Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

Chapter 4: Line Integrals and surface Integrals

In this chapter we consider two generalizations of definite integrals (namely line integrals and
surface integrals) and the four main theorems of vector integral calculus (namely the Fundamental theorem
of Line Integrals, Green’s Theorem, Divergence Theorem and Stokes’ theorem). These theorems are
generalizations of the Fundamental theorem of Calculus.

4.1 Line integrals

Line integrals are generalizations of definite integrals. Definite integrals are defined over
intervals (straight curves) but line integrals can be defined over non-straight curves. A line
integral is also called curve integral, path integral, or curvilinear integral. In some cases it is also called
contour integral (circulation integral). Curve integral would have been the most appropriate term, but
line integral is more widely used.

Line integrals have various forms:

a) line integrals of scalar fields with respect to arc length:  f ( x, y, z)ds

C

b) line integrals of scalar fields with respect to 𝑥, 𝑦 and 𝑧 (aka line integrals in differential forms):

 Mdx ,  Ndy ,  Pdz ,  Mdx  Ndy  Pdz

C C C C

   
c) line integrals of vector fields/work integral/:  F ( x, y, z). dr =  F ( x, y, z).T ds
C C

4.1.1 Line integrals of Scalar Fields with respect to arc length

A. Motivation: mass of a wire (or a thin rod)

Consider a piece of wire having the shape of a smooth curve 𝒞 with end points 𝐴 and 𝐵, lying in xyz-
space. Let 𝑓(𝑥, 𝑦, 𝑧) be the mass per unit length (length density) of the wire at the point 𝑃(𝑥, 𝑦, 𝑧)
on 𝒞. What is the total mass of the wire?

̂𝑃𝑘 be the portion of 𝒞 between the

Let 𝒫: = {𝑃1 , … , 𝑃𝑛 }be a partition of C as shown. Let 𝑃𝑘−1
points Pk-1 and Pk (we call this the kth sub-curve). Let ∆Sk be the length of the kth sub-curve. Let
(xk*, yk*, zk*) be a point on the kth sub-curve. The mass m of the portion of the wire between Pk-1 and
Pk is approximated by

∆mk ≈ f(xk*, yk*, zk*) ∆Sk (1)

Thus the total mass m of the wire is approximated by

m ≈ ∑𝑛𝑘=0 f(xk∗ , yk∗ , zk∗ ) ∆Sk =: Jn (2)

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

If the length || 𝒫|| of the longest sub curve approaches 0 as n tends to infinity, then the total mass
m of the wire is

m =lim𝑛→∞ ∑𝑛𝑘=0 f(xk∗ , yk∗ , zk∗ ) ∆Sk (3)

The limit on the RHS of the above equation can represent many other quantities (not only mass).We
call it the line integral of f along C from A to B with respect to arc length.

B. Definition

Definition 1: Let 𝒞 be a smooth curve of finite length (with end points A and B and oriented from A
to B). Let 𝑓 be a scalar field defined at each point of 𝒞. Let 𝒫 ≔ {𝑃1 , 𝑃2 … 𝑃𝑛 } be a partition of 𝒞 as
shown in the figure. Then the line integral of 𝑓 along 𝒞 with respect to arc length s is denoted and
defined by

∫𝒞 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑠 = lim𝑛→∞ ∑𝑛𝑘=1 𝑓(𝑥𝑘 , 𝑦𝑘 , 𝑧𝑘 ) ∆𝑆𝑘 (4)

where ∆𝑆𝑘 is the length of the portion of C between Pi-1 and P i,

(𝑥𝑘 , 𝑦𝑘 , 𝑧𝑘 ) is any point on 𝒞 between 𝑃𝑘−1 & 𝑃𝑘 i and

Provided that

||𝛥𝑆|| : = max {∆𝑆1,… ∆𝑆𝑛 } tends to 0 as n  

Figure 1: A Partition of a Finite smooth curve

* NB. n   doesn’t imply every S i  0 . But the converse is true. More over S i  0 bi-
implies ||P|| tends to 0.

C. Evaluation:

The basic idea is to convert into definite integral. If C is parameterized by a differentiable vector
valued function

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

r(t) = x(t)i + y(t)j + z(t)k, a ≤ t ≤ b

𝑑𝑠 = ‖𝑟 ′ (𝑡)‖𝑑𝑡 = √[𝑥 ′ (𝑡)]2 + [𝑦 ′ (𝑡)]2 + [𝑧 ′ (𝑡)]2

Thus

∫ 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑠 = ∫ 𝑓(𝑥(𝑡), 𝑦(𝑡), 𝑧(𝑡))√[𝑥 ′ (𝑡)]2 + [𝑦 ′ (𝑡)]2 + [𝑧 ′ (𝑡)]2 𝑑𝑡

𝒞 𝑎

Note: If r(t) = x(t)i + y(t)j + z(t)k, a ≤ t ≤ b is a parameterization of C then, the partition 𝒫 of

C induces a partition {t0,…,tn} of [a, b] and since ∆sk ≈ ||∆rk||≈ r ' (t k ) t k so
*

n n

 f ( x k , y k , z k )s k ≈  f (r (t k )) r ' (t k ) t k
* * * * *

k 1 k 1

The sum in RHS is the Riemann sum of the function g(t): = f (r (t )) r ' (t ) corresponding to the
partition {t0,…,tn} of [a, b].

n n

 f ( x k , y k , z k )s k = lim  f (r (t k )) r ' (t k ) t k

* * * * *
But lim
n  n 
k 1 k 1

Hence


C
f ds =  f (r (t )) r ' (t ) dt
a

Example-1: Evaluate

 f ds
C

Where f = xy4 and C is the right half of the circle x2 + y2 = 16 and have counter clockwise
orientation.

8192
Ans.
5

Question: What if the orientation is ACW?

D. Properties

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

a)  kf  k  f
C C

b)  f g   f g
C C C

c) f ff
C C1 C2
(where C= C1 U C2 and C is smooth)

Further properties:

- If f is continuous at every point on C and if C is smooth, then f

C
exist.

- f
C
is parameterization independent , i.e. any smooth parameterization of C yields the same

value of  f as long as the parameterization of the curve C is traced out exactly once as t
C

increases from a to b.
- f
C
is orientation independent, i.e., if –C is the same curve as C but with opposite

orientation, then

 fds   fds
C C

In view of property (c) above we make the following definition.

Definition 2 (line integral over piecewise smooth curve): Let 𝒞 be a piecewise smooth curve with
smooth pieces 𝒞 1,…, 𝒞 n. We define

 f = f
C C1
+…+ f
Cn

Example-2 (three dimensional): Evaluate

 xyzds
C

Where C is the helix given by r(t): = <cost, sint, 3t>, 0 ≤ t ≤ 4π Ans. -3√10

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

Example-3(piecewise smooth): Evaluate  f ds

C

Where f = 4x3 and C is the union of the following curves.

C1: the line segment with end points (-2, -1) and (0, -1)

C2: part of the curve y = x3 - 1 between (0, -1) and (1, 0)

C3: the line segment with end points (1, 0) and (1, 2)

Ans. -8 + 2/27 (103/2 - 1)

Example 4 (application): Find the mass of the piece of wire described by the curve x2+y2=1
with density function f(x, y) = 3+x+y.

Solution: The curve is the standard unit circle. It can be parameterized by the vector function
r(t)=< cos (t),sin(t)> with 0 ≤ t ≤ π. We have x (t) = cos(t) and y(t)=sin(t), so x'(t)=-sin(t) and
y'(t)= cos (t). The mass is given by the formula

The term in the square root is 1, hence we have

4.1.2 Line Integrals in Differential Form

Definition 3: Let C be a smooth curve (oriented from A to B) of finite length. Let f be a scalar field
𝑃𝑖−1 𝑃𝑖 = ( x i , y i , z i ).
defined at each point of C. Let {P1, P2… P n} be a partition of C and ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
Then we define the line integrals of f(x, y, z) along C from point A to point B with respect to x , y and z as:

 f ( x, y, z )dx  lim
n 
 f ( x , y , z )x
1
i i i i
C

 f ( x, y, z )dy  lim
n 
 f ( x , y , z )y
1
i i i i
C

 f ( x, y, z )dy  lim
n 
 f ( x , y , z )z
1
i i i i
C

The three line integrals often come together and give a line integral in differential form.

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

Definition 4: Let M, N and P be continuous functions of x, y and z. Then we define:

 M dx  Ndy  Pdz   Mdx   Ndy   Pdz

C C C C

Evaluation: The line integrals above are evaluated by expressing x, y, z, dx, dy, dz in terms of t
where t is the parameter in a suitable parameterization of C. The result is an ordinary single integral.

Remark [orientation dependence]: There is one important difference between the integral in
differential form and the line integral of previous subsection. If –C is the same curve as C, but with
opposite orientation then

 M dx  Ndy  Pdz  -  M dx  Ndy  Pdz 

C C

Reason: The integral above changes with changes in x(t),y(t) and z(t). Why?

But changes in x(t),y(t) and z(t) doesn’t affect the arc length element ds. Why?

Example 5: Evaluate  (3  x  y)dy, where C is the standard unit circle oriented

C

a) Counter clockwise (ccw). Ans. 𝜋

b) Clockwise (cw). Ans. −𝜋

Example 6: Find ∫𝐶 𝑦𝑑𝑥 + 𝑧𝑑𝑦 where C is the part of the helix

r(t) = sin t i + cos t j +tk ; 0 < t < 2𝜋

Solution: We have r'(t) = cos t i - sin t j + k so that y dx + z dy = (cos2t - t sin t)dt

2𝜋
This leads us to the integral ∫0 (𝑐𝑜𝑠 2 𝑡 − 𝑡𝑠𝑖𝑛𝑡)𝑑𝑡 with a little bit of effort (using integration by
parts) we get

Example 7: Evaluate ∫𝐶 𝑦𝑑𝑥 + 𝑧𝑑𝑦 + 𝑥𝑑𝑧 where C has the parameterization x = t, y = t2 , z = t3; 0
≤ t ≤1.

Ans. 89/60

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

Example 8: Evaluate ∫𝐶 𝑥𝑑𝑥 + 𝑦𝑑𝑦 + 𝑧 2 𝑑𝑧 when C is the helix parameterized by r( t) = cos( t)

i+ sin( t)j + t k for t in [ 0,2𝜋] .

8𝜋3
Ans.
3

4.1.3 Line Integrals of Vector Fields(with respect to displacement)

A. Motivation (Work done by a variable force )

Let 𝐹 (𝑥, 𝑦, 𝑧) ∶ = 𝑀 𝒊 + 𝑁𝒋 + 𝑃 𝒌 be a force field defined at each point of a region in xyz-space

and let C be a curve in this space with end points A and B. What is the total work done by F in
moving a point from A to B along C?

Note that if a constant force F moves an object through any vector displacement S, then the work
done is given by

W = F. S = ‖𝑭‖‖𝑺‖cos 

If a variable force moves an object through a small vector displacement ∆r, then the small amount of
work ∆w done by the force is the product of the distance traveled ‖∆𝒓‖and the cosine component of
the force. (Because the force is approximately constant on small displacement)

Figure 2: Work done by a constant force F

This implies that ∆w= ‖𝑭‖‖𝑺‖cos  =F(x , y) ·∆r

Subdivide C into n small sections as shown. Let 𝒫: = {P1, …, Pn}be a partition of C as shown. Let
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
∆r k: =P ̂
k−1 Pk . Let (xk*, yk , zk ) be a point on Pk−1 Pk ,i.e., arc Pk−1 Pk . The work ∆Wk done by F
* *

in moving the point from Pk-1 to Pk along C is approximated by

∆W k ≈ F (xk*, yk*, zk*). ∆rk

Thus the total work W is approximated by

W ≈∑𝑛𝑘=0 𝐅(xk∗ , yk∗ , zk∗ ) . ∆rk =: Jn

If the length ‖𝒫‖of the longest sub arc approaches 0 as n tends to infinity, the total work W done by
F is given by:

W = lim𝑛→∞ ∑𝑛𝑘=0 𝐅(xk∗ , yk∗ , zk∗ ) . ∆𝐫𝐤

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

B. Definition:

Definition 5: Let 𝒞 be a smooth curve of finite length with end points A and B and oriented from A
to B. Let F be a vector field defined on C. Let 𝒫 ={P1, P2… P n} be a partition of C as shown in the
̂𝑃𝑘 be the portion of C between Pk-1 and Pk . Then the line integral of F along C (from
figure. Let 𝑃𝑘−1
A to B) is denoted and defined by

 F.dr = lim
n
 F (x , y
k 1
k k , z k )  rk ;
C

where rk := ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ ̂𝑃𝑘 (provided that ‖𝒫‖ tends to 0 as n   ).

Pk−1 Pk and (𝑥𝑘 , 𝑦𝑘 , 𝑧𝑘 ) 𝜖 𝑃𝑘−1

NB. Both F and dr are vectors.

C. Evaluation
b
If r is a parameterization of C, then the line integral of F along C is  F.dr =  F (r (t ))  r ' (t )dt
C a

If F = M i + N j + P k is a vector field and C is the curve with parameterization r = x i + y j + zk, t

in [a, b], then  F.dr =  Mdx  Ndy  Pdz
C C

(i.e., the line integral above and the line integral in differential form are equivalent).

Note that this gives us another method for evaluating line integrals of vector fields.

Justification:

We can rewrite r'(t) dt as

𝑑𝑟 𝑑𝑥 𝑑𝑦 𝑑𝑧
𝑑𝑡
𝑑𝑡 = [ 𝑑𝑡 𝑖 + 𝑑𝑡 𝑗 + 𝑑𝑡 𝑘] 𝑑= dx i + dyj + dzk

So that if

F = Mi + Nj + Pk

Then

F. dr = F. r'(t)dt = M dx + N dy + P dz

Definition 6 (Line Integral of a vector field over piecewise smooth curve): Let C be a piecewise
smooth curve with smooth pieces C1,…, Cn. We define

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

 F.dr =  F
C C1

+…+ F
Cn

D. Properties

a)  F    F (Where C and –C are same curve but have different orientations)

C C

b)
C
 kf  k  f
C

c)  f g   f g
C C C


Example 9: Evaluate F.dr where F(x, y, z) = xz i– yz k and C is the line segment from
C

a) (-1, 2, 0) to (3, 0, 1). Ans. 3

b) (3, 0, 1) to (-1, 2, 0). Ans. -3

Example 10(application): Find the total work done in moving an object along the helix C given by
r( t) = cos( t) i+ sin( t)j + t k for t in [ 0,2𝜋] through the vector field F ( x, y, z) = <y, z2, x2>.

Answer: 2𝜋 + 4𝜋 2

Exercise 4:1

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

4.2 Path independence of Line Integrals: The Fundamental theorem of Line

Integrals

A. Path Dependence of line integrals

Let D be a domain of the vector field F. Let A and B be points in D, C be a curve in D joining A to


B. Then the value of the line integral F (r ).dr generally depends on not only on the end points A
C

and B, but also on the curves joining A and B. However there are exceptions
For example: consider  ydx  xdy (Where C is any curve joining (0, 0) to (1, 1). Consider the 4
C

paths in the diagram.

1 1

1
1
1

1
Figure 3: Four different paths from (0,0) to (1,1)
1

The given line integral is the same along all paths joining (0,0) to (1, 1).

To proceed we need to recall the definitions of some topological notions about curves and regions.

A path/curve/ C is said to be

a) Closed if its initial and terminal points coincide.

b) Simple if does not touch or cross itself.

A (plane or space) region D is said to be

a) Open if it does not contain any of its boundary points.

b) Connected every pair of points in D can be connected by a path in D.
c) Simply connected if any closed path in D can be continuously shrunk till it becomes a point
without leaving D. Otherwise it is called multiply connected.

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

Definition 1: Let D be a connected region in R3. The line integral F (r ).dr is said to be path 
C

independent in D if  F (r ).dr =  F (r ).dr , for any two paths C

C1 C2
1 and C2 in D with the same initial and

terminal points.

B. The Fundamental theorem of Line Integrals

Theorem 1[The Fundamental Theorem of Line Integrals]

a) The line integral  F (r ).dr is independent of path in a region D if and only if there exist a scalar
C

function f defined on D such that F = grad f on D.

b) In this case,  gradf .dr  f ( B)  f ( A)

C

If, for a vector field F, there exists a scalar field f satisfying F = grad f, then F is called a conservative
vector field and f is called a potential function of F.


Notation: we write f (r ).dr =  f  dr
C A

Proof:

Part I. Assume that F = grad f and C is a path from A to B in D.

Let r(t) = < x(t), y(t), z(t) > be a parameterization of C.

 gradf .dr =   f
C C
x , f y , f z    dx, dy, dz 

= f
C
x dx  f y dy  f z dz

 f x x' (t )dt  f y y' (t )dt  f z z ' (t )dt

b
=
a

 [ f x x' (t )  f y y' (t )  f z z ' (t )]dt

b
=
a

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

d

b
= [ f ( x(t ), y(t ), z (t ))]dt [chain rule]
a
dt

= f(x(b),y(b),z(b)) - f(x(a),y(a),z(a))

= f(B) – f(A)

Part II. The proof of the converse is difficult hence it is omitted.

C. Conditions for Path Independence of Line Integrals (or Conservativeness of vector

Fields)

Definition 2:

a) Let M, N and P be scalar fields. An expression of the type

𝑀 𝑑𝑥 + 𝑁𝑑𝑦 + 𝑃 𝑑𝑧

is called a differential form.

b) If in particular, there is a scalar field f such that = 𝑓𝑥 , 𝑁 = 𝑓𝑦 , 𝑃 = 𝑓𝑧 , the expression is
called exact differential.

We now define some topological notions which we need in the sequel.

Theorem 2[Test for exactness of differential forms or test for conservativeness of v.f]:

Let F = Mi + Nj + Pk be a vector field whose components have continuous first order partial
derivatives. If the domain of F is simply connected, then Mdx + Ndy + Pdz is exact (i.e., F is
conservative) if and only if

My = Nx, Nz = Py , My=Px

Proof:

Part I. If F is conservative

There exists a scalar field f such that F = <fx, fy, fz>

This implies that M = fx, N = fy and P = fz. Hence My = fxy = fyx = Nx .

The rest can be proved analogously.

Part II. This requires Stokes theorem hence it is omitted.

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

Theorem 3 (Sufficient and necessary Condition for path Independence)

Consider the following four statements:

1) F =Mi + Nj + Pk is conservative vector field.

2)  F (r ).dr
C
is independent of path.

3)  F (r ).dr
C
=0 for every closed oriented curve C lying in the domain of F. 
(i.e., F .dr  0 )
C

*4) curl F = 0 (i.e. , F. dr is an exact differential)

Then

a) (1)  (2)  (3)

b) (1)  (4)

c) If the domain of F is a simply connected region, then (1)  (2)  (3)  (4)

Example 11: Show that ∫𝐶(𝑦 2 − 6𝑥𝑦 + 6)𝑑𝑥 + (2𝑥𝑦 − 3𝑥 2 )𝑑𝑦 is path independent in ℝ2 and
evaluate from (-1, 0) to (3, 4).

Ans. -36

Example 12: Show that ∫𝐶 (𝑦 + 𝑦𝑧)𝑑𝑥 + (𝑥 + 3𝑧 2 + 𝑥𝑧)𝑑𝑦 + (9𝑦𝑧 2 + 𝑥𝑦 − 1)𝑑𝑧 is path

independent in ℝ2 and evaluate from (-1, 0) to (3, 4).

Ans. 194

Exercise 4.2:

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

4.3 Green’s Theorem and its Applications

Green’s theorem relates the line integral of a two dimensional vector field F= Mi + Nj over the
boundary C of a plane region R with the double integral of the scalar field f: = Nx – My. This
theorem is due to an English self taught mathematician, George Green (1973 - 1841). The theorem is
useful in

a) evaluating line integrals by double integrals

b) evaluating double integrals by line integrals
c) solving some geometrical and physical problems

Orientation of the boundary curve of a region: The boundary C of a plane region R is said to
have positive orientation or positive direction if R is kept to its left as we walk on C in this direction.

Notations:


C
: Integration over a simple closed curve C


C
: Integration over a simple closed curve C in the positive direction


C
: Integration over a simple closed curve C in the negative direction

Theorem 1 [Green’s Theorem]: Let R be a closed bounded region in the xy-plane whose boundary
C is a piecewise smooth curve. Let M, N, My and Nx be continuous everywhere in some domain
containing R. Then

 Mdx  Ndy   ( N
C R
x  M y )dA

Proof:

Case 1: If R is a simple region (i.e., R is both vertically and horizontally simple)

Thus there exists functions g1(x), g2 (x), h1(x), h2(x) and constants a, b, c and d such that

R = {(x, y): a ≤ x ≤ b and g1(x) ≤y≤ g2 (x)} = {(x, y): c ≤ y ≤d and h1(x) ≤ x ≤ h2(x)}

It suffices to prove

i)  Mdx    M
C R
y dA

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

ii)  Ndy   N
C R
x dA

To show (i) we need a parameterization of C. C is the union of the graphs C1 and C2 of g1(x) and g2
(x) for x in [a, b]. If we parameterize C1 and C2 by r1(t) = <t, g1(t)> and r2(t) = <t, g2(t)> , then C1has
the same orientation as C but C2 has opposite orientation. Hence

 Mdx   Mdx   Mdx

C C1 C2
=  M (t , g (t ))dt   M (t , g
C1
1
C2
2 (t )) dt

On the other hand

 g 2 (t )
b 
R y
M dA  a  g ( x) y dx =  C M (t , g1 (t ))dt  C M (t , g 2 (t ))dt
 M dy
 1  1 2

Hence

 Mdx    M
C R
y dA

Similarly we can show

 Ndy   N
C R
x dA

Case 2: If R is not a simple region. (Exercise)

Example 13: Use Green’s theorem to evaluate

 (x  y 2 )dx  (2 y  x)dy
2

Where C consists of the boundary of the region R in the first quadrant bounded by the graphs of y =
x2 and y = x3. Ans. -11/420

Calculating area with Green’s Theorem

a) If a simple closed curve C in xy-plane and the region R it encloses satisfy the hypotheses of
Green’s theorem, then
1
area(R) =  xdy    ydy  2  xdy  ydx
C C C

b) If a simple closed curve C in the polar coordinate plane and the region R it encloses satisfy the
hypotheses of Green’s theorem, then

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

r d
2
area(R) =
C

Example 14: Use Green’s theorem to find the area of the region R bounded by the ellipse x2/a2 +
y2/b2 = 1. Ans. 𝜋𝑎𝑏

Alternative Forms of Green’s Theorem

Green’s theorem in the plane can be expressed in two other forms/sometimes called vectorial forms
of Green’s theorem/. First we define the terms flow, flux and circulation.

Definition 9: If C is a smooth curve in the domain of a continuous vector field F = <M, N> in the
plane and if n is the outward pointing unit normal vector on C, the flux F across C is the line integral
of F.n with respect to arc length, the scalar component of F in the direction of the outward normal.

Flux in space is defined in section 4.4.

Definition 10: If C is a smooth curve in the domain of a continuous vector field F = <M, N> in the
plane and if T is the unit tangent vector on C, the flow along C from its initial point to terminal point is
the line integral of F.T with respect to arc length, the scalar component of F in the direction of T.

Circulation in space will be defined in section 4.5.

Thus flux is the integral of the normal component of F and circulation is the integral of the tangential
component of F.

⃗ 𝑑𝑠
Flux Integral: ∫𝐶 𝐹 ∙ 𝑇 (The Integral of tangential component of F wrt arc length)

Circulation Integral: ∫𝐶 𝐹 ∙ 𝑛⃗𝑑𝑠 (The Integral of normal component of F wrt arc length)

Green’s Theorem in Vector Form I [Divergence Theorem in the Plane]

The outward flux of a field F = Mi + Nj across a simple closed curve C equals the double integral of
div F over the region R enclosed by C.

This theorem is also called outward flux , normal form of Greens theorem or divergence form of
Greens theorem.

Green’s Theorem in Vector Form II [Stokes’ Theorem in the Plane]

The ccw circulation of a field F = Mi + Nj around a simple closed curve C equals the double integral
of the k-component of curl F over the region R enclosed by C.

This theorem is also called ccw circulation, curl form or tangential form of Greens theorem.

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Exercise 4.3

1. Very Green’s theorem for the vector field 𝐹 (𝑥, 𝑦) ≔ 𝑦𝒊 + 3𝑥𝒋 and the path given by 𝑥 2 +
𝑦 2 = 4 on 𝑥𝑦-plane.
1
⃗⃗⃗⃗ where 𝐹 ≔ 〈𝑥𝑦, 𝑥 2 +
2. (line integral by double integral)Use Green’s theorem to evaluate ∮𝐶 𝐹 . 𝑑𝑟
2

𝑥𝑦〉and C is the upper half of the ellipse 𝑥 2 + 4𝑦 2 = 1 and the X-axis on the interval [−1,1]
oriented positively.
[𝑑𝑥−𝑑𝑦]
3. (line integral by double integral )Calculate the integral ∮𝐶 using Green’s theorem. The
𝑥+𝑦

contour C is the boundary of the square with the vertices 𝐴(1,0), 𝐵(0,1), 𝐷(−1,0), 𝐸(0, −1).
4. (double integral by line integral)Use Green’s theorem to evaluate
5. (Geometric Application): Use Green’s Theorem to calculate the area of the region R bounded by
the asteroid: 𝑥 = acos3 𝑡 , 𝑦 = asin3 𝑡 , 0 ≤ 𝑡 ≤ 2𝜋.
6. Evaluate ∮𝐶 𝑥𝑦𝑑𝑥 + 𝑥 2 𝑑𝑦 , where C is the rectangle with vertices (0, 0), (3, 0), (3, 1), (0, 1)
oriented counter-clockwise.
2
7. Compute ∮𝐶 [2𝑦 + √1 + 𝑥 5 ]𝑑𝑥 + [5𝑥 + 𝑒 𝑦 ]𝑑𝑦 ; where 𝐶 ∶= 𝑥 2 + 𝑦 2 = 4.
8. (Green’s theorem with multiple boundary components):

4.4 Surface Integrals

4.4.1 Preparation for Surface Integrals
A. Parametric Equations of surfaces
A surface Σ in 𝑥𝑦𝑧 −space can be described by

a) a Cartesian equation of the form 𝑓 (𝑥, 𝑦, 𝑧) = 0 or

b) a vector equation of the form 𝒓(𝑢, 𝑣) = x(u, v) i + y(u, v)j + y(u, v)k (where (u ,v ) is a point in
some region on uv-plane).In this case 𝒓 is called a parameterization of the surface and R is called
the parameter domain. The vector equation above can be expressed as a system of three scalar
equations:
𝑥 = 𝑥(𝑢, 𝑣)
𝑦 = 𝑦(𝑢, 𝑣)
𝑧 = 𝑧(𝑢, 𝑣)
{(𝑢, 𝑣) ∈ 𝑅 ⊆ ℝ2

z
Σ

r (u, v)
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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

x
v R

(u, v)

Example 1(Parameterizations of some common surfaces):

a) (Sphere): The equation 𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑅 2 represents the sphere with center (0, 0, 0) and
radius 𝑅. In spherical coordinates
𝑥 = 𝑅 𝑠𝑖𝑛 𝜙 𝑐𝑜𝑠𝜃, 𝑥 = 𝑅 𝑠𝑖𝑛 𝜙 𝑠𝑖𝑛 𝜃 𝑎𝑛𝑑 𝑧 = 𝑅𝑐𝑜𝑠 𝜙.
The values of 𝜃 ranges from 0 to 2𝜋 and the values of 𝜙 ranges from 0 to 𝜋. If we let u = 𝜃
and v= 𝜙 we get the set of equations :
𝑥 = 𝑎 𝑐𝑜𝑠𝑢 sin 𝑣
{ 𝑦 = 𝑎 𝑠𝑖𝑛𝑢 sin 𝑣
𝑧 = 𝑎 𝑐𝑜𝑠𝑣
Thus, a parameterization of the sphere with radius 𝑅 and center at the origin is:

⃗⃗⃗⃗⃗⃗⃗⃗ = 𝑅 𝑐𝑜𝑠 𝑢 sin 𝑣 𝒊 + 𝑅 sin 𝑢 𝑠𝑖𝑛 𝑣 𝒋 + 𝑅 𝑐𝑜𝑠 𝑣 𝒌; R: 0 ≤ u ≤2𝜋, 0 ≤ v ≤ π.

𝑟(𝑡)

b) (Any surface of the form z = f(x, y)): A parameterization of the surface z = f(x, y) with domain
R is :
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝑟(𝑢, 𝑣) = 𝑢 𝒊 + 𝑣 𝒋 + 𝑓(𝑢, 𝑣)𝒌 ; (𝑢, 𝑣) 𝜖 𝑅.
In particular, a parameterization of the plane ax + by + c z = d is ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝑟(𝑢, 𝑣) = 𝑢 𝑖 + 𝑣𝑗 +
𝑑−𝑎𝑢−𝑏𝑣)
( )𝑘 (if 𝑐 ≠ 0).
𝑐
Surfaces of the form y= f(x, z)) and x = f(y, z)) can be parameterized in analogous ways.
c) (Cylinder given by ): A parameterization of the cylinder 𝑥 2 + 𝑦 2 = 𝑎2 ; −1 ≤ 𝑧 ≤ 1 is:
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝑟(𝑢, 𝑣) = 𝑎 𝑐𝑜𝑠 𝑢𝒊 + 𝑎 sin 𝑢 𝒋 + 𝑣 𝒌; R: 0 ≤ u ≤ 2π, -1 ≤ v ≤ 1.
d) (Cone): A parameterization of the cone z =√𝑥 2 + 𝑦 2 ; 0 ≤ 𝑧 ≤ ℎ is:
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝑟(𝑢, 𝑣) = 𝑢 𝑐𝑜𝑠 𝑣𝒊 + 𝑢𝑠𝑖 𝑛 𝑣𝒋 + 𝑢𝒌; 𝑅: 0 ≤ 𝑢 ≤ ℎ, 0 ≤ 𝑣 ≤ 2𝜋.
e) (Surfaces of revolution): If the graph z = f(x), a ≤ x ≤ b is rotated about the z-axis then the
resulting surface has a parameterization:
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝑟(𝑢, 𝑣) = 𝑢 𝑐𝑜𝑠 𝑣𝒊 + 𝑢𝑠𝑖 𝑛 𝑣𝒋 + 𝑓(𝑢)𝒌; 0 ≤ 𝑢 ≤ 𝑏, 0 ≤ 𝑣 ≤ 2𝜋.

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

𝑥2 𝑦2 𝑧2
f) (Ellipsoid): A parameterization of the ellipsoid + + = 1 is:
𝑎2 𝑏2 𝑐2
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝑟(𝑢, 𝑣) = 𝑎 𝑐𝑜𝑠 𝑢 sin 𝑣 𝑖 + 𝑐 𝑠𝑖𝑛 𝑢 𝑠𝑖𝑛 𝑣 𝑗 + 𝑐𝑐𝑜𝑠𝑣 𝑘; R: 0 ≤ u ≤2𝜋, 0 ≤ v ≤ π.
g) (Torus): A torus is a 3-D figure formed by rolling a rectangle into a cylinder and bending the
cylinder until its bases meet. Equivalently we can obtain it by rotating a circle C in a
plane about a line perpendicular to the plane. The parameterization of the torus
obtained by rotating the circle with center (R, 0, 0) and radius a about y axis is:
r (u, v) = ((R + a cos u) cos v) i + ((R + r cos u) sin v) j + ((a sin u) k

Figure 4: Torus

B. Normal vectors and Surface area

Theorem 1 [Normal vector of a Parametric Surface]: Let
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝑟(𝑢, 𝑣)=〈𝑥(𝑢, 𝑣), 𝑦(𝑢, 𝑣), 𝑧(𝑢, 𝑣)〉; (𝑢, 𝑣) 𝜖 𝑅 ⊆ ℝ2 be a parameterization of a surface, Σ. Define

𝜕𝑥 𝜕𝑦 𝜕𝑧
𝑟⃗⃗⃗𝑢 (𝑢, 𝑣): = 〈𝜕𝑢 (𝑢, 𝑣), 𝜕𝑢 (𝑢, 𝑣), 𝜕𝑢 (𝑢, 𝑣)〉 ,

𝜕𝑥 𝜕𝑦 𝜕𝑧
𝑟𝑣 (𝑢, 𝑣): = 〈𝜕𝑣 (𝑢, 𝑣), 𝜕𝑣 (𝑢, 𝑣), 𝜕𝑣 (𝑢, 𝑣)〉 and
⃗⃗⃗

⃗ := 𝑟⃗⃗⃗𝑢 (𝑢, 𝑣) × 𝑟⃗⃗⃗𝑣 (𝑢, 𝑣)

𝑁

If 𝑁 ⃗ , then 𝑁
⃗ ≠0 ⃗ is normal to the surface Σ at 𝑟(𝑢, 𝑣).

Example 2 [normal vector to a parametric surface]: Find the surface unit normal 𝑛⃗and the
equation of the tangent plane to the cylinder 𝑟⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
(𝑢, 𝑣) = 〈3 𝑐𝑜𝑠 (𝑢), 3 𝑠𝑖𝑛 (𝑢), 𝑣〉 𝑎𝑡 𝑟(𝜋, 2) =
(3, 0, 2).

Answer: 𝑛⃗ = 〈−1,0,0〉 and the equation of the tangent plane is 𝑥 = −1.

Example 3[Equation of tangent plane to a surface]: Find an equation of the tangent

plane to the parametric surface given by: 𝑟(𝑢, 𝑣)= 〈𝑢 + 𝑣, 3𝑢2 , 𝑢 − 𝑣〉at the point(2, 3, 0).

Answer: 3𝑥 − 𝑦 + 3𝑧 = 3.

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Theorem 2[Surface Area of a Parametric Surface]: The surface area S of the surface Σ with
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
parameterization 𝑟(𝑢, 𝑣)=〈𝑥(𝑢, 𝑣), 𝑦(𝑢, 𝑣), 𝑧(𝑢, 𝑣)〉; (𝑢, 𝑣) 𝜖 𝑅 ⊆ ℝ2 is given by

S = ∬R‖r⃗⃗⃗u (u, v) × ⃗⃗⃗

rv (u, v)‖dA;

(provided that Σ is traced exactly once as (u, v) ranges over all points of R.).

Example 4[surface area of a parametric surface]: Find the area of the parametric surface given by
: 𝑟(𝑢, 𝑣)= 〈𝑢𝑣, 𝑢 + 𝑣, 𝑢 − 𝑣〉;𝑢2 + 𝑣 2 ≤ 1.
𝜋
Answer: 3
(6√6 − 2 √2)

C. Topology of Surfaces: Smoothness and Orientation

Definition 1 [smooth and piecewise smooth surface]: A surface 𝒮 with parameterization

⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝑟(𝑢, 𝑣)=〈𝑥(𝑢, 𝑣), 𝑦(𝑢, 𝑣), 𝑧(𝑢, 𝑣)〉; (𝑢, 𝑣) 𝜖 𝑅 ⊆ ℝ2 is said to be

a) smooth if 𝑟⃗⃗⃗𝑢 (𝑢, 𝑣) 𝑎𝑛𝑑 ⃗⃗⃗

𝑟𝑣 (𝑢, 𝑣) are continuous and 𝑟⃗⃗⃗𝑢 (𝑢, 𝑣) × 𝑟⃗⃗⃗𝑣 (𝑢, 𝑣) is never zero on
the parameter domain.
b) piecewise smooth if it is a union of finitely many smooth surfaces
Intuitively speaking, a surface is smooth if there is no sharp point on the surface. For instance a cone
has a sharp point, hence it is not smooth.

Example 5 [smooth and piecewise smooth surfaces]: Sphere is smooth and cube is piecewise
smooth.

Definition 2 [orientable surface, oriented surface and orientation of a surface]: A surface

⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝒮 with parameterization 𝑟(𝑢, 𝑣)=〈𝑥(𝑢, 𝑣), 𝑦(𝑢, 𝑣), 𝑧(𝑢, 𝑣)〉; (𝑢, 𝑣) 𝜖 𝑅 ⊆ ℝ2 is said to be:

a) Orientable (two sided) if there exists a continuous vector field 𝑛⃗ such that 𝑛⃗ (x, y, z) is normal
to the surface and ‖𝑛⃗ (𝑥, 𝑦, 𝑧)‖=1 for every (x, y, z) ∈ S. In short, S is orientable if there exists a
continuous unit normal field defined on 𝒮. Thus there are exactly two sets of normal vectors to an
orientable surface 𝒮, namely {𝑛⃗ (x, y, z) |(x, y, z) ∈ S} and {−𝑛⃗ (x, y, z) |(x, y, z) ∈ S}. A particular
choice of these two sets is called an orientation of 𝒮. (Read also what we mean by orientation of a piecewise
smooth surface).

b) oriented if one direction of flow through 𝒮 has been chosen as positive direction. Often a
surface is oriented by choosing a normal vector filed that indicates the required direction.

Roughly speaking, a surface is orientable if it has two sides (like inside and outside, left side and right
side, upper side and lower side, etc). If a surface is orientable it can be can be penetrated in two
opposite directions (that is, there are two opposite directions of flow through the surface).

Example 6 (orientable and non-orientable surface):

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

Most surfaces we encounter in the physical world are orientable. Spheres, planes, and tori are
orientable, for example. Every smooth closed surface is orientable. But Möbius strips, Roman
surfaces, real projective planes, and Klein bottles are non-orientable.

Figure 5: Mobius Strip

Definition 3 [simple and closed surfaces]: A surface 𝒮 is said to be:

a) simple if it never touches or intersects itself.
b) closed if it encloses a solid.

Definition 4(orientation of boundary of a surface): Let 𝒮be an oriented non-closed surface with
⃗ . The
surface normal/orientation/ 𝑛⃗ . Let 𝒞 be the boundary curve of S with unit tangent vector 𝑇
⃗ is called the positive direction (orientation) of C if 𝑛⃗ × 𝑇
direction of 𝑇 ⃗ points into S.

Informally, the positive direction of 𝒞 is the direction a person has to walk on 𝒞 in such a way that
his/her head points in the direction of the surface normal while keeping the surface to the left.
4.4.2 Surface Integrals of a Scalar Field
The surface integrals we consider are the following types:

a) surface integral of scalar field:  f ( x, y, z)dydz


b) Surface integrals in differential forms:
- Reduced Surface integrals:  Mdydz ,  Ndxdz ,  Pdxdy
  
- Surface integral in differential form:  Mdydz +  Ndxdz +  Ndxdy
  
c) Surface integral of vector field (Flux Integral): ∬Σ ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝐹(𝑥, 𝑦, 𝑧) ∙ 𝑛⃗ 𝑑𝑠

A. Motivation: Mass of a Surface with given variable density

Let Σ be a surface in xyz space. Suppose that the density per unit area of the surface is given at each
point (x, y, z) of  by the function f(x, y, z). What is the mass of the surface? Let {Σ 1 , …, Σn}

be a partition of  and S k be the area of Σk. Let (x* k, y* k, z* k ) be any point in Σk . Then f (x*

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

k, y* k, z* k) is the density of f at (x* k, y* k, z* k ) . Then, if S k is very, very small, then the mass m
k of the subsurface S k of  can be approximated as

m k  f(x k, y k, z k) S k

(Because density is mass per unit area, for a two dimensional body and because the density g is
almost constant on S k , as S k is almost a point by assumption). The total mass M of the surface
is approximated as

n
M 1
f(x* k, y* k, z* k) S k

If n   as Δ𝑆  0 , then

n
M= lim
n

1
f(x* k, y* k, z* k ) S k (1)

The limit in (1) can be used to express not only mass but other quantities. We call it the surface
integral of f over  and we denote it by

 f ( x, y, z )ds .


n
Hence  f ( x, y, z )ds = lim
n
 f(x k, y k, z k) S k
 1

NB. If Δ𝑆  0 , then n   .But the converse is not necessarily true.

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

B. Definition of surface Integral of a scalar field

Definition 5: Let  be a finite, smooth and simple surface in xyz space. Let 𝑓 (𝑥, 𝑦, z) be a
continuous scalar field defined at each point of Σ . Let {Σ1, Σ2…, Σ n} be a partition of Σ as shown
in the figure. Then the surface integral of f over Σ (with respect to surface area) is denoted and defined
by

 f ( x, y, z )ds = lim
n
 f(x*k, y*k, z*k ) S k
 1

Figure 6: A Partition of a surface

(Where S k is the area of Σ𝑘 , (x*k, y*k , z*k) is any point in Σ k, and max {Δ𝑆𝑘 }𝑛𝑘=1 tends to 0 as n
  .)

C. Evaluation: The basic idea is to convert into double integral.

⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
If 𝑟(𝑢, 𝑣)=〈𝑥(𝑢, 𝑣), 𝑦(𝑢, 𝑣), 𝑧(𝑢, 𝑣)〉, (𝑢, 𝑣) 𝜖 𝑅 is a parameterization of Σ and if f(x, y, z) is
a scalar filed defined on Σ, then

 f ( x, y, z)ds   f ( x(u, v), y(u, v), z(u, v)) r u  rv dA

 R

Example 7 (surface integral over the graph of z =f(x, y)): Compute the surface integral

∬Σ(𝑥𝑦 + 𝑧)𝑑𝑠, where S is that part of the plane x + y+ z = 2 in the first octant.

Answer: 2√3.

The following properties may help in evaluating surface integrals.

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

D. Properties

a) ∬S kf = k ∬S f
b) ∬S f + g = ∬S f + ∬S g
c) ∬S f − g = ∬S f − ∬S g
d) ∬S f = ∬S f + ∬S f
1 ∪S2 1 2

Example 8 (surface integral of a scalar field over a piecewise smooth surface): Evaluate
∬Σ(𝑥 + 𝑦)𝑑𝑠, where ∑ is the cube with vertices (0,0,0), (1,0,0), (1,1,0), (1,0,1), (1,1,1) and (0,1,1)

Ans. 6

Example 9: Evaluate ∬Σ 𝑧𝑑𝑠

Where ∑ is the upper half of the sphere centered at origin and with radius 2. Ans. 8π

Example 10: Evaluate ∬Σ 𝑦𝑑𝑠

Where ∑ is the portion of the cylinder x2 + y2 = 3 that lies between z=0 and z=6. Ans. 0

Example 11: Evaluate∬Σ(𝑦 + 𝑧)𝑑𝑠, where ∑ is the surface whose side is the cylinder x2 + y2 = 3
and whose bottom is the disk x2 + y2 ≤ 3 in the xy-plane and whose top is the plane 𝑧 = 4 − 𝑦..
𝜋
Answer. 2 [29√3 + 24√2 ]

4.4.3 Surface Integral of Vector fields(Flux Integral)

Definition 6:

Let Σ be an orientable finite surface. Let 𝐹 = 𝑀𝒊 + 𝑁𝒋 + 𝑃𝒌 be a vector field defined at each

point of Σ . Let 𝑛⃗ be an orientation of Σ . Let {Σ1, Σ2…, Σ n} be a partition of Σ as shown in the
figure. Then we denote the surface integral of F over 𝛴 in the direction of 𝑛⃗ by


 F .nds or  F . ds and define it as

 
n

 F .nds  lim n

 F (x *
k 1
k , y *k , z *k ).nS k
S

where

for each k=1,…,n , ∆Sk is the surface area of Σ𝑘 ,

(x*k ,y*k, z*k )  Σ𝑘 and

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

Provide that ‖Δ𝑆‖ ≔ max{∆S1 , ∆S2 , , … , ∆Sn }→0 as n→∞.

* Thus flux is a scalar quantity, defined as the surface integral of the component of a vector field
perpendicular to the surface at each point.

The integral can be interpreted physically as the rate of flow of F through S.

Evaluation

Let ∑ be a smooth surface with parameterization

𝑟 (𝑢, 𝑣) = 𝑥(𝑢, 𝑣) 𝑖 + 𝑦(𝑢, 𝑣)𝑗 + 𝑧(𝑢, 𝑣)𝑘

⃗⃗⃗ over ∑ in the direction of the unit vector

The surface integral of a vector field 𝐹

𝑛⃗ can be computed using

 F .nds   F (r (u, v)).r u  rv dA

 R

Where

𝑅 is the projection of ∑ onto uv-plane.

𝑑𝐴 is a differential area element on uv-plane and

𝑑𝑠 is differential surface element [𝑑𝐴 is the projection of 𝑑𝑠 onto uv-plane].

Example 12: Evaluate  F.nds

S

Where 𝐹 = 〈𝑥 2 , 0, 3𝑦 2 〉

S: portion of the plane 𝑥 + 𝑦 + 𝑧 = 1 in the first octant

n is chosen so that its k component is positive.

Ans. 1/3

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

Figure 7: Portion of the Plane x+y+z=1 in the first octant

Example 13: Evaluate  F.nds where 𝐹 ≔ 𝑧𝒌 and S is the unit sphere centered at the origin.
S

4.4.4 Surface Integrals in Differential Forms

Definition 7: Let the vector field 𝑛⃗ be an orientation of a smooth finite surface S. Let α, β and 𝛾 be
the angles between 𝑛⃗ and i, j and k respectively. Let 𝑓 be a scalar field. We define

a. ∬𝑆 𝑓𝑑𝑥𝑑𝑦 = ∬𝑆 𝑓𝑐𝑜𝑠𝛾𝑑𝑠

b. ∬𝑆 𝑓𝑑𝑦𝑑𝑧 = ∬𝑆 𝑓𝑐𝑜𝑠𝛼𝑑𝑠

c. ∬𝑆 𝑓𝑑𝑧𝑑𝑥 = ∬𝑆 𝑓𝑐𝑜𝑠𝛽𝑑𝑠

Caution: 𝑑𝑥𝑑𝑦 ≠ 𝑑𝑦𝑑𝑥

Definition 8: Let 𝐹 = 𝑀𝒊 + 𝑁𝒋 + 𝑃𝒌 be a vector field defined at each point of a smooth surface

S of finite area. We define:

∬𝑆 𝑀𝑑𝑦𝑑𝑧 + 𝑁𝑑𝑧𝑑𝑥 + 𝑃𝑑𝑥𝑑𝑦 = ∬𝑆 𝑓𝑑𝑥𝑑𝑦 + ∬𝑆 𝑓𝑑𝑦𝑑𝑧+ ∬𝑆 𝑓𝑑𝑧𝑑𝑥

Evaluation: ∬𝑆 𝑓𝑑𝑥𝑑𝑦 = ∬𝑆 𝑓(𝑥(𝑢, 𝑣), 𝑦(𝑢, 𝑣), 𝑧(𝑢, 𝑣))𝑛. 𝑘𝑑𝐴

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

Exercise 4.4:

1. Find a parameterization of the surface given by the paraboloid 𝑧 = 𝑥 2 + 𝑦 2 and below the
plane 𝑧 = 4.
2. Find the surface area of the surface given by the parameterization 𝑟(𝑠, 𝑡) ≔ 〈−𝑠 − 𝑡, 𝑠 +
2𝑡, 2𝑠 + 𝑡〉 where 0 ≤ 𝑠 ≤ 1 and 0 ≤ 𝑡 ≤ 1.
3. Evaluate the surface integral ∬𝑆 𝑓𝑑𝑠 where 𝑓(𝑥, 𝑦, 𝑧) = 𝑥 2 over the unit sphere 𝑥 2 + 𝑦 2 +
𝑧 2 = 1.
⃗⃗⃗⃗ where 𝐹 (𝑥, 𝑦, 𝑧) = 〈𝑧 2 , 𝑥, −3𝑧〉 and S is the surface
4. Evaluate the surface integral ∬𝑆 𝐹 . 𝑑𝑠
cut from the parabolic cylinder 𝑦 = 𝑥 2 with −1 ≤ 𝑥 ≤ 1 by the planes 𝑧 = 0 and 𝑧 = 2.
5. Parameterize the hyperboloid 𝑥 2 − 4𝑦 2 + 𝑧 2 = 1.
6. Find a parametric representation for the part of the plane z = x + 3 that lies inside the
cylinder x2 + y2 = 1.
7. Parameterize the upper hemisphere of the sphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 9.
8. Find a parameterization of the portion of the sphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 1.between the
1 1
planes z=− 2 &𝑧 = 2.
9. Find a unit vector with positive z component which is normal to the surface 𝑧 = 𝑥 4 𝑦 +
𝑥𝑦 2 at the point (1,1,2) on the surface.
10. Find the equation of the tangent plane to the surface given by :
𝑟(𝑢, 𝑣) ≔ 〈𝑢, 2𝑣 2 , 𝑢2 + 𝑣〉at the point (2,2,3).

11. For the cylinder of radius 3 and height 5 given by x2 + y2 = 32 and 0  z  5, let the charge
density be proportional to the distance from the xy-plane. Find the total charge on the
cylinder. Ans. 75𝜋
12. [Mass problem]: Find the mass of the lamina that is the portion of 𝑥 2 + 𝑦 2 − 3𝑧 − 1 =
9
0 inside 𝑥 2 + 𝑦 2 = 9/4. The density is δ(x, y, z) = (4 + 𝑥 2 + 𝑦 2 ).

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

4.5 Divergence theorem or Stokes Theorem

A. Divergence Theorem
Divergence theorem relates triple integral over a solid with surface integral over the
bounding surface of the solid. It was stated first by Gauss; hence it is also called Gauss’s theorem. It can
be viewed as a generalization of Green’s theorem; it is analogous to the divergence form of Green’s
Theorem.
Divergence theorem is useful for evaluating triple integrals via surface integrals or vice versa.
It is also useful in the study of physical problems like heat flow, fluid flow, potential theory and
electro-magnetic theory.
Definition 1: The flux of a three-dimensional vector field F across an oriented surface S in the
direction of unit normal vector n is

Flux = ∬𝑆 𝑭. 𝒏𝑑𝑠

The definition of the flux of a two dimensional field across a curve is given in section 4.3.

Theorem 1 [Divergence Theorem]

Let

i) D be a closed and bounded three dimensional region in xyz-space with piecewise smooth
and orientable boundary ∑ that is oriented outward.

ii) ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝐹(𝑥, 𝑦, 𝑧) = 𝑀(𝑥, 𝑦, 𝑧)𝑖 + 𝑁(𝑥, 𝑦, 𝑧)𝑗 + 𝑃(𝑥, 𝑦, 𝑧)𝑘 be a vector field for which M,N and P
are continuous and have continuous first order partial derivatives in a three dimensional region
containing D.Then

∬ 𝐹 . 𝑛⃗𝑑𝑠 = ∭ 𝑑𝑖𝑣(𝐹 )𝑑𝑣

Σ 𝐷

(In words: The flux of a vector field 𝐹 across a closed oriented surface ∑ in the direction of the
surface outward unit normal field 𝑛⃗ equals the triple integral of div (𝐹 ) over the region D enclosed by
the surface).

Partial Proof:

Case 1: If T is a simple region (i.e., the intersection of T with every line parallel to the coordinate
planes is a point or a single segment)

Case 2: if T is divisible to finite number of simple regions

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

Case 3: The general case

Example 1[using triple integral to evaluate surface integral]: Use Divergence theorem to evaluate

∬ 𝐹 ∙ ⃗⃗⃗
𝑛 𝑑𝑠
Σ

where 𝐹 (𝑥, 𝑦, 𝑧) = 𝑥𝒊 − 2𝑦 2 𝒋 + 𝑧 2 𝒌 , Σ is the boundary surface of the solid bounded by the

cylinder 𝑥 2 + 𝑦 2 = 4 and the planes z=0 and z=3 and 𝑛 ⃗⃗⃗ is the unit surface normal vector field
of Σ that is directed outwards.

Solution: ∬Σ 𝐹 ∙ 𝑛
⃗⃗⃗ 𝑑𝑠 = ∭𝐷 𝑑𝑖𝑣(𝐹 )𝑑𝑣 = ∭𝐷(1 − 4𝑦 + 2𝑧)𝑑𝑣 = 48𝜋

Example 2: Verify divergence theorem for the problem of Example 1.

Solution: It follows from the solution of Example 47 that ∭𝐷 𝑑𝑖𝑣(𝐹 )𝑑𝑣 = 48𝜋. It remains to
show that ∬Σ 𝐹 ∙ 𝑛 ⃗⃗⃗ 𝑑𝑠= 48𝜋, without using divergence theorem. Observe that Σ is a piecewise
smooth surface bounding a cylinder. It is composed of two circular regions and a circular
cylindrical surface. The circular faces form the upper and lower faces of the cylinder. Let Σ1, Σ2
& Σ3 be respectively the upper face, the lateral face and the lower face of Σ. Each of these
surfaces can be parameterized as follows:

𝛴1: x = u, y = v, z = 3; (u, v)𝜖 R where R is the disk in uv-plane with center (0,0) and radius 2

𝛴2: x= 2cos u, y = 2 sin u , z = v ; (u, v)𝜖 R where R is the rectangle in uv-plane bounded by
u = 0, u=2𝜋, v = 0 and v=2𝜋

𝛴3: x = u, y = v, z = 0; (u, v)𝜖 R where R is the disk in uv-plane with center (0,0) and radius 2

The outward unit surface normal vector to 𝛴1 and 𝛴3 is the standard unit vector k and a set of
𝑟 ×𝑟 〈−2𝑠𝑖𝑛𝑢,2𝑐𝑜𝑠𝑢,0〉×〈0,0,1〉
unit normal vectors to 𝛴2: is given by n =± ‖𝑟𝑢 ×𝑟𝑣 ‖ =± ‖〈−2𝑠𝑖𝑛𝑢,2𝑐𝑜𝑠𝑢,0〉×〈0,0,1〉‖ = ±
𝑢 𝑣
〈−𝑐𝑜𝑠𝑢, −𝑠𝑖𝑛𝑢, 0〉. For u =0 , the unit vector n has to be the standard unit vector i ; in view of
the assumption that the unit vector has to be outward. Thus n = 〈𝑐𝑜𝑠𝑢, 𝑠𝑖𝑛𝑢, 0〉.

⃗⃗⃗ 𝑑𝑠 = ∬𝑅〈𝑢, −2𝑣 2 , 9〉 ∙ 〈0,0,1〉 𝑑𝐴 = 36𝜋. Likewise we obtain ∬Σ 𝐹 ∙ ⃗⃗⃗

∬Σ 𝐹 ∙ 𝑛 𝑛 𝑑𝑠 =0 and
1 2

∬Σ 𝐹 ∙ 𝑛
⃗⃗⃗ 𝑑𝑠 = 12𝜋.
3

Thus ∬Σ 𝐹 ∙ 𝑛
⃗⃗⃗ 𝑑𝑠 = ∬Σ 𝐹 ∙ 𝑛
⃗⃗⃗ 𝑑𝑠 + ∬Σ 𝐹 ∙ ⃗⃗⃗
𝑛 𝑑𝑠 ∬Σ 𝐹 ∙ ⃗⃗⃗
𝑛 𝑑𝑠 = 36𝜋+0+ 12𝜋= 48𝜋.
1 2 3

The divergence theorem can also be used to evaluate triple integrals by turning them into surface
integrals. This depends on finding a vector field whose divergence is equal to the given function.

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

Example 3[using surface integral to compute triple integral]: Let E be the region defined by
𝑥 2 + 𝑦 2 + 𝑧 2 ≤ 1. Use the divergence theorem to evaluate

∭ 𝑧 2 𝑑𝑣
𝐸

4𝜋
Answer: 15

B. Stokes’sTheorem
Stokes’s Theorem relates surface integral over a surface with line integral over the boundary
curve of the surface (assuming that the surface is not a closed surface).It was appeared first in the
form of an examination question at Cambridge University. It can be viewed as a generalization of
Green’s theorem; it is analogous to the curl form of Green’s Theorem.
Stokes’s theorem is useful in computing line integrals using surface integrals or vice versa. It can
also be used in solving certain physical problems like heat conduction.
Since we will be working in three dimensions, we need to discuss what it means for a boundary
curve of a surface to be oriented positively. Let S be an oriented surface with unit normal vector n
and let C be the boundary of S. Then, we say C is positively oriented if its orientation follows the right
hand rule, that is if your right hand curls around n in the direction of C's orientation, then your
thumb will be pointing in the direction of n.

Definition 2(circulation): The circulation of a three-dimensional vector field F around an oriented

closed path C in the direction of unit tangent vector T is

Circulation =

In short, circulation is the line integral of a vector field around a closed path. The definition of
circulation of a two dimensional vector filed is given in section 4.3.

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

Theorem 2(Stokes's Theorem):

Let

i) S be an oriented surface with unit normal vector n and C be the positively oriented boundary
of S.

ii) F be a vector field defined with continuous first order partial derivatives in a three
dimensional region containing S.

Then

∫ 𝑭. 𝒅𝒓 = ∬ 𝑐𝑢𝑟𝑙 (𝑭). 𝒏𝑑𝑠

𝐶 𝑆

(In words: the line integral of the tangential component of F along C equals the surface integral of
the normal component of curl (F) over S.)

Note: The surface integral in Stokes’s theorem is taken over a closed surface hence it is called a
closed surface integral. Closed surface integral is sometimes denoted using the integral symbol

Example 4 (Using Surface Integral to Compute Line Integral):

Let S be the part of the plane z = 4 - x - 2y in first octant and with upward pointing unit normal
vector. Use Stokes' theorem to find

Where F = yi + zj – xyk and C is the boundary of S with positive orientation.

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

Solution: First notice that without Stokes' theorem, we would have to parameterize three different
line segments. Instead we can find this with just one double integral. We have

and n dS = i + 2j + k

So that Curl F. n dS = 1 + x + 2y - 1 = x 2y

We integrate

Example 5(Using Line Integral to Compute Surface Integral): Use Stokes’ Theorem to evaluate

∬𝑆 𝑐𝑢𝑟𝑙(𝑭) . 𝒏𝑑𝑠 where F = 𝑥 2 𝒊 − 3𝑥𝑦𝒋 + 𝑥 3 𝑦 3 𝒌 and S is the part z = 5-x2 –y2 above the plane z
= 1. Assume that S is oriented upwards.

Answer: 0

Example 6: Verify Stokes's Theorem for F = y2i - xj + 5zk if S is the paraboloid z = x2 + y2 with the
circle x2 + y2 = 1 as its boundary.

Answer: ∬𝑆 𝑐𝑢𝑟𝑙(𝑭) . 𝒏𝑑𝑠 = −𝜋 and ∫𝐶 𝐹. 𝑑𝑟 = −𝜋.

Example 7: Suppose S1 and S2 are two oriented surfaces that share C as boundary. What can you
say about ∬𝑆1 𝑐𝑢𝑟𝑙(𝐹). 𝑛𝑑𝑠 and ∬𝑆2 𝑐𝑢𝑟𝑙(𝐹). 𝑛𝑑𝑠 ?

Example 8: Let S be a solid sphere. Show that ∫∫ (∇XF)·n dS = 0 by using

a) Divergence theorem
b) Stokes’s theorem

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Math 331: Applied mathematics III Tadesse Bekeshie(PhD)

Exercise 4.5:

1. Verify the divergence theorem for the vector field ⃗⃗⃗

𝐹 = 𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌 over the sphere 𝒮
of radius a centered at the origin .
2. Verify the divergence theorem for the vector field ⃗⃗⃗
𝐹 = 𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌 𝒮 where 𝒮 is the
boundary of the solid bounded by the coordinate planes and the plane 2𝑥 + 2𝑦 + 2𝑧 = 6 .
3. Verify the divergence theorem for the vector field ⃗⃗⃗
𝐹 = 𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌 over the sphere 𝒮
of radius a centered at the origin .
4. Verify divergence theorem for the vector field 𝐹 = (4𝑥)𝒊 − (2𝑦2)𝒋 + (𝑧2)𝒌 taken over the
region bounded by 𝑥 2 + 𝑦 2 = 4, 𝑧 = 0 and 𝑧 = 3.
⃗⃗⃗⃗ , where S is the sphere 𝑥 2 + 𝑦 2 +
5. Use Divergence theorem to evaluate∬𝑆 3𝑥𝑖 + 2𝑦𝑗. 𝑑𝑠
𝑧 2 = 9.

6. Verify Stokes’s theorem for the given vector field 𝐹 and surface Σ where 𝐹 (𝑥, 𝑦, 𝑧): =

2𝑦𝑖 − 𝑥𝑗 + 𝑧𝑘⃗; 𝛴: 𝑥2 + 𝑦2 + 𝑧2 = 1, 𝑧 ≥ 0.
2
7. Use Stokes theorem to compute the flux of 𝐹 through the upper half ellipsoid 𝑥 2 + 𝑦9 +
𝑧2
4
= 1, 𝑧 ≥ 0 , oriented by its outward normal.

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