VECTOR ANALYSIS

Contents
1. Vector algebra
• Addition, Subtraction, & Multiplication of Vectors.
2. Orthogonal coordinate system
• Cartesian, cylindrical, & spherical coordinates.
3. Vector calculus
• Differentiation and integration of vectors; line,
surface and volume integrals; “del” operator,
gradient, divergence, and curl operations.

Vector Analysis 2
Vector Addition and Subtraction
• A vector A can be written as:
– A=aAA
– Where A is the magnitude of A and has the
unit and dimension.
– A=|A|
– aA is a dimensionless unit vector with a unity
magnitude having the direction of A.
– aA=A / |A| = A / A A=|A|

A=aAA
Vector Analysis 3
Vector Addition and Subtraction
• Two vectors A and B can be added in two
ways. C=A+B
– Parallelogram rule C

B
A

– Head to tail rule

C B

Vector Analysis 4
Vector Addition and Subtraction
• Vector addition obeys the commutative
and associative laws
– Commutative law
• A+B=B+A
– Associative law
• A + (B + C) = (A + B) + C

Vector Analysis 5
Vector Addition and Subtraction
• Vector subtraction is defined in following
way:
A – B = A + (-B)
Where –B has the same magnitude as B but
the direction is opposite to that of B.
-B = (-aB)B B

-B A-B

Vector Analysis 6
 Product of Vectors
• Multiplication of a vector A by a scalar k
changes only the magnitude of A by a
factor k.
B
kA = aA (kA)
θAB
A
• Scalar or Dot Product:
A . B = AB cos θAB BcosθAB A

Vector Analysis 7
 Product of Vectors
A . A = A2
A=√A.A

– Commutative law
A.B=B.A
– Distributive law
A . (B + C) = A . B + A . C

Vector Analysis 8
 Product of Vectors
• Vector or Cross Product
AxB
A x B = an|AB sin θAB|
B

BsinθAB
a θAB
– Cross Product is not commutative n A
BxA=-AxB
– Cross Product obeys the distributive law
A x (B + C) = A x B + A x C
– Cross Product is not associative
A x (B x C) = (A x B) x C

Vector Analysis 9
 Product of Vectors
• Product of Three Vectors
– Scalar Triple Product
A . (B x C) = B . (C x A) = C . (A x B)
A . (B x C) = -A . (C x B)
= -B . (A x C)
= -C . (B x A)

Vector Analysis 10
 Product of Vectors
Magnitude is equal
BxC
to volume of the
parallelepiped formed θ2
A

by vectors A, B, and C. θ1
C

B
Base area is |B x C| = |BC sin θ1|
Height is |A cos θ2|
Hence the volume is |ABC sin θ1 cos θ2|

Vector Analysis 11
 Product of Vectors
• Vector Triple Product
Back-cab rule
A x (B x C) = B(A . C) – C(A . B)
A|| B(A|| . C)

C
B
θ2
θ1 -C(A|| . B)
A┴
aD D
A = A|| + A┴
A┴ x (B x C) = 0 as both are parallel.
We are left with D = A|| x (B x C)
Vector Analysis 12
 Product of Vectors
– Figure shows the plane containing B, C, A||.
– D also lies in the same plane and is normal to A||.
– Magnitude of (B x C) is BC sin (θ1 – θ2).
– Magnitude of A|| x (B x C) is A||BC sin (θ1 – θ2).
D = D . aD = A||BC sin (θ1 – θ2)
= (B sin θ1)(A||C cos θ2) - (C sin θ2) (A||B cos θ1)
= [B(A|| . C) – C(A|| . B)] . AD
– It is not guaranteed that quantity inside the brackets is
equal to D; as it may contain a vector that is normal to
D; ie parallel to A||. Hence

Vector Analysis 13
 Product of Vectors
B(A|| . C) – C(A|| . B) = D + kA||
Multiplying both sides by A||; we get
(A|| . B)(A|| . C) – (A|| . C)(A|| . B) = A||. D + kA||2
0 = A||. D + kA||2
Since A||. D = 0 ( as D is normal to A||), so k = 0
Hence
D = B(A|| . C) – C(A|| . B)
This proves the Back-Cab rule.
A|| . C = A . C and A|| . B = A . B

Vector Analysis 14
 Division of Vectors
• Division of Vectors is not defined
• Expressions such as k/A and B/A are
meaningless.

Vector Analysis 15
Orthogonal Coordinate Systems
• We need position of the source and the location
of this point in coordinate system to determine
the Electric Field at a certain point in space.
• In three dimensional space a point can be
located as the intersection of three surfaces u1,
u2, u3.
• If these three surfaces are perpendicular to one
another; we have the Orthogonal Coordinate
System.

Vector Analysis 16
Orthogonal Coordinate Systems
• Let au1, au2, and au3 be the unit vectors called the
Base Vectors in the three coordinate system;
then in a general right handed, orthogonal,
curvilinear coordinate system:
• au1 x au2 = au3,
• au2 x au3 = au1,
• au3 x au1 = au2.
• Above three equations are not all independent,
as the specification of one automatically implies
the other two

Vector Analysis 17
Orthogonal Coordinate Systems
• au1 . au2 = au2 . au3 = au3 . au1 = 0

• au1 . au1 = au2 . au2 = au3 . au3 = 1

• A vector A can be written as:

• A = au1Au1 + au2Au2 + au3Au3

• Magnitude of vector A is
• A = |A| = (Au12+Au22+Au32)1/2.

Vector Analysis 18
Orthogonal Coordinate Systems
• EXAMPLE:
– Given three vectors A, B, and C, obtain the
expressions of:
(a) A . B (b) A x B (c) C . (A x B) in the
orthogonal curvilinear coordinate system
(u1,u2,u3).

Vector Analysis 19
Orthogonal Coordinate Systems
• SOLUTION:
A = au1Au1 + au2Au2 + au3Au3
B = au1Bu1 + au2Bu2 + au3Bu3
C = au1Cu1 + au2Cu2 + au3Cu3

a) A . B = (au1Au1 + au2Au2 + au3Au3) . (au1Bu1 +

au2Bu2 + au3Bu3)
= Au1Bu1 + Au2Bu2 + Au3Bu3

Vector Analysis 20
Orthogonal Coordinate Systems
b) A x B = (au1Au1 + au2Au2 + au3Au3) x (au1Bu1 +
au2Bu2 + au3Bu3)
= au1(Au2Bu3 – Au3Bu2) + au2(Au3Bu1 – Au1Bu3)
+ au3(Au1Bu2 – Au2Bu1)
au1 au2 au3
= Au1 Au2 Au3
Bu1 Bu2 Bu3

Vector Analysis 21
Orthogonal Coordinate Systems
c) C . (A x B)
= Cu1(Au2Bu3 – Au3Bu2) + Cu2(Au3Bu1 – Au1Bu3)
+ Cu3(Au1Bu2 – Au2Bu1)
Cu1 Cu2 Cu3
= Au1 Au2 Au3
Bu1 Bu2 Bu3

Vector Analysis 22
Orthogonal Coordinate Systems
• Differential change in length corresponds to the
change in one of the coordinates and a factor is
needed for such a change.
dli = hi dui, (i = 1, 2, or 3)
Where hi is called metric coefficient and may
itself be a function of ui
• e.g: In a two coordinate system (u1, u2) = (r, Ø) a
differential change dØ (=du2) in Ø (=u2)
corresponds to a differential length change dl2 =
rdØ (h2 = r = u1) in the aØ (=au2) direction.

Vector Analysis 23
Orthogonal Coordinate Systems
• A directed differential length change in an
arbitrary direction can be written as vector
sum of component length changes;
dl = au1 dl1 + au2 dl2 + au3 dl3
dl = au1 (h1 du1) + au2 (h2 du2) + au3 (h3 du3)
Magnitude of dl is
dl = [(dl1)2 + (dl2)2 + (dl3)2]1/2
= [(h1 du1)2 + (h2 du2)2 + (h3 du3)2]1/2

Vector Analysis 24
Orthogonal Coordinate Systems
• The differential volume formed by differential
coordinate changes du1, du2, and du3 in
directions au1, au2, and au3 respectively is (dl1
dl2 dl3), or
dv = h1h2h3 du1du2du3

• In order to express the current or flux flowing

through a differential area, cross-sectional area
perpendicular to the current or flux is to be used
ds = ands
Vector Analysis 25
Orthogonal Coordinate Systems
• Let current density J is not perpendicular
to a differential area ds, the current dI,
flowing through ds must be the component
of J normal to the area, multiplied by the
area.
dI = J . ds
=J . ands

Vector Analysis 26
Orthogonal Coordinate Systems
• In general orthogonal curvilinear
coordinate system the differential area ds1
normal to the unit vector au1 is:
• ds1 = dl2 dl3
• ds1 = h2h3du2du3
• Similarly differential areas normal to
vectors au2 and au3 are respectively
• ds2 = h1h3du1du3
• ds3 = h1h2du1du2
Vector Analysis 27
Orthogonal Coordinate Systems
• Main orthogonal coordinate systems are:

• Cartesian (or Rectangular) Coordinates

• Cylindrical Coordinates
• Spherical Coordinates

Vector Analysis 28
 Cartesian Coordinates
• (u1, u2, u3) = (x, y, z) z=z1 plane
• Point P(x1, y1, z1) is
Intersection of three
y=y1 plane
Planes x = x1, y = y1,
z = z1
Base vectors are ax,
ay, az in the
respective Directions. X=x1 plane

Vector Analysis 29
 Cartesian Coordinates
• Base vectors satisfy following relations:
ax x ay = az,
ay x az = ax,
az x ax = ay.
• Position vector to point P P(x1, y1, z1) is:
OP = axx1 + ayy1 + azz1.
• A vector A can be written as:
A = axAx + ayAy + azAz.
Vector Analysis 30
 Cartesian Coordinates
• The dot product of two vectors A and B is:
A . B = AxBx + AyBy + AzBz
• The cross product of A and B is:
AxB=
ax(AyBz-AzBy) + ay(AzBx-AxBz) + az(AxBy-AyBx)
ax ay az
= Ax Ay Az
Bx By Bz

Vector Analysis 31
 Cartesian Coordinates
• Since x, y, and z are lengths so all three matric
coefficients are unity ie, h1 = h2 = h3 = 1. The
expressions for differential length, differential
area, and differential volume are:
dl = axdx + aydy + azdz.
dsx = dydz,
dsy = dxdz,
dsz = dxdy.
dv = dxdydz
Vector Analysis 32
Cartesian Coordinates

dsx =dydz dsz = dxdy

z
Dsy = dxdz
dz

y
o
dx
dy
x
A differential volume in Cartesian Coordinates

Vector Analysis 33
Cartesian Coordinates

Vector Analysis 34
 Cartesian Coordinates
• EXAMPLE: Given A = ax5 – ay2 + az, find
the expression of a unit vector B such
that:
a) B||A
b) B┴A, if B lies in the xy-plane.
• SOLUTION:
– Let B = axBx + ayBy + azBz. We know that
– B = (Bx2 + By2 + Bz2)1/2 = 1

Vector Analysis 35
 Cartesian Coordinates
a) B||A requires B x A = 0, hence we have
-2Bz – By = 0,
Bx – 5Bz = 0,
5By + 2Bx = 0.
Solving above equations along with magnitude
equation; we get:
Bx = 5/√30, By = -2/√30, Bz = 1/√30
Therefore
B = (ax5 – ay2 + az)/√30

Vector Analysis 36
 Cartesian Coordinates
b) B┴A requires B . A = 0, hence we
have 5Bx – 2By = 0.
Bz = 0, since B lies in the xy-plane
Solution of above equation along with
magnitude equation yields:
Bx = 2/√29, By = 5/√29
Hence
B = (ax2 + ay5)/√29
Vector Analysis 37
 Cartesian Coordinates
• EXAMPLE:
– (a) Write the expression of the vector going
from point P1(1,3,2) to point P2(3,-2,4) in
Cartesian coordinates.
– (b) What is the length of this line?
• SOLUTION:

Vector Analysis 38
 Cartesian Coordinates
From Figure, we see P2(3,-2,4) z

P1P2 = OP2 – OP1

= (ax3-ay2+az4) –
P1(1,3,2)
(ax+ay3+az2)
= ax2 – ay5 + az2 x y

The length of the line is

P1P2 =|P1P2| = √22 + (-5)2 + 22
= √33
Vector Analysis 39
 Cylindrical Coordinates
• (u1, u2, u3) = (r, Ø, z)
• Point P(r1, Ø1, z1) is z

the intersection of a z=z1 plane

cylindrical surface r1 az
r=r1, a half plane aØ
ar
containing the z axis
and making an angle o z1
x1
Ø=Ø1 with the xz-r=r1 cylinder y1
y
plane, and a plane Ø1

parallel to xy plane at x Ø=Ø1 plane

z=z1.
Vector Analysis 40
 Cylindrical Coordinates
• Angle Ø is measured from +ve x-axis, and
base vector aØ is tangential to the
cylindrical surface.
• Following right handed relations apply.
ar x aØ = az
aØ x az = ar
az x ar = aØ

Vector Analysis 41
Cylindrical Coordinates
   
ar  ar  1, ar  ar  0,
   
a  a  1, a  a  0,
   
a z  a z  1, a z  a z  0,
 
ar  a  0,
 
a  a z  0,
 
a z  ar  0,

Vector Analysis 42
 Cylindrical Coordinates
• A vector in cylindrical coordinates is
written as:
A = arAr + aØAØ + azAz
• Dot and cross product of two vectors in
cylindrical coordinates follow the equations
as discussed on slides 20,21.
• Two of the coordinats, r and z (u1 and u3)
are lengths; hence h1 = h3 = 1.
Vector Analysis 43
 Cylindrical Coordinates
• However Ø is an angle requiring a metric
co-efficient h2 = r to convert dØ to dl2.
• General expression for a differential length
in cylindrical coordinates is then:
dl = ardr + aØrdØ + azdz
• Expressions for differential areas and
differential volume are:
dsr = r dØ dz,
Vector Analysis 44
 Cylindrical Coordinates
• dsØ = dr dz,
• dsz = r dr dØ,

• dv = r dr dØ dz.

Vector Analysis 45
 Cylindrical Coordinates
• A vector given in cylindrical coordinates
i.e A = arAr + aØAØ + azAz
can be transformed into Cartesian coordinates ie
A = axAx + ayAy + azAz.
• Z component remains un-altered.
• To find Ax, we equate dot product of above both
expressions of A with ax. Thus:
Ax = A . ax = arAr . ax + aØAØ . ax
• az . ax = 0, hence Az disappears.
Vector Analysis 46
 Cylindrical Coordinates
From figure:
ar . ax = cos Ø
aØ . ax = cos(π/2 + Ø)
= - sin Ø
Ax = Ar cos Ø – AØ sin Ø aØ

Similarly Ay = A . ay
= arAr . ay + aØAØ . ay ar
ar . ay = cos(π/2 - Ø)
= sin Ø
aØ . ay = cos Ø
Ay = Ar sin Ø + AØ cos Ø
Vector Analysis 47
Cylindrical Coordinates
 
ar cos  sin  0 ax
 
a   sin  cos  0 ay
 
az 0 0 1 az

Vector Analysis 48
 Cylindrical Coordinates
• Conversion Matrix is:
Ax cosØ -sinØ 0 Ar
Ay = sinØ cosØ 0 AØ
Az 0 0 1 Az
 Ar   cos  sin  0  Ax 
 A    sin  cos   
0  Ay  
  
 Az   0 0 1  Az 
Vector Analysis 49
 Cylindrical Coordinates
• Conversions formulas are:
Cartesian Cylindrical
x = r cos Ø r = √ x2 + y2
y = r sin Ø Ø= tan-1 y/x
z=z z=z

Vector Analysis 50
 Cylindrical Coordinates
• EXAMPLE: The cylindrical coordinates of
an arbitrary point P in the z = 0 plane are
(r, Ø, 0). Find the unit vector that goes
from a point z = h on z-axis toward P.
• SOLUTION:
QP = OP – OQ
= (arr) – (azh)
aQP = QP/|QP|
= (1/√r2 + h2) (arr – azh)

Vector Analysis 51
 Cylindrical Coordinates
• EXAMPLE: Express the vector
A = ar(3cosØ) – aØ2r + az5 in cartesian
coordinates:
Ax cosØ -sinØ 0 3cosØ
Ay = sinØ cosØ 0 -2r
Az 0 0 1 5

• A = ax (3cos2 Ø + 2r sin Ø) + ay (3sin Øcos Ø

– 2r cos Ø) + az 5
Vector Analysis 52
 Cylindrical Coordinates
• Cos Ø = x / √x2 + y2
• Sin Ø = y / √x2 + y2
• Therefore:
A = ax (3x2/(x2 + y2) + 2y)
+ ay (3xy/(x2 + y2) - 2x)
+ az 5

Vector Analysis 53
 Spherical Coordinates
• (u1, u2, u3) = (R, θ, Ø)
• Point P(R1, θ1, Ø1) is the intersection of a
spherical surface centered at the origin
with a radius R=R1, a right circular cone
with it’s apex at the origin, it’s axis
coincides with the + z-axis and having a
half angle θ=θ1 and a half plane containing
the z axis and making an angle Ø=Ø1 with
the xz-plane.

Vector Analysis 54
Spherical Coordinates

Vector Analysis 55
 Spherical Coordinates
• The base vector aR at P is radial from the
origin and is quite different from ar in
cylindrical coordinates, as the latter is
perpendicular to the z-axis. The base
vector aθ lies in the Ø=Ø1 plane and is
tangential to the spherical surface,
whereas the base vector aØ is the same as
in the cylindrical coordinates.

Vector Analysis 56
Spherical Coordinates
   1
a a
   1
R R

a a
   1
a a
   0
a a
   0
R

a a 
   0
a a R

Vector Analysis 57
 Spherical Coordinates
• For a right-handed system we have
aR x aθ = aØ,
aθ x aØ= aR,
aØ x aR = aθ
• Spherical coordinates are important for
problems involving point sources and
regions with spherical boundaries.
• Spherical coordinates are used in solving
antenna problems in the far field.
Vector Analysis 58
 Spherical Coordinates
• A vector in spherical coordinates is written
as:
A = aRAR + aθAθ + aØAØ.
• Expressions for dot and cross products of
two vectors in spherical coordinates are
similar to those shown on slide 31.
• In spherical coordinates only R (u1) is a
length. The other two coordinates θ and Ø
(u2 and u3) are angles.
Vector Analysis 59
Spherical Coordinates

Vector Analysis 60
 Spherical Coordinates
• Metric coefficients h2=R and h3=R sinθ are
required to convert dθ and dØ into dl2 and
dl3 respectively.
• From equation on page 24 the general
expression for differential length is:
• dl = aR dR + aθR dθ + aØR sinθ dØ

Vector Analysis 61
 Spherical Coordinates
• Differential areas and differential volume
resulting from differential changes dR, dθ,
dØ are:
• dsR = R2 sinθ dθ dØ,
• dsθ = R sinθ dR dØ,
• dsØ = R dR dθ,

• dv = R2 sinθ dR dθ dØ.
Vector Analysis 62
 Spherical Coordinates
• A vector in spherical coordinates can be
transformed into Cartesian coordinates as:

x = R sinθ cosØ,
y = R sinθ sinØ,
z = R cosθ,

Vector Analysis 63
 Spherical Coordinates
• Cartesian coordinates can be converted to
spherical coordinates as:
R = √x2 + y2 + z2,
θ = tan-1 √(x2 + y2)/z,
Ø = tan-1 y/x

Vector Analysis 64
 Spherical Coordinates
   
aR  a x  sin  cos  , a  a x   sin  ,
   
aR  a y  sin  sin  , a  a y  cos  ,
   
aR  a z  cos  , a  a z  0,
 
a  a x  cos  cos  ,
 
a  a y  cos  sin  ,
 
a  a z   sin  ,  
aR sin  cos  sin  sin  cos  a x
 
a  cos  cos  cos  sin   sin  a y
 
a  sin  cos  0 az
Vector Analysis 65
 Spherical Coordinates
• Transformation of Vector

   
A  AR aR  A a  A a ,
       
Ax  A  a x  AR aR  a x  A a  a x  A a  a x ,
 AR sin  cos   A cos  cos   A sin  .

Vector Analysis 66
Spherical Coordinates
Ax sin  cos  cos  cos   sin  AR
Ay  sin  sin  cos  sin  cos  A
Az cos   sin  0 A

AR sin  cos  sin  sin  cos  Ax

A  cos  cos  cos  sin   sin  Ay
A  sin  cos  0 Az

Vector Analysis 67
 Spherical Coordinates
• Example: The position of a point P in
spherical coordinates is (8, 120, 330).
Specify it’s location (a) in Cartesian
coordinates (b) in cylindrical coordinates.

• Solution: Coordinates of the point P are

R=8, θ=120°, Ø=330°.

Vector Analysis 68
 Spherical Coordinates
• a) Let us use the equations on page 60.
• x = 8 sin120° cos330° = 6,
• y = 8 sin120° sin330° = -2√3,
• z = 8 cos120° = -4.
• Hence the location of point is P(6, -2√3, -4).
• And the position vector is:
• OP = ax6 – ay2√3 – az4.

Vector Analysis 69
 Spherical Coordinates
• b) The cylindrical coordinates of point P
can be obtained by applying equations on
page: 48; but these can also be calculated
directly from the spherical coordinates by
using following equations:
– r = R sinθ,
– Ø = Ø,
– z = R cos θ.
• Hence we get the point P(4√3, 330, -4).
Vector Analysis 70
 Spherical Coordinates
• Position vector in cylindrical coordinates is:
OP = ar4√3 – az4
• We note that position vector does not contain
Ø=330°; however exact direction of ar depends on
Ø.
• In spherical coordinates position vector contains
only one term:
OP = aR8.
• Here the direction of ar changes with the θ and Ø
coordinates of point P.

Vector Analysis 71
 Spherical Coordinates
• Example: Convert the vector A = aRAR +
aθAθ + aØAØ into Cartesian coordinates.
• Solution: In this problem we want to write
A in the form of A = axAx + ayAy + azAz.
1) We assume that the expression of the given
vector A holds for all points of interest and
that all three given components AR, Aθ, and
AØ may be functions of coordinate variables.
2) At a given point AR, Aθ, and AØ will have
definite numerical values, but these values
Vector Analysis 72
 Spherical Coordinates
that determine the direction of A will, in general, be
entirely different from the coordinate values of the
point.
• Taking dot product of A with ax, we get:
Ax = A . ax
= ARaR . ax + Aθaθ . ax + AØaØ . ax
• aR . ax, aθ . ax, and aØ . ax yield respectively, the
component of unit vectors aR, aθ, and aØ in the
direction of ax, we find from fig on page 57 and
equations on page 60:
Vector Analysis 73
 Spherical Coordinates
• aR . ax = sinθ cosØ = x/√(x2 + y2 + z2)
• aθ . ax = cosθ cosØ = xz/ √((x2 + y2)
(x2 + y2 + z2))
• aØ . ax = - sinØ = -y/√(x2 + y2)
• Thus Ax = AR sinθ cosØ + Aθ cosθ cosØ
- AØ sinØ
• = ARx/√(x2 + y2 + z2) + Aθxz/√((x2 + y2)
(x2 + y2 + z2)) - AØy/√(x2 + y2)

Vector Analysis 74
 Spherical Coordinates
• Similarly Ay = AR sinθ sinØ + Aθ cosθ sinØ
+ AØ cosØ
• = ARy/√(x2 + y2 + z2) + Aθyz/√((x2 + y2)
(x2 + y2 + z2)) + AØx/√(x2 + y2)
• AZ = AR cosθ + Aθ sinθ
= ARz/√(x2 + y2 + z2) - Aθ√(x2 + y2) /
√(x2 + y2 + z2))

Vector Analysis 75
 Spherical Coordinates
• Example: Assuming that a cloud of
electrons confined in a region between two
spheres of radii 2 and 5 cm has a charge
density of -3x10-8 cos2Ø / R4 C/m3.

• Solution: ρ = -3x10-8 cos2Ø / R4,

Q = ∫ ρdv.
Q = 0∫2π 0∫π 0.02∫0.05 ρR2 sinθ dR dθ dØ
Vector Analysis 76
 Spherical Coordinates
• Q = -3x10-8 0∫2π 0∫π 0.02∫0.05 (1/R2) cos2Ø
sinθ dR dθ
dØ
• = -3x10-8 0∫2π 0∫π(-1/0.05 + 1/0.02) cos2Ø
sinθdθ dØ
• = -0.9x10-6 0∫2π (-cosθ)0|π cos2ØdØ
• = -1.8x10-6 (Ø/2 + (sin2Ø)/4)0|2π
• = -1.8π (µC).
Vector Analysis 77
 Integrals Containing Vector
Functions
 Fdv
V

• Above integral can be evaluated as the sum of

three scalar integrals by first resolving the vector
F into it’s three components in the appropriate
coordinate system.
• dv represents the differential volume.
• This is the triple integral over three dimensions
shown in a shorthand way.

Vector Analysis 78
 Integrals Containing Vector
Functions
 Vdl
C

•This integral is a scalar function of space.

•dl shows the differential increment of length.
•C is the path of integration.
P1
•If the integral is from P1 to P2, we write 
P2
Vdl

•If it is for a closed path C, we write CVdl

Vector Analysis 79
 Integrals Containing Vector
Functions
• In Cartesian coordinates Integral can be written
as:

 Vdl   V ( x, y, z)[a dx  a dy  a dz 
C C
x y z

 Vdl  a  V ( x, y, z)dx  a  V ( x, y, z)dy  a  V ( x, y, z)dz

C
x
C
y
C
z
C

•Three integrals on right hand side are ordinary

scalar integrals. These can be evaluated for a
given V(x,y,z) around a path C.

Vector Analysis 80
 Integrals Containing Vector
Functions
• Example: Evaluate the integral O r 2 dr, where
P

r  x  y from the origin to the point P(1,1):

2 2
___________, 2

• a) Along the direct path OP.

• b) Along the path OP1P, and
• c) Along the path OP2P.

Vector Analysis 81
 Integrals Containing Vector
Functions
• Solution:
• a) Along the direct path OP

P 2 2 2
 r dr  ar  r dr  ar
2 2
O 0 3
2 2
 ( a x cos 45  a y sin 45)
3
2 2
 ax  a y
3 3
Vector Analysis 82
 Integrals Containing Vector
Functions
• Solution:
• b) Along the path OP1P
P P1 P
 (x  y )dr a y  y dy  a x  ( x  1)dx
2 2 2 2
O O P1
1
1 31 1 3
 a y y  a x ( x  x)
3 0 3 0

4 1
 ax  a y .
3 3
Vector Analysis 83
 Integrals Containing Vector
Functions
• Along the path OP2P
P P2 P
 (x  y )dr a x  x dx  a y  ( y  1)dy
2 2 2 2
O O P2
1
1 31 1 3
 ax x  a y ( y  y)
3 0 3 0

1 4
 ax  a y .
3 3
Vector Analysis 84
 Integrals Containing Vector
Functions
C
F  dl
•Above is a line integral, in which integrand
represents the component of F along the path of
integration.
•If F is a force, the integral is work done by the
force in moving an object from a point P1 to P2
along a specified path C.
•If F is replaced by E, then the integral is work
done by electric field Vector
in moving
Analysis
a unit charge from
85
P1 to P2.
 Integrals Containing Vector
Functions
• Example: Given F  ax xy  a y 2 x , evaluate the
scalar line integral  F  dl , along the quarter
B

A
circle shown in figure.

Vector Analysis 86
 Integrals Containing Vector
Functions
• Solution:
• a) In Cartesian coordinates:
F  dl  xydx  2 xdy
x 2  y 2  9(0  x, y  3)
B 0 3
 F  dl   x 9  x dx  2  9  y 2 dy
2
A 3 0
3
1 3 0
 y
  (9  x ) 2
2
  y 9  y 2  9 sin 1
3 3  3
0

 9(1  )
2
Vector Analysis 87
 Integrals Containing Vector
Functions
• Solution:
• b) In cylindrical coordinates:
 Ar   cos  sin  0  Ax 
 A    sin  cos  0  Ay 
  
 Az   0 0 1  Az 
 Fr   cos  sin  0  xy 
 F    sin  cos  0  2 x 
  
 Fz   0 0 1  0
F  ar ( xy cos   2 x sin  )  a ( xy sin   2 x cos  )
Vector Analysis 88
 Integrals Containing Vector
Functions
• Path of integration is along a quarter-circle of a
radius 3. There is no change in r or z along the
path (dr=0 and dz=0); hence equation dl = ardr +
aØrdØ + azdz simplifies to:
dl  a 3d
F  dl  3( xy sin   2 x cos  ) d
B 
 F  dl    3(9 sin 2  cos   6 cos 2  ) d
2
A 0

 9(sin     sin  cos  ) 0 2
3


 9(1  ).
2
Vector Analysis 89
 Integrals Containing Vector
Functions
 A  ds
s
• This is a surface integral. It is actually a double
integral over two dimensions.
• The integral measures the flux of the vector field
A flowing through the area S.
• Vector differential surface element ds=ands has
a magnitude ds and the direction shown by an.
• The conventions for the +ve direction of ds are
as follows:
Vector Analysis 90
 Integrals Containing Vector
Functions
• If the surface of integration S is a closed surface
enclosing a volume, then the +ve direction of an
is always is the outward direction.
• Positive direction of an depends on the location
of ds.
• Further closed surface integral requires a small
circle added over the integration sign.

 A  ds   A  a ds.
s s
n

Vector Analysis 91
 Integrals Containing Vector
Functions
• If S is an open surface, the +ve direction of an
depends on the direction in which the perimeter
of the open surface is traversed.
• Acc to right hand rule if the fingers follows the
direction of travel around the perimeter then the
thumb points in the direction of +ve an.
• Again the +ve direction of an depends on the
location of ds.

Vector Analysis 92
 Integrals Containing Vector
Functions
• Example: Given F  ar k1 r  az k2 z , evaluate the
scalar surface integral s F  ds over the surface
of a closed cylinder about the z-axis specified
by z=±3 and r=2.
•Solution: The specified
surface of integration is
that of closed cylinder as
shown. It has three
surfaces: The top face,
the bottom face, and the
side wall. Vector Analysis 93
 Integrals Containing Vector
Functions
 F  ds   F  a ds
s
n

 F .a ds  
n F .an ds   F .an ds
topface bottomface sidewall

•Where an is a unit vector normal outwards from

the respective surfaces.
•Three integrals on the right side can be
evaluated separately.

Vector Analysis 94
 Integrals Containing Vector
Functions
• a) Top face z = 3, an = az,

F  an  k 2 z  3k 2 ,
ds  rdrd ;
2 2
TopFace
F  an ds  
0  3k rdrd  12k
0
2 2

Vector Analysis 95
 Integrals Containing Vector
Functions
• b) Bottom Face: z = -3, an = -az,

F  an  k 2 z  3k 2 ,
ds  rdrd ;
2 2
BottomFace
F  an ds  
0  3k rdrd  12k
0
2 2

Vector Analysis 96
 Integrals Containing Vector
Functions
• C) Side Wall r = 2, an = ar,

k1 k1
F  an   ,
r 2
ds  rddz  2ddz;
3 2
SideWall
F  an ds  
3 0  k1ddz  12k1.

Vector Analysis 97
 Integrals Containing Vector
Functions
• Therefore

 F  ds  12k
s
2  12k 2  12k1
 12 (k1  2k 2 )
•This surface integral gives the net outward flux
of the vector F through the closed cylindrical
surface.

Vector Analysis 98
 Gradient of a Scalar Field
• We encounter scalar and vector fields that are
functions of four variables: (t, u1, u2, u3).
• Method is required for describing the space rate
of change of a scalar field at a given time.
• Consider a scalar function of space coordinates
V(u1, u2, u3) which represents say, the
temperature distribution in a building, the altitude
of a mountainous terrain, or the electric potential
in a region

Vector Analysis 99
 Gradient of a Scalar Field
• Magnitude of V depends on the position of the
point in space, but it may be constant along
certain lines or surfaces as shown in figure two
surfaces having constant magnitudes V1 and
V1+dV.
• Point P1 is on the surface V1; P2 is the
corresponding point on surface V1+dV along the
normal vector dn; and P3 is a point close to P2
along another vector dl ≠ dn.

Vector Analysis 100

 Gradient of a Scalar Field
•For the same change dV in V the space rate of
change, dV/dl, is greatest along dn as dn is the
shortest distance b/w the two surfaces.

•Since the magnitude

of dV/dl depends on
the direction of dl,
dV/dl is a directional
derivative

Vector Analysis 101

 Gradient of a Scalar Field
• “We define the vector that represents both
the magnitude and the direction of the
maximum space rate of increase of a
scalar as the gradient of that scalar.”
dV
gradV  an .
dn
It is customary to employ operator del, represented by thesymbol
 and write V in place of gradV
dV
V  a n
dn
Vector Analysis 102
 Gradient of a Scalar Field
• We have assumed that dV is +ve if increase in
V; if dV is –ve (a decrease in V from P1 to P2) ,
 V will be –ve in an direction.
• Directional derivative along dl is
dV dV dn dV
  cos 
dl dn dl dn
dV
 an  al  (V )  al
dn
•This equation states that the space rate of increase
of V in the al direction is equal to the projection of
the gradient of V in that direction
Vector Analysis 103
 Gradient of a Scalar Field
• We can also write
dV  (V )  dl ,
Where dl  al dl , Now dV is the total differential of V as a result of
a change in position (from P1 to P3 ); hence it can be expressed in
terms of the differential changes in coordinates :
V V V
dV  dl1  dl2  dl3 ,
l1 l2 l3
•Where dl1, dl2, and dl3 are the components of the
vector differential displacement dl in a chosen
coordinate system.
Vector Analysis 104
 Gradient of a Scalar Field
• In terms of general orthogonal coordinates (u1,
u2, u3), dl is:
dl  au1dl1  au 2 dl2  au 3 dl3
 au1 (h1du1 )  au 2 (h2 du2 )  au 3 (h3 du3 )
dV can be written as dot product of two
vectors as follows :
V V V
dV  (au1  au 2  au 3 )  (au1dl1  au 2 dl2  au 3dl3 )
l1 l2 l3
V V V
 (au1  au 2  au 3 )  dl
l1 l2 l3
Vector Analysis 105
 Gradient of a Scalar Field
• Comparing above equation with the equation on
top of slide 104.
V V V
V  au1  au 2  au 3
l1 l2 l3
V V V
V  au1  au 2  au 3
h1u1 h2 u 2 h3u3

• Above equation is useful for computing gradient of

a scalar, when the scalar is given as a function of
space coordinates.

Vector Analysis 106

 Gradient of a Scalar Field
• In Cartesian coordinates, (u1, u2, u3) = (x, y, z)
and h1 = h2 = h3 = 1, hence we have:
V V V
V  a x  ay  az
x y z
  
V  ( a x  ay  a z )V
x y z
It is convenient to consider  in Cartesian coordinates
as a vector differential operator.
  
  ax  ay  az
x y z
Vector Analysis 107
 Gradient of a Scalar Field
• We see that we can define  in general
orthogonal coordinates as:
  
  (au1  au 2  au 3 )
h1u1 h2u2 h3u3

Vector Analysis 108

 Gradient of a Scalar Field
• Example: The Electrostatic field intensity E is
derivable as the –ve gradient of a scalar electric
potential V; that is, E= -  V. Determine E at the
point (1, 1, 0) if
x y
a) V  V e sin ,
4
b) V  E R cos 

Vector Analysis 109

 Gradient of a Scalar Field
• We use Cartesian Coordinates for part (a) and
spherical coordinates for part (b) to solve E= - V.
• a)    x y
E  [ a x  ay  az ]E e sin
x y z 4
y  y
 ( a x sin  ay cos ) E e  x .
4 4 4
 E
Thus E (1,1,0)  ( a x  a y )   a E E ,
4 2
1 2
where E  E (1  ),
2 16
1 
aE  (a x  a y ).
1  ( 2
16) 4
Vector Analysis 110
 Gradient of a Scalar Field
b) E  [a   a 
 a

]E R cos 

R R R sin 
R

 (a R cos   a sin  ) E .

In view of following equation:
AZ = AR cosθ + Aθ sinθ
= ARz/√(x2 + y2 + z2) - Aθ√(x2 + y2) /
√(x2 + y2 + z2))
the result of above converts to E = -azEo in
Cartesian coordinates.

Vector Analysis 111

 Gradient of a Scalar Field
• This is not surprising, as careful examination
of the given V reveals that EoRcosθ is infact
equal to Eoz. Hence in Cartesian coordinates:


E  V  az ( E z )  az E
z

Vector Analysis 112

 Divergence of a Vector Field
• Flux lines or streamlines are directed lines or
curves indicating at each point the direction of
the vector field.
• Magnitude of the field at a point is depicted
either by the density or by the length of the
directed lines in the vicinity of the point.
• This figure shows that the field in the
region A is stronger than that in region
B, as there is higher density of equal
length directed lines in region A.
Vector Analysis 113
 Divergence of a Vector Field
• This figure indicate a radial field
that is strongest in the region
closest to the point q and
decreasing arrow lengths show
the weaker field away from the
charge q.
• This figure depicts a uniform field.

• Vector Field Strength is measured by the

number of flux lines passing through a unit
surface normal to the vector.
Vector Analysis 114
 Divergence of a Vector Field
• The flux of vector field is analogous to the flow of
an incompressible fluid such as water.
• Net +ve divergence indicates the presence of a
source of fluid inside the volume.
• Net –ve divergence indicates the presence of
sink inside the volume.
• In the uniform field, there is an equal amount of
inward and outward flux going through any
closed volume containing no source or sink,
resulting in a zero divergence.
Vector Analysis 115
 Divergence of a Vector Field
• We define the divergence of a vector field at a
point, (abbreviated div A) as the net outward flux
of A per unit volume as the volume about the
point tends to zero:

divA  lim v  0 s
A  ds
v
• The numerator represents the net outward flux,
is an integral over the entire surface S that
bounds the volume
Vector Analysis 116
 Divergence of a Vector Field
• Div A is a scalar quantity whose magnitude may
vary from point to point.
• Consider a differential volume of sides Δx, Δy,
and Δz centered about a point P(xo, yo, zo) in the
field of a vector A; and we wish to find div A at
the point (xo, yo, zo).
• Since the differential volume has six faces, the
surface integral can be decomposed into six
parts.

Vector Analysis 117

 Divergence of a Vector Field
  A  ds
s A  ds 
 frontface backface rightface leftface topface bottomface
    

On the front face

 frontface
A  ds  A frontface  S frontface  A frontface  a x (yz )
x
 Ax ( x  , y , z )yz
2

Vector Analysis 118

 Divergence of a Vector Field
x
• The quantity Ax ( x 
can be expanded as a
, y , z )
2
Taylor series about its value at (xo, yo, zo), as
follows:
x x Ax
Ax ( x  , y , z )  Ax ( x , y , z )   higher _ order _ terms,
2 2 x ( x , y , z )

Where the higher order terms (H.O.T) contain

the factors (Δx/2)2,(Δx/2)3, etc.

Vector Analysis 119

 Divergence of a Vector Field

backface
A  ds  Abackface  S backface  Abackface  ( a x yz )
x
  Ax ( x  , y , z )yz
2
x
The Taylor - series expansion of  Ax ( x  , y , z ) is :
2
x x Ax
Ax ( x  , y , z )  Ax ( x , y , z )   H .O.T
2 2 x ( x , y , z )
Combining equations for front and back face we get :
Ax
[  ] A  ds  (  H .O.T ) xyz.
Frontface Backface x ( x , y , z )

Vector Analysis 120

 Divergence of a Vector Field
• Following the same procedure for the right and
the left faces, where the coordinate changes are
+Δy/2 and –Δy/2, respectively and Δs=ΔxΔz, we
find:
Ay
[  ] A  ds  (  H .O.T ) xyz.
rightface Leftface y ( x , y , z )

Here the H.O.T contains the factors y,(y)2 , For the top and
bottom faces we have :
Az
[  ] A  ds  (  H .O.T ) xyz.
Topface Bottomface z ( x , y , z )

Here the H.O.T contains the factors z, (z)2 .

Vector Analysis 121
 Divergence of a Vector Field
• Now combining the results of all the sides:
Ax Ay Az
 A  ds  (
s x

y
 )
z ( x , y , z )
xyz  higher  order  terms  in  x, y, z.

• Since Δv=ΔxΔyΔz substituting above equation

in div A equation in Cartesian coordinates we
get: Ax Ay Az
divA   
x y z

•The higher order terms vanish as the differential

volume ΔxΔyΔz approaches zero.
Vector Analysis 122
 Divergence of a Vector Field
• Value of div A depends on the position of the
point at which it is evaluated.
• We have dropped notation (xo, yo, zo) in above
equation because it applies to every point at
which A and its partial derivates are defined.
  A  divA
In general orthogonal coordinates (u1 , u2 , u3 ), we have :
1     
 A   (h2 h3 A1 )  (h1h3 A2 )  (h1h2 A3 ).
h1h2 h3  u1 u2 u3 
Vector Analysis 123
 Divergence of a Vector Field
• Example: Find the divergence of the position
vector to an arbitrary point.
• Solution: We will find the solution in Cartesian as
well as in spherical coordinates.
a) Cartesian coordinates:
Expression for a position vector to an arbitrary point
(x, y, z) is:
OP  a x x  a y y  a z z.
x y z
  (OP)     3.
x y z
Vector Analysis 124
Divergence of a Vector Field
b) Spherical coordinates: Here the position
vector is simply:
OP  aR R.
Its divergence in spherical coordinates (R, θ, Ø)
can be obtained from equation on page: 123:
1  1  1 A
 A  2 ( R AR ) 
2
( A sin  ) 
R R R sin   R sin  
Substituting the value of OP in above equation w e get :
  OP  3
Vector Analysis 125
 Divergence of a Vector Field
• Example: The magnetic flux density B outside a
very long current-carrying wire is circumferential
and is inversely proportional to the distance to
the axis of the wire. Find div B.
• Solution: Let the long wire be coincident with the
z-axis in a cylindrical coordinate system. The
problem states that:
k
B  a .
r

Vector Analysis 126

 Divergence of a Vector Field
• The divergence of a vector field in cylindrical
coordinates (r, Ø, z) can be found from equation
on page: 123.
1  1 B Bz
B  (rBr )   .
r r r  z
B  k r , and Br  Bz  0, Hence above equation gives :
  B  0.
• Above vector is not a constant but its divergence is
zero. Hence magnetic flux lines close upon
themselves and there are no sources or sinks. A
divergence less field is called a solenoidal field.

Vector Analysis 127

 Divergence Theorem
• The volume integral of the divergence of a
vector field equals the total outward flux of the
vector through the surface that bounds the
volume; that is,
   Adv  A  ds
V S

• This identity is called the divergence theorem,

also known as Gauss’s theorem.
• The direction of ds is always outward
perpendicular to the surface ds and directed
away from the volume.
Vector Analysis 128
 Divergence Theorem
• For a very small differential volume element Δvj
bounded by a surface sj, the definition of   A in
previous equation gives directly:
(  A) j v j   A  ds.
Sj
• Let an arbitrary volume V, subdivided into many
say N, small differential volumes of which Δvj is
typical as shown in figure.

Vector Analysis 129

 Divergence Theorem
• Combine the contribution of all these differential
volumes to both sides of previous equation:
N  N 
lim v j  0 (  A) j v j   lim v j  0  A  ds 
 j 1   j 1 
Sj

• Left side of above equation is by definition the

volume integral of   A :
N 
lim v j  0 (  A) j v j    (  A)dv
 j 1 
V

Vector Analysis 130

 Divergence Theorem
• The surface integrals on the right side of
equation on the top of page 130 are summed
over all the faces of all the differential volume
elements.
• The contributions from the internal surfaces of
adjacent elements will cancel each other,
because at a common internal surface the
outwards normals of the adjecent elements point
in opposite directions.
• Hence the net contribution is due to only that of
external surface S bounding the volume V.
Vector Analysis 131
 Divergence Theorem
N 
lim v j  0  A  ds    A  ds
 
Sj S
j 1
• The last three equations yield the divergence
theorem.
• Validity of the limiting processes leading to the
proof of the divergence theorem requires that
the vector field A, as well as its first derivatives,
exist and be continuous both in V and on S.
• The Divergence theorem converts a volume
integral of the divergence of a vector to a closed
surface integral of the vector, and vice versa.
Vector Analysis 132
 Divergence Theorem
• Example: Given A=axx2+ayxy+azyz, verify the
divergence theorem over a cube one unit on
each side. The cube is situated in the first octant
of the Cartesian coordinate system with one
corner at the origin.
• Refer to figure. We first
evaluate the surface integral
over the six faces.

Vector Analysis 133

 Divergence Theorem
1. Front face: x=1, ds=axdydz;
1 1
Frontface
A  ds  
0 0  dydz  1
2. Back face: x=0, ds=-axdydz;

 A  ds  0
Frontface

3. Left face: y=0, ds=-aydxdz;

 A  ds  0
Leftface

Vector Analysis 134

 Divergence Theorem
4. Right face: y=1, ds=aydxdz;
1 1 1
Rightface A  ds  0 0 xdxdz  2
5. Top face: z=1, ds=azdxdy;
1 1 1
Topface A  ds  0 0 ydxdy  2
6. Bottom face: z=0, ds=-azdxdy;
 A  ds  0
Bottomface

Vector Analysis 135

 Divergence Theorem
• Adding above six values:
1 1
S A  ds  1  0  0  2  2  0  2
• Now the divergence of A is:
 2  
  A  ( x )  ( xy )  ( yz )  3x  y
x y z
• Hence: 1 1 1
   Adv     (3x  y)dxdydz  2
V 0 0 0

• Results are same; so divergence theorem is

therefore verified.
Vector Analysis 136
 Divergence Theorem
• Example: Given F=aRkR, determine whether the
divergence theorem holds for the shell region
enclosed by spherical surfaces at R=R1 and
R=R2(R2>R1) centered at the origin, as shown in
figure:
• Solution: Here the
region has two
surfaces at R=R1 and
R=R2.

Vector Analysis 137

 Divergence Theorem
• At the outer surface: R=R2, ds=aRR22sinθdθdØ;
2 2

OuterSurface
F  ds  
0  0
( KR2 ) R22 sin dd  4kR23 .

• At the inner surface: R=R1, ds=-aRR12sinθdθdØ;

2 

InnerSurface
F  ds   
0 0
( KR1 ) R12 sin dd  4kR13 .
Adding the two results, we have :

    
3 3
F ds 4 k ( R2 R1)
S

Vector Analysis 138

 Divergence Theorem
• To find the volume integral, we first determine
•   F for an F that has only FR component:
1  1 
 F  2 ( R FR )  2
2
(kR3 )  3k
R R R R
• Since   F is a constant, its volume integral
equals the product of the constant and the
volume. The volume of the shell region between
the two spherical surfaces with radii R1 and R2
is 4 ( R23  R13 ) .
3
Vector Analysis 139
 Divergence Theorem
• Therefore:   Fdv  (  F )V  4k ( R 3  R 3 ),

V
2 1

• This is the same result as in surface integral.

• This example shows that the divergence
theorem holds even when the volume has holes
inside.

Vector Analysis 140

 Curl of a Vector Field
• There is a kind of source called Vortex Source,
which causes a circulation of a vector field
around it.
• The net circulation of a vector field around a
closed path is defined as the scalar line integral
of the vector over the path. We have:
Circulatio n of A around contour C   A  dl
C
• The physical meaning of circulation depends on
what kind of field the vector A represents.

Vector Analysis 141

 Curl of a Vector Field
• If A is a force acting on an object, its circulation
will be the work done by the force in moving the
object once around the contour.
• If A represents an Electric Field Intensity, then
the circulation will be an Electromotive Force
around the closed path.
• The familiar phenomenon of water whirling down
a sink drain is an example of a vortex sink
causing a circulation of fluid velocity.
• A circulation of A may exist even when div A=0.
Vector Analysis 142
 Curl of a Vector Field
• As circulation is a line integral of a dot product,
its value obviously depends on the orientation of
the contour C relative to the vector A.
• To define a point function, which is the measure
of the strength of a vortex source, we must make
C very small and orient it in such a way that the
circulation is a maximum. We define:
curl A    A  lim s  0
1
s C
 
an  A  dl .
max

Vector Analysis 143

 Curl of a Vector Field
• The curl of a vector field A, denoted by curl A
or   A , is a vector whose magnitude is the
maximum net 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 net circulation
maximum.
• Normal to an area can point in two
opposite directions, we stick to the
right hand rule that when fingers
follow the direction of dl, the thumb points to
the an direction Vector Analysis 144
 Curl of a Vector Field
• Curl A is a vector point function its component in
any other direction au is au  (  A) , which can be
determined from the circulation per unit area
normal to au as the area approaches zero.
1
(  A)u  au  (  A)  lim su  0 (  A  dl )
su Cu
• Here the direction of the line integration is
around the contour Cu bounding the area Δsu
and the direction au follow the right hand rule.

Vector Analysis 145

 Curl of a Vector Field
• Let us find the three components of   A in
Cartesian coordinates. Differential rectangular
area parallel to the yz-plane and having sides Δy
and Δz is drawn about a typical point P(xo, yo,
zo). We have au=ax and
Δsu = ΔyΔz, and the
contour Cu consist of the
four sides 1,2,3, and 4.
Thus:

Vector Analysis 146

 Curl of a Vector Field
1
(  A)  lim yz  0 (  A  dl ).
yz sides1, 2,3, 4
•In Cartesian coordinates A=axAx+ayAy+azAz. The
contribution of the four sides to the line integral are
as follows:
Side _ 1
y
dl  a z z , A  dl  Az ( x , y  , z )z ,
2
y
where Az ( x , y  , z ) can be expanded as a Taylor series :
2
y y Az
Az ( x , y  , z )  Az ( x , y , z )   H .O.T
2 2 y ( x , y , z )
Vector Analysis 147
 Curl of a Vector Field
• Where H.O.T (higher order terms) contain the
factors (Δy)2, (Δy)3, etc. Thus:

 y Az 

side_1 A  dl   z   

A ( x , y , z ) 
2 y ( x , y , z )
 H . O .T z.

 
Side _ 3
y
dl   a z z , A  dl  Az ( x , y  , z )z , where :
2
y y Az
Az ( x , y  , z )  Az ( x , y , z )   H .O.T
2 2 y ( x , y , z )

 y Az 

side_ 3 A  dl   z   

A ( x , y , z ) 
2 y
 H .O.T (  z ).

 ( x , y , z ) 
Vector Analysis 148
 Curl of a Vector Field
• Combining equations of side 1 and side 3 we
have:
Az
 A  dl (
sides1&3 y
 H .O.T ) yz.
( x , y , z )

H.O.T in above equation still contain powers of y.Similarly it can

be shown that :
Ay
 A  dl (
sides2& 4 z
 H .O.T ) yz.
( x , y , z )

Vector Analysis 149

 Curl of a Vector Field
Substituting above equations in the equation on the top of page 147
and noting that the H.O.T tend to zero as y  0, we obtain the x -
component of   A :
Az Ay
(  A) x  
y z
A close examinatio n of above equation will reveal a cyclic order in
x, y, and z and enable us to write y - and z - components of   A.
Az Ay Ax Az Ay Ax
  A  ax (  )  ay (  )  az (  ).
y z z x x y

Vector Analysis 150

 Curl of a Vector Field
•   A can be remembered easily by arranging it
in the determinantal form in the manner of the
cross product.
ax ay az
  
 A 
x y z
Ax Ay Az

Vector Analysis 151

 Curl of a Vector Field
• The expression for   A in general orthogonal
curvilinear coordinates (u1, u2, u3) is as below:
au1h1 au 2 h2 au 3h3
1   
 A 
h1h2 h3 u1 u2 u3
h1 A1 h2 A2 h3 A3
• The expression of   Acylindrical and spherical
coordinates can be easily obtained from above
equation by using the appropriate u1, u2, and u3
and their metric coefficients h1, h2, and h3.

Vector Analysis 152

 Curl of a Vector Field
• Example: Show that   A = 0 if
a) A = aØ(k/r) in cylindrical coordinates.
b) A= aRf(R) in spherical coordinates, where
f(R) is any function of the radial distance R.
• Solution:
a) In cylindrical coordinates the following apply:
(u1, u2, u3) = (r, Ø, z); h1 = 1, h2 = r, h3 = 1.
We have:

Vector Analysis 153

Curl of a Vector Field
ar a r az
1   
 A  ,
r r  z
Ar rA Az
which yields for the given A,
ar a r az
1   
 A   0.
r r  z
0 k 0
Vector Analysis 154
 Curl of a Vector Field
a) In spherical coordinates the following apply:
(u1, u2, u3) = (R, θ, Ø); h1 = 1, h2 = R, h3 = R
sinθ. Hence:
aR a R a R sin 
1   
 A  2 ,
R sin  R  
AR RA R sin A

Vector Analysis 155

 Curl of a Vector Field
• And, for the given A,
aR a R a R sin 
1   
 A  2 0
R sin  R  
f ( R) 0 0

• A curl-free vector field is called an Irrotational or

a Conservative field.

Vector Analysis 156

 Stokes’s Theorem
curl A    A  lim s  0
1
s C

an  A  dl .
max

• For a very small differential area Δsj bounded by

a contour Cj, the definition of   A in above
equation leads to:

(  A) j  (s j )   A  dl
Cj
• For an arbitrary surface S, we can subdivide it
into many, say N, small differential areas. Figure
on next page shows such a scheme with Δsj as a
typical differential element
Vector Analysis 157
 Stokes’s Theorem
• Left side of above equation is the flux of the
vector   A through the area Δsj. Adding the
contributions of all differential areas to the flux,
we have:

N
lim s j  0 (  A) j  (s j )
j 1

  (  A)  ds
S

Vector Analysis 158

 Stokes’s Theorem
• Now we sum up the line integrals around the
contours of all the differential elements
represented by the right side of equation on
page 157.
• Since the common parts of the contours of two
adjacent elements is traversed in opposite
directions by two contours, the net contribution
of all the common parts in the interior to the total
line integral is zero, and only the contribution
from the external contour C bounding the entire
area S remains after the summation:
Vector Analysis 159
 Stokes’s Theorem
N
lim s j  0 (  A  dl )   A  dl.
Cj C
j 1

Combining previous two equations, we obtain Stokes' s theorem.

 (  A)  ds  A  dl
S C

• The Stokes’s theorem states that the surface

integral of the curl of a vector field over an open
surface is equal to the closed line integral of the
vector along the contour bounding the surface.

Vector Analysis 160

 Stokes’s Theorem
• As with the divergence theorem, the validity of the
limiting processes leading to Stokes’s theorem
requires that the vector field A, as well as its first
derivatives, exist and be continuous both on S and
along C.
• Stokes’s theorem converts a surface integral of the
curl of a vector to a line integral of the vector and
vice versa.
• Like the divergence theorem, Stokes’s theorem is
an important identity in vector analysis, and we use
it frequently in estabilishing other theorems and
relations in electromagnetics.
Vector Analysis 161
 Stokes’s Theorem
• If the surface integral of   A is carried over a
closed surface, there will be no surface
bounding external contour, and previous
equation tells us that:

 (  A)  ds  0, for any closed surface S.

S

• The geometry in figure on page 158 is chosen

deliberately to emphasize the fact that a
nontrivial application of Stokes’s theorem always
implies an open surface with a rim.
Vector Analysis 162
 Stokes’s Theorem
• Example: Given F=axxy-ay2x, verify Stokes’s
theorem over a quarter circular disk with a
radius 3 in the first quadrant as shown in figure.
• Solution: Let us first find the
surface integral of   F
ax ay az
  
 F   a z (2  x),
x y z
xy  2x 0
Vector Analysis 163
 Stokes’s Theorem
• Therefore
3 9 y 2
 (  F )  ds   
S 0 0
(  F )  (a z dxdy )
3 9 y 2 
    (2  x)dx dy
0
0 
3 1 
   2 9  y 2  (9  y 2 )dy
0
 2 
3
 y 9 y  3
   y 9  y 2  9 sin 1  y  
 3 2 6 0
 
 91  .
 2
Vector Analysis 164
 Stokes’s Theorem
• For the line integral around ABOA, we have
already evaluated the part around the arc from A
to B in example on page 86-89.
From B to O : x  0, and F  dl  F  (a y dy )  2 xdy  0.
From O to A : y  0, and F  dl  F  (a x dx)  xydx  0.
Hence as per example on page 86 - 89
B  
ABOA F  dl  A F  dl  91  2 ,
• Hence Stokes’s Theorem is verified.
Vector Analysis 165
 Two Null Identities
• Identity 1: 
  V  0 
“The curl of the gradient of any scalar field is
identically zero.”
• As per Stokes’s theorem:
   V  ds   V   dl
S C

However as per equation on the top of page 104

 V   dl   dV  0
C C

• The combination of above two equations states

that the surface integral of   V  over any
surface is zero.
Vector Analysis 166
 Two Null Identities
• A converse statement of Identity 1 can be made
as follows:
“If a vector field is curl-free, then it can be
expressed as the gradient of a scalar field.”
• Let a vector field be E. Then, if   E  0 , we can
define a scalar field V such that:
E  V
• The –ve sign is unimportant as far as Identity 1
is concerned, as it is used in a future concept.

Vector Analysis 167

 Two Null Identities

• Identity II     A  0 
“ The divergence of the curl of any vector field is
identically zero.”
• Taking volume integral of above equation on the
left side and applying divergence theorem:
V     Adv  S   A ds
• Let us choose the arbitrary volume V enclosed
by a surface S in figure on next page. The
closed surface S can be split into two open
surfaces S1 and S2 connected by a common
boundary that has been drawn twice as C1 and
C2.
Vector Analysis 168
 Two Null Identities
• We than apply Stokes’s
theorem to surface S1
bounded by C1 and
surface S2 bounded by
C2, and we write the
right side of above
equation as:
   A  ds     A  an1ds     A  an 2 ds
S S1 S2

  A  dl   A  dl
C1 C2

• The normals an1 and an2 to surfaces S1 and S2

are outward normals and their relation with the
Vector Analysis 169
 Two Null Identities
path directions of C1 and C2 follow the right hand
rule.
• As contours C1 and C2 are one and the same
common boundary between S1 and S2, the two
line integrals on the right side of above equation
traverse the same path in opposite direction.
Their sum is therefore zero, and the volume
integral of     A on the left side of equation
on slide 168 vanishes.
• As this is true for any arbitrary volume , the
integrand itself must be zero, as indicated by the
Identity II.
Vector Analysis 170
 Two Null Identities
• A converse statement of Identity II is:
• “If a vector field is divergence-less, then it can
be expressed as the curl of another vector field.”
• Let the vector field be   B  0, we can define a
vector field A such that:
B    A.
• A divergence-less field is also called a
solenoidal field. Solenoidal fields are not
associated with flow sources or sinks.
• The net outward flux of a solenoidal field through
any closed surface is zero, and the flux lines
close upon themselves.
Vector Analysis 171
 Helmholtz’s Theorem
• We may classify vector fields in accordance with
their being solenoidal and / or irrotational.
1. Solenoidal and irrotational if:
  F  0 and   F  0.
e.g: A static electric field in a charge free region.
2. Solenoidal but not irrotational if:
  F  0 and   F  0.
e.g: A steady magnetic field in a current carrying
conductor.

Vector Analysis 172

 Helmholtz’s Theorem
3. Irrotational but not solenoidal if:
  F  0 and   F  0.
e.g: A static electric field in a charged region.
4. Neither solenoidal nor irrotational if:
  F  0 and   F  0.
e.g: An electric field in a charged medium with a
time varying magnetic field.
• The most general vector field then has both a
nonzero divergence and a nonzero curl, and
can be considered as the sum of a solenoidal
field and an irrotational field.
Vector Analysis 173
 Helmholtz’s Theorem
• Helmholtz’s theorem states that:
“A vector field (vector point function) is determined
to within an additive constant if both its divergence
and its curl are specified everywhere.”
• In an unbounded region we assume that both the
divergence and the curl of the vector field vanish at
infinity.
• If a vector field is confined within a region bounded
by a surface, then it is determined if its divergence
and curl throughout the region, as well as the
normal component of the vector over the bounding
surface are given.
Vector Analysis 174
 Helmholtz’s Theorem
• Here we assume that the vector function is
single-valued and that its derivatives are finite
and continuous.
• We remind that the divergence of a vector is a
measure of the strength of the flow source and
that the curl of a vector is a measure of the
strength of the vortex source.
• When the strength of both the flow source and
vortex source are specified, we expect that the
vector field will be determined.
Vector Analysis 175
 Helmholtz’s Theorem
• We can decompose a general vector field F into
an irrotational part Fi and a solenoidal part Fs:
F  Fi  Fs ,
  Fi  0

  Fi  g
   Fs  0

  Fs  G
Where g and G are assumed to be known. We have :
  F    Fi  g
  F    Fs  G.
Vector Analysis 176
 Helmholtz’s Theorem
• Helmholtz’s theorem asserts that g and G are
specified, the vector function F is determined.
• The fact that Fi is irrotational enables us to
define a scalar (potential) function V, in view of
Identity-I discussed earlier.
Fi  V
• Similarly Identity-II and equations on previous
page allow the definition of a vector (potential)
function A such that:
Fs    A.
Vector Analysis 177
 Helmholtz’s Theorem
• Hence according to Helmholtz’s theorem that a
general vector function F can be written as the
sum of the gradient of a scalar function and the
curl of a vector function.

F  V    A.

Vector Analysis 178

 Helmholtz’s Theorem
• Example: Given a vector function:
F  ax 3 y  c1 z   a y c2 x  2 z   az c3 y  z .
a) Determine the constants c1, c2, and c3 if F is
irrotational.
b) Determine the scalar potential function V whose –ve
gradient equals F.

Vector Analysis 179

 Helmholtz’s Theorem
• Solution:
a) For F to be irrotational   F  0; that is:
ax ay az
  
 F 
x y z
3 y  c1 z c2 x  2 z  c3 y  z 
 a x (c3  2)  a y c1  a z (c2  3)  0.
Each component of   F must vanish. Hence :
c1  0, c 2  3, c3  2.

Vector Analysis 180

 Helmholtz’s Theorem
b) Since F is irrotational, it can be expressed as the
negative gradient of a scalar function V; that is:
V V V
F  V  a x  ay  az
x y z
 a x 3 y  a y 3 x  2 z   a z 2 y  z .
Three equations are obtained :
V
 3 y,
x
V
 3 x  2 z ,
y
V
 2y  z
z
Vector Analysis 181
 Helmholtz’s Theorem
• Integrating above three equations with respect
to x, y, and z respectively; we get:
V  3xy  f1  y, z ,
V  3xy  2 yz  f 2  x, z ,
z2
V  2 yz   f 3 ( x, y ).
2
• Examination of above three equations enable us
to write the scalar potential function
2
as:
z
V  3xy  2 yz 
2
• Addition of any constant would still make V an
answer.
Vector Analysis 182