### z - KL University

```Coordinate Systems
To understand the Electromagnetics, we must know basic vector algebra and
coordinate systems. So let us start the coordinate systems.
COORDINATE SYSTEMS
• RECTANGULAR or Cartesian
• CYLINDRICAL
• SPHERICAL
Choice is based on
symmetry of problem
Examples:
Sheets - RECTANGULAR
Wires/Cables - CYLINDRICAL
Spheres - SPHERICAL
Cylindrical Symmetry
Spherical Symmetry
Visualization (Animation)
Orthogonal Coordinate Systems:
1. Cartesian Coordinates
z
P(x,y,z)
Or
Rectangular Coordinates
P (x, y, z)
y
x
z
z
P(r, Φ, z)
2. Cylindrical Coordinates
P (r, Φ, z)
X=r cos Φ,
Y=r sin Φ,
Z=z
3. Spherical Coordinates
x
y
z
θ r
P (r, θ, Φ)
X=r sin θ cos Φ,
Y=r sin θ sin Φ,
Z=z cos θ
r
Φ
x
Φ
P(r, θ, Φ)
y
z
θ r
x
Φ
z
P(r, θ, Φ)
Cartesian Coordinates
P(x, y, z)
P(x,y,z)
y
x
y
Spherical Coordinates
P(r, θ, Φ)
z
Cylindrical Coordinates
P(r, Φ, z)
z
P(r, Φ, z)
x
Φ
r
y
Cartesian coordinate system
Z
dz
dy
dx
P(x,y,z)
Y
X
• dx, dy, dz are infinitesimal
displacements along X,Y,Z.
• Volume element is given by
dv = dx dy dz
• Area element is
da = dx dy or dy dz or dxdz
• Line element is
dx or dy or dz
Ex: Show that volume of a cube
of edge a is a3.
a
V 
a
a
 dv 
 dx  dy  dz  a
v
0
0
0
3
Cartesian Coordinates
Differential quantities:
Length:

d l  xˆ dx  yˆ dy  zˆ dz
Area:

d s x  xˆ dydz

d s y  yˆ dxdz

d s z  zˆ dxdy
Volume:
dv  dxdydz
AREA INTEGRALS
• integration over 2 “delta” distances
dy
dx
Example:
y
7
AREA =
6
  dy
3
2
6
dx
2
Note that: z = constant
3
7
x
= 16
Cylindrical coordinate system
(r,φ,z)
Z
Z
r
φ
X
Y
Cylindrical
Spherical
polarcoordinate
coordinatesystem
system
(r,φ,z)
Z
dz
r dφ
dr
φ
Y
dφ
r
dr
X
φ is azimuth angle
r dφ
• dr is infinitesimal displacement
along r, r dφ is along φ and
dz is along z direction.
• Volume element is given by
dv = dr r dφ dz
• Limits of integration of r, θ, φ
are
0<r<∞ , 0<z <∞ , o<φ <2π
Ex: Show that Volume of a
height ‘H’ is π R2H .
Volume of a Cylinder of radius ‘R’
and Height ‘H’
V 
 dv   r dr d  dz
v
2
R

H
 rdr  d   dz
0
0
0
  R H
2
Try yourself:
1) Surface Area of Cylinder = 2πRH .
2) Base Area of Cylinder (Disc)=πR2.
Cylindrical Coordinates: Visualization of Volume element
Differential quantities:
Length element:

d l  aˆ r dr  aˆ  rd   aˆ z dz
Area element:

d s r  aˆ r rd  dz

d s   aˆ  drdz

d s z  aˆ z rdrd 
Volume element:
dv  r dr d  dz
Limits of integration of r, θ, φ are 0<r<∞ , 0<z <∞ , o<φ <2π
Spherically Symmetric problem
(r,θ,φ)
Z
θ
r
Y
φ
X
Spherical polar coordinate system (r,θ,φ)
• dr is infinitesimal displacement
along r, r dθ is along θ and
r sinθ dφ is along φ direction.
P(r, θ, φ)
Z
• Volume element is given by
dr
dv = dr r dθ r sinθ dφ
P
r cos θ
r dθ
• Limits of integration of r, θ, φ
θ r
are
Y 0<r<∞ , 0<θ <π , o<φ <2π
Ex: Show that Volume of a
r sinθ dφ
φ
r sinθ
sphere of radius R is 4/3 π R3 .
X
θ is zenith angle( starts from +Z reaches up to –Z) ,
φ is azimuth angle (starts from +X direction and lies in x-y plane only)
Volume of a sphere of radius ‘R’
V 
 dv   r
2
dr sin  d  d 
v

R

r
0

2
dr
2
 sin  d   d 
0
R
0
3
3
. 2 . 2

4
 R
3
3
Try Yourself:
1)Surface area of the sphere= 4πR2 .
Spherical Coordinates: Volume element in space
Points to remember
System
Coordinates
dl1
dl2
dl3
Cartesian
Cylindrical
Spherical
x,y,z
r, φ,z
r,θ, φ
dx
dr
dr
dy
rdφ
rdθ
dz
dz
r sinθdφ
• Volume element : dv = dl1 dl2 dl3
• If Volume charge density ‘ρ’ depends only on ‘r’:
Q 
  dv    4 r dr
2
v
l
Ex: For Circular plate: NOTE
Area element da=r dr dφ in both the
coordinate systems (because θ=900)
Quiz: Determine
a) Areas S1, S2 and S3.
b) Volume covered by these surfaces.
S3
Z
Solution :
2
h
1
r
ii ) S 2 
0
h
 dr  dz
0
 rh
S2
S1
0
2 r
iii ) S 3 
  dr .rd  
1 0
r
2
2
h 2 r
b)V 
Height is h,
1     2
r
a ) i ) S 1   rd   dz  rh ( 2   1 )
   dr .rd  .dz
0 1 0

( 2   1 )
r
Y
dφ
2
2
( 2   1 ) h
X
Vector Analysis
• What about A.B=?, AxB=? and AB=?
• Scalar and Vector product:
A.B=ABcosθ
Scalar
or
(Axi+Ayj+Azk).(Bxi+Byj+Bzk)=AxBx+AyBy+AzBz
AxB=ABSinθ n
Vector
n
(Result of cross product is always
perpendicular(normal) to the plane
of A and B
B
A
The Del Operator
• Gradient of a scalar function is a
vector quantity.
• Divergence of a vector is a scalar
quantity.
• Curl of a vector is a vector
quantity.
f
. A
 xA
Vector
Fundamental theorem for
divergence and curl
• Gauss divergence
theorem:
 (  .V ) dv   V .da
v
s
Conversion of volume integral to surface integral and vice verse.
• Stokes curl theorem
 (  x V ). da   V .dl
s
l
Conversion of surface integral to line integral and vice verse.
Operator in Cartesian Coordinate System
T ˆ T ˆ T ˆ
i 
j
k
x
y
z
as
gradT: points the direction of maximum increase of the
function T.
T 
Divergence:
Curl:
 V 
V x
x

V y
y

Vz
z
where
V  V x iˆ  V y ˆj  V z kˆ
  V V y   V
V z  ˆ  V y V x  ˆ
x
z
ˆ
i  
k
  V  



 j  

z   z
x   x
 y 
 y
Operator in Cylindrical Coordinate System
Volume Element:
dv  rdrd  dz
T 
T
r
1 
ˆ 
r
 rV r  
1 T ˆ
T
ˆz
 
r 
z
1  V
 V 
Curl:
  1 V V   V V
z
 rˆ   r  z
  V  

z   z
r
 r 
r r
r 

Vz
Divergence:
z
V  V r rˆ  V ˆ  V z zˆ
Vr
ˆ 1  
    rV   

 r  r

 ˆz


Operator In Spherical Coordinate System
T 
T
r
rˆ 
1 T ˆ
1
T ˆ
 

r 
r sin   
1   r Vr 
2
Divergence:
 V 
r
Curl:

 V 
2
r

1
  sin  V 
r sin 

 
sin  V     V 

r sin    

1
V r
1

 
rV   
r   r


ˆ




1
 V
r sin   

1  1 V r

rV 
 rˆ  

r  sin   
r

V  V r rˆ V ˆ  V ˆ
ˆ

Basic Vector Calculus
  (F  G )  G    F  F    G
     0,
   F  0
  (  F )   (  F )   F
2
Divergence or Gauss’ Theorem
The divergence theorem states that the total outward flux of a
vector field F through the closed surface S is the same as the
volume integral of the divergence of F.



   F dV   F  d S
V
S
Closed surface S, volume V,
outward pointing normal
Stokes’ Theorem
Stokes’s theorem states that the circulation of a vector field F around a
closed path L is equal to the surface integral of the curl of F over the
open surface S bounded by L




   F   d S   F  d l
S
 
d S  n dS
Oriented boundary L
L

n
```