Coor

Coordinate Systems

COORDINATE SYSTEMS

• RECTANGULAR or Cartesian

• CYLINDRICAL

• SPHERICAL

Choice is based on symmetry of problem

Examples:

Sheets - RECTANGULAR

Wires/Cables - CYLINDRICAL

Spheres - SPHERICAL

To understand the Electromagnetics, we must know basic vector algebra and

coordinate systems. So let us start the coordinate systems.

Cylindrical Symmetry Spherical Symmetry

Visualization (Animation)

Orthogonal Coordinate Systems:

3. Spherical Coordinates

2. Cylindrical Coordinates

1. Cartesian Coordinates

P (x, y, z)

P (r, θ, Φ)

P (r, Φ, z)

x

y

zP(x,y,z)

Φ

z

rx

y

z

P(r, Φ, z)

θ

Φ

r

z

yx

P(r, θ, Φ)

Rectangular Coordinates

Or

X=r cos Φ,

Y=r sin Φ,

Z=z

X=r sin θ cos Φ,

Y=r sin θ sin Φ,

Z=z cos θ

Cartesian Coordinates

P(x, y, z)

Spherical Coordinates

P(r, θ, Φ)

Cylindrical Coordinates

P(r, Φ, z)

x

y

zP(x,y,z)

Φ

z

rx

y

z

P(r, Φ, z)

θ

Φ

r

z

yx

P(r, θ, Φ)

Cartesian coordinate system

• 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.

P(x,y,z)

X

Y

Z

3

000

adzdydxdvVaa

v

a

dx

dy

dz

Cartesian Coordinates

Differential quantities:

Length:

Area:

Volume:

dzzdyydxxld ˆˆˆ

dxdyzsd

dxdzysd

dydzxsd

z

y

x

ˆ

ˆ

ˆ

dxdydzdv

AREA INTEGRALS

• integration over 2 “delta” distances

dx

dy

Example:

x

y

2

6

3 7

AREA = 7

3

6

2

dxdy = 16

Note that: z = constant

Cylindrical coordinate system

(r,φ,z)

X

Y

Z

r

φ

Z

Spherical polar coordinate system

• 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

Cylinder of radius ‘R’ and

height ‘H’ is π R2H .φ is azimuth angle

Cylindrical coordinate system

(r,φ,z)

X

Y

Z

rφ

r dφ

dz

dr

r dφ

dr

dφ

Volume of a Cylinder of radius „R‟

and Height „H‟

HR

dzdrdr

dzddrrdvV

R H

v

2

0

2

0 0

Try yourself:

1) Surface Area of Cylinder = 2πRH .

2) Base Area of Cylinder (Disc)=πR2.

Differential quantities:

Length element:

Area element:

Volume element:

dzardadrald zrˆˆˆ

rdrdasd

drdzasd

dzrdasd

zz

rr

ˆ

ˆ

ˆ

dzddrrdv

Limits of integration of r, θ, φ are 0<r<∞ , 0<z <∞ , o<φ <2π

Cylindrical Coordinates: Visualization of Volume element

Spherically Symmetric problem

(р,θ,φ)

X

Y

Z

r

φ

θ

d

d

Spherical Coordinates: Volume element in space

Spherical polar coordinate system (р,θ,φ)

• d р is infinitesimal displacement along р, р dθ is along θ and r sinθ dφ is along φ direction.

• Volume element is given by

dv = dр .рdθ. р sinθ dφ

• Limits of integration of р, θ, φare

0< р <∞ , 0<θ <π , o<φ <2π

Ex: Show that Volume of a sphere of radius R is 4/3 π R3 .

P(r, θ, φ)

X

Y

Z

r

φ

θ

drP

r dθ

r sinθ dφ

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)

r cos θ

r sinθ

Volume of a sphere of radius „R‟

33

0 0

2

0

2

2

3

42.2.

3

sin

sin

RR

dddr

dddrdvV

R

v

Try Yourself:

1)Surface area of the sphere= 4πR2 .

Points to remember

System Coordinates dl1 dl2 dl3

Cartesian x,y,z dx dy dz

Cylindrical r, φ,z dr rdφ dz

Spherical r,θ, φ dr rdθ r sinθdφ

• Volume element : dv = dl1 dl2 dl3• If Volume charge density ‘ρ’ depends only on ‘r’:

Ex: For Circular plate: NOTE

Area element da=r dr dφ in both the

coordinate systems (because θ=900)

drrdvQv l

24

Quiz: Determine

a) Areas S1, S2 and S3.

b) Volume covered by these surfaces.

Radius is r,

Height is h,

X

Y

Z

r

dφ

S1S2

S3

21

hr

dzrddrVb

rrddrSiii

rhdzdrSii

rhdzrdSia

Solution

h r

r

r h

h

)(2

..)

)(2

.3)

2)

)(1))

:

12

2

0 0

12

2

0

0 0

12

0

2

1

2

1

2

1

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(Result of cross product is always

perpendicular(normal) to the plane

of A and BA

B

n

Gradient, Divergence and Curl

• 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 Vector

xA

A

.

The Del Operator

Fundamental theorem for

divergence and curl

• Gauss divergence

theorem:

• Stokes curl theorem

v s

daVdvV .).(

s l

dlVdaVx .).(

Conversion of volume integral to surface integral and vice verse.

Conversion of surface integral to line integral and vice verse.

Gradient:

gradT: points the direction of maximum increase of the

function T.

Divergence:

Curl:

Operator in Cartesian Coordinate System

kz

Tj

y

Ti

x

TT ˆˆˆ

y zxV VV

Vx y z

ky

V

x

Vj

x

V

z

Vi

z

V

y

VV xyzxyz ˆˆˆ

kVjViVV zyxˆˆˆ

where

as

Operator in Cylindrical Coordinate System

Volume Element:

Gradient:

Divergence:

Curl:

dzrdrddv

zz

TˆT

rr

r

TT

1

1 1 z

r

V VV rV

r r r z

zV

rVrr

ˆr

V

z

Vr

z

VV

rV rzrz

11

zVVrVV zr ˆˆˆ

Operator In Spherical Coordinate System

Gradient :

Divergence:

Curl:

ˆT

sinrˆT

rr

r

TT

11

2

2

sin1 1 1

sin sin

rr V VVV

r r r r

ˆVrV

rr

ˆrVr

V

sinrr

VVsin

sinrV

r

r

1

111

ˆˆˆ VVrVV r

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.

Closed surface S, volume V,

outward pointing normal

Basic Vector Calculus

2

( )

0, 0

( ) ( )

F G G F F G

F

F F F

Divergence or Gauss’ Theorem

SV

SdFdVF

dSnSd

Oriented boundary L

n

Stokes’ Theorem

S L

ldFSdF

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