Applied III Chapter 4(1).pdf

1
Chapter 4
Calculus of Vector fields
1. Introduction
2. Vector Calculus
3. Divergence and gradient of a vector
4. Line Integral
5. Volume Integral
6. Surface Integral
7. Green’s Theorem

2
Continued
9. Divergence Theorem (Gauss’ Theorem)
10. Stokes’ Theorem

3
1. Introduction
Definition 2.1 (Scalar and vector)
Vector is a directed quantity, one with
both magnitude and direction.
For instance acceleration, velocity, force
Scalar is a quantity that has magnitude
but not direction.
For instance mass, volume, distance

4
We represent a vector as an arrow from the
origin O to a point A.
The length of the arrow is the magnitude of
the vector written as or .
O
A
or
O
A
OA
a
a
OA

5
Basic Vector System
Unit vectors , ,
• Perpendicular to each other
• In the positive directions
of the axes
• have magnitude (length) 1

6
Magnitude of vectors
Let P = (x, y, z). Vector is defined by
with magnitude (length)
OP = = + +
p x i y j z k
= [ ]
x, y, z
OP = p
OP = = + +
p x y z
2 2 2

7
Calculation of Vectors
1. Vector Equation
Two vectors are equal if and only if the
corresponding components are equals
1 2 3 1 2 3
1 1 2 2 3 3
Let and .
Then
, ,
a a i a j a k b b i b j b k
a b a b a b a b
= + + = + +
=  = = =

8
2. Addition and Subtraction of Vectors
3. Multiplication of Vectors by Scalars
k
b
a
j
b
a
i
b
a
b
a )
(
)
(
)
( 3
3
2
2
1
1 
+

+

=

k
b
j
b
i
b
b )
(
)
(
)
(
then
scalar,
a
is
If
3
2
1 




+
+
=

9
Example
Given 5 3 and 4 3 2 . Find
p i j k q i j k
= + - = - +
a p q
) +
)
b p q
-
) 2 10
d q p
-
c p
) Magnitude of vector

10
Vector Products
1 2 3 1 2 3
~ ~ ~ ~ ~ ~
~ ~
If and ,
a a i a j a k b b i b j b k
= + + = + +
1 1 2 2 3 3
~ ~
a b a b a b a b
 = + +
1) Scalar Product (Dot product)
2) Vector Product (Cross product)
     
~ ~
~
1 2 3
~ ~
1 2 3
2 3 3 2 1 3 3 1 1 2 2 1
~ ~
~
i j k
a b a a a
b b b
a b a b i a b a b j a b a b k
 =
= - - - + -
~
~
~
~
and
between
angle
the
is
,
cos
|
||
|
.
or b
a
b
a
b
a 

=

11
3) Application of Multiplication of Vectors
a) Given 2 vectors and , projection onto
is defined by
b) The area of triangle
~ ~
1
.
2
A a b
= 
a b a b
a
b
b
a
compb a
~ ~
~
~ ~
~
.
comp
| |
| . |
length ( )
| |
b
a b
a
b
a b
l
b
=
=

12
c) The area of parallelogram
d) The volume of tetrahedrone
e) The volume of parallelepiped
a
b
a b
x
A =
a b
c
3
2
1
3
2
1
3
2
1
6
1
c
c
c
b
b
b
a
a
a
=
6
1
=
V a . b c
x
a b
c
3
2
1
3
2
1
3
2
1
c
c
c
b
b
b
a
a
a
=
=
V a . b c
x

13
Example
.
and
between
angle
the
and
,
.
determine
,
2
and
3
2
Given
~
~
~
~
~
~
~
~
~
~
~
~
~
~
b
a
b
a
b
a
k
j
i
b
k
j
i
a

+
+
=
-
+
=

2. Vector Calculus
Limits and Continuity of Vectors
A vector function v(t) of a real variable t is said
to have the limit l as t approaches t0, if v(t) is
defined in some neighborhood of t0 (possibly
except at t0) and
Then we write
Here, a neighborhood of t0 is an interval (segment)
on the t-axis containing t0 as an interior point
(not as an endpoint).
14
0
lim ( ) 0.
t t
t

- =
v l
0
lim ( ) .
t t
t

=
v l

Continuity
• A vector function v(t) is said to be
continuous at t = t0 if it is defined in some
neighborhood of t0 (including at t0 itself!)
and
• If we introduce a Cartesian coordinate
system, we may write
15
1 2 3 1 2 3
( ) ( ), ( ), ( ) ( ) ( ) ( ) .
t v t v t v t v t v t v t
= = + +
 
 
v i j k

Continued
Then v(t) is continuous at t0 if and only if
its three components are continuous at t0.
Derivative of a Vector Function
• A vector function v(t) is said to be
differentiable at a point t if the
following limit exists:
16
0
( ) ( )
( ) lim .
t
t t t
t
t
 
+  -
 =

v v
v

Continuid
• This vector v’(t) is called the derivative of
v(t)
• In components with respect to a given
Cartesian coordinate system,
17

Continuid
• Hence the derivative v’(t) is obtained by
differentiating each component separately.
For instance, if v = [t, t2, 0], then v’ = [1, 2t, 0].
18
1 2 3
( ) ( ), ( ), ( ) .
t v t v t v t
   
=  
 
v

Continuid
Properties
and in particular
Partial Derivatives of a Vector Function
Suppose that the components of a vector
function 19
( )
( )
c c
 
=
  
+ = +
v v
u v u v
( )  
= +
u v u v u v
( )  
= +
u × v u × v u × v
( ) ( ) ( ) ( ).
   
= + +
u v w u v w u v w u v w

Continuid
• are differentiable functions of n variables t1,
… , tn. Then the partial derivative of v with
respect to tm is denoted by ∂v/∂tm and is
defined as the vector function
• Similarly, second partial derivatives are;
20
= = + +
 
 
1 2 3 1 2 3
, ,
v v v v v v
v i j k
3
1 2
.
m m m m
v
v v
t t t t

 

= + +
   
v
i j k

Continuid
21
2
2 2
2
3
1 2
.
l m l m l m l m
v
v v
t t t t t t t t

 

= + +
       
v
i j k

22
Example 1


=
=
+
-
=
2
~
2
~
~
~
~
2
~
hence
5
2
3
If
du
A
d
du
A
d
k
j
u
i
u
A

23
Example 2
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.

24
Solution
• The parameter is t, and the position vector is
• The velocity is given by
• The acceleration is
.
)
5
(
)
3
(
)
3
4
(
)
(
~
2
3
~
2
~
~
k
t
t
j
t
t
i
t
t
r +
+
+
+
+
=
.
)
10
3
(
)
3
2
(
4
~
2
~
~
~
k
t
t
j
t
i
dt
r
d
+
+
+
+
=
.
)
10
6
(
2
~
~
2
~
2
k
t
j
dt
r
d
+
+
=

25
• At t = 1, the velocity of the particle is
and the magnitude of the velocity is
.
13
5
4
))
1
(
10
)
1
(
3
(
)
3
)
1
(
2
(
4
)
1
(
~
~
~
~
2
~
~
~
k
j
i
k
j
i
dt
r
d
+
+
=
+
+
+
+
=
.
210
13
5
4
)
1
( 2
2
2
~
=
+
+
=
dt
r
d

26
• At t = 1, the acceleration of the particle is
and the magnitude of the acceleration is
.
16
2
)
10
)
1
(
6
(
2
)
1
(
~
~
~
~
2
~
2
k
j
k
j
dt
r
d
+
=
+
+
=
.
65
2
16
2
)
1
( 2
2
2
~
2
=
+
=
dt
r
d

27
Example 3
~
~
2
~
2
~
~
2
~
2
~
~
2
~
2
~
~
~
~
~
2
~
~
2
~
~
2
3
~
2
~
2
~
6
,
2
6
,
6
4
,
2
6
,
3
4
3
then
)
(
)
2
(
3
If
i
v
u
v
F
v
u
F
k
i
u
v
F
k
u
j
u
F
k
v
j
i
uv
v
F
k
u
j
u
i
v
u
F
k
v
u
j
v
u
i
uv
F
=



=



+
=


+
=


+
-
=


+
+
=


+
+
-
+
=

28
Exercise






=



=



=


=


=


=


+
+
-
+
=
u
v
F
v
u
F
v
F
u
F
v
F
u
F
k
v
u
j
v
u
i
v
u
F
~
2
~
2
2
~
2
2
~
2
~
~
~
2
3
~
3
~
2
~
,
,
,
then
)
3
(
)
3
(
2
If

29
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 F( ) ( ) ( ) ( )
then
( ) ( )
( ) ( ) .
b b
a a
b b
a a
u f u i g u j h u k
F u du f u du i
g u du j h u du k
= + +
=
+ +
 
 

30
Example
.
80
2
42
]
[
]
5
[
]
2
[
4
)
5
2
(
)
4
3
(
.
calculate
,
4
)
5
2
(
)
4
3
(
If
~
~
~
~
3
1
4
~
3
1
2
~
3
1
2
3
3
1 ~
3
3
1 ~
3
1 ~
2
3
1 ~
3
1 ~
~
3
~
~
2
~
k
j
i
k
t
j
t
t
i
t
t
k
dt
t
j
dt
t
i
dt
t
t
dt
F
dt
F
k
t
j
t
i
t
t
F
+
-
=
+
-
+
+
=
+
-
+
+
=
+
-
+
+
=





Answer

31
Exercise
.
2
7
3
2
4
7
)
4
(
2
)
3
(
.
calculate
,
)
4
(
2
)
3
(
If
~
~
~
1
0 ~
1
0 ~
2
1
0 ~
3
1
0 ~
1
0 ~
~
~
2
~
3
~
k
j
i
k
dt
t
j
dt
t
i
dt
t
t
dt
F
dt
F
k
t
j
t
i
t
t
F
-
+
=
=
=
-
+
+
+
=
-
+
+
+
=







Answer

Curves, Arc Length and Tangent
 In our day to day life, a curve is defined as a line
which is not straight. But, in math, a curve can also be
straight or it is also called continuous line. Curves are
either open or closed.
 Open curve is defined as a curve whose ends do not
meet. Example is parabola, hyperbola.
 Closed curves are curves whose ends are joined.
Closed curves do not have end points. Examples
of closed curves are ellipse and circle.
32

• The application of vector calculus to
geometry is a field known as differential
geometry.
• Bodies that move in space form paths that
may be represented by curves C. This and
other applications show the need for
parametric representations of C with
parameter t, which may denote time or
something else
33

A typical parametric representation is given by
(1) r(t) = [x(t), y(t), z(t)] = x(t)i + y(t)j + z(t)k
34

here t is the parameter and x, y, z are Cartesian
coordinates, that is, the usual rectangular
coordinates.
The use of parametric representations has key
advantages over other representations that involve
projections into the xy-plane and xz-plane or involve
a pair of equations with y or with z as independent
variable. The projections look like this:
(2) y = f(x), z = g(x).
The advantages of using (1) instead of (2) are that, in
(1), the coordinates x, y, z all play an equal role, that
is, all three coordinates are dependent variables.
Moreover, the parametric representation (1) induces
an orientation on C.
35

This means that as we increase t, we travel along the
curve C in a certain direction. The sense of increasing
t is called the positive sense on C. The sense of
decreasing t is then called the negative sense on C,
given by (1).
• Example: Straight Line
A straight line L through a point A with position
vector a in the direction of a constant vector b (see in
the figure below)
can be represented parametrically in the form
(3) r(t) = a + tb = [a1 + tb1, a2 + tb2, a3 + tb3].
36

Parametric representation of a straight line
A plane curve is a curve that lies in a plane in space.
A curve that is not plane is called a twisted curve.
A simple curve is a curve without multiple points,
that is, without points at which the curve intersects
or touches itself. 37

Circle and helix are simple curves. Figures below shows
curves that are not simple.
An arc of a curve is the portion between any two points
of the curve. For simplicity, we say “curve” for curves as
well as for arcs.
Curves with multiple points
38

Tangent to a Curve
The next idea is the approximation of a curve by straight
lines, leading to tangents and to a definition of length.
Tangents are straight lines touching a curve. The tangent
to a simple curve C at a point P of C is the limiting
position of a straight line L through P and a point Q of C
as Q approaches P along C. Let us formalize this
concept. If C is given by r(t), and P and Q correspond to
t and t + Δt, then a vector in the direction of L is
(4)
39

In the limit this vector becomes the derivative
(5)
provided r(t) is differentiable, as we shall assume from
now on. If r′(t) ≠ 0 we call r′(t) a tangent vector of C at
P because it has the direction of the tangent. The
corresponding unit vector is the unit tangent vector
(see figure below)
(6)
Note that both r′ and u point in the direction of
increasing t. Hence their sense depends on the
orientation of C. It is reversed if we reverse the
orientation.
40
0
1
( ) lim ( ) ( ) ,
t
t t t t
t
 
 = +  -
 
 

r r r
1
.

=

u r
r

continued
It is now easy to see that the tangent to C at P is given
by
q(w) = r + wr' (Fig below).
This is the sum of the position vector r of P and a
multiple of the tangent vector r′ of C at P. Both vectors
depend on P. The variable w is the parameter in (7).
41

Length of a Curve
If r(t) has a continuous derivative r′, it can be shown
that the sequence l1, l2, … has a limit, which is
independent of the particular choice of the
representation of C and of the choice of subdivisions.
This limit is given by the integral
l is called the length of C, and C is called rectifiable
42

Arc Length s of a Curve
The length of a curve C is a constant, a positive
number. But if we replace the fixed b in the previous
integral with a variable t, the integral becomes a
function of t, denoted by s(t) and called the arc length
function or simply the arc length of C. Thus
Here the variable of integration is denoted by
because t is now used in the upper limit.
43
( ) .
t
a
d
s t dt
dt
 
  
= =
 
 

r
r r r
l
l
t

Arc Length(Continued)
Linear Element ds.
It is customary to write
dr = [dx, dy, dz] = dx i + dy j + dz k
and
ds2 = dr • dr = dx2 + dy2 + dz2.
ds is called the linear element of C.
Arc Length as Parameter.
The use of s in instead of an arbitrary t simplifies
various formulas. For the unit tangent vector we
simply obtain
44

Arc Length(Continued)
u(s) = r′(s)
Indeed, |r′(s)| = (ds/ds) = 1 shows that r′(s) is a unit
vector.
45

46
Del Operator Or Nabla (Symbol )
• Operator  is called vector differential operator,
defined as
.
~
~
~ 









+


+


=
 k
z
j
y
i
x

47
3. Gradient of Scalar Functions
• If f x,y,z is a scalar function of three variables
and f is differentiable, the gradient of f is
defined as
.
grad
~
~
~
k
z
j
y
i
x 

+


+


=

=
f
f
f
f
f
function
vector
a
is
*
function
scalar
a
is
*
f
f


48
Example
z
xy
yz
x
xyz
z
x
z
y
xyz
z
xy
yz
x
z
xy
yz
x
2
2
2
2
3
2
2
2
3
2
2
3
2
2
2
3
2
2
3
z
2
y
2
x
hence
,
Given
(1,3,2).
P
at
grad
determine
,
If
+
=


+
=


+
=


+
=
=
+
=
f
f
f
f
f
f
Solution

49
.
72
32
84
.
))
2
(
)
3
)(
1
(
2
)
2
)(
3
(
)
1
(
3
(
)
)
2
)(
3
)(
1
(
2
)
2
(
)
1
((
)
)
2
(
)
3
(
)
2
)(
3
)(
1
(
2
(
have
we
(1,3,2),
P
At
.
)
2
3
(
)
2
(
)
2
(
z
y
x
Therefore,
~
~
~
~
2
2
2
~
2
3
2
~
2
2
3
~
2
2
2
~
2
3
2
~
2
2
3
~
~
~
k
j
i
k
j
i
k
z
xy
yz
x
j
xyz
z
x
i
z
y
xyz
k
j
i
+
+
=
+
+
+
+
+
=

=
+
+
+
+
+
=


+


+


=

f
f
f
f
f

50
Exercise
(1,2,3).
P
point
at
grad
determine
,
If 3
2
3
=
+
=
f
f z
xy
yz
x

51
Solution
.
110
111
126
(1,2,3),
P
At
Grad
z
y
x
then
,
Given
~
~
~
3
2
3
k
j
i
z
xy
yz
x
+
+
=

=
=

=

=


=


=


+
=
f
f
f
f
f
f
f





52
Properties of gradient
If A and B are two scalars, then
)
(
)
(
)
(
)
2
)
(
)
1
A
B
B
A
AB
B
A
B
A

+

=


+

=
+


53
Directional Derivative
.
of
direction
in the
r
unit vecto
a
is
which
,
where
.
is
of
direction
in the
of
derivative
l
Directiona
~
~
~
~
~
~
r
d
r
d
r
d
a
grad
a
ds
d
a
=
= f
f
f

54
Example 4.9
.
4
3
2
vector
the
of
direction
in the
)
1
,
2
,
1
(
point
at the
2
of
derivative
l
directiona
the
Compute
~
~
~
~
2
2
2
k
j
i
A
yz
xy
z
x
-
+
=
-
+
+
=
f

55
Solution
Directional derivative of f in the direction of
.
)
2
(
)
4
(
)
2
2
(
hence
,
2
Given
~
2
~
2
~
2
2
2
2
k
yz
x
j
z
xy
i
y
xz
yz
xy
z
x
+
+
+
+
+
=

+
+
=
f
f
.
and
where
.
~
~
~
~
~
~
~
A
A
a
k
z
j
y
i
x
grad
grad
a
ds
d
=


+


+


=

=
=
f
f
f
f
f
f
f
~
a

56
.
29
)
4
(
3
2
then
,
4
3
2
given
Also,
.
3
9
6
.
))
1
)(
2
(
2
)
1
((
)
)
1
(
)
2
)(
1
(
4
(
)
)
2
(
2
)
1
)(
1
(
2
(
(1,2,-1),
At
2
2
2
~
~
~
~
~
~
~
~
~
2
~
2
~
2
=
-
+
+
=
-
+
=
-
+
=
-
+
+
-
+
+
+
-
=

A
k
j
i
A
k
j
i
k
j
i
f

57
.
470462
.
9
29
51
)
3
(
29
4
)
9
(
29
3
)
6
(
29
2
)
3
9
6
.(
29
4
29
3
29
2
.
Then,
.
29
4
29
3
29
2
Therefore,
~
~
~
~
~
~
~
~
~
~
~
~
~

=
-






-
+






+






=
-
+






-
+
=

=
-
+
=
=
k
j
i
k
j
i
a
ds
dφ
k
j
i
A
A
a
f

58
Unit Normal Vector
Equation f (x, y, z) = constant is a surface equation.
Since f (x, y, z) = constant, the derivative of f is
zero; i.e.
.
90
0
cos
0
cos
grad
0
grad
.
~
~

=

=

=

=
=



f
f
f
r
d
r
d
d

59
• This shows that when f (x, y, z) = constant,
• Vector grad f =  f is called normal vector to the
surface f (x, y, z) = constant
.
~
r
d
grad 
f
y
ds
grad f
z
x

60
Unit normal vector is denoted by
Example
Calculate the unit normal vector at (-1,1,1)
for 2yz + xz + xy = 0.
.
~ f
f


=
n

61
Solution
Given 2yz + xz + xy = 0. Thus
.
6
1
1
4
and
2
)
1
2
(
)
1
2
(
)
1
1
(
(-1,1,1),
At
.
)
2
(
)
2
(
)
(
~
~
~
~
~
~
~
~
~
=
+
+
=

+
+
=
-
+
-
+
+
=

+
+
+
+
+
=

f
f
f
k
j
i
k
j
i
k
x
y
j
x
z
i
y
z
)
2
(
6
1
6
2
is
vector
normal
unit
The
~
~
~
~
~
~
k
j
i
k
j
i
n
~
+
+
=
+
+
=


=

f
f

62
4. Divergence and gradient of a Vector
.
.
)
.(
.
as
defined
is
of
divergence
the
,
If
~
~
~
~
~
~
~
~
~
~
~
~
~
~
~
z
a
y
a
x
a
A
A
div
k
a
j
a
i
a
k
z
j
y
i
x
A
A
div
A
k
a
j
a
i
a
A
z
y
x
z
y
x
z
y
x


+


+


=

=

+
+










+


+


=

=
+
+
=

63
Example
.
13
)
3
)(
2
(
2
)
3
)(
1
(
)
2
)(
1
(
2
(1,2,3),
point
At
.
2
2
.
(1,2,3).
point
at
determine
,
If
~
~
~
~
~
2
~
~
2
~
=
+
-
=
+
-
=


+


+


=

=
+
-
=
A
div
yz
xz
xy
z
a
y
a
x
a
A
A
div
A
div
k
yz
j
xyz
i
y
x
A
z
y
x
Answer

64
Exercise
.
114
(3,2,1),
point
At
.
(3,2,1).
point
at
determine
,
If
~
~
~
~
~
3
~
2
~
2
3
~
=
=
=


+


+


=

=
-
+
=


A
div
z
a
y
a
x
a
A
A
div
A
div
k
yz
j
z
xy
i
y
x
A
z
y
x
Answer

65
Remarks
.
called
is
vector
,
0
If
function.
scalar
a
is
but
function,
vector
a
is
~
~
~
~
vector
solenoid
A
A
div
A
div
A
=

66
Curl of a Vector
~ ~
~ ~
~
~ ~
~ ~ ~ ~
~ ~
~ ~
~
~ ~
If ( , , ) ( , , ) ( , , ) , the curl of is defined by
( )
.
A f x y z i g x y z j h x y z k A
curl A A
i j k f i g j hk
x y z
i j k
curl A A
x y z
f g h
= + +
= 
 
  
= + +  + +
 
  
 
  
 =  =
  

67
Example
.
)
2
,
3
,
1
(
at
determine
,
)
(
)
(
If
~
~
2
~
2
2
~
2
2
4
~
-
-
+
+
-
=
A
curl
k
yz
x
j
y
x
i
z
x
y
A

68
Solution
.
)
4
2
(
)
2
2
(
)
(
)
(
)
(
)
(
)
(
)
(
~
3
~
2
~
2
~
2
2
4
2
2
~
2
2
4
2
~
2
2
2
2
2
2
2
2
4
~
~
~
~
~
k
y
x
j
z
x
xyz
i
z
x
k
z
x
y
y
y
x
x
j
z
x
y
z
yz
x
x
i
y
x
z
yz
x
y
yz
x
y
x
z
x
y
z
y
x
k
j
i
A
A
curl
-
+
+
-
-
-
=








-


-
+


+






-


-
-


-








+


-
-


=
-
+
-






=


=

69
.
106
8
2
)
)
3
(
4
)
1
(
2
(
))
2
(
)
1
(
2
)
2
)(
3
)(
1
(
2
(
)
2
(
)
1
(
(1,3,-2),
At
~
~
~
~
3
~
2
~
2
~
k
j
i
k
j
i
A
curl
-
-
=
-
+
-
+
-
-
-
-
-
=
Exercise 4.5
.
)
3
,
2
,
1
(
point
at
determine
,
)
(
)
(
If
~
~
2
2
~
2
2
~
2
2
3
~
A
curl
k
yz
x
j
z
x
i
z
y
xy
A -
+
+
-
=

70
Answer
.
26
12
15
(1,2,3),
At
.
)
2
3
2
(
)
2
2
(
)
2
(
~
~
~
~
~
2
2
~
2
2
~
2
2
~
k
j
i
A
curl
k
yz
xy
x
j
z
y
xyz
i
z
z
x
A
curl
+
+
-
=
+
-
+
+
-
-
-
-
=
Remark
function.
vector
a
also
is
and
function
vector
a
is
~
~
A
curl
A

71
5. 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
O
B
~
~
r
d
r+
~
r

72
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
.
where
~
~
~
~
~
k
dz
j
dy
i
dx
r
d
r
d
V
c
+
+
=


73
Example
(3,2,1).
B
to
(0,0,0)
A
from
along
find
then
,
,
2
,
3
by
given
is
curve
a
and
z
If
~
3
2
2
=
=
=
=
=
=
 C
r
d
V
u
z
u
y
u
x
C
xy
V
c

74
Solution
.
1
,
1
,
2
2
,
3
3
(3,2,1),
B
At
.
0
,
0
,
0
2
,
0
3
(0,0,0),
A
At
.
3
4
3
And,
.
12
)
(
)
2
)(
3
(
z
Given
3
2
3
2
~
2
~
~
~
~
~
~
8
3
2
2
2
=

=
=
=
=
=

=
=
=
=
+
+
=
+
+
=
=
=
=
u
u
u
u
u
u
u
u
k
du
u
j
du
u
i
du
k
dz
j
dy
i
dx
r
d
u
u
u
u
xy
V

75
 
.
11
36
5
24
4
11
36
5
24
4
36
48
36
)
3
4
3
)(
12
(
~
~
~
~
1
0
11
~
1
0
10
~
1
0
9
1
0 ~
10
1
0
~
9
1
0
~
8
1
0 ~
2
~
~
8
~
k
j
i
k
u
j
u
i
u
k
du
u
j
du
u
i
du
u
k
du
u
j
udu
i
du
u
r
d
V
u
u
B
A
+
+
=






+






+
=
+
+
=
+
+
=

 



=
=

76
Exercise
.
11
768
144
5
384
(4,3,2).
B
to
(0,0,0)
A
from
curve
the
along
calculate
,
2
,
3
,
4
by
given
is
curve
the
and
If
~
~
~
~
~
2
3
2
2
k
j
i
r
d
V
C
r
d
V
u
z
u
y
u
x
C
yz
x
V
B
A
c
+
+
=
=
=
=
=
=
=


Answer

77
Vector Field, Integral
Let a vector field
and
The scalar product is written as
.
)
).(
(
.
~
~
~
~
~
~
~
~
dz
F
dy
F
dx
F
k
dz
j
dy
i
dx
k
F
j
F
i
F
r
d
F
z
y
x
z
y
x
+
+
=
+
+
+
+
=
~
~
~
~
k
F
j
F
i
F
F z
y
x +
+
=
.
~
~
~
~
k
dz
j
dy
i
dx
r
d +
+
=
~
~
. r
d
F

78
.
.
by
given
is
B
point
another
A to
point
a
from
curve
the
along
of
integral
line
then the
,
curve
the
along
is
field
vector
a
If
~
~
~
~



 +
+
=
c
z
c
y
c
x
c
dz
F
dy
F
dx
F
r
d
F
C
F
C
F

79
Example
.
y
2
y
if
,
2
,
4
curve
the
along
(4,2,1)
B
to
(0,0,0)
A
from
.
Calculate
~
~
~
2
~
3
2
~
~
k
z
j
xz
i
x
F
t
z
t
y
t
x
r
d
F
c
-
+
=
=
=
=
=
=


80
Solution
.
3
4
4
And
.
4
4
32
)
(
)
2
(
2
)
(
)
4
(
)
2
(
)
4
(
2
y
Given
~
2
~
~
~
~
~
~
~
5
~
4
~
4
~
3
2
~
3
~
2
2
~
~
~
2
~
k
dt
t
j
dt
t
i
dt
k
dz
j
dy
i
dx
r
d
k
t
j
t
i
t
k
t
t
j
t
t
i
t
t
k
yz
j
xz
i
x
F
+
+
=
+
+
=
-
+
=
-
+
=
-
+
=

81
.
)
12
16
128
(
12
16
128
)
3
)(
4
(
)
4
)(
4
(
)
4
)(
32
(
)
3
4
4
)(
4
4
32
(
.
Then
7
5
4
7
5
4
2
5
4
4
~
2
~
~
~
5
~
4
~
4
~
~
dt
t
t
t
dt
t
dt
t
dt
t
dt
t
t
tdt
t
dt
t
k
dt
t
j
dt
t
i
dt
k
t
j
t
i
t
r
d
F
-
+
=
-
+
=
-
+
+
=
+
+
-
+
=
.
1
,
1
,
2
2
,
4
4
(4,2,1),
B
at
and,
.
0
,
0
,
0
2
,
0
4
(0,0,0),
A
At
3
2
3
2
=

=
=
=
=
=

=
=
=
=
t
t
t
t
t
t
t
t

82
.
30
23
26
2
3
3
8
5
128
2
3
3
8
5
128
)
12
16
128
(
.
1
0
8
6
5
1
0
7
5
4
~
~
=
-
+
=






-
+
=
-
+
=
 

=
=
t
t
t
dt
t
t
t
r
d
F
t
t
B
A

83
Exercise
.
168
61
7
.
.
3
,
2
,
curve
on the
(1,2,3)
B
to
(0,0,0)
A
from
.
calculate
,
3
If
~
~
3
2
~
~
~
2
~
~
2
~
=
=
=
=
=
=
+
-
=


r
d
F
t
z
t
y
t
x
r
d
F
k
z
x
j
yz
i
xy
F
B
A
c
Answer

84
* Double Integral (Revisions)*
2
2
Given ( , ) 4 in region bounded
by a straight line 0, and 2.
Find ( , ) in both order integrals.
( , ) 4 unit .
R
R
f x y y R
x y x y
f x y dA
f x y dA
= -
= = =
=


Example
Answer

85
2
2
Using double integral, find the area of a region
bounded by 5 and 3.
1
The area of the region 4 unit .
2
y x y x
= - = +
=
Example
Answer

86
2 2
Evaluate a solid which is bounded by
16 and 2.
Stated ( , , ) as integral in order .
z x y z
f x y z dV dzdydx
= - - =

Example

87
2
Describe ( , , ) as integral in order
if is a solid which is bounded by
0, , and 4 2 .
f x y z dV
dzdydx S
z z x y x
= = = -

Example

88
6. Volume Integral
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

 =
V
V
Fdxdydz
FdV

89
Example
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
x
y
3
O
2
1

90
  
 = = =
=
2
0
3
0
1
0
2
z y x
V
xdxdydz
FdV
Solution
6
]
[
3
3
]
[
2
1
.
2
2
1
2
2
2
2
0
2
0
3
0
2
0
2
0
3
0
1
0
2
0
3
0
2
=
=
=
=
=






=


 
 
=
=
= =
= =
z
dz
dz
y
dydz
dydz
x
z
z
z y
z y

91
If V is a closed region and , vector field in region
V, Volume integral of V is
~
F
~
F
   
=
V
x
x
y
y
z
z
dzdydx
F
dV
F
2
1
2
1
2
1 ~
~

92
Evaluate , where V is a region bounded by
x = 0, y = 0, z = 0 and 2x + y + z = 2, and also
given
V
dV
F
~
~
~
~
2 k
y
i
z
F +
=
Example

93
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)
Solution

94
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
z
x
y
2
O
2
1
2x + y + z = 2
y = 2 (1 - x)

95
Therefore,
   
=
-
=
-
-
=
=
V x
x
y
y
x
z
dzdydx
F
dV
F
1
0
)
1
(
2
0
)
1
(
2
0 ~
~
 

-
=
-
-
=
=
+
=
)
1
(
2
0
)
1
(
2
0 ~
~
1
0
)
2
(
x
y
y
x
z
x
dzdydx
k
y
i
z
~
~ 3
1
3
2
k
i+
=



96
Example
Evaluate where
and V is region bounded by z = 0, z = 4 and
x2 + y2 = 9
V
dV
F
~ ~
~
~
~
2
2 k
y
j
z
i
F +
+
=
x
z
f
y

4 -
3
3

97
f
 cos
=
x f
 sin
=
y z
z =
z
d
ρdρd
dV f
=
; ; ;
where
,
3
0 
  ,
2
0 
f 
 4
0 
 z
Using polar coordinate of cylinder,

98
  +
+
=
V V
dxdydz
k
y
j
z
i
dV
F )
2
2
(
~
~
~
~
Therefore,
  
= = =
+
+
=
4
0
2
0
3
0 ~
~
~
)
sin
2
2
(
z
k
j
z
i

f 

f
 dz
d
d f

~
~
144
72 j
i 
 +
=



100
7. Surface Integral
Scalar Field, V Integral
If scalar field V exists on surface S, surface
integral V of S is defined by
 
=
S S
dS
n
V
S
Vd
~
~
where
S
S
n


=
~

101
Example
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
S
S
Vd
~
Solution
Given S : x2 + y2 = 4 , so grad S is
~
~
~
~
~
2
2 j
y
i
x
k
z
S
j
y
S
i
x
S
S +
=


+


+


=


102
Also,
4
4
2
2
)
2
(
)
2
( 2
2
2
2
=
=
+
=
+
=
 y
x
y
x
S
Therefore,
)
(
2
1
4
2
2
~
~
~
~
~
j
y
i
x
j
y
i
x
S
S
n +
=
+
=


=
Then,
  +






=
S S
dS
j
y
i
x
xyz
dS
n
V )
(
2
1
~
~
~
 +
= dS
j
z
xy
i
yz
x )
(
2
1
~
2
~
2

103
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,
So, we will use polar coordinate of cylinder to find
the surface integral.
.
2
4 =
=

x
z
f
y
2
2
3
O

104
Polar Coordinate for Cylinder
cos 2cos
sin 2sin
ρ
x
y
z z
dS d dz
 f f
 f f
f
= =
= =
=
=
where
2
0

f 
 3
0 
 z
(1st octant) and

105
Using polar coordinate of cylinder,
f
f
f
f
f
f
f
f
cos
sin
8
)
(
)
sin
2
)(
cos
2
(
sin
cos
8
)
sin
2
(
)
cos
2
(
2
2
2
2
2
2
z
z
z
xy
z
z
yz
x
=
=
=
=
From


 =
+
=
S
S
S
S
Vd
dS
j
z
xy
i
yz
x
dS
n
V
~
~
2
~
2
~
)
(
2
1

106
  
= =
+
=
S
z
dzd
j
z
i
z
S
Vd 2
0
3
0 ~
2
~
2
~
)
2
)(
cos
sin
8
sin
cos
8
(
2
1 
f
f
f
f
f
f
3
2 2 2 2
2
0 ~ ~ 0
2 2
2
0 ~ ~
1 1
8 cos sin sin cos
2 2
9 9
8 cos sin sin cos
2 2
z i z j d
i j d


f f f f f
f f f f f
 
= +
 
 
 
= +
 
 


2 2
2
0 ~ ~
3 3 2
~ ~
0
~ ~
9
8 cos sin sin cos
2
cos sin sin cos
36
3( sin ) 3(cos )
12( )
i j d
i j
i j


f f f f f
f f f f
f f
 
=  +
 
 
 
= +
 
-
 
= +

Therefore,

107
2
2 2
~
,
9
0 2
S
If V is a scalar field whereV xyz evaluate
V d S for surface S that region bounded by x y
between z and z in the first octant.
=
+ =
= =

Exercise
~ ~
: 24( )
Answer i j
+

108
If vector field defeated on surface S, surface
integral of S is defined as

 =
S
S
dS
n
F
S
d
F .
.
~
~
~
~
~
F
~
F
~
where
S
n
S

=


109
Example
~ ~ ~
~
2 2 2
~ ~
Vector field 2 defeated on surface
: 9 and bounded by 0, 0, 0 in
the first octant.
Evaluate . .
S
F y i j k
S x y z x y z
F d S
= + +
+ + = = = =


110
Solution
2 2 2
Given : 9 is bounded by 0, 0,
0 in the1st octant. This refer to sphere with center
at (0,0,0) and radius, 3, in the1st octant.
S x y z x y
z
r
+ + = = =
=
=
x
z
y
3
3
3
O

111
~ ~
~
~ ~
~
2 2 2
2 2 2
So, grad is
2 2 2 ,
and
(2 ) (2 ) (2 )
2
2 9 6.
S
S S S
S i j k
x y z
x i y j z k
S x y z
x y z
  
 = + +
  
= + +
 = + +
= + +
= =

112
~ ~
~ ~
~ ~ ~ ~
~ ~
Therefore,
. .
1
( 2 ) ( )
3
1
( 2 ) .
3
S S
S
S
F d S F ndS
y i j k x i y j z k dS
xy y z dS
=
 
= + + + +
 
 
= + +
 


).
(
3
1
6
2
2
2
~
~
~
~
~
~
~
k
z
j
y
i
x
k
z
j
y
i
x
S
S
n
+
+
=
+
+
=


=


113
Using polar coordinate of sphere,
2
sin cos 3sin cos
sin sin 3sin sin
cos 3cos
sin 9sin
where 0 , .
2
x r
y r
z r
dS r d d d d
 f  f
 f  f
 
  f   f

 f
= =
= =
= =
= =
 

114
f



f

f
f

f



f

f

f

 
 
f 
f 
d
d
d
d
S
d
F
S
]
cos
sin
sin
sin
2
cos
sin
sin
3
[
9
]
sin
9
[
]
cos
3
)
sin
sin
3
(
2
)
sin
sin
3
)(
cos
sin
3
[(
3
1
.
2
0 0
3
0 0
~
~
2 2
2 2
+
+
=
+
+
=

 
  
= =
= =






+
=
4
3
1
9




115
Exercise
octant.
1
in the
0
and
0
,
0
,
4
by
bounded
region
the
of
surface
a
is
and
2
where
,
on
Evaluate
st
2
2
2
~
~
~
~
~
~
=
=
=
=
+
+
+
+
=


z
y
x
z
y
x
S
k
y
j
z
i
x
F
S
S
d
F
S
:8 1
6
Answer

 
+
 
 

116
8. 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),
where S is the area of C.

 +
=










-


c
S
dy
Q
dx
P
dy
dx
y
P
x
Q
)
(

117
Example
2 2
2 2
Prove Green's Theorem for
[( ) ( 2 ) ]
which has been evaluated by boundary that defined as
0, 0 4 in the first quarter.
c
x y dx x y dy
x y and x y
+ + +
= = + =

y
2
x
2
C3
C2
C1
O
x2 + y2 = 22
Solution

118
1 1
2 2
2 2
1 2 3
1
2 2
2
2
0
2
3
0
Given [( ) ( 2 ) ] where
and 2 . We defined curve
as , .
i) For : 0, 0 0 2
( ) ( ) ( 2 )
1 8
.
3 3
c
c c
x y dx x y dy
P x y Q x y c
c c and c
c y dy and x
Pdx Qdy x y dx x y dy
x dx
x
+ + +
= + = +
= =  
 
+ = + + +
 
=
 
= =
 
 

 


119
2 2
2
ii) For : 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,
2cos , 2sin , 0
c x y
x y
 
+ =
= =
2
2sin , 2cos .
dx d dy d


   
 
 = - =

120
 
 
.
4
4
8
sin
4
2
sin
2
cos
8
)
cos
sin
8
2
cos
2
2
sin
8
(
)
cos
sin
8
cos
4
sin
8
(
)]
cos
2
))(
sin
2
(
2
cos
2
((
)
sin
2
)(
)
sin
2
(
)
cos
2
((
[
)
2
(
)
(
)
(
2
2
2
2
2
2
0
2
0
0
2
2
2
0
2
2
-
=
+
+
-
=
+
+
+
=
+
+
+
-
=
+
+
-
=
+
+
-
+
=
+
+
+
=
+

































d
d
d
d
dy
y
x
dx
y
x
Qdy
Pdx
c
c

121
3 3
3
2 2
0
2
0
2
2
iii) : 0, 0, 0 2
( ) ( ) ( 2 )
2
4.
8 16
( ) ( 4) 4 .
3 3
c c
c
For c x dx y
Pdx Qdy x y dx x y dy
y dy
y
Pdx Qdy  
= =  
 
+ = + + +
 
=
 
=  
= -
 + = + - - = -
 



122
b) Now, we evaluate
where 1 2 .
Again,because this is a part of the circle,
we shall integrate by using polar coordinate of plane,
cos , sin
where
S
Q P
dxdy
x y
Q P
and y
x y
x r y r
 
 
 
-
 
 
 
 
= =
 
= =

0 r 2, 0 .
2
and dxdy dS r dr d

 
    = =

123
.
3
16
cos
3
16
2
sin
3
16
2
sin
3
2
2
1
)
sin
2
1
(
)
2
1
(
2
2
2
2
0
0
2
0
0
3
2
0
2
0
-
=






+
=






-
=






-
=
-
=
-
=










-




 


=
=
= =
















d
d
r
r
d
dr
r
r
dy
dx
y
dy
dx
y
P
x
Q
r
S
S

124
Therefore,
( )
16
.
3
Green's Theorem has been proved.
C S
Q P
Pdx Qdy dxdy
x y
LHS RHS

 
 
+ = -
 
 
 
= -
=

 

125
9. Divergence Theorem (Gauss’ Theorem)
If S is a closed surface including region V in
vector field
.
.
~
~
~ 
 =
S
V
S
d
F
dV
F
div
~
F
~
y
x z
f
f f
div F
x y z

 
= + +
  

126
Example
2
~ ~ ~
~
2 2
Prove Gauss' Theorem for vector field,
2 in the region bounded by
planes 0, 4, 0, 0 4
in the first octant.
F x i j z k
z z x y and x y
= + +
= = = = + =

127
Solution
x
z
y
2
2
4
O
S3
S4
S2
S1
S5

128
1
2
3
4
2 2
5
~
~ ~
For this problem, the region of integration is bounded
by 5 planes :
: 0
: 4
: 0
: 0
: 4
To prove Gauss' Theorem, we evaluate both
. ,
The answer should be the same.
V
S
S z
S z
S y
S x
S x y
div F dV
and F d S
=
=
=
=
+ =



129
2
~ ~ ~ ~
~
2
~
~
1) We evaluate . Given 2 .
So,
( ) (2) ( )
1 2 .
Also, (1 2 ) .
The region is a part of the cylinder. So, we integrate by using
polar c
V
V V
div F dV F x i j z k
div F x z
x y z
z
div F dV z dV
= + +
  
= + +
  
= +
= +

 
oordinate of cylinder ,
; sin ;
where 0 2, 0 , 0 4.
2
x = cos y z z
dV d d dz
z
 f  f
  f

 f
= =
=
     

130
 
2
2
2
2
2
2
2 4
0 0 0
2
2 4
0
0 0
2
0 0
2 2
0
0
0
0
~
Therefore,
(1 2 ) (1 2 )
[ ]
(20 )
[10 ]
(40)
40
20 .
20 .
V z
V
z dV z dzd d
z z d d
d d
d
d
div F dV






f 
f 
f 
f
f
  f
  f
  f
 f
f
f


= = =
= =
= =
=
=
+ = +
= +
=
=
=
=
=
 =
   
 
 




131
1
~ ~
~ ~
1
~ ~
~ ~ ~
~
~ ~ ~ ~
~
~ ~
2) Now, we evaluate . . .
i) : 0, ,
2 0
. ( 2 ).( ) 0
. 0.
S S
S
F d S F ndS
S z n k dS rdrd
F x i j k
F n x i j k
F ndS

=
= = - =
 = + +
 = + - =
 =
 


132
2
2
2
~ ~
2
~ ~ ~ ~ ~
~ ~
~ ~ ~ ~ ~
~
2
2
0 0
~ ~
ii) : 4, ,
2 (4) 2 16
. ( 2 16 ).( ) 16.
Therefore for , 0 r 2, 0
2
. 16
16 .
S r
S z n k dS rdrd
F x i j k x i j k
F n x i j k k
S
F ndS rdrd







= =
= = =
 = + + = + +
 = + + =
   
 =
=
=
  

133
3
3
~ ~
2
~ ~ ~
~
2
~ ~ ~ ~
~ ~
3
2 4
0 0
~ ~
iii) : 0, ,
2
. ( 2 ).( )
2.
Therefore for S , 0 2, 0 4
. ( 2)
16.
S x z
S y n j dS dxdz
F x i j z k
F n x i j z k j
x z
F ndS dzdx
= =
= = - =
 = + +
 = + + -
= -
   
 = -
=
= -
  

134
4
4
~ ~
2 2
~ ~ ~ ~
~ ~
2
~ ~ ~ ~
~
~ ~
iv) : 0, ,
0 2 2
. (2 ).( ) 0.
. 0.
S
S x n i dS dydz
F i j z k j z k
F n j z k i
F ndS
= = - =
 = + + = +
 = + - =
 =


135
2 2
5
5 5
~ ~
~
5 ~
~
5
~ ~
5
v) : 4,
2 2 4
2 2
4
1
( ).
2
By using polar coordinate of cylinder :
cos , sin ,
where for :
2, 0 , 0 4, 2
2
S x y dS d dz
S x i y j and S
x i y j
S
n
S
x i y j
x y z z
S
z dS d dz
 f
 f  f

 f f
+ = =
 = +  =
+

 = =

= +
= = =
=     =

136
.
4
16
)
2
)(
sin
)(cos
2
(
.
).
sin
(cos
2
2.
kerana
;
sin
2
cos
2
)
sin
(
)
cos
(
2
1
2
1
2
1
2
1
).
2
(
.
5
2
0
4
0
2
~
~
2
2
2
2
~
~
~
2
~
~
~
~

f
f
f
f
f

f
f
f

f


f
+
=
=
+
=

+
=
=
+
=
+
=
+
=






+
+
+
=

  
= =

S z
dz
d
dS
n
F
y
x
j
y
i
x
k
z
j
i
x
n
F

137
1 2 3 4 5
~ ~ ~ ~ ~ ~
~ ~ ~ ~ ~ ~
~ ~
Finally,
. . . . . .
0 16 16 0 16 4
20 .
. 20 .
Gauss' Theorem has been proved.
S S S S S S
S
F d S F d S F d S F d S F d S F d S
F d S
LHS RHS
 


= + + + +
= + - + + +
=
 =
=

     


138
10. Stokes’ Theorem
If is a vector field on an open surface S and
boundary of surface S is a closed curve C,
therefore
  
=

S c
r
d
F
S
d
F
curl
~
~
~
~
~
F
~ ~
~
~ ~
x y z
i j k
curl F F
x y z
f f f
  
=   =
  

139
Example
Surface S is the combination of
2 2
~ ~ ~
~
i) part of the cylinder 9 0
and 4 0.
ii) half of the circle with radius 3 at 4, and
iii) 0
, prove Stokes' Theorem
for this case.
a x y between z
z for y
a z
plane y
If F z i xy j xz k
+ = =
= 
=
=
= + +

140
Solution
2 2
1
2
3
We can divide surface S as
S : x y 9 0 z 4 y 0
S : z 4, half of the circle with radius 3
S : y 0
for and
+ =   
=
=
z
y
x
3
4
O
S3
C2
S2
C1
S1
3

141
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 first.
Given
~ ~
S
curl F d S


~
~
~
~
k
xz
j
xy
i
z
F +
+
=

142
So,
~
~
~
~
~
~
~
~
~
)
1
(
)
(
)
(
)
(
)
(
)
(
)
(
k
y
j
z
k
z
y
xy
x
j
xz
x
z
z
i
xy
z
xz
y
xz
xy
z
z
y
x
k
j
i
F
curl
+
-
=










-


+








-


+










-


=






=

143
By integrating each part of the surface,
2 2
1
1
~ ~
2 2
1
2 2
( ) : 9,
2 2
(2 ) (2 )
2 6
i For surface S x y
S x i y j
and S x y
x y
+ =
 = +
 = +
= + =

144
)
(
3
1
6
2
2
~
~
~
~
1
1
~
j
y
i
x
j
y
i
x
S
S
n +
=
+
=


=
and
).
1
(
3
1
3
1
3
1
)
1
(
~
~
~
~
~
~
z
y
j
y
i
x
k
y
j
z
n
F
curl
-
=






+






 +
-
=

Then ,

145
By using polar coordinate of cylinder ( because
is a part of the cylinder),
9
: 2
2
1 =
+ y
x
S
cos , sin ,
3, 0 0 4.
x y z z
dS d dz
where
dan z
 f  f
 f
 f 
= = =
=
=    

146
Therefore,
  
~ ~
1
(1 )
3
1
sin 1
3
sin (1 ) ; 3
curl F n y z
z
z because
 f
f 
 = -
= -
= - =
Also, dz
d
dS f
3
=

147
 
 
1 1
~ ~
~ ~
4
0 0
4
0
0
4
0
3 sin (1 )
3 (1 ) cos
3 (1 )(1 ( 1))
24
S S
z
curl F d S curl F n dS
z d dz
z dz
z dz

f

f f
f
= =
  = 
= -
= - -
= - - -
= -
 
 



148
(ii) For surface , normal vector unit to the
surface is
By using polar coordinate of plane ,
4
:
2 =
z
S
.
~
~
k
n =

 d
dr
r
dS
dan
z
r
y =
=
= 4
,
sin
0 r 3 and 0 .
where  
   

149
 
2 2
~ ~ ~ ~
~
~ ~
~ ~
3
0 0
3
2
0 0
(1 )
sin
( sin )( )
sin
18
S S
r
r
curl F n z j y k k
y r
r rdrd
r d dr





 
 
= =
= =
 
  = - + 
 
 
= =
  = 
=
=
=
 
 
 

150
(iii) For surface S3 : y = 0, normal vector unit
to the surface is
dS = dxdz
The integration limits :
.
~
~
j
n -
=
3 3 0 4
x and z
-    
So,
1
)
(
)
)
1
((
~
~
~
~
~
-
=
-

+
-
=

z
j
k
y
j
z
n
F
curl

151
3 3
1 2 3
~ ~
~ ~
3 4
3 0
~ ~ ~ ~
~ ~ ~ ~
Then,
. .
( 1)
24.
. . . .
24 18 24
18.
S S
x z
S S S S
z dzdx
curl F d S curl F d S curl F d S curl F d S
=- =
=
= -
=
=
 = + +
= - + +
=
 
 
   

152
~ ~
1
1
Now, we evaluate . for each pieces of the curve C.
i) is a half of the circle.
Therefore, integration for will be more easier if we use
polar coordinate for plane with radius
C
F d r
C
C

3, that is
3cos , 3sin dan z 0
where 0 .
r
x y
 
 
=
= = =
 

153
~ ~ ~
~
~
~
~ ~
~
~ ~
(3cos )(3sin )
9sin cos
and
3sin 3cos .
F z i xy j xzk
j
j
dr dx i dy j dzk
d i d j
 
 
   
 = + +
=
=
= + +
= - +

154
1
2
~ ~
2
0
~ ~
3
0
From here,
. 27sin cos .
. 27sin cos
9cos
18.
C
F d r d
F d r d


  
  

=
 =
 
= -
 
=
 

155
2
2
~ ~ ~
~
~
~ ~
ii) Curve is a straight line defined as
, 0 z 0, where 3 3.
Therefore,
0.
. 0.
C
C
x t y and t
F z i xy j xz k
F d r
= = = -  
= + +
=
 =


156
1 2
~ ~ ~ ~ ~ ~
~ ~ ~
~
. . .
18 0
18.
We already show that
. .
Stokes' Theorem has been proved.
C C C
S C
F d r F d r F d r
curl F d S F d r
 = +
= +
=
=

  
 

Line Integral Independent of Path
• Suppose that C is a smooth curve given
by Also suppose that f is a
function whose gradient vector is continuous
on C. Then,
• Notice the strong resemblance of this
theorem to the usual FTC. Some sort
• of definite integral is equal to the value of a
function at one point subtracted from the
157
( ), .
r t a t b
 
f

 
   
 
.
C
f dr f r b f r a
 = -


Continued
value of a function at another point. The role
of an anti derivative for F is played by the
scalar function f, and the role of the
endpoints of the interval [a, b] are played by
the endpoints of C, r(a), r(b).
• One consequence of this theorem is that if
F is conservative on R 2 , then the value of
a line integral of F along C does not
actually depend on the path taken by C,
but only on the endpoints of C. That is, a
phenomenon like that which we observed
in the previous set of examples can never
158

Continued
take place for a conservative vector field. We say
that a line integral of F over C is independent of
path if its values only depend on the endpoints of
C. In this terminology, the line integral of a
conservative vector field is independent of path for
all curves C.
• A special case of the above case occurs when
we calculate the line integral of F along a curve
C whose starting and endpoints are the same. If
C is such a curve, we call C a closed curve, and
the line integral of a conservative vector field
along a closed curve C is equal to
159

Continued
160
 
   
  0
C
Fdr f r b f r a
= - =


Continued
since r(b) = r(a). Therefore, the line integral of a
conservative vector field along a closed path is always
equal to 0, regardless of the shape of the closed path.
Example.
Let and let C, be the
parameterization of C,0 ≤ t ≤ 1. Calculate the line
integral of F along C.
161
 
,
x x
F ye e
=    
17 3
,cos
r t t
t 
=

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