Chapter 2
Vector Calculus
1. Elementary
2. Vector Product
3. Differentiation of Vectors
4. Integration of Vectors
5. Del Operator or Nabla (Symbol )
6. Polar Coordinates
1
Chapter 2 Continued
7. Line Integral
8.
Volume Integral
9.
Surface Integral
10. Green’s Theorem
11. Divergence Theorem (Gauss’ Theorem)
12. Stokes’ Theorem
2
2.1 Elementary Vector Analysis
Definition 2.1 (Scalar and vector)
Scalar is a quantity that has magnitude
but not direction.
For instance mass, volume, distance
Vector is a directed quantity, one with
both magnitude and direction.
For instance acceleration, velocity, force
3
We represent a vector as an arrow from the
origin O to a point A.
A
O
OA
A
or
a
O
The length of the arrow is the magnitude of
the vector written as OA or a .
4
2.1.1 Basic Vector System
Unit vectors , ,
• Perpendicular to each other
• In the positive directions
of the axes
• have magnitude (length) 1
5
Define a basic vector system and form a
right-handed set, i.e
6
2.1.2 Magnitude of vectors
Let P = (x, y, z). Vector OP =
p
is defined by
OP = p = x i + y j + z k
= [x, y, z]
with magnitude (length)
OP = p =
x +y +z
2
2
2
7
2.1.3 Calculation of Vectors
1. Vector Equation
Two vectors are equal if and only if the
corresponding components are equals
Let a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k .
Then
a = b  a1 = b1 , a2 = b2 , a3 = b3
8
2. Addition and Subtraction of Vectors
a  b = (a1  b1 )i + (a2  b2 ) j + (a3  b3 )k
3. Multiplication of Vectors by Scalars
If  is a scalar, then
 b = (b1 )i + (b2 ) j + (b3 )k
9
Example 2.1
Given p = 5i + j - 3k and q = 4i - 3j + 2k . Find
a) p + q
b) p - q
c) Magnitude of vector p
d) 2q - 10p
10
2.2
Vector Products
If a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k ,
~
~
~
~
~
~
~
~
1) Scalar Product (Dot product)
a  b = a1b1 + a2b2 + a3b3
~
~
or a . b =| a || b | cos  ,  is the angle between a and b
~
~
~
~
2) Vector Product (Cross product)
i
j k
~
~
a  b = a1 a2
~
~
b1
b2
~
a3
b3
=  a2b3 - a3b2  i -  a1b3 - a3b1  j +  a1b2 - a2b1  k
~
~
~
11
3) Application of Multiplication of Vectors
a) Given 2 vectors
is defined by
comp b a =
a
and
b , projection a onto b
a .b
~
a
~
|b|
~
length (l ) =
| a .b |
~
~
b
compb a
|b|
~
b) The area of triangle
1
A = a b .
2 ~ ~
b
a
12
c) The area of parallelogram
A=
a
a xb
b
d) The volume of tetrahedrone
1
V=
6
a1
a . b xc
=
1
b1
6
c1
a2
a3
b2
b3
c2
c3
a
b
c
e) The volume of parallelepiped
a1
a2
a3
V = a . b x c = b1
c1
b2
b3
c2
c3
a b
c
13
Example 2.3
Given a = 2 i + 3 j - k and b = i + 2 j + k ,
~
~
~
~
~
~
~
~
determine a . b, a  b and the angle between a and b .
~
~
~
~
~
~
14
2.4 Vector Differential Calculus
• Let A be a vector depending on parameter u,
A(u ) = ax (u ) i + a y (u ) j + az (u ) k
~
~
~
~
• The derivative of A(u) is obtained by
differentiating each component separately,
dA
da y
dax
daz
=
i+
j+
k
du
du ~ du ~ du ~
~
15
• The nth derivative of vector A(u ) is given by
~
dn A
n
n
n
d
a
d
a
d
az
y
x
~
=
i+
j+
k.
n
n ~
n
n ~
du
du
du ~ du
dn A
• The magnitude of
du
n
d A
du
~
n
~
n
is
2
 d a x   d a y   d az 
=  n  +  n  +  n 
 du   du   du 
n
2
n
n
2
16
Example 2.4
If A = 3u 2 i - 2u j + 5 k
~
~
~
~
hence
dA
~
du
d2 A
du
~
2
=
=
17
Example 2.5
The position of a moving particle at time t is given
by x = 4t + 3, y = t2 + 3t, z = t3 + 5t2. Obtain
• The velocity and acceleration of the particle.
• The magnitude of both velocity and acceleration
at t = 1.
18
Solution
• The parameter is t, and the position vector is
r (t ) = (4t + 3) i + (t 2 + 3t ) j + (t 3 + 5t 2 ) k .
~
~
~
~
• The velocity is given by
dr
~
dt
= 4 i + (2t + 3) j + (3t 2 + 10t ) k .
~
~
~
• The acceleration is
d2 r
dt
~
2
= 2 j + (6t + 10) k .
~
~
19
• At t = 1, the velocity of the particle is
d r (1)
~
dt
= 4 i + (2(1) + 3) j + (3(1)2 + 10(1)) k
~
~
~
= 4 i + 5 j + 13 k .
~
~
~
and the magnitude of the velocity is
d r (1)
~
dt
= 42 + 52 + 132
= 210 .
20
• At t = 1, the acceleration of the particle is
d 2 r (1)
~
dt
2
= 2 j + (6(1) + 10) k
~
~
= 2 j + 16 k .
~
~
and the magnitude of the acceleration is
d 2 r (1)
~
dt 2
= 22 + 162
= 2 65.
21
2.4.1 Differentiation of Two Vectors
If both A(u ) and B(u ) are vectors, then
~
~
dA
d
a)
(c A) = c ~
du ~
du
dA dB
d
b)
( A+ B) = ~ + ~
du ~ ~
du du
dB dA
d
c)
( A . B) = A . ~ + ~ .B
~ du
du ~ ~
du ~
dB dA
d
d)
( A  B) = A  ~ + ~  B
~
du ~ ~
du du ~
22
2.4.2 Partial Derivatives of a Vector
•
If vector A depends on more than one
~
parameter, i.e
A(u1 , u 2 ,  , u n ) = a x (u1 , u 2 ,  , u n ) i
~
~
+ a y (u1 , u 2 ,  , u n ) j
~
+ a z (u1 , u 2 ,  , u n ) k
~
23
•
Partial derivative of A with respect to
~
given by
u1 is
A
a y
a x
a z
=
i+
j+
k,
u1 u1 ~ u1 ~ u1 ~
~
2 A
 2a y
2

a

az
x
~
=
i+
j+
k
u1u2 u1u2 ~ u1u2 ~ u1u2 ~
2
e.t.c.
24
Example 2.6
If F = 3uv 2 i + (2u 2 - v) j + (u 3 + v 2 ) k
~
~
~
~
then
F
~
u
F
~
v
2 F
v
~
2
= 3v 2 i + 4u j + 3u 2 k ,
~
~
~
= 6uv i - j + 2v k ,
~
~
= 6u i + 2 k ,
~
~
2 F
u
~
2 F
~
uv
=
~
2
= 4 j + 6u k ,
~
~
2 F
~
vu
= 6v i
~
25
Exercise 2.1
If F = 2u 2 v i + (3u - v 3 ) j + (u 3 + 3v 2 ) k
~
~
~
~
then
F
~
u
2 F
= ,
~
2
= ,
~
= ,
u
2 F
uv
F
~
v
2 F
=
~
2
=
~
=
v
2 F
vu
26
2.5 Vector Integral Calculus
• The concept of vector integral is the same as
the integral of real-valued functions except
that the result of vector integral is a vector.
If A(u ) = a x (u ) i + a y (u ) j + a z (u ) k
~
~
~
~
then

b
b
A(u ) du =  a x (u ) du i
a ~
a
~
b
b
+  a y (u ) du j +  a z (u ) du k .
a
~
a
~
27
Example 2.7
If F = (3t 2 + 4t ) i + (2t - 5) j + 4t 3 k ,
~
~
calculate

3
~
~
F dt.
1
~
Answer

3
1
F dt =
~

3
1
3
3
(3t + 4t ) dt i +  (2t - 5) dt j +  4t 3 dt k
2
~
1
~
1
~
= [t 3 + 2t 2 ]13 i + [t 2 - 5t ]13 j + [t 4 ]13 k
~
~
~
= 42 i - 2 j + 80 k .
~
~
~
28
Exercise 2.2
If F = (t 3 + 3t ) i + 2t 2 j + (t - 4) k ,
~
~
calculate
1

~
~
F dt.
0 ~
Answer
1

F dt =
0 ~
1

0
1
1
(t + 3t ) dt i +  2t dt j +  (t - 4) dt k
3
~
0
2
~
0
~
=
=
7
2
7
= i + j- k.
4~ 3 ~ 2 ~
29
2.6 Del Operator Or Nabla (Symbol )
• Operator  is called vector differential operator,
defined as


 
 =  i +
j + k .
 x ~ y ~ z ~ 
30
2.6.1 Grad (Gradient of Scalar Functions)
• If  x,y,z is a scalar function of three variables
and  is differentiable, the gradient of  is
defined as



grad  =  =
i+
j+
k.
x ~ y ~ z ~
* is a scalar function
*  is a vector function
31
Example 2.8
If  = x 2 yz 3 + xy2 z 2 , determine grad  at P = (1,3,2).
Solution
Given  = x 2 yz 3 + xy2 z 2 , hence

= 2 xyz3 + y 2 z 2
x

= x 2 z 3 + 2 xyz2
y

= 3 x 2 yz 2 + 2 xy2 z
z
32
Therefore,



 =
i+
j+
k
x ~ y ~ z ~
= (2 xyz3 + y 2 z 2 ) i + ( x 2 z 3 + 2 xyz2 ) j
~
~
+ (3 x 2 yz 2 + 2 xy2 z ) k .
~
At P = (1,3,2), we have
 = (2(1)(3)( 2) 3 + (3) 2 (2) 2 ) i + ((1) 2 (2) 3 + 2(1)(3)( 2) 2 ) j
~
~
+ (3(1) 2 (3)( 2) 2 + 2(1)(3) 2 (2)) k .
~
= 84 i + 32 j + 72 k .
~
~
~
33
Exercise 2.3
If  = x 3 yz + xy2 z 3 ,
determine grad  at point P = (1,2,3).
34
Solution
Given  = x 3 yz + xy2 z 3 , then

=
x

=
y

=
z
 Grad  =  = 
At P = (1,2,3),
 = 126 i + 111 j + 110 k .
~
~
~
35
2.6.1.1 Grad Properties
If A and B are two scalars, then
1) ( A + B ) = A + B
2) ( AB) = A(B ) + B(A)
36
2.6.2 Directional Derivative
Directiona l derivative of  in the direction of a is
~
d
= a . grad
ds ~
dr
where a = ~ ,
~
dr
~
which is a unit vecto r in the direction of d r .
~
37
Example 2.9
Compute the directiona l derivative of  = x 2 z + 2 xy2 + yz 2
at the point (1,2,-1) in the direction of the vector
A = 2i +3 j - 4k.
~
~
~
~
38
Solution
Directional derivative of  in the direction of
d
= a . grad
ds ~
a
~
A



where grad =  =
i+
j+
k and a = ~ .
~
x ~ y ~ z ~
A
~
Given  = x 2 z + 2 xy2 + yz 2 , hence
 = (2 xz + 2 y 2 ) i + (4 xy + z 2 ) j + ( x 2 + 2 yz ) k .
~
~
~
39
At (1,2,-1),
 = (2(1)( -1) + 2(2) 2 ) i + (4(1)( 2)
~
+ (-1) 2 ) j + ((1) 2 + 2(2)( -1)) k .
~
~
= 6 i + 9 j - 3k .
~
~
~
Also, given A = 2 i + 3 j - 4 k , then
~
~
~
~
A = 2 2 + 3 2 + ( -4 ) 2
~
= 29 .
40
Therefore, a =
~
A
~
A
=
2
3
4
i+
jk.
29 ~
29 ~
29 ~
~
dφ
Then,
= a . 
ds ~
3
4 
 2
=
i+
jk  .(6 i + 9 j - 3 k )
~
~
29 ~
29 ~  ~ ~
 29
4 
 2 
 3 

=
(6) + 
(9) +  (-3)
29 
 29 
 29 

51
=
 9.470462.
29
41
2.6.3 Unit Normal Vector
Equation  (x, y, z) = constant is a surface equation.
Since  (x, y, z) = constant, the derivative of  is
zero; i.e.
d = d r .grad  = 0
~
 d r grad  cos  = 0
~
 cos  = 0
  = 90.
42
• This shows that when  (x, y, z) = constant,
grad   d r .
~
y
grad 
ds
x
z
• Vector grad  =   is called normal vector to the
surface  (x, y, z) = constant
43
Unit normal vector is denoted by

n=
.
~

Example 2.10
Calculate the unit normal vector at (-1,1,1)
for 2yz + xz + xy = 0.
44
Solution
Given 2yz + xz + xy = 0. Thus
 = ( z + y ) i + ( 2 z + x ) j + ( 2 y + x ) k .
~
~
~
At (-1,1,1),  = (1 + 1) i + (2 - 1) j + (2 - 1) k
~
~
~
= 2 i+ j + k
~
~
~
and  = 4 + 1 + 1 = 6 .
 The unit normal vector is

n=
=
~

2 i + j+ k
~
~
6
~
1
=
(2 i + j + k )
6 ~ ~ ~
45
2.6.4 Divergence of a Vector
If A = a x i + a y j + a z k , the divergence of A is
~
~
~
~
~
defined as
div A = . A
~
~


 
=  i +
j + k .(a x i + a y j + a z k )
~
~
~
 x ~ y ~ z ~ 
a x a y a z
 div A = . A =
+
+
.
~
~
x
y
z
46
Example 2.11
If A = x 2 y i - xyz j + yz 2 k ,
~
~
~
~
determine div A at point (1,2,3).
~
Answer
a x a y a z
+
+
div A = . A =
~
~
z
y
x
= 2 xy - xz + 2 yz.
At point (1,2,3),
div A = 2(1)( 2) - (1)(3) + 2(2)(3)
~
= 13.
47
Exercise 2.4
If A = x 3 y 2 i + xy2 z j - yz 3 k ,
~
~
~
~
determine div A at point (3,2,1).
~
Answer
a x a y a z
div A = . A =
+
+
~
~
x
y
z
=
At point (3,2,1),
div A = 
~
= 114.
48
Remarks
A is a vector function, but div A is a scalar function.
~
~
If div A = 0, vector A is called solenoid vector.
~
~
49
2.6.5 Curl of a Vector
If A = a x i + a y j + a z k , the curl of A is defined by
~
~
~
~
~
curl A =   A
~
~


 
=  i +
j + k   (a x i + a y j + a z k )
~
~
~
 x ~ y ~ z ~ 
i
j
k
~

 curl A =   A =
~
~
x
ax
~

y
ay
~

.
z
az
50
Example 2.12
If A = ( y 4 - x 2 z 2 ) i + ( x 2 + y 2 ) j - x 2 yz k ,
~
~
~
~
determine curl A at (1,3,-2).
~
51
i
Solution
curl A =   A =
~
~
j
~
~

x

y
y4 - x2 z 2
x2 + y2
k
~

z
- x 2 yz
 
 2
2
2 
=  (- x yz ) - ( x + y )  i
z
 y
~
 4


2
-  (- x yz ) - ( y - x 2 z 2 )  j
z
 x
~
 2
 4
2
2 2 
+  ( x + y ) - ( y - x z )  k
y
 x
~
= - x 2 z i - (-2 xyz + 2 x 2 z ) j + (2 x - 4 y 3 ) k .
~
~
~
52
At (1,3,-2),
curl A = -(1) 2 (-2) i - (-2(1)(3)( -2) + 2(1) 2 (-2)) j
~
~
~
+ (2(1) - 4(3) 3 ) k
~
= 2 i - 8 j - 106 k .
~
~
~
Exercise 2.5
If A = ( xy3 - y 2 z 2 ) i + ( x 2 + z 2 ) j - x 2 yz 2 k ,
~
~
~
~
determine curl A at point (1,2,3).
~
53
Answer
curl A = (- x 2 z 2 - 2 z ) i - (-2 xyz2 + 2 y 2 z ) j
~
~
~
+ (2 x - 3 xy2 + 2 yz 2 ) k .
~
At (1,2,3), curl A = -15 i + 12 j + 26 k .
~
~
~
~
Remark
A is a vector function and
~
curl A is also a vector function.
~
54
2.7 Polar Coordinates
• Polar coordinate is used in calculus to
calculate an area and volume of small
elements in easy way.
• Lets look at 3 situations where des Cartes
Coordinate can be rewritten in the form of
Polar coordinate.
55
2.7.1 Polar Coordinate for Plane (r, θ)
x = r cos 
y
ds
d
y = r sin 
dS = r dr d

x
56
2.7.2 Polar Coordinate for Cylinder (, , z)
x =  cos 
z
ds
y =  sin 
z=z
dv
z

y

dS =  d dz
dV =  d d dz
x
57
2.7.3 Polar Coordinate for Sphere (r, , 
z
x = r sin  cos 
r


y = r sin  sin 
z = r cos 
y
dS = r 2 sin  d d
x
dV = r 2 sin  dr d d
58
Example 2.13 (Volume Integral)
Calculate

F dV where F = 2 i + 2 z j + y k
V ~
~
~
~
~
and V is a space bounded by z = 0, z = 4
and x 2 + y 2 = 9.
z
4 -

3
x

3
y
59
Solution
Since it is about a cylinder, it is easier if we use
cylindrical polar coordinates, where
x =  cos  , y =  sin  , z = z , dV =  d d dz
where 0    3, 0    2 , 0  z  4.
60
2.8 Line Integral
Ordinary integral  f (x) dx, we integrate along
the x-axis. But for line integral, the integration is
along a curve.
 f (s) ds =  f (x, y, z) ds
A
B
r
~
O
r+ d r
~
~
61
2.8.1 Scalar Field, V Integral
If there exists a scalar field V along a curve C,
then the line integral of V along C is defined by
Vdr
c
~
where d r = dx i + dy j + dz k .
~
~
~
~
62
Example 2.14
If V = xy2 z and a curve C is given by
x = 3u ,
then find
y = 2u 2 , z = u 3 ,
Vdr
c
~
along C
from A = (0,0,0) to B = (3,2,1).
63
Solution
Given V = xy2 z
= (3u )( 2u 2 ) 2 (u 3 ) = 12u 8 .
And,
d r = dx i + dy j + dz k
~
~
~
~
= 3 du i + 4u du j + 3u 2 du k .
~
~
~
At A = (0,0,0), 3u = 0, 2u 2 = 0, u 3 = 0,
 u = 0.
At B = (3,2,1), 3u = 3, 2u 2 = 2, u 3 = 1,
 u = 1.
64
u =1
B
  V d r =  (12u8 )( 3du i + 4udu j + 3u 2du k )
A
~
u =0
1
~
~
~
1
1
=  36u du i +  48u du j +  36u10du k
8
0
~
 
9
0
~
1
0
~
1
 24 10 
 36 11 
= 4u i +  u  j +  u  k
~
5
 0 ~  11  0 ~
24
36
= 4i +
j+
k.
~
5 ~ 11 ~
9 1
0
65
Exercise 2.6
If V = x 2 yz 2 and the curve C is given by
x = 4u,
calculate
Vdr
c
~
y = 3u 3 , z = 2u 2 ,
along the curve C
from A = (0,0,0) to B = (4,3,2).
Answer
384
768
A V d r~ = 5 ~i +144 ~j + 11 k~ .
B
66
2.8.2 Vector Field, F Integral
~
Let a vector field
F = Fx i + Fy j + Fz k
~
and
~
d r = dx i + dy j + dz k .
~
~
The scalar product
~
~
~
~
F . d r is written as
~
~
F . d r = ( Fx i + Fy j + Fz k ).( dx i + dy j + dz k )
~
~
~
~
~
~
~
~
= Fx dx + Fy dy + Fz dz.
67
If a vector field F is along the curve C ,
~
then the line integral of F along the curve C
~
from a point A to another point B is given by
 F. d r =  F
c ~
~
c
x
dx +  Fy dy +  Fz dz.
c
c
68
Example 2.15
 F . d r from
Calculate
c ~
~
A = (0,0,0) to B = (4,2,1)
along the curve x = 4t , y = 2t 2 , z = t 3 if
F = x 2 y i + xz j - 2 yz k .
~
~
~
~
69
Solution
Given F = x 2 y i + xz j - 2 yz k
~
~
~
~
= (4t ) 2 (2t 2 ) i + (4t )(t 3 ) j - 2(2t 2 )(t 3 ) k
~
~
~
= 32t 4 i + 4t 4 j - 4t 5 k .
~
And
~
~
d r = dx i + dy j + dz k
~
~
~
~
= 4 dt i + 4t dt j + 3t 2 dt k .
~
~
~
70
Then
F . d r = (32t 4 i + 4t 4 j - 4t 5 k )( 4 dt i + 4t dt j + 3t 2 dt k )
~
~
~
~
~
~
~
~
= (32t 4 )( 4dt ) + (4t 4 )( 4tdt ) + (-4t 5 )(3t 2 dt )
= 128t 4 dt + 16t 5 dt - 12t 7 dt
= (128t 4 + 16t 5 - 12t 7 ) dt.
At A = (0,0,0), 4t = 0, 2t 2 = 0, t 3 = 0,
 t = 0.
and, at B = (4,2,1), 4t = 4, 2t 2 = 2, t 3 = 1,
 t = 1.
71
t =1
B
  F . d r =  (128t + 16t - 12t )dt
A ~
~
4
5
7
t =0
1
128 5 8 6 3 8 
=
t + t - t 
3
2 0
 5
128 8 3
=
+ 5
3 2
23
= 26 .
30
72
Exercise 2.7
If F = xy2 i - yz j + 3 x 2 z k ,
~
calculate
~
~
~
 F. d r
c ~
~
from A = (0,0,0) to B = (1,2,3) on the
curve x = t , y = 2t 2 , z = 3t 3 .
Answer

B
A
61
F .d r = 7
.
~
~
168
73
* Double Integral *
Example 2.16
Given f ( x, y ) = 4 - y 2 in region R bounded
by a straight line x = 0, y = x and y = 2.
Find
 f ( x, y) dA in both order integrals.
R
Answer

f ( x, y ) dA = 4 unit 2 .
R
74
Example 2.17
Using double integral, find the area of a region
bounded by y = 5 - x 2 and y = x + 3.
1
Answer The area of the region = 4 unit 2 .
2
75
Example 2.18
Evaluate a solid which is bounded by
z = 16 - x 2 - y 2 and z = 2.
Stated

f ( x, y, z ) dV as integral in order dz dy dx.
76
Example 2.19
Describe

f ( x, y, z ) dV as integral in order
dz dy dx if S is a solid which is bounded by
z = 0, z = x, and y = 4 - 2 x.
2
77
2.9 Volume Integral
2.9.1 Scalar Field, F Integral
If V is a closed region and F is a scalar field in
region V, volume integral F of V is
 FdV = 
V
V
Fdxdydz
78
Example 2.20
Scalar function F = 2 x defeated in one cubic that
has been built by planes x = 0, x = 1, y = 0, y = 3,
z = 0 and z = 2. Evaluate volume integral F of the
cubic.
z
2
O
3
y
1
x
79
Solution
2
3
1
z =0
y =0
x =0
 FdV =   
V
2 xdxdydz
1
x 
= 2     dydz
z =0 y =0
 2 0
2
3
1
= 2 
dydz
z =0 y =0 2
3
1 2
= 2.  [ y ] dz
0
2 z =0
2
=
3
2
2
2
3
dz
=
3
[
z
]
0 = 6

z =0
80
2.9.2 Vector Field, F Integral
~
If V is a closed region and
F, vector field in region
~
V, Volume integral F of V is
~

F dV = 
V ~
x2
x1
y2
z2
y1
z1
 
F dzdydx
~
81
Example 2.21
Evaluate

F dV, where V is a region bounded by
V ~
x = 0, y = 0, z = 0 and 2x + y + z = 2, and also
given
F = 2z i + y k
~
~
~
82
Solution
If x = y = 0, plane 2x + y + z = 2 intersects z-axis at z = 2.
(0,0,2)
If x = z = 0, plane 2x + y + z = 2 intersects y-axis at y = 2.
(0,2,0)
If y = z = 0, plane 2x + y + z = 2 intersects x-axis at x = 1.
(1,0,0)
83
z
2
2x + y + z = 2
O
x
1
2
y
y = 2 (1 - x)
We can generate this integral in 3 steps :
1. Line Integral from x
= 0 to x = 1.
2. Surface Integral from line y
= 0 to line y = 2(1-x).
3. Volume Integral from surface z = 0 to surface
2x + y + z = 2 that is z = 2 (1-x) - y
84
Therefore,

1
F dV = 

2 (1- x )
x =0 y =0
V ~
1
=
x =0


2 (1- x )
y =0
2 (1- x ) - y
z =0

F dzdydx
~
2 (1- x ) - y
z =0
(2 z i + y k ) dzdydx
~
~


2
1
= i+ k
3~ 3~
85
Example 2.22
Evaluate

F dV where F = 2 i + 2 z j + y k
~
V ~
~
~
~
and V is region bounded by z = 0, z = 4 and
x2 + y2 = 9
z
4 -

3

3
y
x
86
Using polar coordinate of cylinder,
x =  cos  ; y =  sin  ; z = z ;
dV = ρdρddz
where
0    3, 0    2 , 0  z  4
87
Therefore,

F dV =  (2 i + 2 z j + y k )dxdydz
V ~
=
4
z =0
V
2
3
 
=0
=0
~
~
~
(2 i + 2 z j +  sin  k )  dddz
~
~
~


= 72 i + 144 j
~
~
88
Exercise 2.8
89
2.10 Surface Integral
2.10.1 Scalar Field, V Integral
If scalar field V exists on surface S, surface
integral V of S is defined by
 Vd S =  V n dS
S
where
~
S
~
S
n=
~
S
90
Example 2.23
Scalar field V = x y z defeated on the surface
S : x2 + y2 = 4 between z = 0 and z = 3 in the
first octant.
Evaluate
 Vd S
S
~
Solution
Given S : x2 + y2 = 4 , so grad S is
S S
S
S =
i+
j+ k = 2x i + 2 y j
~
x ~ y ~ z ~
~
91
Also,
S = (2 x)2 + (2 y )2 = 2 x 2 + y 2 = 2 4 = 4
Therefore,
S
n=
=
~
S
2x i + 2 y j
~
~
4
1
= ( x i + y j)
2 ~
~
Then,
1
S V n~ dS = S xyz 2 ( x i~ + y ~j )dS
1
=  ( x 2 yz i + xy2 z j )dS
~
2
~
92
Surface S : x2 + y2 = 4 is bounded by z = 0 and z = 3
that is a cylinder with z-axis as a cylinder axes and
radius,  = 4 = 2.
So, we will use polar coordinate of cylinder to find
the surface integral.
z
3
O

2
y
2
x
93
Polar Coordinate for Cylinder
x =  cos  = 2 cos 
y =  sin  = 2sin 
z=z
dS = ρ d dz
where 0   

2
(1st octant) and
0 z 3
94
Using polar coordinate of cylinder,
x yz = (2 cos  ) (2 sin  ) z = 8z cos  sin 
2
2
2
xy z = (2 cos  )( 2 sin  ) ( z ) = 8z sin  cos 
2
2
2
From
1
2
2
V
n
dS
=
(
x
yz
i
+
xy
z j )dS =  Vd S
S ~

S
~
~
2S
~
95
Therefore,
1 2 3
2
2
Vd
S
=
(
8
z
cos

sin

i
+
8
z
sin
 cos  j )( 2)dzd
S ~ 2  =0 z =0
~
~

2
0
1 2 2
1 2 2

z
cos

sin

i
+
z
sin

cos

j
d
 2

~ 2
~ 0

2
0
9 2
9 2

 2 cos  sin  i~ + 2 sin  cos  ~j  d
= 8
= 8
3

9 2 2
= 8   cos  sin  i + sin 2  cos  j  d
~
2 0 
~

 cos3  sin  sin 3  cos 
= 36 
i+
 3( - sin  ) ~ 3(cos  )
= 12( i + j )
~

2
j
~
0
~
96
Exercise 2.9
If V is a scalar field where V = xyz 2 , evaluate

S
V d S for surface S that region bounded by x 2 + y 2 = 9
~
between z = 0 and z = 2 in the first octant.
Answer : 24( i + j )
~
~
97
2.10.2 Vector Field,
F Integral
~
If vector field F defeated on surface S, surface
~
integral F of S is defined as
~

F . d S =  F . n dS
S ~
~
S ~
~
S
where n =
~
S
98
Example 2.24
Vector field F = y i + 2 j + k defeated on surface
~
~
~
~
S : x 2 + y 2 + z 2 = 9 and bounded by x = 0, y = 0, z = 0 in
the first octant.
Evaluate

F .d S .
S ~
~
99
Solution
Given S : x 2 + y 2 + z 2 = 9 is bounded by x = 0, y = 0,
z = 0 in the 1st octant. This refer to sphere with center
at (0,0,0) and radius, r = 3, in the 1st octant.
z
3
O
3
y
3
x
100
So, grad S is
S
S
S
S =
i+
j+
k
x ~ y ~ z ~
= 2x i + 2 y j + 2z k ,
~
~
~
and
S = (2 x ) 2 + (2 y ) 2 + (2 z ) 2
= 2 x2 + y2 + z2
= 2 9 = 6.
101
S
n=
=
~
S

2x i + 2 y j + 2z k
~
~
~
6
1
= ( x i + y j + z k ).
~
3 ~
~
Therefore,

F . d S =  F . n dS
S ~
S ~
~
=
S
~
1
( y i + 2 j + k )   (x i + y j + z k ) dS
~
~
~
~
~
 3 ~
1
=  (xy + 2 y + z ) dS .
3 S
102
Using polar coordinate of sphere,
x = r sin  cos  = 3sin  cos 
y = r sin  sin  = 3sin  sin 
z = r cos  = 3cos 
dS = r 2 sin  d d = 9 sin  d d
where 0   ,  

2
.
103
1 2 2
  F . d S =   [(3 sin  cos  )(3 sin  sin  )
S ~
~
3  =0  =0
+ 2(3 sin  sin  ) + 3 cos  ][9 sin  ]d d
= 9

2


2
 =0  =0
[3 sin 3  sin  cos 
+ 2 sin 2  sin  + sin  cos  ]dd


3 

= 9 1 +

4 

104
Exercise 2.9
Evaluate
 F  d S on S , where F = x i + 2 z j + y k
S
~
~
~
~
~
~
and S is a surface of the region bounded by
x 2 + y 2 + z 2 = 4, x = 0, y = 0 and z = 0 in the 1st octant.


Answer : 8  + 1
6

105
2.11 Green’s Theorem
If c is a closed curve in counter-clockwise on
plane-xy, and given two functions P(x, y) and
Q(x, y),
 Q P 
S  x - y  dx dy = c( P dx + Q dy)
where S is the area of c.
106
Example 2.25
Prove Green's Theorem for

c
[( x 2 + y 2 )dx + ( x + 2 y )dy ]
which has been evaluated by boundary that defined as
x = 0, y = 0 and x 2 + y 2 = 4 in the first quarter.
Solution
y
x2 + y2 = 22
2
C2
C3
O
C1
2
x
107
Given

c
[( x 2 + y 2 )dx + ( x + 2 y )dy ] where
P = x 2 + y 2 and Q = x + 2 y. We defined curve c
as c1 , c2 and c3.
i) For c1 : y = 0, dy = 0 and 0  x  2
2
2

(
Pdx
+
Qdy
)
=
(
x
+
y
)dx + ( x + 2 y )dy 
c1
c1 
2
=  x 2dx
0
2
1 3 8
= x  = .
 3 0 3
108
ii) For c2 : x 2 + y 2 = 4 ,in the first quarter from (2,0) to (0,2).
This curve actually a part of a circle.
Therefore, it's more easier if we integrate by using polar
coordinate of plane,
x = 2 cos  , y = 2sin  , 0   

 dx = -2sin  d , dy = 2 cos  d .
2
109

c2

( Pdx + Qdy ) =  ( x 2 + y 2 )dx + ( x + 2 y )dy
c2


=  [(( 2 cos  ) 2 + (2 sin  ) 2 )( -2 sin  d )
2
0
+ (( 2 cos  + 2(2 sin  ))( 2 cos  d )]

=  (-8 sin  + 4 cos 2  + 8 sin  cos  )d
2
0

=  (-8 sin  + 2 + 2 cos 2 + 8 sin  cos  )d
2

0
= 8 cos  + 2 + sin 2 + 4 sin 
2


2
0
= -8 +  + 4 =  - 4.
110
iii) For c3 : x = 0, dx = 0, 0  y  2
2
2

(
Pdx
+
Qdy
)
=
(
x
+
y
)dx + ( x + 2 y )dy 
c3
c3 
0
=  2 y dy
2
2 0
=  y 
2
= -4.
8
16
  ( Pdx + Qdy ) = + ( - 4) - 4 =  - .
c
3
3
111
 Q P 
S  x - y  dxdy
Q
P
where
= 1 and
= 2 y.
x
y
Again,because this is a part of the circle,
b) Now, we evaluate
we shall integrate by using polar coordinate of plane,
x = r cos  , y = r sin 
where 0  r  2, 0   

2
and
dxdy = dS = r dr d .
112
 Q P 
S  x - y  dx dy = S (1 - 2 y ) dx dy
=

2

2
 =0 r =0
(1 - 2 r sin  ) r dr d
2
1 2 2 3

2
=   r - r sin   d
 =0 2
3

0

16


=   2 - sin   d
=0
3



2

16

2
= 2 + cos  
3

0
16
= - .
3
113
Therefore,
 Q P 
 C ( Pdx + Qdy ) = S  x - y  dx dy
16
= - .
3
LHS = RHS
 Green's Theorem has been proved.
114
2.12 Divergence Theorem (Gauss’ Theorem)
If S is a closed surface including region V in
vector field F
~
 div F dV =  F . d S.
V
~
S ~
~
f x f y f z
div F =
+
+
~
x
y
z
115
Example 2.26
Prove Gauss' Theorem for vector field,
F = x i + 2 j + z 2 k in the region bounded by
~
~
~
~
planes z = 0, z = 4, x = 0, y = 0 and x 2 + y 2 = 4
in the first octant.
116
Solution
z
4
S2
S4
O
2
S3
x
2
S5
y
S1
117
For this problem, the region of integration is bounded
by 5 planes :
S1 : z = 0
S2 : z = 4
S3 : y = 0
S4 : x = 0
S5 : x 2 + y 2 = 4
To prove Gauss' Theorem, we evaluate both  div F dV
V
and

~
F.d S,
S ~
~
The answer should be the same.
118
1) We evaluate

V
div F dV . Given F = x i + 2 j + z 2 k .
~
~
~
~
~
So,



( x ) + (2) + ( z 2 )
x
y
z
= 1 + 2 z.
div F =
~
Also,

V
div F dV =  (1 + 2 z ) dV .
~
V
The region is a part of the cylinder. So, we integrate by using
polar coordinate of cylinder ,
x =  cos ;
y =  sin  ; z = z
dV =  d  d dz
where 0    2, 0   

2
, 0  z  4.
119
Therefore,

V
(1 + 2 z ) dV = 

2
2
 
4
 =0  =0 z =0
=

2

2
 =0  =0
=

2
 =0
=

2
 =0
=

2
 =0
2

=0
(1 + 2 z )  dzd  d
 [ z + z 2 ]04 d  d
(20 ) d  d
[10 2 ]02 d
(40) d
= 40 02

= 20 .
  div F dV = 20 .
V
~
120
2) Now, we evaluate
i) S1 :

F . d S =  F . n dS .
S ~
S ~
~
~
z = 0, n = - k , dS = rdrd
~
~
 F = x i + 2 j + 0k
~
~
~
~
 F . n = ( x i + 2 j ).( - k ) = 0
~


~
~
~
~
F . n dS = 0.
S1 ~
~
121
z = 4, n = k , dS = rdrd
ii) S2 :
~
~
 F = x i + 2 j + (4) 2 k = x i + 2 j + 16 k
~
~
~
~
~
~
~
 F . n = ( x i + 2 j + 16 k ).( k ) = 16.
~
~
~
~
~
Therefore for S2 , 0  r  2, 0   


F . n dS = 
S2 ~

2

2
 =0 r =0
~
~

2
16 rdrd
=
= 16 .
122
y = 0, n = - j, dS = dxdz
iii) S3 :
~
~
 F = x i + 2 j + z2 k
~
~
~
~
 F . n = ( x i + 2 j + z 2 k ).( - j )
~
~
~
~
~
~
= -2.
Therefore for S3 , 0  x  2, 0  z  4


F . n dS = 
S3 ~
~
2
x =0

4
z =0
( -2) dzdx
=
= -16.
123
x = 0, n = - i , dS = dydz
iv) S4 :
~
~
 F = 0 i + 2 j + z2 k = 2 j + z2 k
~
~
~
~
~
~
 F . n = (2 j + z 2 k ).( - i ) = 0.
~


~
~
~
~
F . n dS = 0.
S4 ~
~
124
v) S5 :
x 2 + y 2 = 4, dS =  d  dz
S5 = 2 x i + 2 y j and
~
S 5
n=
=
~
S 5
S 5 = 4
~
2x i + 2 y j
~
~
4
1
( x i + y j ).
2 ~
~
By using polar coordinate of cylinder :
=
x =  cos  , y =  sin  , z = z
where for S5 :
 = 2, 0   

2
, 0  z  4, dS = 2d dz
125
1
1

 F . n = ( x i + 2 j + z k ). x i + y j 
~ ~
~
~
~
2 ~ 2 ~
1 2
= x +y
2
1
= (  cos  ) 2 + (  sin  )
2
= 2 cos2  + 2 sin  ; kerana  = 2.
2
= 2(cos 2  + sin  ).


F . n dS = 
S5 ~
~

2

4
 =0 z =0
( 2)(cos 2  + sin  )( 2) d dz
=
= 16 + 4 .
126
Finally,

F.d S =  F.d S + F.d S + F.d S + F.d S + F.d S
S ~
S1 ~
~
~
S2 ~
~
S3 ~
~
S4 ~
~
S5 ~
~
= 0 + 16 - 16 + 0 + 16 + 4
= 20 .


F . d S = 20 .
S ~
~
LHS = RHS
 Gauss' Theorem has been proved.
127
2.13 Stokes’ Theorem
If F is a vector field on an open surface S and
~
boundary of surface S is a closed curve c,
therefore
 curl F  d S =  F  d r
S
~
~
i
~

curl F =   F =
~
~
x
fx
c~
j
~
k
~
~

y
fy

z
fz
128
Example 2.27
Surface S is the combination of
i) a part of the cylinder x 2 + y 2 = 9 between z = 0
and z = 4 for y  0.
ii) a half of the circle with radius 3 at z = 4, and
iii) plane y = 0
If F = z i + xy j + xz k , prove Stokes' Theorem
~
~
~
~
for this case.
129
Solution
z
S3
4
S2
S1
C2
O
x
3 C
1
3
y
We can divide surface S as
S1 : x 2 + y 2 = 9 for 0  z  4 and y  0
S 2 : z = 4, half of the circle with radius 3
S3 : y = 0
130
We can also mark the pieces of curve C as
C1 :
Perimeter of a half circle with radius 3.
C2 :
Straight line from (-3,0,0) to (3,0,0).
Let say, we choose to evaluate
Given

S
curl F  d S first.
~
~
F = z i + xy j + xz k
~
~
~
~
131
So,
i
~

curl F =
~
x
z
j
~

y
xy
k
~

z
xz

 



=  ( xz) - ( xy)  i +  ( z ) - ( xz)  j
z
x
~
 y
 ~  z



+  ( xy) - ( z )  k
y  ~
 x
= (1 - z ) j + y k
~
~
132
By integrating each part of the surface,
(i )
For surface S1 : x 2 + y 2 = 9,
S1 = 2 x i + 2 y j
~
and
S1 =
~
(2 x ) 2 + (2 y ) 2
= 2 x2 + y2 = 6
133
Then ,
S1
n=
=
~
S1
2x i + 2 y j
~
~
6
1
= ( x i + y j)
3 ~
~
and
1
1



curl F  n =  (1 - z ) j + y k    x i + y j 
~ ~
~
~

3 ~ 3 ~
1
= y (1 - z ).
3
134
By using polar coordinate of cylinder ( because
S1 : x 2 + y 2 = 9 is a part of the cylinder),
x =  cos  , y =  sin  , z = z
dS =  d dz
where
 = 3, 0     dan 0  z  4.
135
Therefore,
1
curl F  n = y (1 - z )
~ ~
3
1
=   sin  1 - z 
3
= sin  (1 - z ) ; because  = 3
Also,
dS = 3 d dz
136


S1
curl F  d S =
~
~

curl F  n dS
S1
~
= 3
4
z =0
~

  sin  (1 - z )  d dz
=0
= 3 (1 - z )  - cos   0 dz
4
0
4
= 3 (1 - z )(1 - ( -1))dz
0
= -24
137
(ii) For surface
surface is
S2 : z = 4
, normal vector unit to the
n = k.
~
~
By using polar coordinate of plane ,
y = r sin  , z = 4 dan dS = r dr d
where 0  r  3 and 0     .
138


curl F  n =  (1 - z ) j + y k   k
~ ~
~
~ 
~

= y = r sin 


S2
curl F  d S =
~
~
=
=

S2
curl F  n dS
3
~

 
r =0
3

 
r =0
=0
=0
~
( r sin  )( rdrd )
r 2 sin  d dr
= 18
139
(iii) For surface S3 : y = 0, normal vector unit
to the surface is n = - j .
~
~
dS = dxdz
The integration limits : -3  x  3
and
0 z4
So,
curl F  n = ((1 - z ) j + y k )  ( - j )
~
~
~
~
~
= z -1
140
Then,

S3
curl F . d S =  curl F . n dS
~
S3
~
=
3
~

4
x =-3 z =0
~
( z - 1) dzdx
=
= 24.


S
curl F . d S =  curl F . d S +
~
~
S1
~
~

S2
curl F . d S +
~
~

S3
curl F . d S
~
~
= -24 + 18 + 24
= 18.
141
Now, we evaluate

F . d r for each pieces of the curve C.
C ~
~
i) C1 is a half of the circle.
Therefore, integration for C1 will be more easier if we use
polar coordinate for plane with radius r = 3, that is
x = 3cos  ,
y = 3sin 
dan z = 0
where 0     .
142
 F = z i + xy j + xz k
~
~
~
~
= (3cos  )(3sin  ) j
~
= 9sin  cos  j
~
and
dr = dx i + dy j + dz k
~
~
~
= -3sin  d i + 3cos  d j .
~
~
143
From here,
F . d r = 27sin  cos  d .
2
~


~

F . d r =  27sin  cos 2  d
C1 ~
~
0

=  -9 cos  
0
= 18.
3
144
ii) Curve C2 is a straight line defined as
x = t,
y = 0 and
z = 0,
where - 3  t  3.
Therefore, F = z i + xy j + xz k
~
~
~
~
= 0.
~


F . d r = 0.
C2 ~
~
145


F.d r =
C ~
~

F.d r +
C1 ~
~

F.d r
C2 ~
~
= 18 + 0
= 18.
We already show that

S
curl F . d S =
~
~

F.d r
C ~
~
 Stokes' Theorem has been proved.
146