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Note on Vector Calculus
1
...
Vector Product

3
...
Integration of Vectors
5
...
Polar Coordinates
1

7
...


Volume Integral

9
...
Green’s Theorem
11
...
Stokes’ Theorem

2

Elementary Vector Analysis
Definition (Scalar and vector)

Scalar is a quantity that has magnitude
but not direction
...

For instance acceleration, velocity, force
3

We represent a vector as an arrow from the
origin O to a point A
...

4

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
...
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

Calculation of Vectors
1
...

Then
a = b  a1 = b1 , a2 = b2 , a3 = b3

8

2
...
Multiplication of Vectors by Scalars

If  is a scalar, then
 b = (b1 )i + (b2 ) j + (b3 )k

9

Example 1

Given p = 5i + j - 3k and q = 4i - 3j + 2k
...
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 |
~

~

b

compb a

|b|
~

b) The area of triangle

1
A = a b
...
b x c = b1
6
6
c1

a2
b2

b

b3

c2

a

a3
c3

c

e) The volume of parallelepiped
V = a
...
b, a  b and the angle between a and b
...

n
n ~
n
n ~
du
du
du ~ du
dn A
~
n

• The magnitude of

is

du

n

d A
du

~
n

2

 d ax   d a y   d az 
=  n  + n  + n 
 du   du   du 

 

 
n

2

n

n

2

16

Example 3
If A = 3u 2 i - 2u j + 5 k
~

~

~

~

hence
dA
~

du
d2 A
du

~
2

=
=

17

Example 4
The position of a moving particle at time t is given
by x = 4t + 3, y = t2 + 3t, z = t3 + 5t2
...

• The magnitude of both velocity and acceleration
at t = 1
...

~

~

~

~

• The velocity is given by

dr
~

dt

= 4 i + (2t + 3) j + (3t 2 + 10t ) 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
...

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
...

21

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
~ du
du ~ ~
du ~
dB dA
d
d)
( A  B) = A  ~ + ~  B
~
du ~ ~
du du ~
22

Partial Derivatives of a Vector
•

If vector A depends on more than one
~

parameter, i
...
t
...

24

Example 5
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 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

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
...

a

~

a

~

27

Example 6
If F = (3t 2 + 4t ) i + (2t - 5) j + 4t 3 k ,
~

~

~

~

3

calculate  F dt
...

~

~

~

28

Exercise 2
If F = (t 3 + 3t ) i + 2t 2 j + (t - 4) k ,
~

~

calculate

1



~

~

F dt
...

4~ 3 ~ 2 ~
29

Del Operator Or Nabla (Symbol )
• Operator  is called vector differential operator,
defined as



 
= i+
j
 x ~ y ~ + z k 
...

x ~ y ~ z ~

* is a scalar function
*  is a vector function
31

Example 7
If  = x 2 yz 3 + xy 2 z 2 , determine grad  at P = (1,3,2)
...

~

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
...

~

~

~

33

Exercise 3

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

~

~

~

35

Grad Properties
If A and B are two scalars, then

1) ( A + B) = A + B
2) ( AB ) = A(B) + B(A)

36

Directional Derivative
Directional derivative of  in the direction of a is
~

d
= a
...

~

37

Example 8
Compute the directiona l derivative of  = x 2 z + 2 xy 2 + yz 2
at the point (1,2,-1) in the direction of the vector
A = 2i +3 j - 4k
...
grad
ds ~

a
~

A



where grad =  =
i+
j+
k and a = ~
...

~

~

~

39

At (1,2,-1),
 = (2(1)(-1) + 2(2) 2 ) i + (4(1)(2)
~

+ (-1) 2 ) j + ((1) 2 + 2(2)(-1)) k
...

~

~

~

Also, given A = 2 i + 3 j - 4 k , then
~

~

~

~

A = 2 2 + 32 + (-4) 2
~

= 29
...

29 ~
29 ~
29 ~

~

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

29
41

Unit Normal Vector
Equation  (x, y, z) = constant is a surface
equation
...
e
...
grad  = 0
~

 d r grad  cos  = 0
~

 cos  = 0
  = 90
...

~

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=

...

44

Solution
Given 2yz + xz + xy = 0
...

~

~

~

At (-1,1,1),  = (1 + 1) i + (2 - 1) j + (2 - 1) k
~

~

~

= 2 i+ j + k
~

~

~

and  = 4 + 1 + 1 = 6
...
A
~

~



 
= i +
j
i
j
 x ~ y ~ + z k 
...
A =
+
+

...

~

Answer
a x a y a z
div A = 
...

At point (1,2,3),
div A = 2(1)(2) - (1)(3) + 2(2)(3)
~

= 13
...

~

Answer

a x a y a z
div A = 
...

48

Remarks
A is a vector function, but div A is a scalar function
...

~

~

49

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
i
j
 x ~ y ~ + z k   (a x ~ + a y ~ + a z k )

~
~


i
j
k
~


 curl A =   A =
~
~
x
ax

~


y
ay

~



...

~

51

i

Solution

~

curl A =   A =
~

j

~

~


x

k


y

y4 - x2 z 2

x2 + y2

~


z
- x 2 yz

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

~

~

~

Exercise 5
If A = ( xy 3 - y 2 z 2 ) i + ( x 2 + z 2 ) j - x 2 yz 2 k ,
~

~

~

~

determine curl A at point (1,2,3)
...

~

At (1,2,3), curl A = -15 i + 12 j + 26 k
...

~

54

Polar Coordinates
• Polar coordinate is used in calculus to
calculate an area and volume of small
elements in easy way
...

55

Polar Coordinate for Plane (r, θ)

y
ds

d

x = r cos 
y = r sin 
dS = r dr d



x
56

Polar Coordinate for Cylinder ( ,, z)
x =  cos 

z
ds
dv

y =  sin 
z=z

z



y



dS =  d dz
dV =  d d dz

x
57

Polar Coordinate for Sphere (r,  ,
z
r





x = r sin  cos 
y = r sin  sin 
z = r cos 
y

dS = r 2 sin  d d
x

dV = r 2 sin  dr d d
58

Example (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
...


60

Line Integral
Ordinary integral

 f (x) dx, we integrate along

the x-axis
...


 f (s) ds =  f (x, y, z) ds
A

B

r
~

O

r+ d r
~

~

61

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
...


63

Solution
Given V = xy 2 z
= (3u )(2u 2 ) 2 (u 3 ) = 12u 8
...

~

~

~

At A = (0,0,0), 3u = 0, 2u 2 = 0, u 3 = 0,
 u = 0
...


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
...

Answer
384
768
i
A V d r = 5 ~ +144 ~j + 11 k
...

~

~

The scalar product

~

~

~

~

F
...
d r = ( Fx i + Fy j + Fz k )
...


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
...

c

c

68

Example
Calculate

 F
...

~

~

~

~

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
...

~

~

~

70

Then
F
...

At A = (0,0,0), 4t = 0, 2t 2 = 0, t 3 = 0,
 t = 0
...


71

t =1

B

  F
...

30

72

Exercise
If F = xy 2 i - yz j + 3x 2 z k ,
~

~

~

~

calculate  F
...


Answer



B

A

61
F
...

~
~
168

73

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

 FdV = 
V

V

Fdxdydz

78

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
...

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
...

(0,0,2)
If x = z = 0, plane 2x + y + z = 2 intersects y-axis at y = 2
...

(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
...


2
...


3
...


Evaluate

 Vd S
S

~

Solution
Given S : x2 + y2 = 4 , so grad S is

S
S S
j+ k = 2x i + 2 y j
i+
S =
~
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 + xy 2 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
...


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 = 8 z cos  sin 
2

2

2

xy 2 z = (2 cos  )(2 sin  )2 ( z ) = 8 z sin 2  cos 
From

1
V n dS =  ( x 2 yz i + xy 2 z j )dS =  Vd S
S ~
S
~
~
2S
~

95

Therefore,

1  3
Vd S =  2  (8 z cos 2  sin  i + 8 z sin 2  cos  j )(2)dzd
 ~ 2  =0 z =0
~
~
S

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

cos  sin  i + sin  cos  j  d
2
~ 2
~


= 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
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
...
d S =  F
...

Evaluate



F
...


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
...

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
...

~
3 ~
~
Therefore,



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
...


103

1 2 2
  F
...




Answer : 8  + 1
6


105

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
...

Solution

y
x 2 + y 2 = 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
...

i) For c1 : y = 0, dy = 0 and 0  x  2
( Pdx + Qdy ) =  ( x 2 + y 2 )dx + ( x + 2 y )dy 
c1

c1 
2

=  x 2dx
0

2

1 3 8
= x  =
...

This curve actually a part of a circle
...


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
...

8
16
  ( Pdx + Qdy ) = + ( - 4) - 4 =  -
...

where
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
...

3

113

Therefore,
 Q P 
 C ( Pdx + Qdy ) = S  x - y  dx dy


16
= -
...


114

Divergence Theorem (Gauss’ Theorem)
If S is a closed surface including region V in
vector field F
~



V

div F dV =  F
...

~

S ~

~

f x f y f z
div F =
+
+
~
x
y
z

115

Example
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
...
d S,

S ~

~

The answer should be the same
...
Given F = x i + 2 j + z 2 k
...


div F =
~

Also,



V

div F dV =  (1 + 2 z ) dV
...
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
...

  div F dV = 20
...
d S =  F
...


S ~

S ~

~

~

z = 0, n = - k , dS = rdrd
~

~

 F = x i + 2 j + 0k
~

~

~

~

 F
...
( - k ) = 0
~





~

~

~

~

F
...


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
...
(k ) = 16
...
n dS = 

S2 ~

~


2



2

 =0 r =0

~


2

16 rdrd

=
= 16
...
n = ( x i + 2 j + z 2 k )
...

Therefore for S3 , 0  x  2, 0  z  4




F
...


123

x = 0, n = - i , dS = dydz

iv) S4 :

~

~

 F = 0 i + 2 j + z2 k = 2 j + z2 k
~

~

~

~

~

~

 F
...
( - i ) = 0
...
n dS = 0
...

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
...
 x i + y j 
~ ~
~
~
~
2 ~ 2 ~
1 2
= x +y
2
1
= (  cos  ) 2 + (  sin  )
2
= 2 cos 2  + 2 sin  ; kerana  = 2
...





F
...


126

Finally,



F
...
d S + F
...
d S + F
...
d S

S ~

S1 ~

~

~

S2 ~

~

S3 ~

~

S4 ~

~

S5 ~

~

= 0 + 16 - 16 + 0 + 16 + 4
= 20
...
d S = 20
...


127

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
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
...


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
...


Let say, we choose to evaluate

Given

 curl F  d S
S

~

~

first
...

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
...


(ii) For surface
surface is

~

, normal vector unit to the

~

By using polar coordinate of plane ,

y = r sin  , z = 4 dan dS = r dr d

where 0  r  3 and 0    
...

~

~

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
...
n dS
~

S3

~

=

3

~



4

x =-3 z =0

~

( z - 1) dzdx

=
= 24
...
d S =  curl F
...
d S +
~

~



S3

curl F
...


141

Now, we evaluate



F
...


C ~

~

i) C1 is a half of the circle
...


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
...
d r = 27sin  cos  d
...
d r =  27sin  cos  d

C1 ~

~

2

0



=  -9 cos  

0
= 18
...

Therefore, F = z i + xy j + xz k
~

~

~

~

= 0
...
d r = 0
...
d r =

C ~

~



F
...
d r

C2 ~

~

= 18 + 0
= 18
...
d S =
~

~



F
...


146



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