1. VECTOR CALCULUS
1.10 GRADIENT OF A SCALAR
1.11 DIVERGENCE OF A VECTOR
1.12 DIVERGENCE THEOREM
1.13 CURL OF A VECTOR
1.14 STOKES’S THEOREM
1.15 LAPLACIAN OF A SCALAR

2. 1.10 GRADIENT OF A
SCALAR
Suppose is the temperature at ,
and is the temperature at
as shown.
( )zyxT ,,1 ( )zyxP ,,1
2P( )dzzdyydxxT +++ ,,2

3. The differential distances are the
components of the differential distance
vector :
dzdydx ,,
zyx dzdydxd aaaL ++=
Ld
However, from differential calculus, the
differential temperature:
dz
z
T
dy
y
T
dx
x
T
TTdT
∂
∂
+
∂
∂
+
∂
∂
=−= 12
GRADIENT OF A SCALAR (Cont’d)

4. But,
z
y
x
ddz
ddy
ddx
aL
aL
aL
•=
•=
•=
So, previous equation can be rewritten as:
Laaa
LaLaLa
d
z
T
y
T
x
T
d
z
T
d
y
T
d
x
T
dT
zyx
zyx
•





∂
∂
+
∂
∂
+
∂
∂
=
•
∂
∂
+•
∂
∂
+•
∂
∂
=
GRADIENT OF A SCALAR (Cont’d)

5. The vector inside square brackets defines the
change of temperature corresponding to a
vector change in position .
This vector is called Gradient of Scalar T.
Ld
dT
GRADIENT OF A SCALAR (Cont’d)
For Cartesian coordinate:
zyx
z
V
y
V
x
V
V aaa
∂
∂
+
∂
∂
+
∂
∂
=∇

6. GRADIENT OF A SCALAR (Cont’d)
For Circular cylindrical coordinate:
z
z
VVV
V aaa
∂
∂
+
∂
∂
+
∂
∂
=∇ φρ
φρρ
1
For Spherical coordinate:
φθ
φθθ
aaa
∂
∂
+
∂
∂
+
∂
∂
=∇
V
r
V
rr
V
V r
sin
11

7. EXAMPLE
10
Find gradient of these scalars:
yxeV z
cosh2sin−
=
φρ 2cos2
zU =
φθ cossin10 2
rW =
(a)
(b)
(c)

8. SOLUTION TO EXAMPLE 10
(a) Use gradient for Cartesian coordinate:
z
z
y
z
x
z
zyx
yxe
yxeyxe
z
V
y
V
x
V
V
a
aa
aaa
cosh2sin
sinh2sincosh2cos2
−
−−
−
+=
∂
∂
+
∂
∂
+
∂
∂
=∇

9. SOLUTION TO EXAMPLE 10
(Cont’d)
(b) Use gradient for Circular cylindrical
coordinate:
z
z
zz
z
UUU
U
a
aa
aaa
φρ
φρφρ
φρρ
φρ
φρ
2cos
2sin22cos2
1
2
+
−=
∂
∂
+
∂
∂
+
∂
∂
=∇

10. SOLUTION TO EXAMPLE 10
(Cont’d)
(c) Use gradient for Spherical coordinate:
φ
θ
φθ
φθ
φθφθ
φθθ
a
aa
aaa
sinsin10
cos2sin10cossin10
sin
11
2
−
+=
∂
∂
+
∂
∂
+
∂
∂
=∇
r
r
W
r
W
rr
W
W

11. 1.11 DIVERGENCE OF A VECTOR
Illustration of the divergence of a vector
field at point P:
Positive
Divergence
Negative
Divergence
Zero
Divergence

12. DIVERGENCE OF A VECTOR
(Cont’d)
The divergence of A at a given point P
is the outward flux per unit volume:
v
dS
div s
v ∆
•
=•∇=
∫
→∆
A
AA lim
0

13. DIVERGENCE OF A VECTOR
(Cont’d)
What is ??∫ •
s
dSA Vector field A at
closed surface S

14. Where,
dSdS
bottomtoprightleftbackfronts
•








+++++=• ∫∫∫∫∫∫∫ AA
And, v is volume enclosed by surface S
DIVERGENCE OF A VECTOR
(Cont’d)

15. For Cartesian coordinate:
z
A
y
A
x
A zyx
∂
∂
+
∂
∂
+
∂
∂
=•∇ A
For Circular cylindrical coordinate:
( ) z
AA
A z
∂
∂
+
∂
∂
+
∂
∂
=•∇
φρ
ρ
ρρ
φ
ρ
11
A
DIVERGENCE OF A VECTOR
(Cont’d)

16. For Spherical coordinate:
( ) ( )
φθθ
θ
θ
φθ
∂
∂
+
∂
∂
+
∂
∂
=•∇
A
r
A
r
Ar
rr
r
sin
1sin
sin
11 2
2
A
DIVERGENCE OF A VECTOR
(Cont’d)

17. EXAMPLE
11
Find divergence of these vectors:
zx xzyzxP aa += 2
zzzQ aaa φρφρ φρ cossin 2
++=
φθ θφθθ aaa coscossincos
1
2
++= r
r
W r
(a)
(b)
(c)

18. 18
(a) Use divergence for Cartesian
coordinate:
SOLUTION TO EXAMPLE 11
( ) ( ) ( )
xxyz
xz
zy
yzx
x
z
P
y
P
x
P zyx
+=
∂
∂
+
∂
∂
+
∂
∂
=
∂
∂
+
∂
∂
+
∂
∂
=•∇
2
02
P

19. (b) Use divergence for Circular cylindrical
coordinate:
( )
( ) ( ) ( )
φφ
φρ
φρ
φρ
ρρ
φρ
ρ
ρρ
φ
ρ
cossin2
cos
1
sin
1
11
22
+=
∂
∂
+
∂
∂
+
∂
∂
=
∂
∂
+
∂
∂
+
∂
∂
=•∇ Q
z
z
z
z
QQ
Q z
SOLUTION TO EXAMPLE 11
(Cont’d)

20. SOLUTION TO EXAMPLE 11
(Cont’d)
(c) Use divergence for Spherical coordinate:
( ) ( )
( ) ( )
( )
φθ
θ
φθ
φθ
θθ
θ
φθθ
θ
θ
φθ
coscos2
cos
sin
1
cossin
sin
1
cos
1
sin
1sin
sin
11
2
2
2
2
=
∂
∂
+
∂
∂
+
∂
∂
=
∂
∂
+
∂
∂
+
∂
∂
=•∇ W
r
r
rrr
W
r
W
r
Wr
rr
r

21. It states that the total outward flux of
a vector field A at the closed surface S
is the same as volume integral of
divergence of A.
∫∫ •∇=•
VV
dVdS AA
1.12 DIVERGENCE THEOREM

22. EXAMPLE
12
A vector field exists in the region
between two concentric cylindrical surfaces
defined by ρ = 1 and ρ = 2, with both cylinders
extending between z = 0 and z = 5. Verify the
divergence theorem by evaluating:
ρρ aD 3
=
→
∫ •
S
dsD
∫ •∇
V
DdV
(a)
(b)

23. SOLUTION TO EXAMPLE 12
(a) For two concentric cylinder, the left side:
topbottomouterinner
S
d DDDDSD +++=•∫
Where,
πφρ
φρρ
π
φ
ρ
ρρ
π
φ
ρ
ρρ
10)(
)(
2
0
5
0
1
4
2
0
5
0
1
3
−=•−=
−•=
∫ ∫
∫ ∫
= =
=
= =
=
z
z
inner
dzd
dzdD
aa
aa

24. πφρ
φρρ
π
φ
ρ
ρρ
π
φ
ρ
ρρ
160)(
)(
2
0
5
0
2
4
2
0
5
0
2
3
=•=
•=
∫ ∫
∫ ∫
= =
=
= =
=
z
z
outer
dzd
dzdD
aa
aa
∫ ∫
∫ ∫
= =
=
= =
=
=•=
=−•=
2
1
2
0
5
3
2
1
2
0
0
3
0)(
0)(
ρ
π
φ
ρ
ρ
π
φ
ρ
ρφρρ
ρφρρ
z
ztop
z
zbottom
ddD
ddD
aa
aa
SOLUTION TO EXAMPLE 12
Cont’d)

25. Therefore
π
ππ
150
0016010
=
+++−=•∫ SD
S
d
SOLUTION TO EXAMPLE 12
Cont’d)

26. SOLUTION TO EXAMPLE 12
Cont’d)
(b) For the right side of Divergence Theorem,
evaluate divergence of D
( ) 23
4
1
ρρρ
ρρ
=
∂
∂
=•∇ D
So,
πρ
φρρρ
π
φ
π
φ ρ
150
4
5
0
2
0
2
1
4
5
0
2
0
2
1
2
=























=
=•∇
=
=
=
= = =
∫ ∫ ∫∫∫∫
z
r
z
dzdddVD

27. 1.13 CURL OF A VECTOR
The curl of vector A is an axial
(rotational) vector whose magnitude is
the maximum circulation of A per unit
area tends to zero and whose direction
is the normal direction of the area
when the area is oriented so as to
make the circulation maximum.

28. maxlim
0
a
A
AA n
s
s s
dl
Curl










∆
•
=×∇=
∫
→∆
Where,
CURL OF A VECTOR (Cont’d)
dldl
dacdbcabs
•







+++=• ∫∫∫∫∫ AA

29. CURL OF A VECTOR (Cont’d)
The curl of the vector field is concerned
with rotation of the vector field. Rotation
can be used to measure the uniformity
of the field, the more non uniform the
field, the larger value of curl.

30. For Cartesian coordinate:
CURL OF A VECTOR (Cont’d)
zyx
zyx
AAA
zyx ∂
∂
∂
∂
∂
∂
=×∇
aaa
A
z
xy
y
xz
x
yz
y
A
x
A
z
A
x
A
z
A
y
A
aaaA 





∂
∂
−
∂
∂
+



∂
∂
−
∂
∂
−





∂
∂
−
∂
∂
=×∇

31. z
z
AAA
z
φρ
φρ
ρ
φρ
ρ
ρ ∂
∂
∂
∂
∂
∂
=×∇
aaa
A
1
( )
z
zz
AA
z
AA
z
AA
a
aaA






∂
∂
−
∂
∂
+






∂
∂
−
∂
∂
−





∂
∂
−
∂
∂
=×∇
φρ
ρ
ρ
ρφρ
ρφ
φ
ρ
ρ
φ
1
1
For Circular cylindrical coordinate:
CURL OF A VECTOR (Cont’d)

32. CURL OF A VECTOR (Cont’d)
For Spherical coordinate:
( ) φθ
φθ
θ
φθθ
ArrAA
rr
r
r
sin
sin
1
2
∂
∂
∂
∂
∂
∂
=×∇
aaa
A
( ) ( )
( )
φ
θ
θ
φθφ
θ
φθφθ
θ
θ
a
aaA






∂
∂
−
∂
∂
+






∂
∂
−
∂
∂
+





∂
∂
−
∂
∂
=×∇
r
r
r
A
r
rA
r
r
rAA
r
AA
r
)(1
sin
11sin
sin
1

33. EXAMPLE
13
zx xzyzxP aa += 2
zzzQ aaa φρφρ φρ cossin 2
++=
φθ θφθθ aaa coscossincos
1
2
++= r
r
W r
(a)
(b)
(c)
Find curl of these vectors:

34. SOLUTION TO EXAMPLE 13
(a) Use curl for Cartesian coordinate:
( ) ( ) ( )
( ) zy
zyx
z
xy
y
xz
x
yz
zxzyx
zxzyx
y
P
x
P
z
P
x
P
z
P
y
P
aa
aaa
aaaP
22
22
000
−−=
−+−+−=






∂
∂
−
∂
∂
+



∂
∂
−
∂
∂
−





∂
∂
−
∂
∂
=×∇

35. (b) Use curl for Circular cylindrical coordinate
( )
( )
( )
( ) ( ) z
z
z
zz
zz
z
z
y
Q
x
QQ
z
Q
z
QQ
aa
a
aa
aaaQ
φρρφ
ρ
φρρ
ρ
ρφ
ρ
ρ
ρρφρ
ρ
φρ
ρφ
φ
ρ
ρ
φ
cos3sin
1
cos3
1
00sin
11
3
2
2
−++−=
−+
−+





−
−
=






∂
∂
−
∂
∂
+





∂
∂
−
∂
∂
+





∂
∂
−
∂
∂
=×∇
SOLUTION TO EXAMPLE 13
(Cont’d)

36. (c) Use curl for Spherical coordinate:
( ) ( )
( )
( ) ( ) ( )
φ
θ
φ
θ
θ
φθφ
θ
θ
φθ
θ
φ
θ
θφ
φθ
θ
θθ
θ
θ
φθφθ
θ
θ
a
aa
a
aaW










∂




∂
−
∂
∂
+










∂
∂
−
∂




∂
+





∂
∂
−
∂
∂
=






∂
∂
−
∂
∂
+






∂
∂
−
∂
∂
+





∂
∂
−
∂
∂
=×∇
22
2
cos
)cossin(1
cos
cos
sin
11cossinsincos
sin
1
)(1
sin
11sin
sin
1
r
r
r
r
r
rr
r
r
r
W
r
rW
r
r
rWW
r
WW
r
r
r
r
r
SOLUTION TO EXAMPLE 13
(Cont’d)

37. SOLUTION TO EXAMPLE 13
(Cont’d)
( ) ( )
a
aa
a
aa
φ
θ
φ
θ
θφ
θ
φ
θ
θ
θ
φθ
θφθθ
θ
sin
1
cos2
cos
sin
sin
2cos
sin
cossin2
1
cos0
1
sinsin2cos
sin
1
3
2






++
+





+=






++
−++=
r
rr
r
r
r
r
r
r
r
r

38. 1.14 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 that A and curl
of A are continuous on S.
( )∫∫ •×∇=•
SL
dSdl AA

39. STOKE’S THEOREM (Cont’d)

40. EXAMPLE
14
By using Stoke’s Theorem, evaluate
for
∫ • dlA
φρ φφρ aaA sincos +=
→

41. EXAMPLE 14 (Cont’d)

42. SOLUTION TO EXAMPLE 14
Stoke’s Theorem,
( )∫∫ •×∇=•
SL
dSdl AA
where, andzddd aS ρφρ=
Evaluate right side to get left side,
( ) zaA φρ
ρ
sin1
1
+=×∇

43. SOLUTION TO EXAMPLE 14
(Cont’d)
( ) ( )
941.4
sin1
1
0
0
60
30
5
2
=
+=•×∇ ∫ ∫∫
= =
aA z
S
dddS ρφφρρ
ρφ ρ

44. EXAMPLE
15
Verify Stoke’s theorem for the vector field
for given figure by evaluating:φρ φφρ aaB sincos +=
→
(a) over the
semicircular contour.
∫ • LB d
(b) over the
surface of semicircular
contour.
( )∫ •×∇ SB d

45. SOLUTION TO EXAMPLE 15
(a) To find∫ • LB d
∫∫∫∫ •+•+•=•
321 LLL
dddd LBLBLBLB
Where,
( ) ( )
φφρρφρ
φρρφφρ φρφρ
dd
dzddd z
sincos
sincos
+=
++•+=• aaaaaLB

46. So
20
2
1
sincos
2
0
2
0
0
0
0,0
2
01
=+





=








+








=•
=
=
=
==
=
∫∫∫
ρ
φφρ
ρ
φφρρφρLB
zz
L
ddd
( ) 4cos20
sincos
0
0,2
0
0
2
22
=−+=








+








=•
=
==
=
=
=
∫∫∫
π
φ
ρ
π
φρ
φ
φφρρφρLB
zz
L
ddd
SOLUTION TO EXAMPLE 15
(Cont’d)

47. 20
2
1
sincos
0
2
2
00,0
0
23
=+





−=








+








=•
=
=
=
==
=
∫∫∫
r
zz
L
ddd
ρ
φφρρφρ
π
πφφρ
LB
SOLUTION TO EXAMPLE 15
(Cont’d)
Therefore the closed integral,
8242 =++=•∫ LB d

48. SOLUTION TO EXAMPLE 15
(Cont’d)
(b) To find ( )∫ •×∇ SB d
( )
( ) ( ) ( ) ( )
( ) ( )
( ) ( ) ( )
z
z
z
zz
a
aaa
a
aa
aaB






+=
+++=






∂
∂
−
∂
∂
+






∂
∂
−
∂
∂
+





∂
∂
−
∂
∂
=
+×∇=×∇
ρ
φ
φρφ
ρ
φρ
φ
φρ
ρρ
ρ
φρφ
φρ
φφρ
φρ
φρ
φρ
1
1sin
sinsin
1
00
cossin
1
0cossin0
1
sincos

49. SOLUTION TO EXAMPLE 15
(Cont’d)
Therefore
( )
8
2
1
cos
1sin
1
1sin
0
2
0
2
0
2
0
0
2
0
=




















+−=
+=
•











+=•×∇
=
=
= =
= =
∫ ∫
∫ ∫∫∫
π
φ
ρ
π
φ ρ
π
φ ρ
ρρφ
φρρφ
φρρ
ρ
φ aaSB
dd
ddd zz

50. 1.15 LAPLACIAN OF A SCALAR
The Laplacian of a scalar field, V
written as:
V2
∇
Where, Laplacian V is:






∂
∂
+
∂
∂
+
∂
∂
•





∂
∂
+
∂
∂
+
∂
∂
=
∇•∇=∇
zyxzyx
z
V
y
V
x
V
zyx
VV
aaaaaa
2

51. For Cartesian coordinate:
2
2
2
2
2
2
2
z
V
y
V
x
V
V
∂
∂
+
∂
∂
+
∂
∂
=∇
For Circular cylindrical coordinate:
2
22
2
2 11
z
VVV
V
∂
∂
+
∂
∂
+





∂
∂
∂
∂
=∇
φρρ
ρ
ρρ
LAPLACIAN OF A SCALAR (Cont’d)

52. LAPLACIAN OF A SCALAR (Cont’d)
For Spherical coordinate:
2
2
22
2
2
2
2
sin
1
sin
sin
11
φθ
θ
θ
θθ
∂
∂
+






∂
∂
∂
∂
+





∂
∂
∂
∂
=∇
V
r
V
rr
V
r
rr
V

53. EXAMPLE
16
Find Laplacian of these scalars:
yxeV z
cosh2sin−
=
φρ 2cos2
zU =
φθ cossin10 2
rW =
(a)
(b)
(c)
You should try this!!

54. SOLUTION TO EXAMPLE 16
yxeV z
cosh2sin22 −
−=∇
02
=∇ U
( )θ
φ
2cos21
cos102
+=∇
r
W
(a)
(b)
(c)

55. Yea, unto God belong all things in the
heavens and on earth, and enough is God
to carry through all affairs
Quran:4:132
END