Upload
alban-long
View
220
Download
1
Tags:
Embed Size (px)
Citation preview
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
= +
= +=
=