1

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

2

Chapter 2 Continued

7. Line Integral

8. Volume Integral

9. Surface Integral

10. Green’s Theorem

11. Divergence Theorem (Gauss’ Theorem)

12. Stokes’ Theorem

3

2.1 Elementary Vector Analysis

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

orO

A

O Aa

aOA

5

2.1.1 Basic Vector System

Unit vectors , ,

• Perpendicular to each other• In the positive directions of the axes• have magnitude (length) 1

6

Define a basic vector system and form a right-handed set, i.e

7

2.1.2 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 z2 2 2

8

2.1.3 Calculation of Vectors

1. Vector Equation

Two vectors are equal if and only if the corresponding components are equals

332211

321321

, ,

Then

. and Let

babababa

kbjbibbkajaiaa

====

++=++=

9

2. Addition and Subtraction of Vectors

3. Multiplication of Vectors by Scalars

kbajbaibaba )()()( 332211 ++=

kbjbibb )()()(

thenscalar, a is If

321

++=

10

Example 2.1

Given 5 3 and 4 3 2 . Findp i j k q i j k= + - = - +

a p q) +

) b p q-

) 2 10 d q p-

c p) Magnitude of vector

11

2.2 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 theis ,cos||||.or bababa =

12

3) Application of Multiplication of Vectors

a) Given 2 vectors and , projection onto is defined by

b) The area of triangle

~ ~

1.

2A a b=

a b a b

a

b

ba

compb a||

|.|)(length

||

.comp

~

~~

~

~~

b

bal

b

baab

=

=

13

c) The area of parallelogram

d) The volume of tetrahedrone

e) The volume of parallelepiped

aba bxA =

a b

c321

321

321

6

1

ccc

bbb

aaa

=6

1=V a . b cx

a b

c321

321

321

ccc

bbb

aaa

==V a . b cx

14

Example 2.3

. and between angle theand ,. determine

,2and32Given

~~~~~~

~~~~~~~~

bababa

kjibkjia

++=-+=

15

2.4 Vector Differential Calculus

• Let A be a vector depending on parameter u,

• The derivative of A(u) is obtained by differentiating each component separately,

~~~

~ kdu

daj

du

dai

du

da

du

Adzyx ++=

~~~~)()()()( kuajuaiuauA zyx ++=

16

• The nth derivative of vector is given by

• The magnitude of is

)(~

uA

.~~~

~ kdu

adj

du

adi

du

ad

du

Adnz

n

ny

n

nx

n

n

n

++=

222

~

+

+

=

nz

n

ny

n

nx

n

n

n

du

ad

du

ad

du

ad

du

Ad

n

n

du

Ad~

17

Example 2.4

=

=

+-=

2~

2

~

~~~

2

~

hence

523 If

du

Ad

du

Ad

kjuiuA

18

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.

19

Solution

• The parameter is t, and the position vector is

• The velocity is given by

• The acceleration is

.)5()3()34()(~

23

~

2

~~kttjttittr +++++=

.)103()32(4~

2

~~

~ kttjtidt

rd++++=

.)106(2~~

2~

2

ktjdt

rd++=

20

• At t = 1, the velocity of the particle is

and the magnitude of the velocity is

.1354

))1(10)1(3()3)1(2(4)1(

~~~

~

2

~~

~

kji

kjidt

rd

++=

++++=

.210

1354)1(

222~

=

++=dt

rd

21

• At t = 1, the acceleration of the particle is

and the magnitude of the acceleration is

.162

)10)1(6(2)1(

~~

~~2

~

2

kj

kjdt

rd

+=

++=

.652

162)1(

222

~

2

=

+=dt

rd

22

2.4.1 Differentiation of Two Vectors If both and are vectors, then)(

~uA )(

~uB

~

~~

~~~

~

~~

~~~

~~

~~

~

~

)()

..).()

)()

)()

Bdu

Ad

du

BdABA

du

dd

Bdu

Ad

du

BdABA

du

dc

du

Bd

du

AdBA

du

db

du

AdcAc

du

da

+=

+=

+=+

=

23

2.4.2 Partial Derivatives of a Vector

• If vector depends on more than

one parameter, i.e

~A

~21

~21

~2121

~

),,,(

),,,(

),,,(),,,(

kuuua

juuua

iuuuauuuA

nz

ny

nxn

+

+

=

24

• Partial derivative of with respect to is

given by

e.t.c.

~A

~21

2

~21

2

~21

2

21

~

2

~1~1

~11

~ ,

kuu

aj

uu

ai

uu

a

uu

A

ku

aj

u

ai

u

a

u

A

zyx

zyx

+

+

=

+

+

=

1u

25

Example 2.6

~

~

2

~

2

~~2~

2

~~2~

2

~~~

~

~

2

~~

2~

~

23

~

2

~

2

~

6 ,26

,64 ,26

,343

then

)()2(3 If

ivuv

F

vu

Fkiu

v

F

kuju

Fkvjiuv

v

F

kujuivu

F

kvujvuiuvF

=

=

+=

+=

+-=

++=

++-+=

26

Exercise 2.1

=

=

=

=

=

=

++-+=

uv

F

vu

F

v

F

u

F

v

F

u

F

kvujvuivuF

~

2

~

2

2~

2

2~

2

~~

~

23

~

3

~

2

~

,

,

,

then

)3()3(2 If

27

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.

.)()(

)()(

then

)()()()( If

~~

~~

~~~~

kduuajduua

iduuaduuA

kuajuaiuauA

b

a z

b

a y

b

a x

b

a

zyx

++

=

++=

28

Example 2.7

.80242

][]5[]2[

4)52()43(

.calculate

,4)52()43( If

~~~

~

31

4

~

31

2

~

31

23

3

1 ~

33

1 ~

3

1 ~

23

1 ~

3

1 ~

~

3

~~

2

~

kji

ktjttitt

kdttjdttidtttdtF

dtF

ktjtittF

+-=

+-++=

+-++=

+-++=

Answer

29

Exercise 2.2

.2

7

3

2

4

7

)4(2)3(

. calculate

,)4(2)3( If

~~~

1

0 ~

1

0 ~

21

0 ~

31

0 ~

1

0 ~

~~

2

~

3

~

kji

kdttjdttidtttdtF

dtF

ktjtittF

-+=

==

-+++=

-+++=

Answer

30

2.6 Del Operator Or Nabla (Symbol )

• Operator is called vector differential operator, defined as

.~~~

+

+

= kz

jy

ix

31

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~~~k

zj

yi

x

+

+

==

function vector a is *

functionscalar a is *

32

Example 2.8

zxyyzx

xyzzx

zyxyz

zxyyzx

zxyyzx

222

232

223

2232

2232

23z

2y

2x

hence ,Given

(1,3,2).Pat grad determine , If

+=

+=

+=

+=

=+=

Solution

33

.723284

.))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),PAt

.)23(

)2()2(

zyx

Therefore,

~~~

~

222

~

232

~

223

~

222

~

232

~

223

~~~

kji

k

ji

kzxyyzx

jxyzzxizyxyz

kji

++=

++

+++=

=

++

+++=

+

+

=

34

Exercise 2.3

(1,2,3).Ppoint at grad determine

, If 323

=+=

zxyyzx

35

Solution

.110111126(1,2,3),PAt

Gradz

y

x

then,Given

~~~

323

kji

zxyyzx

++====

=

=

=

+=

36

2.6.1.1 Grad Properties

If A and B are two scalars, then

)()()()2

)()1

ABBAAB

BABA

+=+=+

37

2.6.2 Directional Derivative

. ofdirection in ther unit vecto a iswhich

, where

.

is ofdirection in the of derivative lDirectiona

~

~

~

~

~

~

rd

rd

rda

gradads

d

a

=

=

38

Example 2.9

.432

vector theofdirection in the )1,2,1(point at the

2 of derivative ldirectiona theCompute

~~~~

222

kjiA

yzxyzx

-+=-

++=

39

Solution

Directional derivative of in the direction of

.)2()4()22(

hence ,2Given

~

2

~

2

~

2

222

kyzxjzxyiyxz

yzxyzx

+++++=

++=

. and where

.

~

~

~~~~

~

A

Aak

zj

yi

xgrad

gradads

d

=

+

+

==

=

~a

40

.29

)4(32

then,432given Also,

.396

.))1)(2(2)1(())1(

)2)(1(4())2(2)1)(1(2(

(1,2,-1),At

222

~

~~~~

~~~

~

2

~

2

~

2

=

-++=

-+=

-+=

-++-+

++-=

A

kjiA

kji

kj

i

41

.470462.929

51

)3(29

4)9(

29

3)6(

29

2

)396.(29

4

29

3

29

2

. Then,

.29

4

29

3

29

2Therefore,

~~~~~~

~

~~~

~

~

~

=

-

-+

+

=

-+

-+=

=

-+==

kjikji

ads

dφ

kjiA

Aa

42

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.

.90

0cos

0cosgrad

0grad.

~

~

=

=

=

==

rd

rdd

43

• This shows that when (x, y, z) = constant,

• Vector grad = is called normal vector to the surface (x, y, z) = constant

.~rdgrad

y

ds

grad

z

x

44

Unit normal vector is denoted by

Example 2.10

Calculate the unit normal vector at (-

1,1,1) for 2yz + xz + xy = 0.

.~

=n

45

Solution

Given 2yz + xz + xy = 0. Thus

.6114 and

2

)12()12()11((-1,1,1),At

.)2()2()(

~~~

~~~

~~~

=++=

++=

-+-++=

+++++=

kji

kji

kxyjxziyz

)2(6

1

6

2

is vector normalunit The

~~~

~~~kji

kjin~

++=++

=

=

46

2.6.4 Divergence of a Vector

..

).(

.

as defined

is of divergence the, If

~~

~~~~~~

~~

~~~~~

z

a

y

a

x

aAAdiv

kajaiakz

jy

ix

AAdiv

AkajaiaA

zyx

zyx

zyx

+

+

==

++

+

+

=

=

++=

47

Example 2.11

.13

)3)(2(2)3)(1()2)(1(2

(1,2,3),point At

.22

.

(1,2,3).point at determine

, If

~

~~

~

~

2

~~

2

~

=

+-=

+-=

+

+

==

+-=

Adiv

yzxzxy

z

a

y

a

x

aAAdiv

Adiv

kyzjxyziyxA

zyx

Answer

48

Exercise 2.4

.114

(3,2,1),point At

.

(3,2,1).point at determine

, If

~

~~

~

~

3

~

2

~

23

~

=

=

=

+

+

==

-+=

Adiv

z

a

y

a

x

aAAdiv

Adiv

kyzjzxyiyxA

zyxAnswer

49

Remarks

.called is vector ,0 If

function.scalar a is but function, vector a is

~~

~~

vectorsolenoidAAdiv

AdivA

=

50

2.6.5 Curl of a Vector

.

)(

by defined is of curl the,If

~~~

~~

~~~~~~

~~

~~~~~

zyx

zyx

zyx

aaa

zyx

kji

AAcurl

kajaiakz

jy

ix

AAcurl

AkajaiaA

==

++

+

+

=

=

++=

51

Example 2.12

.)2,3,1(at determine

,)()( If

~

~

2

~

22

~

224

~

-

-++-=

Acurl

kyzxjyxizxyA

52

Solution

.)42()22(

)()(

)()(

)()(

~

3

~

2

~

2

~

22422

~

2242

~

222

222224

~~~

~~

kyxjzxxyzizx

kzxyy

yxx

jzxyz

yzxx

iyxz

yzxy

yzxyxzxy

zyx

kji

AAcurl

-++---=

-

-+

+

-

--

-

+

--

=

-+-

==

53

.10682

))3(4)1(2(

))2()1(2)2)(3)(1(2()2()1(

(1,3,-2),At

~~~

~

3

~

2

~

2

~

kji

k

jiAcurl

--=

-+

-+-----=

Exercise 2.5

.)3,2,1(point at determine

,)()( If

~

~

22

~

22

~

223

~

Acurl

kyzxjzxizyxyA -++-=

54

Answer

.261215 (1,2,3),At

.)232(

)22()2(

~~~~

~

22

~

22

~

22

~

kjiAcurl

kyzxyx

jzyxyzizzxAcurl

++-=

+-+

+----=

Remark

function. vector a also is

andfunction vector a is

~

~

Acurl

A

55

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.

56

2.7.1 Polar Coordinate for Plane (r, θ)

x

ds

y

d

ddrrdS

ry

rx

===

sin

cos

57

2.7.2 Polar Coordinate for Cylinder (, , z)

dzdddV

dzddS

zz

y

x

==

===

sin

cos

x

y

z

dv

z

ds

58

2.7.3 Polar Coordinate for Sphere (r,

dddrrdV

ddrdS

rz

ry

rx

sin

sin

cos

sinsin

cossin

2

2

=

=

===

y

x

r

z

59

Example 2.13 (Volume Integral)

.9 and

4,0by bounded space a is and

22 where Calculate

22

~~~~~

=+

==

++=

yx

zzV

kyjziFdVFV

x

z

y

4 -

3 3

60

Solution

Since it is about a cylinder, it is easier if we use cylindrical polar coordinates, where

.40,20,30 where

,,sin,cos

====

z

dzdddVzzyx

61

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

O

B

~~rdr+

~r

62

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

.where~~~~

~

kdzjdyidxrd

rdVc

++=

63

Example 2.14

(3,2,1).B to(0,0,0)A from

along findthen

,,2,3

by given is curve a and z If

~

32

2

==

===

=

CrdV

uzuyux

CxyV

c

64

Solution

.1

,1,22,33 (3,2,1),BAt

.0

,0,02,03 (0,0,0),AAt

.343

And,

.12)()2)(3(

zGiven

32

32

~

2

~~

~~~~

8322

2

=====

=====

++=

++===

=

u

uuu

u

uuu

kduujduuidu

kdzjdyidxrd

uuuu

xyV

65

.11

36

5

244

11

36

5

244

364836

)343)(12(

~~~

~

1

0

11

~

1

0

10

~

1

09

1

0 ~

101

0~

91

0~

8

1

0 ~

2

~~

8

~

kji

kujuiu

kduujduuiduu

kduujuduiduurdVu

u

B

A

++=

+

+=

++=

++=

=

=

66

Exercise 2.6

.11

768144

5

384

(4,3,2).B to(0,0,0)A from

curve thealong calculate

,2,3,4

by given is curve theand If

~~~~

~

23

22

kjirdV

C rdV

uzuyux

CyzxV

B

A

c

++=

==

===

=

Answer

67

2.8.2 Vector Field, Integral

Let a vector field

and

The scalar product is written as

.

)).((. ~~~~~~~~

dzFdyFdxF

kdzjdyidxkFjFiFrdF

zyx

zyx

++=

++++=

~F

~~~~kFjFiFF zyx ++=

.~~~~kdzjdyidxrd ++=

~~. rdF

68

. .

bygiven is Bpoint another A topoint a from

curve thealong of integral line then the

, curve thealong is field vector a If

~~

~

~

++=c zc yc xc

dzFdyFdxFrdF

CF

CF

69

Example 2.15

.y2y

if,2,4curve thealong

(4,2,1)B to(0,0,0)A from .Calculate

~~~

2

~

32

~~

kzjxzixF

tztytx

rdFc

-+=

===

==

70

Solution

.344

And

.4432

)()2(2)()4()2()4(

2yGiven

~

2

~~

~~~~

~

5

~

4

~

4

~

32

~

3

~

22

~~~

2

~

kdttjdttidt

kdzjdyidxrd

ktjtit

kttjttitt

kyzjxzixF

++=

++=

-+=

-+=

-+=

71

.)1216128(

1216128

)3)(4()4)(4()4)(32(

)344)(4432(.

Then

754

754

2544

~

2

~~~

5

~

4

~

4

~~

dtttt

dttdttdtt

dttttdttdtt

kdttjdttidtktjtitrdF

-+=

-+=

-++=

++-+=

.1

,1,22,44 (4,2,1),Bat and,

.0

,0,02,04 (0,0,0),AAt

32

32

=====

=====

t

ttt

t

ttt

72

.30

2326

2

3

3

8

5

128

2

3

3

8

5

128

)1216128(.

1

0

865

1

0

754

~~

=

-+=

-+=

-+= =

=

ttt

dttttrdFt

t

B

A

73

Exercise 2.7

.168

617.

.3,2, curve

on the (1,2,3)B to(0,0,0)A from

. calculate

,3 If

~~

32

~~

~

2

~~

2

~

=

===

==

+-=

rdF

tztytx

rdF

kzxjyzixyF

B

A

c

Answer

74

* Double Integral *

.unit 4),(

integrals.order both in ),( Find

.2 and,0 linestraight aby

bounded region in 4),(Given

2

2

=

===-=

R

R

dAyxf

dAyxf

yxyx

Ryyxf

Answer

2.16 Example

75

.unit 2

14region theof area The

.3 and5by bounded

region a of area thefind integral, double Using

2

2

=

+=-=

Answer

2.17 Example

xyxy

76

.order in integral as ),,( Stated

.2 and16

by bounded is which solid a Evaluate

22

dxdydzdVzyxf

zyxz

=--=

2.18 Example

77

.24 and,,0

by bounded is which solid a is if

orderin integral as ),,( Describe

2 xyxzz

Sdxdydz

dVzyxf

-===

2.19 Example

78

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

=VV

FdxdydzFdV

79

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

x

y 3 O

2

1

80

= = ==

2

0

3

0

1

02

z y xVxdxdydzFdV

Solution

6][33

][2

1.2

2

12

22

20

2

0

3

0

2

0

2

0

3

0

1

0

2

0

3

0

2

===

=

=

=

=

=

= =

= =

zdz

dzy

dydz

dydzx

z

z

z y

z y

81

2.9.2 Vector Field, Integral ~F

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

zdzdydxFdVF

2

1

2

1

2

1 ~~

82

Evaluate , where V is a region bounded by

x = 0, y = 0, z = 0 and 2x + y + z = 2, and also

given

V dVF~

~~~2 kyizF +=

Example 2.21

83

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

84

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

y2O

2

1

2x + y + z = 2

y = 2 (1 - x)

85

Therefore,

=

-

=

--

==

V x

x

y

yx

zdzdydxFdVF

1

0

)1(2

0

)1(2

0 ~~

-

=

--

==+=

)1(2

0

)1(2

0 ~~

1

0)2(

x

y

yx

zxdzdydxkyiz

~~ 3

1

3

2ki+=

86

Example 2.22

Evaluate where

and V is region bounded by z = 0, z = 4

and x2 + y2 = 9

V dVF~ ~~~~

22 kyjziF ++=

x

z

y

4 -

3 3

87

cos=x sin=y zz =zdρdρddV =

; ; ;

where

,30 ,20 40 z

Using polar coordinate of cylinder,

88

++=V V

dxdydzkyjzidVF )22(~~~~

Therefore,

= = =++=

4

0

2

0

3

0 ~~~)sin22(

zkjzi

dzdd

~~14472 ji +=

89

Exercise 2.8

90

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

=S S

dSnVSVd~~

where

S

Sn

=~

91

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 S SVd~

Solution

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

~~~~~22 jyixk

z

Sj

y

Si

x

SS +=

+

+

=

92

Also,

4422)2()2( 2222 ==+=+= yxyxS

Therefore,

)(2

1

4

22

~~

~~

~jyix

jyix

S

Sn +=

+=

=

Then,

+

=

S SdSjyixxyzdSnV )(

2

1

~~~

+= dSjzxyiyzx )(2

1

~

2

~

2

93

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.

.24 ==

x

z

y

2

2

3

O

94

Polar Coordinate for Cylinder

cos 2cos

sin 2sin

ρ

x

y

z z

dS d dz

= == ===

where2

0 30 z(1st octant) and

95

Using polar coordinate of cylinder,

cossin8)()sin2)(cos2(

sincos8)sin2()cos2(222

222

zzzxy

zzyzx

==

==

From

=+=S

SS

SVddSjzxyiyzxdSnV~~

2

~

2

~)(

2

1

96

= =+=

Sz

dzdjzizSVd 2

0

3

0 ~

2

~

2

~)2)(cossin8sincos8(

2

1

3

2 2 2 22

0 ~ ~ 0

2 22

0 ~ ~

1 18 cos sin sin cos

2 2

9 98 cos sin sin cos

2 2

z i z j d

i j d

= +

= +

2 22

0 ~ ~

3 3 2

~ ~0

~ ~

98 cos sin sin cos

2

cos sin sin cos36

3( sin ) 3(cos )

12( )

i j d

i j

i j

= +

= + - = +

Therefore,

97

2

2 2

~

,

9

0 2S

If V is a scalar field where V xyz evaluate

V d S for surface S that region bounded by x y

between z and z in the first octant.

=

+ =

= =

Exercise 2.9

~ ~: 24( )Answer i j+

98

2.10.2 Vector Field, Integral

If vector field defeated on surface S, surface

integral of S is defined as

=SS

dSnFSdF ..~~~~

~F

~F

~F

~where

Sn

S

=

99

Example 2.24

~ ~ ~~

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

= + +

+ + = = = =

100

Solution

2 2 2Given : 9 is bounded by 0, 0,

0 in the 1st octant. This refer to sphere with center

at (0,0,0) and radius, 3, in the 1st octant.

S x y z x y

z

r

+ + = = ==

=

x

z

y

3

3

3

O

101

~ ~~

~ ~~

2 2 2

2 2 2

So, grad is

2 2 2 ,

and

(2 ) (2 ) (2 )

2

2 9 6.

S

S S SS i j k

x y z

x i y j z k

S x y z

x y z

= + +

= + +

= + +

= + +

= =

102

~ ~~ ~

~ ~ ~ ~~ ~

Therefore,

. .

1( 2 ) ( )

3

1( 2 ) .

3

S S

S

S

F d S F n dS

y i j k x i y j z k dS

xy y z dS

=

= + + + +

= + +

).(3

1

6

222

~~~

~~~

~

kzjyix

kzjyix

S

Sn

++=

++=

=

103

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

= == == =

= =

104

dd

dd

SdFS

]cossinsinsin2

cossinsin3[9

]sin9[]cos3)sinsin3(2

)sinsin3)(cossin3[(3

1.

2

0 0

3

0 0~~

2 2

2 2

++

=

++

=

= =

= =

+=

4

319

105

Exercise 2.9

octant. 1 in the 0 and 0,0,4

by boundedregion theof surface a is and

2 where,on Evaluate

st222

~~~~~~

====++

++=

zyxzyx

S

kyjzixFSSdFS

: 8 16

Answer +

106

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),

where S is the area of c.

+=

-

cSdyQdxPdydx

y

P

x

Q)(

107

Example 2.25

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.

cx y dx x y dy

x y and x y

+ + +

= = + =

y

2

x 2

C3

C2

C1O

x2 + y2 = 22 Solution

108

1 1

2 2

2 2

1 2 3

1

2 2

2 2

0

23

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

+ + +

= + = +

= =

+ = + + +

=

= =

109

2 22ii) 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

=- =

110

.448

sin42sin2cos8

)cossin82cos22sin8(

)cossin8cos4sin8(

)]cos2))(sin2(2cos2((

)sin2)()sin2()cos2(([

)2()()(

2

2

2

2

22

02

0

0

2

22

0

22

-=++-=+++=

+++-=

++-=

++

-+=

+++=+

d

d

d

d

dyyxdxyxQdyPdxcc

111

3 3

3

2 2

0

2

02

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

= =

+ = + + +

=

= =-

+ = + - - = -

112

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 Pdxdy

x y

Q Pand y

x y

x r y r

-

= =

= =

0 r 2, 0 .2

and dxdy dS r dr d = =

113

.3

16

cos3

162

sin3

162

sin3

2

2

1

)sin21(

)21(

2

2

2

2

0

0

2

00

32

0

2

0

-=

+=

-=

-=

-=

-=

-

=

=

= =

d

drr

ddrrr

dydxydydxy

P

x

Q

r

SS

114

Therefore,

( )

16.

3

Green's Theorem has been proved.

C S

Q PPdx Qdy dx dy

x y

LHS RHS

+ = -

= -

=

115

2.12 Divergence Theorem (Gauss’ Theorem)

If S is a closed surface including region V in

vector field

..~~~ =

SVSdFdVFdiv

~F

~

yx zff f

div Fx y z

= + +

116

Example 2.26

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

= + +

= = = = + =

117

Solution

x

z

y

2

2

4

O

S3

S4

S2

S1

S5

118

1

2

3

4

2 25

~

~ ~

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

====

+ =

119

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 zx 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

= ==

120

2

2

2

2

2

2

2 4

0 0 0

2 2 400 0

2

0 0

2 200

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

= = =

= =

= =

=

=

+ = +

= +

=

=

=

=

=

=

121

1

~ ~~ ~

1~ ~

~ ~ ~~

~ ~ ~ ~~

~ ~

2) Now, we evaluate . . .

i) : 0, ,

2 0

. ( 2 ).( ) 0

. 0.

S S

S

F d S F n dS

S z n k dS rdrd

F x i j k

F n x i j k

F n dS

=

= =- =

= + +

= + - =

=

122

2

2

2~ ~

2

~ ~ ~ ~ ~~ ~

~ ~ ~ ~ ~~

2

2

0 0~ ~

ii) : 4, ,

2 (4) 2 16

. ( 2 16 ).( ) 16.

Therefore for , 0 r 2, 02

. 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 n dS rdrd

= =

= = =

= + + = + +

= + + =

=

==

123

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 n dS dzdx= =

= =- =

= + +

= + + -

=-

= -

==-

124

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 n dS

= =- =

= + + = +

= + - =

=

125

2 25

5 5~ ~

~5 ~

~5

~ ~

5

v) : 4,

2 2 4

2 2

4

1( ).

2By using polar coordinate of cylinder :

cos , sin ,

where for :

2, 0 , 0 4, 22

S x y dS d dz

S x i y j and S

x i y jSn

S

x i y j

x y z z

S

z dS d dz

+ = =

= + =

+ = =

= +

= = =

= =

126

.416

)2)(sin)(cos2(.

).sin(cos2

2.kerana ;sin2cos2

)sin()cos(2

12

1

2

1

2

1).2(.

5

2

0

4

0

2

~~

2

2

2

2

~~~

2

~~~~

+==

+=

+=

=+=

+=

+=

+++=

= =

S z

dzddSnF

yx

jyixkzjixnF

127

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

= + + + +

= + - + + +=

=

=

128

2.13 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

rdFSdFcurl~~~~

~F

~ ~~

~ ~

x y z

i j k

curl F Fx y z

f f f

= =

129

Example 2.27

Surface S is the combination of2 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

+ = ==

==

= + +

130

Solution

2 21

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

yx

3

4

O

S3

C2

S2

C1

S1

3

131

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

~ ~Scurl F d S

~~~~kxzjxyizF ++=

132

So,

~~

~

~~

~~~

~

)1(

)()(

)()()()(

kyjz

kzy

xyx

jxzx

zz

ixyz

xzy

xzxyzzyx

kji

Fcurl

+-=

-

+

-

+

-

=

=

133

By integrating each part of the surface,

2 21

1~ ~

2 21

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

+ = = +

= +

= + =

134

)(3

1

6

22

~~

~~

1

1

~jyix

jyix

S

Sn +=

+=

=

and

).1(3

1

3

1

3

1)1(

~~~~~~

zy

jyixkyjznFcurl

-=

+

+-=

Then ,

135

By using polar coordinate of cylinder ( because

is a part of the cylinder), 9: 221 =+ yxS

cos , sin ,

3, 0 0 4.

x y z z

dS d dz

where

dan z

= = ==

=

136

Therefore,

~ ~

1(1 )

31

sin 13sin (1 ) ; 3

curl F n y z

z

z because

= -

= -

= - =

Also, dzddS 3=

137

1 1~ ~~ ~

4

0 0

4

00

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

= =

=

= -

= - -

= - - -

=-

138

(ii) For surface , normal vector unit to the

surface is

By using polar coordinate of plane ,

4:2 =zS.

~~kn =

ddrrdSdanzry === 4,sin

0 r 3 and 0 .where

139

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

curl F d S curl F n dS

r rdrd

r d dr

= =

= =

= - +

= =

=

=

=

=

140

(iii) For surface S3 : y = 0, normal vector unit

to the surface is

dS = dxdz

The integration limits :

.~~jn -=

3 3 0 4x and z-

So,

1

)())1((~~~~~

-=

-+-=

z

jkyjznFcurl

141

3 3

1 2 3

~ ~~ ~

3 4

3 0

~ ~ ~ ~~ ~ ~ ~

Then,

. .

( 1)

24.

. . . .

24 18 24

18.

S S

x z

S S S S

curl F d S curl F n dS

z dzdx

curl F d S curl F d S curl F d S curl F d S

=- =

=

= -

==

= + +

=- + +=

142

~ ~

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

CF d r

C

C

3, that is

3cos , 3sin dan z 0

where 0 .

r

x y

=

= = =

143

~ ~ ~~

~

~

~ ~~

~ ~

(3cos )(3sin )

9sin cos

and

3sin 3cos .

F z i xy j xz k

j

j

dr dx i dy j dz k

d i d j

= + +

=

=

= + +

=- +

144

1

2

~ ~

2

0~ ~

3

0

From here,

. 27sin cos .

. 27sin cos

9cos

18.

C

F d r d

F d r d

=

=

= - =

145

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

= = = - = + +

=

=

146

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

= +

= +=

=