f (x, y, z) dz dA.
Q R g1 (x,y)

ZZZ
Example 11.4.2. Evaluate 6xy dV , where Q is bounded by x = 0, y = 0, z =
Q
0 and 2x + y + z = 4. z
y
4 s
6
6
4 s
J
3 JJ A
 2x + y + z = 4 A
2 J
J A
 A
1 J 2 2x + y = 4

 Os
J R AA
1 J
1 XXX A
s X5
XX 2 3J 4
s 
   R XX XX
2 XXJX As -x
3  XzXy 2 4
4 

+
x
De-Yu Wang CSIE CYUT 198
11.5. TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL
COORDINATES

Solution:
ZZZ ZZ Z g2 (x,y)
f (x, y, z) dV = f (x, y, z) dz dA
Q R g1 (x,y)
ZZ Z 4−2x−y
= 6xy dz dA
R 0
Z 2 Z 4−2x
= 6xy(4 − 2x − y) dy dx
0 0
Z 2   y=4−2x
y2 2y
2
y3
= 6 4x − 2x −x dx
0 2 2 3 y=0
Z 2 Z 0
3
= 8x(2 − x) dx = − 8(2 − u)u3 du (u = 2 − x)
0 2
Z 2 2
3 8u5 4
= 8(2 − u)u du = 4u −
0 5 u=0
64
= .
5

Exercise 11.4.1. Evaluate the triple integrals

ZZZ
(a) (2x + y 2 ) dV , where Q = {(x, y, z)|0 ≤ x ≤ 3, −2 ≤ y ≤ 1, 1 ≤ z ≤ 2}.
Q
ZZZ
(b) (2x − y) dV , where Q = {(x, y, z)|1 ≤ x ≤ 3, 0 ≤ y ≤ 1, 1 ≤ z ≤ 2}.
Q
ZZZ
(c) (2x + 3y + z) dV , where Q is bounded by x = 0, y = 0, z = 0 and
Q
3x − y − z = 4
ZZZ
(d) (x − 3y + z) dV , where Q is bounded by x = 0, y = 0, z = 0 and
Q
−3x + y − z = 3

11.5 Triple Integrals in Cylindrical and Spherical

Coordinates
Definition 11.5.1. Cylindrical Coordinates
Cylindrical coordinates represent a point P in space by (r, θ, z) in which
(a) (r, θ) is the polar coordinate for the vertical projection of P on the xy-plane.

(b) z is the rectangular vertical coordinate.

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11.5. TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL
COORDINATES

The rectangular conversion equations for cylindrical coordinates:

x = r cos θ z
y = r sin θ 6

s
z=z P (r, θ, z)

z
O XX
 X XX
 : X
X θ r  Xz
X X y
  S
+
 x
x y

Example 11.5.1. Write the equations (a) x2 + y 2 = 16 and (b) z 2 = x2 + y 2 in

cylindrical coordinates.
z z
6 6
8 8
6 6
2 2
4 x + y = 16 4 z 2 = x2 + y 2
2 @ 2
@
O @
O
1 XX1XX
2 3
XXX4 5
1 @XX1XX2 3
XXX4 5
2  XX 2  @@ XX
3  z
Xy 3  @ z
Xy
4  4 
 
x+

x +

Solution: (a) x2 + y 2 = 16, r2 = 16, r = ± 4.

(b) z 2 = x2 + y 2 = r2 , z = ± r.

Theorem 11.5.1. Triple integrals in cylindrical coordinates

Suppose that the solid Q as

Q = {(r, θ, z)|(r, θ) ∈ R and k1 (r, θ) ≤ z ≤ k2 (r, θ)},

where k1 (r, θ) ≤ k2 (r, θ), for all (r, θ) in the region R of the xy-plane defined by

R = {(r, θ)|α ≤ β and g1 (θ) ≤ r ≤ g2 (θ)}.

The triple integrals in cylindrical coordinates:

ZZZ Z β Z g2 (θ) Z k2 (r,θ)
f (r, θ, z) dV = f (r, θ, z) dz {z dθ} .
| rdr
Q α g1 (θ) k1 (r,θ)
dV

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11.5. TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL
COORDINATES

z
6

O
X
HHXX
 ee H XX
  HH XXzXy
 e∆θ HH
+
 re
x e r∆θ
e

Z Z √ Z
1 1−x2 2−x2 −y 2
Example 11.5.2. Evaluate the triple integral √
(x2 +
−1 − 1−x2 x2 +y 2
3
y 2 ) 2 dz dy dx in cylindrical coordinates.

Solution:
Z Z √ Z Z Z Z
1 1−x2 2−x2 −y 2 2π 1 2−r 2
3 3
2 2
√
(x + y ) dz dy dx =
2 (r2 ) 2 dz rdr dθ
−1 − 1−x2 x2 +y 2 0 0 r2
Z 2π Z 1Z 2−r 2
= r4 dz dr dθ
r2
Z0 2π Z0 1
= r4 (2 − r2 − r2 ) dr dθ
0 0
Z 2π  5 1
r r7
=2 − dθ
0 5 7 r=0
8π
= .
35

Definition 11.5.2. Spherical Coordinates

Spherical coordinates represent a point P in space by (ρ, φ, θ) in which
p
(a) ρ = x2 + y 2 + z 2 ≥ 0 is the distance from the origin O;

−→
(b) 0 ≤ φ ≤ π is angle from the positive z-axis to the vector OP ;

(c) 0 ≤ θ ≤ 2π the same as the θ in the polar coordinates.

The rectangular conversion equations for spherical coordinates:

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11.5. TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL
COORDINATES
z
6

s
P (ρ, φ, θ)
x = ρ sin φ cos θ
φ  7
y = ρ sin φ sin θ j
ρ z
z = ρ cos φ O
 XXX X
 : XX
X r  Xz
X X y
  θ S
+
 x
x y

Example 11.5.3. Write the equations (a) x2 + y 2 + z 2 = 4 and (b) z 2 = x2 + y 2 in

spherical coordinates.
Solution: (a) x2 + y 2 + z 2 = ρ2 = 4, ρ = 2.
(b) z 2 = (ρ cos φ)2 = x2 + y 2 = (ρ sin φ cos θ)2 + (ρ sin φ sin θ)2
ρ2 cos2 φ = ρ2 sin2 φ(cos2 θ + sin2 θ)
ρ2 cos2 φ = ρ2 sin2 φ,
π 3π
∴ ρ = 0 or cos2 φ = sin2 φ, φ = , .
4 4

Theorem 11.5.2. Triple integrals in spherical coordinates

Suppose that the solid Q as
Q = {(ρ, φ, θ)|α ≤ θ ≤ β, c ≤ φ ≤ d and g1 (θ, φ) ≤ ρ ≤ g2 (θ, φ)}.
The triple integrals in cylindrical coordinates:
ZZZ Z β Z d Z g2 (θ,φ)
f (ρ, φ, θ) dV = f (ρ, φ, θ) ρ2 sin φ dρ dφ dθ .
Q α c g1 (θ,φ)
| {z }
dV
z
6
ρ∆φ
ρ

∆φ
^

O
X
HHXX
 ee H XX
  HH XXzXy
 e∆θ HH
+
 e
r
x e
e r∆θ
Proof:
∆Vk ≈ ∆ρk (ρk ∆φk )(ρk sin φk ∆θk ) = ρ2k sin φk ∆ρk ∆φk ∆θk
dV = ρ2 sin φ dρ dφ dθ

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11.5. TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL
COORDINATES

Z 2 Z √
4−x2 Z √4−x2 −y2
Example 11.5.4. Evaluate the triple integral (x2 + y 2 +
−2 0 0
z 2 ) dz dy dx in spherical coordinates.

Solution:
Z 2 Z √
4−x2 Z √4−x2 −y2 Z π Z π Z 2
2
2 2 2
(x + y + z ) dz dy dx = ρ2 ρ2 sin φ dρ dφ dθ
−2 0 0 0 0 0
Z Z π 2
π 5
2 ρ
= sin φ dφ dθ
0 0 5 ρ=0
Z Z π
π 2 32
= sin φ dφ dθ
0 0 5
Z π
π
32 2
= − cos φ dθ
0 5 φ=0
32π
= .
5

Exercise 11.5.1. Write the given equation in cylindrical coordinates.

2 −y 2
(a) (x − 2)2 + y 2 = 9 (c) z = e−x
p
(b) z = x2 + y 2 (d) y = 2x

Exercise 11.5.2. Evaluate the triple integral in cylindrical coordinates.

Z Z √ Z
1 1−x2 2−x2 −y 2 p
(a) √
x2 + y 2 dz dy dx
−1 − 1−x2 0

Z 1 Z √
1−x2 Z √x2 +y2
(b) √
3z dz dy dx
−1 − 1−x2 0

Z 2 Z √4−y2 Z √8−x2 −y2

(c) √ √ 2 dz dx dy
0 − 4−y 2 x2 +y 2

Exercise 11.5.3. Convert the equation into spherical coordinates.

(a) x2 + y 2 + z 2 = 6 (c) z = 2
p
(b) y = x2 (d) z = − x2 + y 2

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Exercise 11.5.4. Evaluate the triple integral in the spherical coordinate system.
Z 2 Z √4−x2 Z √4−x2 −y2 p
(a) 2 2 2
√ 2 2 x + y + z dz dy dx
−2 0 − 4−x −y

Z 1 Z √
1−x2 Z √1−x2 −y2
3
(b) √ √ (x2 + y 2 + z 2 ) 2 dz dy dx
0 − 1−x2 − 1−x2 −y 2

Z Z √ Z
4 16−x2 0 p
(c) √ x2 + y 2 + z 2 dz dy dx
0 0 − 16−x2 −y 2

11.6 Change of Variables: Jacobians

Definition 11.6.1. Jacobian

(a) For x = g(u, v) and y = h(u, v), the Jacobian is

∂x ∂x
∂(x, y) ∂u ∂v
= .
∂(u, v) ∂y ∂y
∂u ∂v

(b) For x = g(u, v, w), y = (u, v, w) and z = k(u, v, w), the Jacobian is

∂x ∂x ∂x
∂u ∂v ∂w
∂(x, y, z) ∂y ∂y ∂y
= .
∂(u, v, w) ∂u ∂v ∂w
∂z ∂z ∂z
∂u ∂v ∂w

2u
Example 11.6.1. Find the Jacobian of the transformation x = , y = 3uv + v 2 .
v
Solution:
∂x ∂x 2 2u
∂(x, y) − 2 12u
= ∂u ∂v = v v = + 4.
∂(u, v) ∂y ∂y v
3v 3u + 2v
∂u ∂v

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Example 11.6.2. Compute the Jacobian for the spherical-like transformation x =

ρ sin φ cos θ, y = ρ sin φ sin θ and z = ρ cos φ.

Solution:

∂x ∂x ∂x
∂ρ ∂φ ∂θ
sin φ cos θ ρ cos φ cos θ −ρ sin φ sin θ
∂(x, y, z) ∂y ∂y ∂y
= = sin φ sin θ ρ cos φ sin θ ρ sin φ cos θ
∂(ρ, φ, θ) ∂ρ ∂φ ∂θ
cos φ −ρ sin φ 0
∂z ∂z ∂z
∂ρ ∂φ ∂θ
ρ cos φ cos θ −ρ sin φ sin θ
= cos φ
ρ cos φ sin θ ρ sin φ cos θ
sin φ cos θ −ρ sin φ sin θ
+ ρ sin φ
sin φ sin θ ρ sin φ cos θ
= cos φ(ρ2 sin φ cos φ cos2 θ + ρ2 sin φ cos φ sin2 θ)
+ ρ sin φ(ρ sin2 φ cos2 θ + ρ sin2 φ sin2 θ)
=ρ2 sin φ(cos2 φ + sin2 φ) = ρ2 sin φ.

Theorem 11.6.1. Change of Variables

(a) For x = g(u, v) and y = h(u, v),

ZZ ZZ
∂(x, y)
f (x, y) dA = f (g(u, v), h(u, v)) du dv.
R S ∂(u, v)

(b) For x = g(u, v, w), y = (u, v, w) and z = k(u, v, w),

ZZZ ZZZ
∂(x, y, z)
f (x, y, z) dV = f (g(u, v, w), h(u, v, w), k(u, v, w)) du dv dw.
Q S ∂(u, v, w)

Proof: Consider the region ∆A with vertices M (g(u, v), h(u, v)), N (g(u +
∆u, v), h(u + ∆u, v)), P (g(u + ∆u, v + ∆v), h(u + ∆u, v + ∆v)), and Q(g(u, v +
∆v), h(u, v + ∆v)),

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Q P
 
 
 
 
 
 
M N

−−→ −−→
∆x∆y = ∆A ≈ M N × M Q
i j k
−−→ −−→
M N × M Q = g(u + ∆u, v) − g(u, v) h(u + ∆u, v) − h(u, v) 0
g(u, v + ∆v) − g(u, v) h(u, v + ∆v) − h(u, v) 0
i j k
∂x ∂y
∂x ∂y
∂u ∂u ∆u∆vk
≈ ∂u ∆u ∂u ∆u 0 =
∂x ∂y
∂x ∂y
∆v ∆v 0 ∂v ∂v
∂v ∂v
∂(x, y)
= ∆u∆vk
∂(u, v)
−−→ −−→ ∂(x, y)
∆A ≈ M N × M Q ≈ ∆u∆v
∂(u, v)

Example 11.6.3. Use Theorem 11.6.1 to derive the evaluation formula for polar
coordinates:
ZZ ZZ
f (x, y) dA = f (r cos θ, r sin θ)r dr dθ.
R S

Proof:

∂x ∂x
∂(x, y)
= ∂r ∂θ = cos θ −r sin θ = r cos2 θ + r sin2 θ = r
∂(r, θ) ∂y ∂y sin θ r cos θ
ZZ Z∂r
Z ∂θ
∂(x, y)
f (x, y) dA = f (r cos θ, r sin θ) dr dθ
R S ∂(r, θ)
ZZ
= f (r cos θ, r sin θ)r dr dθ.
S

ZZ
Example 11.6.4. Evaluate the integral (x + 2y) dA, where R is the region
R
bounded by the lines y = −1 − 2x, y = 3 − 2x, y = x − 1 and y = x − 3.

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u+v u − 2v
Solution: Let u = y + 2x and v = x − y, then x = and y =
3 3
1 1
∂(x, y) 3 3 1
= 1 −2 = −
∂(u, v) 3
3 3
ZZ Z 3Z 3  
u + v 2(u − 2v) ∂(x, y)
(x + 2y) dA = + du dv
R 1 −1 3 3 ∂(u, v)
Z 3Z 3
1
= (u − v) − du dv
1 −1 3 y
1
Z 3 2
u
3 6 y =x−1
= − uv dv @
3 1 2 @ v =1 y =x−3
u=−1 1 
Z @  v=3
1 3   @ - x
= (4 − 4v) dv  @ 
3 1 @  1 2  3
−1@ R @  
1
3
@   @ y = 3 − 2x
= 4v − 2v 2 v=1
3 @  @
u=3
8 @
−3  @ y = −1 − 2x
=− .
3 @ u = −1

Exercise 11.6.1. Find the Jacobian of the transformation.

(a) x = uev , y = ue−v (c) x = 2uv, y = 3u − v

2u u
(b) x = , y = 3uv + v 2 (d) x = , y = v 2
v v

Exercise 11.6.2. Evaluate the integral

ZZ
(a) (x−2y) dA, where R is the region bounded by y = 3x, y = 3x−4, y = 1−2x
R
and y = 2 − 2x.
ZZ
(b) (x+2y) dA, where R is the region bounded by y = ex , y = ex +2, y = 3−ex
R
and y = 5 − ex .
ZZ
(c) (x + 2y) dA, where R is the region bounded by y = 1 − 2x, y = 3 − 2x, y =
R
x − 1 and y = x − 3.

Exercise 11.6.3. Prove that

ZZ ZZ
(a) f (x, y) dA = f (r cos θ, r sin θ)rdr dθ.
R S
ZZZ ZZZ
(b) f (x, y, z) dV = f (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ)ρ2 sin φ dρ dφ dθ.
Q S

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