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.
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
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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
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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
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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
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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.
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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.
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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:
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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
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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
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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
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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.
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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
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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
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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
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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
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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
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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
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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
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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.
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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.
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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.
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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
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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
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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
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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
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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
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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
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