VECTOR ANALYSIS
Vector Analysis 2Vector Analysis 2Vector Analysis 2
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 3Vector Analysis 3Vector Analysis 3
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 / AA=|A|
A=aAA
Vector Analysis 4Vector Analysis 4Vector Analysis 4
• Two vectors A and B can be added in two
ways. C=A+B
– Parallelogram rule
– Head to tail rule
Vector Addition and Subtraction
A
B
C
A
BC
Vector Analysis 5Vector Analysis 5Vector Analysis 5
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 6Vector Analysis 6Vector Analysis 6
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
A
B
-B A-B
Vector Analysis 7Vector Analysis 7Vector Analysis 7
• Multiplication of a vector A by a scalar k
changes only the magnitude of A by a
factor k.
kA = aA (kA)
• Scalar or Dot Product:
A . B = AB cos θAB
Product of Vectors
A
ABcosθAB
B
θAB
Vector Analysis 8Vector Analysis 8Vector Analysis 8
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 9Vector Analysis 9Vector Analysis 9
Product of Vectors
• Vector or Cross Product
A x B = an|AB sin θAB|
– Cross Product is not commutative
B x A = - A x B
– 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
A
B
BsinθABθABan
A x B
Vector Analysis 10Vector Analysis 10Vector Analysis 10
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 11Vector Analysis 11Vector Analysis 11
Product of Vectors
Magnitude is equal
to volume of the
parallelepiped formed
by vectors A, B, and C.
Base area is |B x C| = |BC sin θ1|
Height is |A cos θ2|
Hence the volume is |ABC sin θ1 cos θ2|
B
C
A
B x C
θ1
θ2
Vector Analysis 12Vector Analysis 12Vector Analysis 12
Product of Vectors
• Vector Triple Product
Back-cab rule
A x (B x C) = B(A . C) – C(A . B)
A = A|| + A┴
A┴ x (B x C) = 0 as both are parallel.
We are left with D = A|| x (B x C)
A||
A┴
CB
B(A|| . C)
-C(A|| . B)
DaD
θ1
θ2
Vector Analysis 13Vector Analysis 13Vector Analysis 13
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 14Vector Analysis 14Vector Analysis 14
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 15Vector Analysis 15Vector Analysis 15
Division of Vectors
• Division of Vectors is not defined
• Expressions such as k/A and B/A are
meaningless.
Vector Analysis 16Vector Analysis 16Vector Analysis 16
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 17Vector Analysis 17Vector Analysis 17
• Let au1, au2, and au3 be the unit vectors called theBase 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 impliesthe other two
Orthogonal Coordinate Systems
Vector Analysis 18Vector Analysis 18Vector Analysis 18
• 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+Au2
2+Au32)1/2.
Orthogonal Coordinate Systems
Vector Analysis 19Vector Analysis 19Vector Analysis 19
• 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).
Orthogonal Coordinate Systems
Vector Analysis 20Vector Analysis 20Vector Analysis 20
• 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
Orthogonal Coordinate Systems
Vector Analysis 21Vector Analysis 21Vector Analysis 21
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
Orthogonal Coordinate Systems
Vector Analysis 22Vector Analysis 22Vector Analysis 22
c) C . (A x B)
= Cu1(Au2Bu3 – Au3Bu2) + Cu2(Au3Bu1 – Au1Bu3)
+ Cu3(Au1Bu2 – Au2Bu1)
Cu1 Cu2 Cu3
= Au1 Au2 Au3
Bu1 Bu2 Bu3
Orthogonal Coordinate Systems
Vector Analysis 23Vector Analysis 23Vector Analysis 23
• 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.
Orthogonal Coordinate Systems
Vector Analysis 24Vector Analysis 24Vector Analysis 24
• 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
Orthogonal Coordinate Systems
Vector Analysis 25Vector Analysis 25Vector Analysis 25
• The differential volume formed by differential
coordinate changes du1, du2, and du3 in
directions au1, au2, and au3 respectively is (dl1dl2 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
Orthogonal Coordinate Systems
Vector Analysis 26Vector Analysis 26Vector Analysis 26
• 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
Orthogonal Coordinate Systems
Vector Analysis 27Vector Analysis 27Vector Analysis 27
• In general orthogonal curvilinearcoordinate system the differential area ds1
normal to the unit vector au1 is:
• ds1 = dl2 dl3• ds1 = h2h3du2du3
• Similarly differential areas normal tovectors au2 and au3 are respectively
• ds2 = h1h3du1du3
• ds3 = h1h2du1du2
Orthogonal Coordinate Systems
Vector Analysis 28Vector Analysis 28Vector Analysis 28
• Main orthogonal coordinate systems are:
• Cartesian (or Rectangular) Coordinates
• Cylindrical Coordinates
• Spherical Coordinates
Orthogonal Coordinate Systems
Vector Analysis 29Vector Analysis 29Vector Analysis 29
Cartesian Coordinates
• (u1, u2, u3) = (x, y, z)
• Point P(x1, y1, z1) is
Intersection of three
Planes x = x1, y = y1,
z = z1
Base vectors are ax,
ay, az in the
respective Directions.
z=z1 plane
y=y1 plane
X=x1 plane
Vector Analysis 30Vector Analysis 30Vector Analysis 30
• 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.
Cartesian Coordinates
Vector Analysis 31Vector Analysis 31Vector Analysis 31
• The dot product of two vectors A and B is:
A . B = AxBx + AyBy + AzBz
• The cross product of A and B is:
A x B =
ax(AyBz-AzBy) + ay(AzBx-AxBz) + az(AxBy-AyBx)
ax ay az
= Ax Ay Az
Bx By Bz
Cartesian Coordinates
Vector Analysis 32Vector Analysis 32Vector Analysis 32
• 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
Cartesian Coordinates
Vector Analysis 33Vector Analysis 33Vector Analysis 33
x
y
z
dx
dy
dz
dsx =dydz dsz = dxdy
Dsy = dxdz
o
Cartesian Coordinates
A differential volume in Cartesian Coordinates
Vector Analysis 34Vector Analysis 34Vector Analysis 34
Cartesian Coordinates
Vector Analysis 35Vector Analysis 35Vector Analysis 35
• 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 + By
2 + Bz2)1/2 = 1
Cartesian Coordinates
Vector Analysis 36Vector Analysis 36Vector Analysis 36
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
Cartesian Coordinates
Vector Analysis 37Vector Analysis 37Vector Analysis 37
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
Cartesian Coordinates
Vector Analysis 38Vector Analysis 38Vector Analysis 38
• 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:
Cartesian Coordinates
Vector Analysis 39Vector Analysis 39Vector Analysis 39
From Figure, we see
P1P2 = OP2 – OP1
= (ax3-ay2+az4) –
(ax+ay3+az2)
= ax2 – ay5 + az2
The length of the line is
P1P2 =|P1P2| = √22 + (-5)2 + 22
= √33
Cartesian Coordinates
P1(1,3,2)
P2(3,-2,4)
x y
z
Vector Analysis 40Vector Analysis 40Vector Analysis 40
• (u1, u2, u3) = (r, Ø, z)
• Point P(r1, Ø1, z1) is
the intersection of a
cylindrical surface
r=r1, a half plane
containing the z axis
and making an angle
Ø=Ø1 with the xz-
plane, and a plane
parallel to xy plane at
z=z1.
Cylindrical Coordinates
z
y
x
r1az
aØ
ar
z1
y1
x1
o
Ø1
Ø=Ø1 plane
z=z1 plane
r=r1 cylinder
Vector Analysis 41Vector Analysis 41Vector Analysis 41
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 42Vector Analysis 42
Cylindrical Coordinates
Vector Analysis 42
,0
,0
,0
,1
,1
,1
rz
z
r
zz
rr
aa
aa
aa
aa
aa
aa
,0
,0
,0
zz
rr
aa
aa
aa
Vector Analysis 43Vector Analysis 43Vector Analysis 43
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 44Vector Analysis 44Vector Analysis 44
• 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,
Cylindrical Coordinates
Vector Analysis 45Vector Analysis 45Vector Analysis 45
• dsØ = dr dz,
• dsz = r dr dØ,
• dv = r dr dØ dz.
Cylindrical Coordinates
Vector Analysis 46Vector Analysis 46Vector Analysis 46
• 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.
Cylindrical Coordinates
Vector Analysis 47Vector Analysis 47Vector Analysis 47
From figure:
ar . ax = cos Ø
aØ . ax = cos(π/2 + Ø)
= - sin Ø
Ax = Ar cos Ø – AØ sin Ø
Similarly Ay = A . ay
= arAr . ay + aØAØ . ay
ar . ay = cos(π/2 - Ø)
= sin Ø
aØ . ay = cos Ø
Ay = Ar sin Ø + AØ cos Ø
Cylindrical Coordinates
ar
aØ
Vector Analysis 48Vector Analysis 48
Cylindrical Coordinates
Vector Analysis 48
z
y
x
z
r
a
a
a
a
a
a
100
0cossin
0sincos
Vector Analysis 49Vector Analysis 49Vector Analysis 49
• Conversion Matrix is:
Ax cosØ -sinØ 0 Ar
Ay = sinØ cosØ 0 AØ
Az 0 0 1 Az
Cylindrical Coordinates
z
y
x
z
r
A
A
A
A
A
A
100
0cossin
0sincos
Vector Analysis 50Vector Analysis 50Vector Analysis 50
• 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
Cylindrical Coordinates
Vector Analysis 51Vector Analysis 51Vector Analysis 51
• EXAMPLE: The cylindrical coordinates ofan arbitrary point P in the z = 0 plane are(r, Ø, 0). Find the unit vector that goesfrom a point z = h on z-axis toward P.
• SOLUTION:
QP = OP – OQ
= (arr) – (azh)
aQP = QP/|QP|
= (1/√r2 + h2) (arr – azh)
Cylindrical Coordinates
Vector Analysis 52Vector Analysis 52Vector Analysis 52
• 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
Cylindrical Coordinates
Vector Analysis 53Vector Analysis 53Vector Analysis 53
• Cos Ø = x / √x2 + y2
• Sin Ø = y / √x2 + y2
• Therefore:
A = ax (3x2/(x2 + y2) + 2y)
+ ay (3xy/(x2 + y2) - 2x)
+ az 5
Cylindrical Coordinates
Vector Analysis 54Vector Analysis 54Vector Analysis 54
Spherical Coordinates
• (u1, u2, u3) = (R, θ, Ø)
• Point P(R1, θ1, Ø1) is the intersection of aspherical surface centered at the originwith a radius R=R1, a right circular conewith it’s apex at the origin, it’s axiscoincides with the + z-axis and having ahalf angle θ=θ1 and a half plane containingthe z axis and making an angle Ø=Ø1 withthe xz-plane.
Vector Analysis 55Vector Analysis 55Vector Analysis 55
Spherical Coordinates
Vector Analysis 56Vector Analysis 56Vector Analysis 56
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 57Vector Analysis 57
0
0
0
1
1
1
aa
aa
aa
aa
aa
aa
R
R
RR
Spherical Coordinates
Vector Analysis 58Vector Analysis 58Vector Analysis 58
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 forproblems involving point sources andregions with spherical boundaries.
• Spherical coordinates are used in solvingantenna problems in the far field.
Vector Analysis 59Vector Analysis 59Vector Analysis 59
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 60Vector Analysis 60Vector Analysis 60
Spherical Coordinates
Vector Analysis 61Vector Analysis 61Vector Analysis 61
• 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
Spherical Coordinates
Vector Analysis 62Vector Analysis 62Vector Analysis 62
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 63Vector Analysis 63Vector Analysis 63
• A vector in spherical coordinates can be
transformed into Cartesian coordinates as:
x = R sinθ cosØ,
y = R sinθ sinØ,
z = R cosθ,
Spherical Coordinates
Vector Analysis 64Vector Analysis 64Vector Analysis 64
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 65Vector Analysis 65
Spherical Coordinates
,cos
,sinsin
,cossin
zR
yR
xR
aa
aa
aa
,sin
,sincos
,coscos
z
y
x
aa
aa
aa
,0
,cos
,sin
z
y
x
aa
aa
aa
z
y
xR
a
a
a
a
a
a
0cossin
sinsincoscoscos
cossinsincossin
Vector Analysis 66Vector Analysis 66
Spherical Coordinates
• Transformation of Vector
.sincoscoscossin
,
,
AAA
aaAaaAaaAaAA
aAaAaAA
R
xxxRRxx
RR
Vector Analysis 67Vector Analysis 67
Spherical Coordinates
A
A
A
A
A
A R
z
y
x
0sincos
cossincossinsin
sincoscoscossin
z
y
xR
A
A
A
A
A
A
0cossin
sinsincoscoscos
cossinsincossin
Vector Analysis 68Vector Analysis 68Vector Analysis 68
• 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°.
Spherical Coordinates
Vector Analysis 69Vector Analysis 69Vector Analysis 69
• 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.
Spherical Coordinates
Vector Analysis 70Vector Analysis 70Vector Analysis 70
• 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).
Spherical Coordinates
Vector Analysis 71Vector Analysis 71Vector Analysis 71
• 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.
Spherical Coordinates
Vector Analysis 72Vector Analysis 72Vector Analysis 72
• 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
Spherical Coordinates
Vector Analysis 73Vector Analysis 73Vector Analysis 73
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:
Spherical Coordinates
Vector Analysis 74Vector Analysis 74Vector Analysis 74
• 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)
Spherical Coordinates
Vector Analysis 75Vector Analysis 75Vector Analysis 75
• 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))
Spherical Coordinates
Vector Analysis 76Vector Analysis 76Vector Analysis 76
• 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
Spherical Coordinates
Vector Analysis 77Vector Analysis 77Vector Analysis 77
• Q = -3x10-80∫
2π 0∫
π0.02∫
0.05 (1/R2) cos2Ø
sinθ dR dθ
dØ
• = -3x10-80∫
2π 0∫
π(-1/0.05 + 1/0.02) cos2Ø
sinθdθ dØ
• = -0.9x10-60∫
2π (-cosθ)0|π cos2ØdØ
• = -1.8x10-6 (Ø/2 + (sin2Ø)/4)0|2π
• = -1.8π (µC).
Spherical Coordinates
Vector Analysis 78Vector Analysis 78
Integrals Containing Vector
Functions
• 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.
VFdv
Vector Analysis 79Vector Analysis 79
Integrals Containing Vector
Functions
CVdl
1
2
P
PVdl
•This integral is a scalar function of space.
•dl shows the differential increment of length.
•C is the path of integration.
•If the integral is from P1 to P2, we write
•If it is for a closed path C, we write CVdl
Vector Analysis 80Vector Analysis 80
Integrals Containing Vector
Functions
• In Cartesian coordinates Integral can be written
as:
Cz
CCyx
C
Czyx
C
dzzyxVadyzyxVadxzyxVaVdl
dzadyadxazyxVVdl
),,(),,(),,(
)[,,(
•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 81Vector Analysis 81
Integrals Containing Vector
Functions
• Example: Evaluate the integral , where
___________, from the origin to the point P(1,1):
• a) Along the direct path OP.
• b) Along the path OP1P, and
• c) Along the path OP2P.
P
Odrr 2
222 yxr
Vector Analysis 82Vector Analysis 82
Integrals Containing Vector
Functions
• Solution:
• a) Along the direct path OP
3
2
3
2
)45sin45cos(3
22
3
222
0
22
yx
yx
P
Orr
aa
aa
adrradrr
Vector Analysis 83Vector Analysis 83
Integrals Containing Vector
Functions• Solution:
• b) Along the path OP1P
.3
1
3
4
)3
1(
3
1
)1()(
1
0
31
0
3
2222
1
1
yx
xy
P
Px
P
Oy
P
O
aa
xxaya
dxxadyyadryx
Vector Analysis 84Vector Analysis 84
• Along the path OP2P
.3
4
3
1
)3
1(
3
1
)1()(
1
0
31
0
3
2222
2
2
yx
yx
P
Py
P
Ox
P
O
aa
yyaxa
dyyadxxadryx
Integrals Containing Vector
Functions
Vector Analysis 85Vector Analysis 85
Integrals Containing Vector
Functions
C
dlF
•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 in moving a unit charge from
P1 to P2.
Vector Analysis 86Vector Analysis 86
Integrals Containing Vector
Functions
• Example: Given , evaluate the scalar line integral , along the quarter circle shown in figure.
xaxyaF yx 2
B
AdlF
Vector Analysis 87Vector Analysis 87
Integrals Containing Vector
Functions
• Solution:
• a) In Cartesian coordinates:
)2
1(9
3sin99)9(
3
1
929
)3,0(9
2
3
0
120
3
23
2
3
0
20
3
2
22
yyyx
dyydxxxdlF
yxyx
xdyxydxdlF
B
A
Vector Analysis 88Vector Analysis 88
Integrals Containing Vector
Functions
• Solution:
• b) In cylindrical coordinates:
)cos2sin()sin2cos(
0
2
100
0cossin
0sincos
100
0cossin
0sincos
xxyaxxyaF
x
xy
F
F
F
A
A
A
A
A
A
r
z
r
z
y
x
z
r
Vector Analysis 89Vector Analysis 89
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 thepath (dr=0 and dz=0); hence equation dl = ardr +aØrdØ + azdz simplifies to:
).2
1(9
)cossin(sin9
)cos6cossin9(3
)cos2sin(3
3
2
0
3
2
0
22
B
AddlF
dxxydlF
dadl
Vector Analysis 90Vector Analysis 90
Integrals Containing Vector
Functions
• 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:
s
dsA
Vector Analysis 91Vector Analysis 91
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.
s
ns
dsaAdsA .
Vector Analysis 92Vector Analysis 92
• If S is an open surface, the +ve direction of andepends on the direction in which the perimeterof the open surface is traversed.
• Acc to right hand rule if the fingers follows thedirection of travel around the perimeter then thethumb points in the direction of +ve an.
• Again the +ve direction of an depends on thelocation of ds.
Integrals Containing Vector
Functions
Vector Analysis 93Vector Analysis 93
Integrals Containing Vector
Functions
• Example: Given , evaluate the
scalar surface integral over the surface
of a closed cylinder about the z-axis specified
by z=±3 and r=2.
zkarkaF zr 21
s
dsF
•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 94Vector Analysis 94
Integrals Containing Vector
Functions
sidewalln
topface bottomfacenn
sn
dsaFdsaFdsaF
dsaFdsF
...
•Where an is a unit vector normal outwards from
the respective surfaces.
•Three integrals on the right side can be
evaluated separately.
Vector Analysis 95Vector Analysis 95
Integrals Containing Vector
Functions
• a) Top face z = 3, an = az,
TopFacen
n
krdrdkdsaF
rdrdds
kzkaF
2
0
2
022
22
123
;
,3
Vector Analysis 96Vector Analysis 96
Integrals Containing Vector
Functions
• b) Bottom Face: z = -3, an = -az,
BottomFacen
n
krdrdkdsaF
rdrdds
kzkaF
2
0
2
022
22
123
;
,3
Vector Analysis 97Vector Analysis 97
Integrals Containing Vector
Functions
• C) Side Wall r = 2, an = ar,
SideWalln
n
kdzdkdsaF
dzddzrdds
k
r
kaF
.12
;2
,2
3
3
2
011
11
Vector Analysis 98Vector Analysis 98
Integrals Containing Vector
Functions
• Therefore
)2(12
121212
21
122
kk
kkkdsFs
•This surface integral gives the net outward flux
of the vector F through the closed cylindrical
surface.
Vector Analysis 99Vector Analysis 99
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 100Vector Analysis 100
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
Vector Analysis 101Vector Analysis 101
Gradient of a Scalar Field
Vector Analysis 101
•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 102Vector Analysis 102
• “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.”
Vector Analysis 102
Gradient of a Scalar Field
dn
dVaV
dn
dVagradV
n
n
gradV of placein V writeand
symbol by the drepresente del,operator employ tocustomary isIt
.
Vector Analysis 103Vector Analysis 103
• 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
Vector Analysis 103
Gradient of a Scalar Field
lln aVaadn
dV
dn
dV
dl
dn
dn
dV
dl
dV
)(
cos
•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 104Vector Analysis 104
• We can also write
Vector Analysis 104
Gradient of a Scalar Field
,
:scoordinatein changes aldifferenti theof terms
in expressed becan it hence );P toP (fromposition in change a
ofresult a as V of aldifferenti total theis dV Now , Where
,)(
3
3
2
2
1
1
31
dll
Vdl
l
Vdl
l
VdV
dladl
dlVdV
l
•Where dl1, dl2, and dl3 are the components of the
vector differential displacement dl in a chosen
coordinate system.
Vector Analysis 105Vector Analysis 105
• In terms of general orthogonal coordinates (u1,
u2, u3), dl is:
Vector Analysis 105
Gradient of a Scalar Field
dll
Va
l
Va
l
Va
dladladlal
Va
l
Va
l
VadV
duhaduhaduha
dladladladl
uuu
uuuuuu
uuu
uuu
)(
)()(
:follows as vectors
twoofproduct dot as written becan dV
)()()(
3
3
2
2
1
1
332211
3
3
2
2
1
1
333222111
332211
Vector Analysis 106Vector Analysis 106
Gradient of a Scalar Field
• Comparing above equation with the equation on top of slide 104.
33
3
22
2
11
1
3
3
2
2
1
1
uh
Va
uh
Va
uh
VaV
l
Va
l
Va
l
VaV
uuu
uuu
• Above equation is useful for computing gradient of
a scalar, when the scalar is given as a function of
space coordinates.
Vector Analysis 107Vector Analysis 107
Gradient of a Scalar Field
• In Cartesian coordinates, (u1, u2, u3) = (x, y, z)
and h1 = h2 = h3 = 1, hence we have:
za
ya
xa
Vz
ay
ax
aV
z
Va
y
Va
x
VaV
zyx
zyx
zyx
operator. aldifferenti vector a as
scoordinateCartesian in consider toconvenient isIt
)(
Vector Analysis 108Vector Analysis 108
Gradient of a Scalar Field
• We see that we can define in general
orthogonal coordinates as:
)(33
3
22
2
11
1uh
auh
auh
a uuu
Vector Analysis 109Vector Analysis 109
Gradient of a Scalar Field
• Example: The Electrostatic field intensity E isderivable as the –ve gradient of a scalar electricpotential V; that is, E= - V. Determine E at thepoint (1, 1, 0) if
cos )
,4
sin )
REVb
yeVVa x
Vector Analysis 110Vector Analysis 110
Gradient of a Scalar Field
• We use Cartesian Coordinates for part (a) and spherical coordinates for part (b) to solve E= - V.
• a)
).4
()16(1
1
,)16
1(2
1
,2
)4
()0,1,1(
.)4
cos44
sin(
4sin][
2
2
yxE
Eyx
x
yx
x
zyx
aaa
EEwhere
EaE
aaEThus
eEy
ay
a
yeE
za
ya
xaE
Vector Analysis 111Vector Analysis 111
Gradient of a Scalar Field
b)
.)sincos(
cos]sin
[
Eaa
RER
aR
aR
aE
R
R
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 112Vector Analysis 112
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:
EazEz
aVE zz
)(
Vector Analysis 113Vector Analysis 113
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 114
Divergence of a Vector Field
Vector Analysis 114
• 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 115
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
Vector Analysis 116
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:
Vector Analysis 116
v
dsAvdivA s
0lim
• The numerator represents the net outward flux,
is an integral over the entire surface S that
bounds the volume
Vector Analysis 117
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
Vector Analysis 118
Divergence of a Vector Field
Vector Analysis 118
zyzyx
xA
zyaASAdsA
dsAdsA
x
xfrontfacefrontfacefrontface
frontface
s bottomfacetopfaceleftfacerightfacebackfacefrontface
),,2
(
)(
facefront On the
Vector Analysis 119
Divergence of a Vector Field
• The quantity can be expanded as a
Taylor series about its value at (xo, yo, zo), as
follows:
Vector Analysis 119
),,2
( zyx
xAx
,__2
),,(),,2
(),,(
termsorderhigherx
AxzyxAzy
xxA
zyx
xxx
Where the higher order terms (H.O.T) contain
the factors (Δx/2)2,(Δx/2)3, etc.
Vector Analysis 120
Divergence of a Vector Field
Vector Analysis 120
.)..(][
:get wefaceback andfront for equations Combining
..2
),,(),,2
(
:is ),,2
( ofexpansion series-Taylor The
),,2
(
)(
),,(
),,(
zyxTOHx
AdsA
TOHx
AxzyxAzy
xxA
zyx
xA
zyzyx
xA
zyaASAdsA
zyx
x
BackfaceFrontface
zyx
xxx
x
x
xbackfacebackfacebackface
backface
Vector Analysis 121
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:
Vector Analysis 121.z)(z, factors thecontains H.O.T theHere
.)..(][
:have wefaces bottom
and topFor the ,y)(y, factors thecontains H.O.T theHere
.)..(][
2
),,(
2
),,(
zyxTOHz
AdsA
zyxTOHy
AdsA
zyx
z
BottomfaceTopface
zyx
y
Leftfacerightface
Vector Analysis 122
Divergence of a Vector Field
• Now combining the results of all the sides:
Vector Analysis 122
.,,)(),,(
zyxintermsorderhigherzyxz
A
y
A
x
AdsA
zyx
zy
s
x
• Since Δv=ΔxΔyΔz substituting above equation
in div A equation in Cartesian coordinates we
get:
z
A
y
A
x
AdivA zyx
•The higher order terms vanish as the differential
volume ΔxΔyΔz approaches zero.
Vector Analysis 123
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.
Vector Analysis 123
.)()()(1
:have we),,,( scoordinate orthogonal generalIn
321
3
231
2
132
1321
321
Ahhu
Ahhu
Ahhuhhh
A
uuu
divAA
Vector Analysis 124
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:
Vector Analysis 124
.3)(
.
z
z
y
y
x
xOP
zayaxaOP zyx
Vector Analysis 125
Divergence of a Vector Field
b) Spherical coordinates: Here the position
vector is simply:
Its divergence in spherical coordinates (R, θ, Ø)
can be obtained from equation on page: 123:
Vector Analysis 125
.RaOP R
3
:get eequation w abovein OP of value thengSubstituti
sin
1)sin(
sin
1)(
1 2
2
OP
A
RA
RAR
RRA R
Vector Analysis 126
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:
Vector Analysis 126
.r
kaB
Vector Analysis 127
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.
Vector Analysis 127
.0
:givesequation above Hence ,0 and ,
.1
)(1
B
BBrkB
z
BB
rrB
rrB
zr
zr
• 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 128
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,
• 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
SV
dsAAdv
Vector Analysis 129
Divergence Theorem
• For a very small differential volume element Δvj
bounded by a surface sj, the definition of in
previous equation gives directly:
• 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
A
Sj
jj dsAvA .)(
Vector Analysis 130
Divergence Theorem
• Combine the contribution of all these differential
volumes to both sides of previous equation:
• Left side of above equation is by definition the
volume integral of :
Vector Analysis 130
N
jSj
j
N
j
jjj dsAvvAv11
0lim)(0lim
A
dvAvAvV
N
j
jjj )()(0lim1
Vector Analysis 131
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
Vector Analysis 132
Divergence Theorem
• 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
S
N
jSj
j dsAdsAv1
0lim
Vector Analysis 133
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.
Vector Analysis 133
• Refer to figure. We first
evaluate the surface integral
over the six faces.
Vector Analysis 134
Divergence Theorem
1. Front face: x=1, ds=axdydz;
2. Back face: x=0, ds=-axdydz;
3. Left face: y=0, ds=-aydxdz;
Vector Analysis 134
Frontface
dydzdsA1
0
1
01
Frontface
dsA 0
Leftface
dsA 0
Vector Analysis 135
Divergence Theorem
4. Right face: y=1, ds=aydxdz;
5. Top face: z=1, ds=azdxdy;
6. Bottom face: z=0, ds=-azdxdy;
Vector Analysis 135
Rightface
xdxdzdsA1
0
1
0 2
1
Topface
ydxdydsA1
0
1
0 2
1
Bottomface
dsA 0
Vector Analysis 136
Divergence Theorem• Adding above six values:
• Now the divergence of A is:
• Hence:
• Results are same; so divergence theorem is
therefore verified.Vector Analysis 136
202
1
2
1001 S dsA
yxyzz
xyy
xx
A
3)()()( 2
V
dxdydzyxAdv1
0
1
0
1
02)3(
Vector Analysis 137
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:
Vector Analysis 137
• Solution: Here the
region has two
surfaces at R=R1 and
R=R2.
Vector Analysis 138
Divergence Theorem
• At the outer surface: R=R2, ds=aRR22sinθdθdØ;
• At the inner surface: R=R1, ds=-aRR12sinθdθdØ;
.4sin)( 3
2
2
2
2
0
2
02 kRddRKRdsF
ceOuterSurfa
)(4
:have weresults, two theAdding
.4sin)(
3
1
3
2
3
1
2
1
2
0 01
RRkdsF
kRddRKRdsF
S
ceInnerSurfa
138Vector Analysis
Vector Analysis 139
Divergence Theorem
• To find the volume integral, we first determine
• for an F that has only FR component:
• Since 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
F
kkRRR
FRRR
F R 3)(1
)(1 3
2
2
2
F
.3
)(4 3
1
3
2 RR
139Vector Analysis
Vector Analysis 140
Divergence Theorem
• Therefore:
• This is the same result as in surface integral.
• This example shows that the divergence
theorem holds even when the volume has holes
inside.
),(4)( 3
1
3
2 RRkVFdvFV
140Vector Analysis
Vector Analysis 141
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:
• The physical meaning of circulation depends on
what kind of field the vector A represents.
Vector Analysis 141
C
dlACcontour aroundA ofn Circulatio
Vector Analysis 142
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
Vector Analysis 143
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:
Vector Analysis 143
.1
0lim curlmax
Cn dlAa
ssAA
Vector Analysis 144
Curl of a Vector Field• The curl of a vector field A, denoted by curl A
or , 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
A
Vector Analysis 145
Curl of a Vector Field
• Curl A is a vector point function its component in
any other direction au is , which can be
determined from the circulation per unit area
normal to au as the area approaches zero.
• 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
)( Aau
)(1
0lim)()(
uC
u
uuu dlAs
sAaA
Vector Analysis 146
Curl of a Vector Field
• Let us find the three components of 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
A
Vector Analysis 147
Curl of a Vector Field
Vector Analysis 147
).(1
0lim)(4,3,2,1
sidesdlA
zyzyA
•In Cartesian coordinates A=axAx+ayAy+azAz. The
contribution of the four sides to the line integral are
as follows:
TOHy
AyzyxAz
yyxA
zy
yxA
zzy
yxAdlAzadl
Side
zyx
zzz
z
zz
..2
),,(),2
,(
:seriesTaylor a as expanded becan ),2
,( where
,),2
,(,
1_
),,(
Vector Analysis 148
Curl of a Vector Field
• Where H.O.T (higher order terms) contain the
factors (Δy)2, (Δy)3, etc. Thus:
Vector Analysis 148
).(..2
),,(
..2
),,(),2
,(
:,),2
,(,
3_
...2
),,(
3_),,(
),,(
1_),,(
zTOHy
AyzyxAdlA
TOHy
AyzyxAz
yyxA
wherezzy
yxAdlAzadl
Side
zTOHy
AyzyxAdlA
sidezyx
zz
zyx
zzz
zz
sidezyx
zz
Vector Analysis 149
Curl of a Vector Field
• Combining equations of side 1 and side 3 we
have:
Vector Analysis 149
.)..(
:shown that be
canit Similarly y. of powerscontain stillequation abovein H.O.T
.)..(
),,(4&2
),,(3&1
zyTOHz
AdlA
zyTOHy
AdlA
zyx
y
sides
zyx
z
sides
Vector Analysis 150
Curl of a Vector Field
Vector Analysis 150
).()()(
A. of components-z and -y write tous enable and z and y, x,
inorder cyclic a reveal illequation w above ofn examinatio closeA
)(
:A ofcomponent
- xobtain the we0,y as zero to tendH.O.T that thenoting and
147 page of topon theequation in the equations above ngSubstituti
y
A
x
Aa
x
A
z
Aa
z
A
y
AaA
z
A
y
AA
xy
zzx
y
yzx
yzx
Vector Analysis 151
Curl of a Vector Field
• can be remembered easily by arranging it
in the determinantal form in the manner of the
cross product.
A
zyx
zyx
AAA
zyx
aaa
A
Vector Analysis 152
Curl of a Vector Field• The expression for in general orthogonal
curvilinear coordinates (u1, u2, u3) is as below:A
332211
321
332211
321
1
AhAhAh
uuu
hahaha
hhhA
uuu
• The expression of cylindrical and sphericalcoordinates can be easily obtained from aboveequation by using the appropriate u1, u2, and u3
and their metric coefficients h1, h2, and h3.
A
Vector Analysis 153
Curl of a Vector Field
• Example: Show that = 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:
A
Vector Analysis 154
Curl of a Vector Field
.0
00
1
A,given for the yieldswhich
,1
kzr
araa
rA
ArAAzr
araa
rA
zr
zr
zr
Vector Analysis 155
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:
,
sin
sin
sin
12
ARRAA
R
RaRaa
RA
R
R
Vector Analysis 156
Curl of a Vector Field
• And, for the given A,
• A curl-free vector field is called an Irrotational or
a Conservative field.
0
00)(
sin
sin
12
RfR
RaRaa
RA
R
Vector Analysis 157
Stokes’s Theorem
• For a very small differential area Δsj bounded bya contour Cj, the definition of in aboveequation leads to:
• For an arbitrary surface S, we can subdivide itinto many, say N, small differential areas. Figureon next page shows such a scheme with Δsj as atypical differential element
.1
0lim curlmax
Cn dlAa
ssAA
A
jC
jj dlAsA )()(
Stokes’s Theorem
• Left side of above equation is the flux of the
vector through the area Δsj. Adding the
contributions of all differential areas to the flux,
we have:
Vector Analysis 158
A
S
N
j
jjj
dsA
sAs
)(
)()(0lim1
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
• 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
CS
N
jC C
j
dlAdsA
dlAdlAsj
)(
theorem.sStokes'obtain weequations, twoprevious Combining
.)(0lim1
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 is carried over a
closed surface, there will be no surface
bounding external contour, and previous
equation tells us that:
• 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
A
S
dsA S. surface closedany for ,0)(
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
Vector Analysis 163
F
),2(
02
xa
xxy
zyx
aaa
F z
zyx
Stokes’s Theorem
• Therefore
Vector Analysis 164
.2
19
62
9
3sin99
)9(2
192
)2(
)()()(
3
0
312
3
0
22
3
0
9
0
3
0
9
0
2
2
yy
yyy
dyyy
dydxx
dxdyaFdsF
y
S
y
z
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.
• Hence Stokes’s Theorem is verified.
Vector Analysis 165
,2
19
89-86 pageon exampleper as
.0)( and 0,y :A toO From
.02)( and 0, x:O toB From
ABOA
B
A
x
y
dlFdlF
Hence
xydxdxaFdlF
xdydyaFdlF
Two Null Identities• Identity 1:
“The curl of the gradient of any scalar field is
identically zero.”
• As per Stokes’s theorem:
• The combination of above two equations states
that the surface integral of over any
surface is zero.Vector Analysis 166
0 V
0
104 page of topon theequation per asHowever
CC
CS
dVdlV
dlVdsV
V
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 , we can
define a scalar field V such that:
• The –ve sign is unimportant as far as Identity 1
is concerned, as it is used in a future concept.
Vector Analysis 167
0 E
VE
Two Null Identities• Identity II
“ 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:
• 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
0 A
SV
dsAdvA
Two Null Identities
Vector Analysis 169
• 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:
21
2121
CC
nS
nSS
dlAdlA
dsaAdsaAdsA
• The normals an1 and an2 to surfaces S1 and S2
are outward normals and their relation with the
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 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
A
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 we can define a
vector field A such that:
• 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
,0 B
.AB
Helmholtz’s Theorem
• We may classify vector fields in accordance with
their being solenoidal and / or irrotational.
1. Solenoidal and irrotational if:
e.g: A static electric field in a charge free region.
2. Solenoidal but not irrotational if:
e.g: A steady magnetic field in a current carrying
conductor.
Vector Analysis 172
.0 and 0 FF
.0 and 0 FF
Helmholtz’s Theorem3. Irrotational but not solenoidal if:
e.g: A static electric field in a charged region.
4. Neither solenoidal nor irrotational if:
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
.0 and 0 FF
.0 and 0 FF
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:
Vector Analysis 176
.
:have Weknown. be toassumed areG and g Where
0
0
,
GFF
gFF
GF
F
gF
F
FFF
s
i
s
s
i
i
si
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.
• Similarly Identity-II and equations on previous
page allow the definition of a vector (potential)
function A such that:
Vector Analysis 177
VFi
.AFs
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.
Vector Analysis 178
.AVF
Helmholtz’s Theorem
• Example: Given a vector function:
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
.23 321 zycazxcazcyaF zyx
Helmholtz’s Theorem
• Solution:
a) For F to be irrotational that is:
Vector Analysis 180
;0 F
.2c ,3c ,0c
:Hence h.must vanis F ofcomponent Each
.0)3()2(
23
321
213
321
cacaca
zyczxczcyzyx
aaa
F
zyx
zyx
Helmholtz’s Theoremb) Since F is irrotational, it can be expressed as the
negative gradient of a scalar function V; that is:
Vector Analysis 181
zyz
V
zxy
V
yx
V
zyazxaya
z
Va
y
Va
x
VaVF
zyx
zyx
2
,23
,3
:obtained are equations Three
.2233
Helmholtz’s Theorem• Integrating above three equations with respect
to x, y, and z respectively; we get:
• Examination of above three equations enable us
to write the scalar potential function as:
• Addition of any constant would still make V an
answer.
Vector Analysis 182
).,(2
2
,,23
,,3
3
2
2
1
yxfz
yzV
zxfyzxyV
zyfxyV
223
2zyzxyV