1

EELE 3331 – Electromagnetic I

Chapter 3

Vector Calculus

Islamic University of Gaza

Electrical Engineering Department

Dr. Talal Skaik

2012

The divergence of A at a given point P is the outward flux per unit

volume as the volume shrinks about P.

∆v→is the volume enclosed by the closed surface S in which P is

located.

The divergence at a given point is a measure of how much the field

diverges from or converges to that point.

2

Divergence of a vector and Divergence Theorem

0

A S

div A= A= lim S

v

d

v

3

Divergence of a vector

Illustration of the divergence of a vector field at P: (a) positive divergence (source point), example: positive charges. (b) negative divergence (sink point), example: negative charges. (c) zero divergence. There is neither sink nor source.

4

Divergence of a vector

5

Divergence of a vector

0 0 0To find the divergence of a vector A at point ( , , ) :

Let the point be enclosed by the differential volume.

A S A SS front back left right top bottom

P x y z

d d

6

Divergence of a vector

0 0 0 0 0

0

Three dimensional Taylor series expansion of about is:

A AA ( , , ) A ( , , )

A + higher order terms

x

x xx x

P P

x

P

A P

x y z x y z x x y yx y

z zz

2

Note: One dimensional Taylor series:

( ) ( ) '

''

2!

Evaluate near point

f x f a f a x a

f ax a

x a

7

Divergence of a vector

0 x 0 0

0 0 0

0 x

For the front side, , S= a ( , ) 2

A S= A , ,2

higher order terms

For the back side, , S= ( a )2

A S= A

xx

Pfront

x

back

dxx x d dy dz y y z z

dx Ad dy dz x y z

x

dxx x d dy dz

d dy dz x

0 0 0, , h.o. terms2

Hence A S A S= h.o. terms

x

P

x

Pfront back

dx Ay z

x

Ad d dx dy dz

x

8

Divergence of a vector

0

By taking similar steps, we obtain:

A S+ A S= higher order terms

A S+ A S= higher order terms

Noting that ,

A S

lim

y

left right P

z

Ptop bottom

S

v

Ad d dx dy dz

y

Aand d d dx dy dz

z

v dx dy dz

d

v

The higher order terms will vanish as 0

yx z

at P

AA A

x y z

v

9

Divergence of a vector 0 0 0Thus the divergence of A at point ( , , ) is:

A= Cartesian

cos sin 0

sin cos 0

0 0 1

sin co cos , sin

yx z

x

y

z z

P x y z

AA A

x y z

A A

Since A A

A A

andx y

2

2

s

1 1A= Cylindrical

and in spherical coordinates:

1 1 1A= sin Spherical

sin sin

z

r

A AA

z

Ar A A

r r r r

2

2

Cartesian

A=

Cylindrical

1 1 A=

Spherical

1 1 1 A= sin

sin sin

yx z

z

r

AA A

x y z

A AA

z

Ar A A

r r r r

10

Divergence of a vector

Divergence Theorem: (Guass-Ostrogradsky)

The total flux of a vector field A through the closed surface S is the

same as the volume integral of the divergence of A.

11

The Divergence Theorem

A S A S v

d dv

* Volume integrals are easier than surface integrals.

Properties of divergence:

produces a scalar field.

A+B A+ B

A A+A ( scalar)V V V V

Explanation: Let the volume v bounded by the surface S subdivided

into a number of small cells. Each cell has a volume ∆vk and bounded

by a surface Sk.

• Since the outward flux to one cell is inward to some neighbouring

cells, there is cancellation on every interior surface.

• As a result, the divergence of the flux density throughout the

volume leads to the same result as determining the net flux crossing the

closing surface.

12

The Divergence Theorem

A S A S v

d dv

2

2 2

(a) P=

0 2

1 1(b) Q=

1 1 = sin cos

2sin cos

x y z

z

P P Px y z

x yz xy xyz xx y z

Q QQ

z

z zz

Determine the divergence of the vector fields:

13

Example 3.6

2 2

x

2

r

P= a + a (b) Q sin a + a + cos a

(c) T (1 / )cos a sin cos a cos a

z zx yz xz z z

r r

14

Example 3.6 - continued

2

r

2

2

2

2

(c) T (1 / )cos a sin cos a cos a

1 1 1T= sin

sin sin

1 1 1= cos sin cos cos

sin sin

10 2 sin cos cos 0

sin

T=2cos cos

r

r r

r T T Tr r r r

rr r r r

rr

2 1

2

0 0

Total flux: G S

flux through top

flux through bottom

flux through sides (curved surface)

For , z=1, S a

G S 10

t b s

S

t

b

s

t z

t

d

d d d

d e d d

15

Example 3.7

-2If G(r)=10 , determine the flux of G out of the

entire surface of the cylinder =1, 0 z 1. Confirm the result

by using the divergence theorem.

z

ze a a

12

2 2

0

2 1

0

0 0

12

0

12 1 22 2 2

0 0 0

10 2 102

For , z=0, S ( a )

G S 10

10 2 102

For , =1, S a

G S 10 10 2 10 12

,

t

b z

b

b

S

zz

S

z

t b

e e

d d d

d e d d

d dz d

ed e dz d e

Thus

0S 16

Example 3.7 - continued

2 2

2

, Since S is a closed surface, we can apply

the divergence theorem:

= G S= G

1 1G

1G 10 10

20

S v

z

z z

z

Alternatively

d dv

G G Gz

e ez

e

220 0

G has no outward flux.

ze

17

Example 3.7 - continued

Fields with zero divergence are called Solenoidal Fields.

(They obey the rule: what goes in, must come out).

Example: magnetic fields.

18

Divergence

The curl of A is a rotational vector whose magnitude is the

maximum circulation of A per unit area as the area tends to

zero, and whose direction is the normal direction of the area

when the area is oriented to make the circulation maximum.

• The area ∆S is bounded by the curve L.

• an is a unit vector normal to the surface ∆S, determined by right

hand rule. The direction of the curl, an, is the axis of rotation. 19

Curl of a vector and Stoke’s Theorem

0

max

A l

curl A= A= lim aLn

S

d

S

0 0 0

0 0 0

To obtain expression for A:

Consider the differential area

in the yz plane.

A l= A l

Taylor series expansion about the centre point ( , , )

( , , ) ( , , )

L ab bc cd da

y y

d d

P x y z

A x y z A x y z

0 0

0

( ) ( )

( ) higher order terms

y y

P P

y

P

A Ax x y y

x y

Az z

z

20

Curl of a vector

0 0 0 0 0

0

y 0

0 0 x

0 0 0

,

( , , ) ( , , ) ( ) ( )

( ) h. o. terms

on side ab, l=dy a , , 2

, , A= a + a a

A l ( , , )2

z zz z

P P

y

P

x y y z z

y

y

ab P

Similarly

A AA x y z A x y z x x y y

x y

Az z

z

dzd z z

x x y y A A A

Adzd dy A x y z

z

21

Curl of a vector

0

0 0 x

0 0 0

0

0 0 x

0 0 0

on side bc, l=dz a , y , 2

, z , A= a + a a

A l ( , , )2

on side cd, l= dy a , z , 2

, y , A= a + a a

A l ( , , )2

z

x y y z z

zz

bc P

y

x y y z z

y

y

c P

dyd y

x x z A A A

dy Ad dz A x y z

y

dzd z

x x y A A A

Adzd dy A x y z

z

d

22

Curl of a vector

0

0 0 x

0 0 0

on side da, l= dz a , y , 2

, z , A= a + a a

A l ( , , ) 2

z

x y y z z

zz

da P

dyd y

x x z A A A

dy Ad dz A x y z

y

23

Curl of a vector

0 0 0

0 0 0

0 0 0

0 0 0

on side ab : A l ( , , )2

on side bc: A l ( , , )2

on side cd: A l ( , , ) 2

on side da : A l ( , , )2

y

y

ab P

zz

bc P

y

y

cd P

zz

d P

Adzd dy A x y z

z

dy Ad dz A x y z

y

Adzd dy A x y z

z

dy Ad dz A x y z

y

0

Noting that S= , Add 4 equations:

A l

lim or curl A

a

y yz zL

xS

dy dz

dA AA A

S y z y z

24

Curl of a vector

The y and z components of curl A are found in similar way:

curl A

curl A

In Cartesian: A=

A= a a

x z

y

y x

z

x y z

x y z

y yxz zx y

A A

z x

A A

x y

a a a

x y z

A A A

A AAA Aor

y z z x x

axz

A

y

25

Curl of a vector

In Cylindrical:

a a a

1A=

1A= a a

1

a

z

z z

z

z

A A A

or

A AA A

z z

A A

26

Curl of a vector

2

In Spherical:

a a sin a

1A=

sin

sin

sin1 1 1A= a a

sin sin

1 a

r

r

rr

r

r r

r r

A rA r A

or

A rAA A

r r r

rA A

r r

27

Curl of a vector

The curl of a vecor field is another vector field.

A+B A B

A B A B B A B A A B

A A A

The divergence of the curl of a vector field vanishes,

that is A =0

The curl of

V V V

the gradient of a scalar field vanishes,

that is 0V 28

Properties of Curl

The curl of field A at point P is a measure of the circulation, or

how much the field curls around P.

29

The curl of field A at point P

Illustration of a curl: (a) curl at P points out of the page, (b) curl at P is zero.

Stoke’s Theorem: The circulation of a vector field A around a closed

path L is equal to the surface integral of the curl of A over the open

surface S bounded by L, provided A and are continuous on S.

30

Stoke’s Theorem

A l A SL S

d d

A

To find direction of dS, use right-

hand rule.

→Fingers in the direction of dl,

thumb indicates direction of dS.

• Surface is subdivided into large number of cells.

• If the kth cell has surface area ∆Sk and is bounded by path Lk.

• There is cancellation on every interior path, so the sum of line

integrals around the Lk’s is the same as the line integrals around the

bounding curve L.

31

Stoke’s Theorem

A l A SL S

d d

2 2

2 2

(a) P= a a

a

P 0 0 a a 0 a

P a a

yz x zx y

y xz

x y z

y z

PP P P

y z z x

P P

x y

x y z x z

x y z x z

32

Example 3.8

Determine the curl of the vector fields:

2

x

2

2

r

(a) P= a + a

(b) Q sin a + a + cos a

(c) T (1 / )cos a sin cos a cos a

z

z

x yz xz

z z

r r

33

Example 3.8 - continued

2

2 2

3

(b) Q sin a + a + cos a

1= a a

1 a

1sin a 0 0 a 3 cos a

1sin a 3 cos a

z

z z

z

z

z

z z

Q QQ QQ

z z

Q Q

zz

z z

2

2

sin1(c) A= a

sin

1 1 1 a

sin

1cos sin sin cos a

sin

1 1 cos cos a

sin

1 sin cos

r

r r

r

A A

r

rA rAA A

r r r r

rr

rr r r

rr r

2

3

cosa

cos2 cos 1sin a a 2cos sin a

sinr

r

r r r

34

Example 3.8 - continued

35

Example 3.9

L

If A cos a +sin , evaluate A. l around the path

shown in the figure. Confirm this by using Stoke's theorem.

a d

30

60

30

60

A l A l

Along ab, 2, l

A l sin

2 cos 3 1

b c d a

L a b c d

b

a

d d

d d a

d d

5

2

52

2

60

30

60

30

Along bc, 30, l

A l cos

21 3cos30

2 4

Along cd, 5, l

A l sin

55 cos 3 1

2

c

b

d

c

d d a

d d

d d a

d d

36

Example 3.9 - continued

A cos a +sin a

2

5

22

5

Along da, 60, l

A l cos

21cos60

2 4

A l

A l 4.941

a

d

L

b c d a

a b c d

d d a

d d

Hence d

d

37

Example 3.9 - continued

A cos a +sin a

Using Stoke's Theorem (because L is a closed path)

A l= A S

S a

1A= a

a

1 a

A= 0 0 a 0 0 a 1 / 1 s

L S

z

z

z

z

d d

d d d

AA

z

A A

z

A A

in a z 38

Example 3.9 - continued

A cos a +sin a

60 5

30 2

52

60

30

2

S a

A= 1/ 1 sin a

A S 1/ 1 sin

27 = cos 3 1 4.941

2 4

z

z

S

d d d

d d d

39

Example 3.9 - continued

A cos a +sin a

22

Assume A in Cartesian coordinates,

A= , , , ,y yz z x x

y yz z x x

yz

A AA A A A

x y z y z x z x y

A AA A A A

x y z y x z z x y

AA

x y x z

22 2 2

0yz x x

AA A A

y x y z z x z y

40

Example 3.10

For a vector field, show that A=0, that is, the

divergence of the curl of any vector is zero.

The Laplacian of a scalar field V, written as , is the

divergence of the gradient of V.

2

2

2 2 22

2 2 2

2 22

2 2 2

In Cartesian, Laplacian

a a a a a a

(scalar)

In Cylindrical coordinates,

1 1

x y z x y z

V V V

V V VV

x y z x y z

V V VV

x y z

V V VV

z

41

Laplacian of a scalar 2V

2 2 22

2 2 2

2 22

2 2 2

2 2

2 2

In Cartesian,

In Cylindrical,

1 1

In Spherical,

1 1sin

sin

V V VV

x y z

V V VV

z

V VV r

r r r r

2

2 2 2

1

sin

V

r

42

Laplacian of a scalar

2

2

For charge free region:

0 (Laplace equation)

will be solved in Chapter 6

Laplacian of a vector:

applies to a vector and returns a vector

For vector A, A= A A

In Cartesian:

V

2 2 2 2

A a a ax x y y z zA A A 43

Laplacian

2 2 22

2 2 2

-

(a)

2 cos2 cosh sin 2 sin h

sin 2 cosh

4 sin 2 cosh sin 2 cosh sin 2 cosh

z z

z

z z z

V V VV

x y z

e x y e x yx y

e x yz

e x y e x y e x y

44

Example 3.11

Find the Laplacian of the following scalar fields:

2

2

(a) sin 2 cosh

(b) z cos2

(c) 10 sin cos

zV e x y

U

W r

2 22

2 2 2

2 2 2

2

2

2 2

2 2

2

2 2 2

1 1(b)

1 12 cos2 4 cos2 0

4 cos2 4 cos 0

1 1(c) sin

sin

1

sin

U U UU

z

U z z

U z z

W WW r

r r r r

W

r

2 10cos 1 2cos2W

r

45

Example 3.11 2 2 (b) z cos2 (c) 10 sin cos U W r

46

Classification of Vector Fields

A=0 A 0 A=0 A 0

A=0 A=0 A 0 A 0

Such field has neither source nor sink of flux.

• Flux lines of A entering any closed surface must leave it.

Examples: magnetic fields, conduction current density under steady

state conditions.

• A solenoidal field A can always be expressed in terms of another

vector F, 47

Classification of Vector Fields

A vector field A is said to be solenoidal (or divergenceless) if:

A=0

A S= A 0S v

d dv

A F, A= F 0

A curl-free vector is irrotational.

Example: electrostatic field.

• An irrotational field A can always be expressed in terms of a

scalar field V,

(Negative sign for physical reasons….Chapter 4) 48

Classification of Vector Fields

A vector field A is said to be irrotational (or conservative) if:

A=0

A l= A S 0L S

d d

A , A= 0V V