2102222 ENGINEERING ELECTROMAGNETICS
Vector Analysis & Vector Calculus
1
Outline
• Vector representation and vector algebra
• Orthogonal coordinate systems
• Differentiation and integration of vectors
• Line, surface, and volume integrals
• “Del” operator ()
• Gradient, divergence, and curl
2
Scalar & Vector
• A quantity is called a scalar if it has only
magnitude.
– Examples: mass, temperature, electric potential,
population
• A quantity is called a vector if it has both
magnitude and direction.
– Examples: velocity, force, electric field intensity
3
Vector Representation
 

• A vector A is represented by A  a A A

where A  | A | is the magnitude

a A is a unit vector specifying the direction
Note: A can be real or complex!!
• A vector can be represented graphically by an arrow.
4
Equality of Vectors
 
A  aA A
 
B  aB B
 


A  B where A  B and a A  aB
Note: They may be displaced in space!!
5
Vector Addition
• Head-to-tail rule
  
ABC
   
ABD  C

D

C

A

B
6
Vector Subtraction
  
ABC

C

B

A


Note: C arrowhead points to that of A
7
Vector Dot Product

B
 
A  B  AB cos  AB
 AB
B cos AB

A


B is projected onto A and multiplied by A
Examples:
 
A  A  A  A  A2  A 
 
ax ax 1
 

A  A  A
8
Example
• Find
 

AB  C

B  AB

C

A
Solution:


B
     AB
 

AB  C
   
 A  B  A  B

2 2
 
A  B  2 A B cos  AB
 A 2  B 2  2AB cos 
9
Vector Dot Product (cont.)
• Vector dot product is the sum of products of their
components on the same direction.
Example:
 
 




A  a x A1  a y A2  a z A3 ; B  a x B1  a y B2  a z B3
 
A  B  A1B1  A2 B2  A3 B3
 
 
 
ax  ax  a y  a y  az  az  1
 
   
ax  a y  ax  az  a y  az  0
10
Vector Cross Product

an
 
A B

B


A

A

B
  
A  B  an A  B sin 
Right-hand rule



an is a unit vector normal to the plane containing A and B
 
 
A  B  ( B  A)
11
Example

az

ax

ay
 
 
  
ax  ax  a y  a y  az  az  0
12
Length & Area Calculation
• Length calculation by dot product

  12
C  C  C 
• Area calculation by cross product
  


A  B  a A A   aB B   an A B sin  AB



where an  a A  aB
 
Area  A  B  A B sin  AB
 
A B

B


A
13
Cartesian Coordinate System
Z
  
• Base vectors: a x , a y , a z


Do not change with position
• Point: P  ( x, y , z )
PP (X
( x, Y
, y, Z, )z )
0
Y
X

• Position vector: OP  x, y, z   ax x  a y y  az z
 


• Vector: A  a x Ax  a y Ay  a z Az

• Vector field: A( x, y , z )
• Scalar field: V ( x, y , z )
14
Vector Differential Line
Z


0
  
d  a d
 
  d
 
Y
  ( x, y, z)
X
 



d   a d  ax dx  a y dy  az dz
15
Vector Differential Surface
 
ds  a n ds

an

d
dz
Example:




ds z  a x dx  a y dy   a z dxdy


an  a z and ds  dxdy
dy
dx
Z   
   d
X
ds
Y

Note: The unit normal vector an is perpendicular
to the plane containing ds
16
Differential Volume



dv  a z d z   [( a x d x )  ( a y d y )]


 a z d z   a z d x d y 
 d xd yd z
17
Cylindrical Coordinate System
  
• Base vectors: ar , a , a z
Change with 
Constant
18
Cylindrical Coordinate (cont.)
Z

PP 
(X,rY, 
, Z,)z


0
X


Z
r
Y
• Point: P  r, , z
•
•



Position vector: OP  l ( r,  , z )  ar r  a z z
 


Vector: A  a r A r  a  A  a z A z

• Vector field: Ar ,  , z  • Scalar field: V r,  , z 
19
Vector Differential Line
Z


0
  
d  a d


  d
  
Y
 (rx,,y,, zz)
X
 



d   a  d  a r dr  a  rd  a z dz
20
Vector Differential Surface
 
ds  a n ds
For example:



ds z  (ar dr )  (a rd )
 
 (ar  a )(rdrd )

 a z rdrd
Note: The unit normal vector perpendicular
to the plane containing ds
21
Differential Volume


dv  a z dz   ds z


 a z dz   a z rdrd 
 rdrddz
22
Spherical Coordinate System
  
• Base vectors: aR , a , a
All change with  and 
• Point: P  (R , ,  )
•

Position vector: OP  l ( R, ,  )  aR R
 


• Vector: A  a R A R  a  A  a  A
•

Vector field: A(R, ,  ) •
Scalar field: V(R , ,  )
23
Vector Differential Line
 


dl  aR dR  a Rd  a R sin d
24
Vector Differential Surface
 
ds  a n ds
For example:
R sin d



ds R  (a Rd )  (a R sin d )
 2
 aR R sin dd
dR

R sin(  d )d
R
Rd

Note: The unit normal vector perpendicular
to the plane containing ds
25
Differential Volume
R sin d
dR

R sin(  d )d
R

Rd


dv  aR dR  dsR


 (aR dR )  (aR R 2 sin  dd )
 R 2 sin  dRd d
26
Summary of Vector Relations
27
Coordinate Transformation Relations
28
Integral Containing Vector Functions
• Line integral (of scalar and vector)

 
 Vdl  F  dl
C
C
• Surface integral (of vector)
 
 A  ds or
S
 
 A  ds
S
• Volume integral (of vector)

 Fdv or
V

 Fdv
V
29
Line Integral of Scalar Function
• General form

 Vdl
C
 Vis a scalar function of space
 dl is a differential increment of length
 C is the path of integration
• Integration path from a point P1 to another point P2
P2

 Vdl
P1
• Integration around a closed path C

 Vdl
C
30
Line Integral of Scalar Function
• In Cartesian coordinates:




 Vdl   V ( x, y, z )ax dx  a y dy  az dz 
C
C



 a x  V ( x, y , z )dx  a y  V ( x, y , z )dy  a z  V ( x, y , z )dz
C
C
C
• In cylindrical coordinates:




 Vdl   arV (r, , z )dr  r  aV (r, , z )d  az  V (r, , z )dz
C
C
C
C
31
Exercise
• Write down the general form of a line integral of

a scalar function  Vdl in spherical coordinates.
C
R sin d
dR

R sin(  d )d
R

Rd
 


dl  aR dR  a Rd  a R sin d
32
Example
P

• Evaluate the integral  r 2dr , where r 2  x 2  y 2
O
from the origin O to the point P(1,1) along the path
OP, the path OP1P, and the path OP2P.
33
Solution
• Along the path OP:
P
2
 
 2 2
2
O r dr  ar 0 r dr  ar 3
2
2 2 


a x cos 45  a y sin 45 
3
 2  2
 ax  a y
3
3

34
Solution
• Along the path OP1P:
P
1
1
O
0
0
 

2
2
2
2
(
x

y
)
d
r

a
y
dy

a
(
x
 1)dx
y
x

1
 1

 ax  x3  x 
0
3
0
 4  1
 ax  a y
3
3
 1
 ay y
3
31
35
Exercise
P

• Evaluate the integral  r dr along the path OP2P.
2
O
36
Example
 
• Evaluate the line integral  F  dl
B
 

where F  a x xy  a y 2 x
along the quarter-circle
shown in the figure.
A

dl
37
Solution
x 2  y 2  9 (0  x, y  3)
• In Cartesian coordinates:

dl
 
F  dl  Fx dx  Fy dy  xydx  2 xdy
3
  0
2
2
F

d
l

x
9

x
dx

2
9

y
dy



B
A
3
1
  9  x
3
0

 
 91  
2

3

2
1 y 
  y 9  y  9 sin ( )
3
3 0

0
2 3/ 2
38
Exercise
 
• Write down the integral  F  dl
B
in cylindrical coordinates.
 

F  a x xy  a y 2 x

dl
A
 Fr   cos  sin  0  Fx 
 F    sin  cos  0  F 
  
 y 
0
1  Fz 
 Fz   0
 


F  ar Fr  a  F  a z Fz
 
dl  a  rd
39
Surface Integral of Vector Function
• It is a double integral over two dimensions.
 
 
 A  ds or  A  ds
S
S

• The integral measures the flux of the vector field A flowing
through the surface S. (Note: The result is a scalar.)

 

A
A  an A cos   at A sin 

ds

at A sin 

an A cos 


a n  at
 
d
s
 an ds

aA
   
A  ds  A  an ds  ( A cos  )ds  A( ds cos  )
40
Closed Surfaces vs. Open Surfaces
41
Example
 

• Given a vector field F  ar (k1 / r )  a z k2 z
where k1 and k2 are constants,
 
evaluate the scalar surface integral  F  ds
over the surface of a closed
cylinder about the z-axis
specified by r = 2 and z =
3.
S
42
Solution
• The cylinder has three surfaces: the top face, the
bottom face, and the side wall.
 
 
 
 
 F  ds   F  ands   F  ands   F  ands
S
top
bottom
side
wall
43
Solution


• Top face: z  3, an  a z
 
F  an  k2 z  3k2
ds  rdrd
2 2
 
 F  ands    3k2rdrd  12k2
top
0 0


• Bottom face: z  3, an  a z
 
ds  rdrd
F  an  k2 z  3k2
2 2
 
 F  ands    3k2rdrd  12k2
bottom
0 0
44
Solution


• Side wall: r  2, an  ar
  k1 k1
ds  rd dz  2d dz
F  an  
r
2
3 2
 
 F  ands    k1d dz  12k1
side
wall
3 0
• The total integral:
 
 F  ds  12k2  12k2  12k1  12 (k1  2k2 )
S
45
Volume Integral of Vector Function

 Fdv or
V

 Fdv
V

• By resolving the vector F into its three components
in the appropriate
coordinate system, the volume

integral of F can be evaluated as the sum of three
scalar integrals.
• In Cartesian coordinates:




F
(
x
,
y
,
z
)
dv

(
a
F
(
x
,
y
,
z
)

a
F
(
x
,
y
,
z
)

a
y y
z Fz ( x , y , z )) dxdydz

 x x
V
V



 a x  Fx ( x, y , z )dxdydz  a y  Fy ( x, y , z )dxdydz  a z  Fz ( x, y , z )dxdydz
V
V
V
46
Exercise

• Write down the integral  Fdv
V
in cylindrical coordinates.
 


F  ar Fr  a  F  a z Fz
dv  rdrd dz
47
Del Operation
48
Gradient of Scalar Field
• The gradient of a scalar is the vector that represents
the magnitude and the direction of the maximum
space rate of increase of that scalar.
 dV
grad V  V  an
dn
 is the “del” operator
dV dV dn dV
dV  



cos  
( an  al )  (V )  al
dl
dn dl dn
dn
49
Gradient of Scalar Field (cont.)


dV  (V )  al dl  (V )  dl
 
 




e.g. dl  a x dx  a y dy  a z dz
dl  au1dl1  au 2dl2  au 3dl3
V
V
V
dV 
dl1 
dl2 
dl3
l1
l2
l3
  V  V  V  


  au1dl1  au 2dl2  au 3dl3 
  au1
 au 2
 au 3
l1
l2
l3 

  V  V  V  
  dl
  au1
 au 2
 au 3
l1
l2
l3 

 V  V  V
V  au1
 au 2
 au 3
l1
l2
l3
50
Gradient in General Orthogonal
Curvilinear Coordinates
 V  V  V
V  au1
 au 2
 au 3
l1
l2
l3
     
  au 1
 au 2
 au 3
l1
l2
l3
 1 
 1 
 1 
 au 1
 au 2
 au 3
h1 u1
h2 u2
h3 u3
li
where hi 
ui
51
Gradient in Principal Coordinates
• In Cartesian coordinates:
     
  ax
 ay
 az
x
y
z
• In cylindrical coordinates:
   1   
  ar  a
 az
r
r 
z
• In spherical coordinates:
1

   1  
  aR
 a
 a
R
R 
R sin  
52
Example

• The electric field intensity E is derivable as the
negative gradient of a scalar electric potential V, i.e.

E  V

Determine E at the point (0, 1, 0) if
x
V  E0e sin(
y
4
)
53
Solution

       
y 
x
E  V   a x
 ay
 a z  E0e sin( ) 
y
z 
4 
 x
y  y
y   x

  a x sin( )  a y cos( )  E0e
4
4
4 


     E0 
E (0, 1, 0)   a x  a y 
 aE E
4 2

Exercise

E  ?; aE  ?
54
Divergence of Vector Field

• The divergence of a vector field A at
 any point is
defined as the net outward flux of A per unit volume
as the volume about the point tends to zero.
 
A  ds



div A    A  lim S
v 0
v
55
Mathematical Representation of Divergence
• Cartesian coordinates:
 Ax Ay Az
 A 


x
y
z
• Cylindrical coordinates:
 1 
1 A Az
 A 
( rAr ) 

r r
r 
z
• Spherical coordinates:
 1 
1

1 A
2
 A  2
( R AR ) 
( A sin  ) 
R R
R sin  
R sin  
56
Example
• Find the divergence of the position
vector to an arbitrary point P.

P
P
Solution
 


• In Cartesian coordinates: P  a x x  a y y  a z z
 Px Py Pz
P 


x
y
z
x y z


 3
x y z
57
Solution
 
• In spherical coordinates: P  aR R
 1 
1

1 P
2
P  2
( R PR ) 
( P sin  ) 
R R
R sin  
R sin  
1 
1 
2
 2
( R  R)  2
( R3 )  3
R R
R R
 

• In cylindrical coordinates: P  ar r  a z z
 1 
1 P Pz
P 
( rPr ) 

r r
r  z
1 
z

(r  r)   3
r r
z
58
Example

• The magnetic flux density B outside a very
long current-carrying wire is circumferential
and is inversely proportional to the distance to
the distance to the axis of the wire.

Find   B
Magnetic flux
Wire
59
Solution
• Let the long wire be coincident with the z-axis in a
cylindrical coordinate system.
  k
B  a
r
where k is a constant
 1 
1 B Bz
B 
( rBr ) 

r r
r 
z
1  k

( )0
r  r
60
Divergence Theorem
• The volume integral of the divergence of a vector
field equals the total outward flux of the vector
through the surface that bounds the volume.

 
   Adv   A  ds
V
Closed surface S
S

A
Volume dv
61
Example
  2 

• Given A  a x x  a y xy  a z yz
Verify the divergence theorem over a cube one unit
on each side, i.e. show that
 

 A  ds     Adv
cube
cube
Given that the cube is situated in
the first octant of the Cartesian
coordinate system with one corner
at the origin.
z
y
x
62
Solution
 
• Front face: x  1, ds  a x dydz
  11 2
2 1 1
A

d
s

x
dydz

x
y0z0 1


front
0 0


• Back face: x  0, ds  a x dydz
1 1
 
2
2 1 1
A

d
s


x
dydz


x
y0z0  0


back
• Right face:
0 0
 
y  1, ds  a y dxdz
1
  11
 x2  1 1
A  ds    xydxdz  y   z 0 

2
 2 0
right
0 0
63
Solution


• Left face: y  0, ds  a y dxdz
1 1
 
x
A  ds     xydxdz  y 

 2
left
0 0
• Top face:
 
z  1, ds  a z dxdy
1 1
2
1
 1
 z 0  0
0
1
 
1
1 y 
A  ds    yzdxdy  z x 0   

 2 0 2
top
0 0
2


• Bottom face: z  1, ds  a z dxdy
1 1
1
 
1 y 
A  ds     yzdxdy   z x 0    0

 2 0
bottom
0 0
2
64
Solution
• Add the surface integral over the six faces:
 
1 1
A  ds  1    2

2 2
cube

• The divergence of A is
 Ax Ay Az
 2


 A 


 ( x )  ( xy )  ( yz )  3x  y
x
y
z x
y
z
1 1 1

   Adv     (3x  y )dxdydz
cube
0 0 0
3 1
1 1 1 2 1
 3 2 1
 y 0 z 0 x  x 0 z 0  y    2
2 0
2 0 2 2
1
1
65
Curl of Vector Field

• The curl of a vector field B describes its rotational
property, or circulation.

• Thecirculation of B is defined as the line integral
of B around a closed contour C:

 
Circulation B   B  dl
C
66
Example
 
• Find the circulation of a uniform field B  a x B0
around the contour abcd.
c
 b



Circulation B   a x B0  a x dx   a x B0  a y dy
a
b
d
a
c
d




  a x B0  a x dx   a x B0  a y dy
 B0 x  B0 x  0
The circulation of a uniform field is zero.
67
Example
 
• Find the circulation of an azimuthal field B  a B0
around the contour C.

 
Circulation B   B  dl
C
2


  a B0  a rd
0
 2rB0

The circulation of B depends on the choice of contour.
68
Curl of Vector Field (cont.)


• The curl of a vector field B is the circulation of B per
unit area, with the area s of the contour C being
oriented such that the circulation is maximum.


1   
  B  curl B  lim
nˆ  B  dl 

s 0 s
 C
 max
69
Mathematic Representation of Curl
• Cartesian coordinates:
   Az Ay    Ax Az    Ay Ax 
  a y 

  A  a x 



  a z 
z 
x 
y 
 z
 y
 x
• Cylindrical coordinates:
   1 Az A    Ar Az  
  a 
  A  ar 


  az
z 
r 
 z
 r 
1 
Ar
 (rA ) 
r  r




• Spherical coordinates:
 
  A  aR
1
R sin 

  1
A   1  1 AR 
AR 
(
A
sin

)


a

(
RA
)

a
(
RA
)


 


  
 
R  sin   R
R  R
 


70
Stokes’s Theorem
• Stokes’s theorem converts the surface integral of the
curl of a vector over an open surface S into a line
integral of the vector along the contour C bounding
the surface S.
 
 
 (  B)  ds   B  dl
S
C
71