Vector Calculus: Divergence, Curl, and Laplacian - Detailed Notes with Examples, Lecture notes of Electronics

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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.

Divergence of a vector and Divergence Theorem

0

A S

div A= A= lim S

v

d

  v



Divergence of a vector

Divergence of a vector

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

Let the point be enclosed by the differential volume.

A S A S

S front back left right top bottom

P x y z

d d

            

Divergence of a vector

 

0 x 0 0

0 0 0

0 x

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

A S= A , ,

higher order terms

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

A S= A

x x front P

x back

dx x x d dy dz y y z z

dx A d dy dz x y z x

dx x x d dy dz

d dy dz x

 

  ^0 ,^0 ,^0  h.o. terms 2

Hence A S A S= h.o. terms

x

P

x

front back P

dx A y z x

A

d d dx dy dz x

 ^  

   

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

top bottom P

S v

A

d d dx dy dz y

A

and d d dx dy dz z

v dx dy dz

d

  v

   

   



The higher order terms will vanish as 0

x y z

at P

A A A

x y z

v

 ^ ^  

 

  ^ 

2 2

Cartesian

A=

Cylindrical

1 1 A=

Spherical

1 1 1 A= sin sin sin

x y z

z

r

A A A

x y z

A (^) A A z

A r A A r r r r

 

 

    

    

         

         

         

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.

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

V A V A+A V ( V scalar)

  ^ ^ ^ 

 

    ^ 

2

2 2

(a) P=

(b) Q=

= sin cos

2sin cos

x y z

z

P P P

x y z

x yz xy xyz x x y z

Q (^) Q Q z

z z z



Determine the divergence of the vector fields:

13

Example 3.

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

x yz xz (^) z z z z

r r

 

 

Example 3.6 - continued

  ^ ^  

  (^)    

2 r 2 2 2 2 (c) T (1 / ) cos a sin cos a cos a

T= sin sin sin

1 1 1 = cos sin cos cos sin sin

1 0 2 sin cos cos 0 sin

T=2 cos cos

r

r r

r T T T r r r r

r r r r r

r r

 

 

   

    

       

   

 

 

 

  (^)  

2 1 2 2

0

2 1 0

0 0

2 1

0

2 1 2 1 2 2 2

(^0 0 )

10 2 10 2

For , z=0, S ( a )

G S 10

10 2 10 2

For , =1, S a

G S 10 10 2 10 1 2

,

t

b z

b b S z z S z

t b

e e

d d d

d e d d

d dz d

e d e dz d e

Thus



 







 

 

  

 

   

  

    

    

 

       

     

   

   

  (^) S  (^016)

Example 3.7 - continued

 

 

   

2 2

2

, Since S is a closed surface, we can apply

the divergence theorem:

= G S= G

1 1 G

1 G 10 10

20

S v

z

z z

z

Alternatively

d dv

G G G z

e e z

e

 



   

          

       



 

2 20 0

G has no outward flux.

z e

  



Example 3.7 - continued

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 L a n

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 A

x x y y

x y

A

z z

z

Curl of a vector