A Crash Course in Vector Calculus
SCALAR AND VECTOR FIELDS
This introductory chapter is a review of mathematical concepts required for the
course. It is assumed that the reader is already familiar with elementary vector
analysis.
Physical quantities that we deal with in electromagnetism can be scalars or vectors.
A scalar is an entity which only has a magnitude. Examples of scalars are mass,
time, distance, electric charge, electric potential, energy, temperature etc.
A vector is characterized by both magnitude and direction. Examples of vectors
in physics are displacement, velocity, acceleration, force, electric field, magnetic
field etc.
A field is a quantity which can be specified everywhere in space as a function of
position. The quantity that is specified may be a scalar or a vector. For instance,
we can specify the temperature at every point in a room. The room may, therefore, be said to be a region of “temperature field” which is a scalar field because
the temperature T (x, y, z) is a scalar function of the position. An example of a
scalar field in electromagnetism is the electric potential.
In a similar manner, a vector quantity which can be specified at every point in a
region of space is a vector field. For instance, every point on the earth may be
considered to be in the gravitational force field of the earth. we may specify the
field by the magnitude and the direction of acceleration due to gravity (i.e. force
per unit mass ) g(x, y, z) at every point in space. As another example consider
flow of water in a pipe. At each point in the pipe, the water molecule has a velocity ~v (x, y, z). The water in the pipe may be said to be in a velocity field. There
~
are several examples of vector field in electromagnetism, e.g., the electric field E,
~ etc.
the magnetic flux density B
1.1 Coordinate Systems :
We are familiar with cartesian coordinate system. For systems exhibiting cylindrical or spherical symmetry, it is convenient to use respectively the cylindrical
and spherical coordinate systems.
1.1.1 Polar Coordinates :
In two dimensions one defines the polar coordinate (ρ, θ) of a point by defining
1
ρ as the radial distance from the origin O and θ as the angle made by the radial
vector with a reference line (usually chosen to coincide with the x-axis of the
cartesian system). The radial unit vector ρ̂ and the tangential (or angular) unit
vector θ̂ are taken respectively along the direction of increasing distance ρ and
that of increasing angle θ respectively, as shown in the figure.
y
^θ
θ
θ
ρ^
ρ
y
θ
x
x
Relationship with the cartesian components are
x = ρ cos θ
y = ρ sin θ
so that the inverse relationships are
q
x2 + y 2
y
θ = tan−1
x
ρ =
By definition, the distance ρ > 0. we will take the range of angles θ to be 0 ≤ θ <
2π (It is possible to define the range to be −π ≤ θ < +π). One has to be careful
in using the inverse tangent as the arc-tan function is defined in 0 ≤ θ < π. If y
is negative, one has to add π to the principal value of θ calculated by the arc - tan
function so that the point is in proper quadrant.
2
Example 1 :
~ has cartesian components Ax and Ay . Write the vector in terms of its
A vector A
radial and tangential components.
Solution :
Let us write
~ = Aρ ρ̂ + Aθ θ̂
A
Since ρ̂ and θ̂ are orthonormal basis vectors ρ̂ · ρ̂ = θ̂ · θ̂ = 1 and ρ̂ · θ̂ = 0. Thus
~ · ρ̂, Aθ = A
~ · θ̂
Ar = A
Note that (see figure) ρ̂ makes an angle θ with the x-axis (ı̂) and π/2 − θ with the
y-axis (̂). Similarly, the unit vector θ̂ makes π/2 + θ with the x-axis and θ with
the y-axis. Thus
~ · ρ̂ =
Aρ = A
=
=
~ · θ̂ =
Aθ = A
Ax ı̂ · ρ̂ + Ay ̂ · ρ̂
Ax cos θ + Ay cos(π/2 − θ)
Ax cos θ + Ay sin θ
Ax ı̂θ̂ + Ay  · θ̂
= Ax cos(π/2 + θ) + Ay cos θ
= −Ax sin θ + Ay cos θ
The Jacobian :
When we transform from one coordinate system to another, the differential element also transform.
For instance, in 2 dimension the element of an area is dxdy but in polar
coordinates the element is not dθdρ
but (ρdθ)dρ. This extra factor ρ is
important when we wish to integrate
a function using a different coordinate system.
ρ dθ
dρ
dθ
ρ
θ
If f (x, y) is a function of x, y we may express the function in polar
coordinates
R
and write it as g(ρ, θ). However, when we evaluate the integral f (x, y)dxdy
R
in polar coordinates, the corresponding integral is g(r, θ)ρdθdρ. In general, if
x = f (u, v) and y = g(u, v), then, in going from (x, y) to (u, v), the differential
3
element dxdy →| J | dudv where J is given by the determinant
J=
∂x
∂u
∂y
∂u
∂x
∂v
∂y
∂v
The differentiations are partial, i.e., while differentiating ∂x/∂u = ∂f (u, v)/∂u,
the variable v is treated as constant. An useful fact is that the Jacobian of the
inverse transformation is 1/J because the detrminant of the inverse of a matrix is
equal to the inverse of the determinant of the original matrix.
Example 2 : Show that the Jacobian of the transformation from cartesian to polar
coordinates is ρ.
Solution :
We have
∂x
∂ρ
∂y
∂ρ
J =
Using x = ρ cos θ and y = ρ sin θ, we have
∂x
∂θ
∂y
∂θ
cos θ −ρ sin θ
J=
sin θ ρ cos θ
=ρ
Exercise :
Show that√the Jacobian of the inverse transformation from polar to cartesian is
1/ρ = 1/ x2 + y 2 .
Example 3 :
find the area of a circle of radius R.
Solution :
Take the origin to be at the centre of the circle and the plane of the circle to be the
ρ − θ plane. Since the area element in the polar coordinates is ρdθdr, the area of
the circle is
" #R
Z 2π
Z R
ρ2
= πR2
dθ
ρdρ = 2π
2 0
0
0
- a very well known result !
Example 4 :
Z
2
2
Find the integral e−(x +y ) dxdy where the region of integration is a unit circle
about the origin.
2
Using polar coordinates the integrand becomes e−ρ . The range of i integration
for ρ is from 0 to 1 and for θ is from 0 to 2π. The integral is given by
I=
Z
0
2π
dθ
Z
0
1
2
e−ρ ρdρ = 2π
4
Z
0
1
2
e−ρ ρdρ
The radial integral is evaluated by substitution w = ρ2 so that ρdρ = dw/2. The
value of the integral is
I = 2π
Z
1
0
1
e−w dw/2 = π[−e−w ]10 = π(1 − )
e
Exercise R: R
Evaluate
xydxdy where the region of integration is the part of the area between
circles of radii 1 and 2 that lies in the first quadrant.
(Ans. 15/8)
Exercise :
R
2
Evaluate the Gaussian integral I = 0∞ e−x dx
[ Hint : The integration cannot be done using cartesian coordinates but is relatively
easy using polar coordinates and propertiesR of definite integrals. By changing the
2
dummy variable x to y, one can write I = 0∞ e−y dy, so that we can write
I2 =
Z
0
∞
Z
∞
0
e−(x
2 +y 2 )
dxdy
Transform this to polar. Range of√integration for ρ is from 0 to ∞ and that of θ is
from 0 to π/2(why ?) Answer : π/2 ]
Differentiation of polar unit vectors with respect to time :
It may be noted that the basis vectors ρ̂ and θ̂, unlike ı̂ and ̂ are not constant
vectors but depend on the position of the point. The time derivative of the unit
vectors are defined as follows
dρ̂
∆ρ̂
= lim
=
dt ∆t→0 ∆t
dθ̂
∆θ̂
= lim
=
dt ∆t→0 ∆t
ρ̂(t + ∆t) − ρ̂(t)
∆t→0
∆t
θ̂(t + ∆t) − θ̂(t)
lim
∆t→0
∆t
lim
One can evaluate the derivatives by laborious process of expressing the unit vectors ρ̂ and θ̂ in terms of constant unit vectors of cartesian system, differentiating
the resulting expressions and finally transform back to the polar form. Alternatively, we can look at the problem geometrically, as shown in the following figure.
5
ρ^(t+dt)
^ (t+dt)
θ
dθ
^ (t)
θ
dθ
ρ^ (t)
dθ
θ
In the figure, the positions of a particle are shown at time t and t + dt. The unit
vectors ρ̂ is shown in red while the unit vector θ̂ is shown in blue. It can be easily
seen by triangle law of addition of vectors that the magnitude of ∆ρ̂ and ∆θ̂ is
1.dθ = dθ. However, as the limit dt → 0, the direction of ∆ρ̂ is in the direction
of θ̂ while that of ∆θ̂ is in the direction of −ρ̂. Thus
dρ̂
dθ
∆ρ̂
θ̂
= lim
=
dt ∆t→0 ∆t
dt
dθ̂
dθ
∆θ̂
= lim
= − ρ̂
dt ∆t→0 ∆t
dt
Now, dθ/dt is the angular velocity of the point, which is usually denoted by ω,
Thus we have,
dρ̂
= ω θ̂
dt
dθ̂
= −ω ρ̂
dt
Exercise :
Obtain the above relationships directly by using the expression for the unit vectors
ρ̂ and θ̂ in terms of the cartesian unit vectors, viz.,
ρ̂ = ı̂ cos θ + ̂ sin θ
θ̂ = −ı̂ sin θ + ̂ cos θ
6
1.1.2 Cylindrical coordinates :
Cylnidrical coordinate system is obtained by extending the polar coordinates by
adding a z-axis along the height of a right circular cylinder. The z-axis of the
coordinate system is same as that in a cartesian system.
In the figure ρ is the distance of the
foot of the perpendicular drawn from
the point to the x − y(ρ, θ) plane.
Note that ρ here is not the distance
z
of the point P from the origin, as is
the case in polar coordinate systems.
^z
(Some texts use r to denote what we
^
are calling as ρ here. However, we
θ
use ρ to denote the distance from the
P
θ
origin to the foot of the perpendicu^ρ
lar to avoid confusion.) In terms of
cartesian coordinates
z
x = ρ cos θ
y = ρ sin θ
z = z
O
x θ
y
y
ρ
P’
so that the inverse relationships are
x
q
x2 + y 2
y
θ = tan−1
x
z = z
Exercise :
Find the cylindrical coordinate of the point 3ı̂ + 4̂ + k̂. [Hint : Determine ρ and
tan θ using above transformation]
(Ans. 5ρ̂ + tan−1 (4/3)θ̂ + k̂)
The line element in the system is given by
ρ =
~ = ρ̂dρ + ρdθθ̂ + dẑ
dl
and the volume element is
dV = ρdθdρdz
The Jacobian of transformation from cartesian to cylindrical is ρ as in the polar
coordinates since z coordinate remains the same.
7
1.2.3 Spherical Polar Coordinates :
Spherical coordinates are useful in dealing with problems which possess spherical symmetry. The independent variables of the system are (r, θ, φ). Here r is the
distance of the point P from the origin. Angles θ and φ are similar to latitudes
and longitudes.
Two mutually perpendicular lines are chosen, taken to coincide with the x-axis
and z-axis of the cartesian system. We take angle θ to be the angle made by the
radius vector (i.e. the vector connecting the origin to P) with the z-axis (the angle
θ is actually complementary to the latitude). The angle φ is the angle between the
x-axis and the line joining the origin to P ′ , the foot of the perpendicular from P
to the x-y plane.
The unit vectors r̂, θ̂ and φ̂ are respectively along the directions of increasing r, θ
and φ.
z
^r
r sin θ
P
^φ
r cos
θ
O
r
z
θ
^
θ
y
φ
y
P’
x
The surface of constant r are spheres of radius r about the centre. the surface of
constant θ is a cone of semi-angle θ about the z-axis. The reference for measuring φ is the x-z plane of the cartesian system. A surface of constant φ is a plane
containing the z-axis which makes an angle φ with the reference plane.
Example 5 :
8
Express unit vectors of spherical coordinate system in terms of unit vectors of
cartesian system.
Solution :
~ ′ by ρ~. The
From the point P drop a perpendicular on to the x-y plane. Denote OP
figure below shows the unit vectors in both the systems. By triangle law of vector
addition,
~ = OP
~ ′ + P~′ P = r sin θρ̂ + r cos θk̂
~r = OP
However, ρ̂ = cos φı̂ + sin φ̂. Substituting this in the expression for ~r, we get on
dividing both sides by the magnitude of r
r̂ = sin θ cos φı̂ + sin θ sin φ̂ + r cos θk̂
z
The unit vector θ̂ is perpendicular to
OP in the direction of increasing θ.
The angle that θ̂ makes with the positive z-axis is π/2 + θ. Hence,
P θ
^k
θ
z
^j
O
Substituting for ρ, we get
^i
θ̂ = cos θ cos φı̂+cos θ sin φ̂−sin θk̂
^
θ
r
π
π
θ̂ = cos( +θ)k̂+sin( +θ)ρ̂ = − sin θk̂+cos θρ̂
2
2
^r
φ
y
φ
φ^
ρ
P’
x
~ is in the direction of increasing angle φ. It is perpenThe azimuthal unit vector φ
dicular to ρ~ and has no z- component. It can be easily seen that
φ̂ = − sin φı̂ + cos φ̂
Transformation from spherical to cartesian :
Using the expression for ~r in terms of cartesian basis, it is seen that
x = r sin θ cos φ
y = r sin θ sin φ
z = r cos θ
and the inverse transformation
r =
q
x2 + y 2 + z 2
9
−1
θ = tan
√
y
x
Range of the variables are as follows :
x2 + y 2
z
= cos−1
z
r
φ = tan−1
0 ≤ r < ∞ 0 ≤ θ ≤ π 0 ≤ φ ≤ 2π
Exercise :
A particle moves along a spherical helix. its position coordinate at time t is given
by
cos t
sin t
t
√
√
x=√
,
y
=
,
z
=
1 + t2
1 + t2
1 + t2
Express the equation
(Ans.
√ of the path in spherical coordinates.
r = 1, cos θ = t/ 1 + t2 φ(t) = t)
z
The differential element of volume
is obtained by constructing a closed
volume by extending r, θ and φ respectively by dr, dθ and dφ. The
length elements in the direction of
r̂ is dr, that along θ̂ rdθ while that
along φ̂ is r sin θdφ (see figure). The
volume element, therefore, is
dφ
r sin θ
r
θ
r sin θ d φ
rdθ
dr
dθ
y
dV = dr·rdθ·r sin θdφ = r 2 sin θdθdφdr
φ
dφ
Thus the Jacobian of transformation
is r 2 sin θ.
x
Example 6:
Find the volume of a solid region in the first octant that is bounded from above by
the sphere x2 + y 2 + z 2 = 9 and from below by the cone x2 + y 2 = 3z 2 .
Solution :
Because of obvious spherical symmetry, the problem is best solved in spherical
polar coordinates. The equation to sphere is r = 3 so that the range of r vaiable
for our solid is from 0 to 3.
10
The equation to the cone x2 + y 2 =
3z 2 becomes r 2 sin2 θ = 3r 2 cos2 θ.
Solving, √the semi-angle of cone is
tan θ = 3 i.e. θ = π/3. Since the
solid is restricted to the first octant,
i.e., (x, y, z ≥ 0), the range of the
azimuthal angle φ is from 0 to π/2.
The volume of the solid is
Z
dV =
Z
r 2 sin θdθdφdr =
Z
π/2
0
dφ
Z
0
π/3
sin θdθ
r3
π
π/3
. [− cos θ]0 .
=
2
3
9π
=
4
"
Z
0
3
r 2 dr
#3
0
Exercise :
Using direct integration find the volume of the first octant bounded by a sphere
x2 + y 2 + z 2 = 9.
1.2 Line and Surface Integrals of a Vector Field :
Since a vector field is defined at every position in a region of space, like a scalar
function it can be integrated and differentiated. However, as a vector field has
both magnitude and direction it is necessary to define operations of calculus to
take care of both these aspects.
1.2.1 Line Integral :
If a vector field F~ is known in a certain region of space, one can define a line
integral of the vector function may be defined as
11
Z
C
~
F~ · dl
where C is the curve along which the
integral is calculated. Like the integral of a scalar function the integral
above is also interpreted as a limit
of a sum. We first divide the curve
C into a large number of infinitisimally small line segments such that
dl θ
F
the vector function is constant (in
magnitude and direction) over each
such line segment. The integrand
is then equal to the product of the
length of the line segment and the
component of F~ along the segment.
The integration is defined as the limit
of the sum of contributions from all
such segments in the same manner as
ordinary integration is defined.
The concept of line integral is very useful in many branches of physics. In mechanics, we define work done by a force F~ in moving an object from an initial
position A to a final position B as the line integral of the force along the curve
joining the end points. Except in the case of conservative forces, the line integral
depends on the actual path along which the particle moves under the force.
Example 7:
A force F~ = zyı̂ + x̂ + z 2 xk̂ acts on a particle that travels from the origin to
the point (1, 2, 3). Calculate the work done if the particle travels (i) along the
path (0, 0, 0) → (1, 0, 0) → (1, 2, 0) → (1, 2, 3) along straightline segments joining each pair of end points (ii) along the straightline joining the initial and final
points.
Solution :
12
Along the path (0, 0, 0) → (1, 0, 0),
~ = ı̂dx and y = 0, z = 0. Since
dl
Fx = 0 along this segment, the integral along C1 is zero. Along the
path C2 joining (1, 0, 0) and (1, 2, 0),
~ = ̂dyand x = 1. Along this path
dl
Fy = x = 1. The integral
Z
2
0
Fy dy =
Z
0
2
(1,2,3)
C3
dy = 2
(1,2,0)
Along the third path connecting
(1, 2, 0) to (1, 2, 3), x = 1, y = 2.
~ = k̂dz and Fz = z 2 x = z 2 . The
dl
line integral is
Z
0
3
z 2 dz =
z3 3
| =9
3 0
C2
(0,0,0)
C1
(1,0,0)
The work done is, therefore, 0 + 2 +
9 = 11.
In order to calculate the work done when the particle moves along the straightline
connecting the initial and final points, we need to write down the equation to the
line in a parametric form. If (x1 , y1 , z1 ) and (x2 , y2, z2 ) are the end points, the
equation is
y − y1
z − z1
x − x1
=
=
=m
x2 − x1
y2 − y1
z2 − z1
Substituting for the coordinates, we get the equation to the line as
x =m
y = 2m
z = 3m
Thus the differential elements are dx = dm, dy = 2dm and dz = 3dm. The line
integral is given by
Z
~ =
F~ · dl
=
Z
Z
0
(Fx dx + Fy dy + Fz dz)
1
(6m2 .dm + m.2dm + 9m3 .3dm)
13
= 2m3 + m2 + 27
m4 1
| = 9.75
4 0
Example 8 :
Find the line integral of F~ = yı̂ + ̂ over an anticlockwise circular loop of radius
1 with the origin as the centre of the circle.
Solution :
y
The length element dl has a magnitude 1.dθ = dθ. Since the unit vector
~ makes an angle of (π/2) + θ
along dl
with the positive x− axis,
dl
θ
R=1
θ
~ = | dl | cos( π + θ)ı̂+ | dl | sin( π + θ)̂
dl
2
2
= − sin θı̂ + cos θ̂
In polar coordinates, x = cos θ and y = sin θ (since the radius is 1). Thus
Z
~ =
F~ · dl
Z
= −
Fx dx + Fy dy
Z
0
2π
2
sin θdθ +
= −π + 0 = −π
Z
0
2π
cos θdθ
Exercise :
A force F~ = xyı̂+ (x2 −z 2 )̂ −xz 2 k̂ acts on a particle. Calculate the work done if
the particle is taken from the point (0, 0, 0) to the point (2, 1, 3) along straight line
segment connecting (0, 0, 0) → (0, 1, 0) → (2, 1, 0) → (2, 1, 3). What would be
the work done if the particle directly moved to the final point along the straightline
connecting to origin.
(Ans. −16, −13.8.)
Exercise :
A vector field is given by
F~ = (2x + 3y)ı̂ + (3x + 2y)̂
14
x
Evaluate the line integral of the field around a circle of unit radius traversed in
clockwise fashion.
(Ans. 6π)
Exercise :
Evaluate the line integral of a scalar function xy along a parabolic path y = x2
connecting the origin to the point (1, 1).
q
[]Hint : Remember that the arc length along a curve is given by (dx)2 + (dy)2.
√
The curve can be parametrized by x = t and y = t2 .
Ans. (25 5 + 1)/120 ]
1.2.3 Surface Integral :
We have seen that an area element can be regarded as a vector with its direction
being defined as the outward normal to the surface. The concept of a surface
integral is related to flow. Suppose the vector field represents the rate at which
water flows at a point in the region of flow. The flow may be measured in cubic
meter of water flowing per square meter of area per second. If an area is oriented
perpendicular to the direction of flow, as shown in the figure to the left, maximum
amount of water would flow through the surface. The amount of water passing
through the area is the flux (measured in cubic meter per second).
V
V
^n
^n
θ
If, on the other hand, the surface is tilted relative to the flow, as shown to the right,
the amount of flux through the area decreases. Clearly, only the part of the area
that is perpendicular to the direction of flow will contribute to the flux.
~ as the dot product of the vector field
We define flux through an area element dS
~ When V~ is parallel to dS,
~ i.e. if the surface is oriented
~ with the area vector dS.
V
perpendicular to the direction of flow, the flux is maximum. On the other hand, a
surface oriented parallel to the flow does not contribute to the flux.
15
Example 9 :
A hemispherical bowl of radius R is oriented such that the circular base is perpendicular to direction of flow. Calculate the flux through the curved surface of the
~ to be constant.
bowl, assumuing the flow vector V
Solution :
Since the curved surface makes different angles at different positions, it is somewhat difficult to calculate the flux through it. However, one can circumvent it by
calculating the flux through the circular base.
V
^n
2
^n
1
^n
1
^n
2
Since the flow vector is constant all over the circular face which is oriented perpendicular to the direction of flow, the flux through the base is −πR2 V . The
minus sign is a result of the fact that the direction of the surface is opposite to
the direction of flow. Thus we may call the flux through the base as inward flux.
Since there is no source or sink of flow field (i.e. there is no accumulation of
water) inside the hemisphere, whatever fluid enters through the base must leave
through the curved face. Thus the outward flux from the curved face is +πR2 V .
We may now generalize the above for a surface over which the field is not uniform
by defining the flux through an area as the sum of contribution to the flux from
infinitisimal area elements which comprises the total area by treating the field to
be uniform over such area elements. Since the flux is a scalar, the surface integral,
defined as the limit of the sum, is also a scalar.
16
If ∆Si is the i−th surface element,
the normal n̂i to which makes an an~i at the
gle θi with the vector field V
position of the element, the total flux
Φ=
X
i
~i · n̂i ∆Si =
V
X
^n
i
θi
∆S i
Vi δSi cos θi
i
In the limit of ∆Si → 0, the sum
above becomes a surface integral
Z
Φ = V~ · dS
S
If the surface is closed, it encloses a volume and we define
Φ=
I
S
~ · dS
V
to be the net outward flux. In terms of cartesian components
Φ=
Z
S
(Vx dydz + Vy dxdz + Vz dxdy)
Example 10 :
~ = xyı̂ + yz̂ + zxk̂. Evaluate the flux through each
A vector field is given by B
face of a unit cube whose edges along the cartesian axes and one of the corners is
at the origin.
Solution :
Consider the base of the cube
(OGCD), which is the x-y plane on
~ = xyı̂.
which z = 0. On this face B
The surface vector n̂ is along −k̂ direction.
Thus on
this surface flux
R
R
~
B · n̂dxdy = xyı̂ · k̂dxdy = 0
since ı̂ · k̂ = 0. For the top surface
(ABFE), z = 1 and n̂ = k̂. The flux
from this surface is
z
E
A
F
B
O
G
y
1
D
C
2
0
0
x
In a similar way one can show that flux from left side (AEOD) is zero while the
contribution from the right side (BFGC) is 1/2. The back face (EFGO) contributes
Z
Vz dxdy =
Z
1
xdx
Z
1
dy =
17
zero while the front face (ABCD) contributes 1/2. The net flux, therefore, is 3/2.
Exercise :
z
~ =
Find the flux of the vector field V
2
Axı̂ + By ̂ through a rectangular
surface in the x-y plane having dimensions a × b. The origin of the coordinate system is at one of the corners of the rectangle and the x-axis
along its length. (Ans. Bab3 /3)
a
O
x
b
y
Example 11 :
Calculate directly the flux through the curved surface of the hemispherical bowl
of Example 10.
Solution :
Use a spherical polar coordinates with the
base of the hemisphere being the x-y plane
and the direction of the vector field as the zaxis. We have seen that an area element on x
the surface is givenR by dS = R2 sin θdθdφ.
~ · n̂ is given by
φ
Thus the flux Φ = V
Φ = | V | R2
Z
0
2
πdφ
Z
π/2
0
sin2 θ
= | V | R .2π
2
2
"
cos θ sin θdθ
#2π
^n
z−axis
O
θ
y
0
= π | V | R2
Exercise :
Find the flux through a hemispherical bowl with its base on the x-y plane and the
origin at the centre of the base. The vector field, in spherical polar coordinates is
~ = r sin θr̂ + θ̂ + φ̂.
V
(Ans. π(1 + π/2))
18
V
Example 12 :
A cylindrical object occupies a volume defined by x2 + y 2 ≤ R2 and 0 ≤ z ≤ h.
Find the flux through each of the surfaces when the object is in a vector field
~ = ı̂x + ̂y + k̂z.
V
Solution :
Because of cylindrical symmetry, it is convenient to work in a cylindrical (ρ, θ, z)
coordinates. The vector field is given by
~ = ρρ̂ + z k̂
V
z
^n
Bottom face has z = 0 and the normal to the face points in −k̂ direction. Thus the flux from this face
R
~ · n̂ = R Vz ρdθdρ = 0.
V
The top face has z = h and n̂ = k̂. h
The flux is
Z
Vz ρdθdr = h
Z
R
0
2
ρdρ
Z
0
2π
ρ
^n
dθ
y
R
= h .2π = πR2 h
2
θ
^n
x
The normal to the curved face is along ρ̂ direction. An area element on the curved
face is Rdθdz Thus the flux from this face is
Z
(Rρ̂ + z k̂) · ρ̂Rdθdz = R
2
Z
2π
0
dθ
Z
0
h
dz = 2πR2 h
The net flux from the faces of the object is 3πR2 h.
Exercise :
~ = 2ρ̂ − 3r θ̂ + zρk̂ through surfaces of a right
Find the flux of the vector field V
cylinder of radius 1 and height 2. The base of the cylinder is in the z = 0 plane
with the origin at the centre of the base.
(Ans. 28π/3)
19
1.3 Gradient of a Scalar Function :
Consider a scalar field such as temperature T (x, y, z) in some region of space. The
distribution of temperature may be represented by drawing isothermal surfaces or
contours connecting points of identical temperatures,
T (x, y, z) = constant
One can draw such contours for different temperatures. If we are located at a point
~r on one of these contours and move away along any direction other than along
the contour, the temperature would change.
The change ∆T in temperature as we
move away from a point P (x, y, z) to
a point Q(x + ∆x, y + ∆y, z + ∆z)
is given by
∆
∂T
∂T
∂T
∆T =
∆x +
∆y +
∆z
∂x
∂y
∂z
T
θ
P
Q
T( r + d r )
where the derivatives in the above
expression are partial derivatives.
If the displacement from the initial position is infinitisimal, we get
dT =
∂T
∂T
∂T
dx +
dy +
dz
∂x
∂y
∂z
Note that the change dT involves a change in temperature with respect to each of
the three directions. We define a vector called the gradient of T , denoted by ∇T
or grad T as
∂T
∂T
∂T
+ ̂
+ k̂
∇T = ı̂
∂x
∂y
∂z
using which, we get
dT = ∇T · (ı̂dx + ̂dy + k̂dz) = ∇T · d~r
20
T( r )
Note that ∇T , the gradient of a scalar T is itself a vector. If θ is the angle between
the direction of ∇T and d~r,
dT =| ∇T || d~r | cos θ = (∇T )r | d~r |
where (∇T )r is the component of the gradient in the direction of d~r. If d~r lies
on an isothermal surface then dT = 0. Thus, ∇T is perpendicular to the surfaces
of constant T . When d~r and ∇T are parallel, cos θ = 1 dT has maximum value.
Thus the magnitude of the gradient is equal to the maximum rate of change of T
and its direction is along the direction of greatest change.
The above discussion is true for any scalar field V . If a vector field can be written
as a gradient of some some scalar function, the latter is called the potential of the
vector field. This fact is of importance in defining a conservative field of force in
mechanics. Suppose we have a force field F~ which is expressible as a gradient
F~ = ∇V
The line integral of F~ can then be written as follows :
Z
i
f
~ =
F~ · dl
=
Z
f
i
Z
f
i
∇V · d~l
dV = Vf − Vi
where the symbols i and f represent the initial and thec final positions and in the
last step we have used an expression for dV similar to that derived for dT above.
Thus the line integral of the force field is independent of the path connecting the
initial and final points. If the initial and final points are the same, i.e., if the particle
is taken through a closed loop under the force field, we have
I
~ =0
F~ · dl
Since the scalar product of force with displacement is equal to the work done by
a force, the above is a statement of conservation of mechanical energy. Because
of this reason, forces for which one can define a potential function are called conservative forces.
Example 13 :
Find the gradient of the scalar function V = x2 y + y 2 z + z 2 x + 2xyz.
21
Solution :
∇V
∂V
∂V
∂V
+ ̂
+ k̂
∂x
∂y
∂z
= 2(xy + zx + yz)(ı̂ + ̂ + k̂)
= ı̂
Exercise : Find the gradients of
(i) xz − x2 y + y 2z 2
(ii)x3 +√y 3 + z 3
(iii) ln x2 + y 2 + z 2 (Ans. (xı̂ + y̂ + k̂)/(x2 + y 2 + z 2 )
Gradient can be expressed in other coordinate systems by finding the length elements in the direction of basis vectors. For example, in cylindrical coordinates the
length elements are dρ, ρdθ and dz along ρ̂, θ̂ and k̂ respectively. The expression
for gradient is
1 ∂V
∂V
∂V
+ θ̂
+ k̂
∇V = ρ̂
∂ρ
ρ ∂θ
∂z
The following facts may be noted regarding the gradient
1. The gradient of a scalar function is a vector
2. ∇(U + V ) = ∇U + ∇V
3. ∇(UV ) = U∇(V ) + V ∇(U)
4. ∇(V n ) = nV n−1 ∇V
Example 14 :
2
2
Find the gradient of V = e−(x +y ) in cylindrical (polar) coordinates.
Solution :
2
In polar variables the function becomes V = e−ρ . Thus
2
∇V
∂e−ρ
= ρ̂
∂ρ
−ρ2
= ρ̂e .(−2)ρ = −2ρ̂ρV
Exercise :
Find the gradient of the function V of Example 15 in cartesian coordinates and
then transform into polar form to verify the answer.
22
Exercise :
√
Find the gradient of the function ln ρ2 + z 2 in cylindrical coordinates.(Ans.(ρ̂ρ+
k̂z)/(ρ2 + z 2 ))
In spherical coordinates the length elements are dr, rdθ and r sin θdφ. Hence the
gradient of a scalar function U is given by
∇V = r̂
∂V
1 ∂V
1 ∂V
+ θ̂
+ k̂
∂r
r ∂θ
r sin θ ∂φ
Exercise :
Find the gradient of V = r 2 cos θ cos φ (Ans. 2r cos θ cos φr̂ − r sin θ cos φθ̂ −
r cos θ sin φφ̂.)
Exercise :
√
A potential function is given in cylidrical coordinates as k/ ρ2 + z 2 Find the
force field it represents and express the field in spherical polar coordinates. (Ans.
−k~r/r 3 )
1.4 Divergence of a Vector Field :
Divergence of a vector field F~ is a measure of net outward flux from a closed
surface S enclosing a volume V , as the volume shrinks to zero.
~ · F~ = lim
divF~ ≡ ∇
∆V →0
H
S
~
F~ · dS
∆V
where ∆V is the volume (enclosed by the closed surface S) in which the point P
at which the divergence is being calculated is located. Since the volume shrinks
to zero, the divergence is a point relationship and is a scalar.
Consider a closed volume V bounded by S. The volume may be mentally broken
into a large number of elemental volumes closely packed together. It is easy to
see that the flux out of the boundary S is equal to the sum of fluxes out of the
surfaces of the constituent volumes. This is because surfaces of boundaries of two
adjacent volumes have their outward normals pointing opposite to each other. The
following figure illustrates it.
23
=
+
We can generalize the above to closely packed volumes and conclude that the flux
out of the bounding surface S of a volume V is equal to the sum of fluxes out of
the elemental cubes. If ∆V is the volume of an elemental cube with ∆S as the
surface, then,
1
~ ∆V
F~ · dS
∆V
→0
∆V
S
∆S
The quantity in the bracket of the above expression was defined as the divergence
of F~ , giving
Z
~ =
F~ · dS
XZ
F~ · dS = lim
Z
Z
S
~ =
F~ · dS
V
Z
divF~ dV
This is known as the Divergence Theorem.
We now calculate the divergence of F~ from an infinitisimal volume over which
variation of F~ is small so that one can retain only the first order term in a Taylor
expansion. Let the dimensions of the volume element be ∆x × ∆y × ∆z and let
the element be oriented parallel to the axes.
Consider the contribution to the flux from the two shaded faces. On these faces,
the normal is along the +̂ and −̂ directions so that the contribution to the flux is
from the y-component of F~ only and is given by
24
z
[Fy (y + dy) − Fy (y)] dxdz
Expanding Fy (y +dy) in a Taylor series and retaining only the fiirst order
term
^n
^n
∂Fy
F( y + dy) = Fy +
dy
∂y
(x,y,z) dz
so that the flux from these two faces
is
dx
∂Fy
∂Fy
dxdydz =
dV
∂y
∂y
(x+dx,y,z)
dy
where dV = dxdydz is the volume x
of the cuboid.
Combining the above with contributions from the two remaining pairs of faces,
the total flux is
!
∂Fx ∂Fy ∂Fz
dV
+
+
∂x
∂y
∂z
Thus
Z
S
~ =
F~ · dS
∂Fx ∂Fy ∂Fz
+
+
∂x
∂y
∂z
Z
!
dV
Comparing with the statement of the divergence theorem, we have
divF~ =
∂Fx ∂Fy ∂Fz
+
+
∂x
∂y
∂z
!
Recalling that the operator ∇ is given by
∇ = ı̂
y
∂Fy
∂Fz
∂Fx
+ ̂
+ k̂
∂x
∂y
∂z
and using F~ = ı̂Fx + ̂Fy + k̂Fz , we can write divF~ = ∇ · F~ .
The following facts may be noted :
1. The divergence of a vector field is a scalar
~ = ∇ · F~ + ∇ · G
~
2. ∇ · (F~ + G)
25
3. ∇ · (φF~ ) = φ∇ · F~ + ∇φ · F~
4. In cylindrical coordinates
1 ∂
∂
1 ∂
(ρFρ ) +
Fθ + Fz
∇ · F~ =
ρ ∂ρ
ρ ∂θ
∂z
5. In spherical polar coordinates
1 ∂
1 ∂
1 ∂
∇ · F~ = 2 (r 2 Fr ) +
(Fθ sin θ) +
Fφ
r ∂r
r sin θ ∂θ
r sin θ ∂φ
6. The divergence theorem is
I
S
~ =
F~ · dS
Z
V
∇ · F~ dV
Example 15 : Divergence of ~r = ı̂x + ̂y + k̂z
Divergence of position vector ~r is very useful to remember.
∇ · ~r =
∂x ∂y ∂z
+
+
=1+1+1=3
∂x ∂y ∂z
One can also calculate easily in spherical coordinate since ~r only has radial component
1 ∂
1
∇ · ~r = 2 (r 2 .r) = 2 .3r 2 = 3
r ∂r
r
Exercise :
Calculate the divergence of the vector field ~r/r 3 using all the three coordinate systems.
(Ans. 0)
Example 16 :
A vector field is given by F~ = 4xzı̂ − y 2 ̂ + yz k̂. Find the surface integral of the
field from the surfaces of a unit cube bounded by planes x = 0, x = 1, y = 0, y =
1, z = 0 and z = 1. Verify that the result agrees with the divergence theorem.
Solution :
Divergence of F~ is
∂Fx ∂Fy ∂Fz
+
+
∂x
∂y
∂z
∂4xz ∂(−y 2 ) ∂yz
=
+
+
∂x
∂y
∂z
= 4z − 2y + y = 4z − y
∇ · F~ =
26
The volume integral of above is
Z
∇ · F~ dV =
Z
0
1
dx
Z
0
1
dy
Z
0
Consider the surface integral from
the six faces individually. For the
face AEOD, the normal is along −̂.
On this face y = 0 so that F~ = 4xzı̂.
Since ı̂ · ̂ = 0, the integrand is zero.
For the surface BFGC, the normal
^
is along ̂ and on this face y = 1. ^n = −j
On this face the vector field is F~ =
4xzı̂ − ̂ + z k̂. The surface integral
is
Z
~ =
F~ · dS
Z
= −
F~ · ̂dxdz
Z
1
0
dx
Z
0
1
1
(4z − y)dxdydz =
z
E (0,0,1)
F
B
A
^n = ^j
G (0,1,0)
O
D
(1,0,0)
dz = −1
3
2
C
x
Consider
the top face (ABFE) for which the normal is k̂ so that the surface integral
R
is Fz dydx. On this face z = 1 and Fz = y. The contribution to the surface
integral from this face is
Z 1
Z 1
1
ydy =
dx
2
0
0
For the bottom face (DOGC) the normal is along −k̂ and z = 0. This gives Fz = 0
so that the integral vanishes.
R
For the face EFGO the normal is along −ı̂ so that the surface integral is − Fx dydz.
On this face x = 0 giving Fx = 0. The surface integral is zero. For the front face
ABCD, the normal is along ı̂ and on this face x = 1 giving Fx = 4z. The surface
integral is
Z
Z
1
0
dy
1
0
4zdz = 2
Adding the six contributions above, the surface integral is 3/2 consistent with the
divergence theorem.
Exercise :
Verify the divergence theorem by calculating the surface integral of the vector
field F~ = ı̂x3 + ̂y 3 + k̂z 3 for the cubical volume of Example 17. (Ans. Surface
27
y
integral has value 3)
Example 17 :
In Example 13 we found that the surface integral of a vector field xı̂ + y̂ + z k̂
over a cylinder of radius R and height h is 3πR2 h. Verify this result using the
divergence theorem.
Solution :
In Example 16 we have seen that the divergence of the field vector is 3. Since the
integrand is constant, the volume integral is 3V = 3πR2 h.
Example 18 :
A vector field is given by F~ = x3 ı̂+y 3̂. Verify Divergence theorem for a cylinder
of radius 2 and height 5. The origin of the coordinate system is at the centre of the
base of the cylinderand z-axis along the axis.
Solution :
The problem is obviously simple in cylindrical coordinates. The divergence can
be easily seen to be 3(x2 + y 2) = 3ρ2 . Recalling that the volume element is
ρdρdθdz, the integral is
Z
div F~ dV =
Z
3ρ2 dV = 3
Z
2
0
ρ3 dρ
Z
2π
0
dθ
Z
5
0
dz = 120π
In order to calculate the surface integral, we first observe that the end faces have
their normals along ±k̂. Since the field does not have any z- component, the
contribution to surface integral from the end faces is zero.
We will calculate the contribution to the surface integral from the curved surface.
Using the coordinate transformation to cylindrical
x = ρ cos θ y = ρ sin θ
and
ı̂ = ρ̂ cos θ − θ̂ sin θ
̂ = ρ̂ sin θ + θ̂ cos θ
k̂ = k̂
Using these
F~ = ρ3 (cos4 θ + sin4 θ)ρ̂ + ρ3 (sin3 θ cos θ − cos3 θ sin θ)θ̂
The area element on the curved surface is Rdθdz ρ̂, where R is the radius. Thus
the surface integral is
Z
Fρ dS = R4
Z
2π
0
(cos4 θ + sin4 θ)dθ
28
Z
0
5
dz
= 16.
3π
.5 = 120π
2
where we have used 02π sin4 θdθ = 02π cos4 θdθ = 3π/4.
Exercise :
In the Exercise following Example 13, we had seen that surface integral of the
~ = 2ρ̂ − 3ρθ̂ + zρk̂ through the surface of a cylinder of radius 1 and
vector field V
height 2 is 28π/3. Re-confirm the same result using divergence theorem.
Example 19 :
A hemispherical bowl of radius 1 lies with its base on the x-y plane and the origin
at the centre of the circular base. Calculate the surface integral of the vector field
F~ = x3 ı̂ + y 3̂ + z 3 k̂ in the hemisphere and verify the divergence theorem.
Solution :
The divergence of F~ is easily calculated
R
R
∂Fx ∂Fy ∂Fz
+
+
∂x
∂y
∂z
3
3
∂x
∂y
∂z 3
=
+
+
∂x
∂y
∂z
2
2
2
= 3(x + y + z ) = 3r 2
∇ · F~ =
where r is the distance from origin. The volume integral over the hemisphere is
conveniently calculated in spherical polar using the volume element r 2 sin θdθdφ.
Since it is a hemisphere with z = 0 as the base, the range of θ is 0 to π/2.
∇ · F~ dV
= 3
Z
= 3×
2
r dV = 3
Z
0
1
4
r dr
6π
1
× 1 × 2π =
5
5
Z
0
π/2
sin θdθ
Z
0
2π
dφ
The surface integrals are calculated conveniently in spherical polar. There is no
contribution to the flux from the base because the outward normal points in the
−z direction but the z-component of the field is zero because the base of the hemisphere is z = 0.
In order to calculate the flux from the curved face we need to express the force
field and the unit vectors in spherical polar coordinates. Using the tranformation
properties given earlier and observing that we only require the radial component
of the vector field since the area element is radially directed. Using R2 sin θdθdφ
as the area element, a bit of laborious algebra gives
29
~ = R5
F~ · dS
+ R5
π/2
Z
0
π/2
Z
0
5
sin θdθ
Z
2π
0
4
cos φdφ + R
cos4 θ sin θdθ
R π/2
Using 0 sin5 θdθ = 8/15 and
seen to give the correct result.
R 2π
0
2π
Z
0
5
Z
0
π/2
5
sin θdθ
Z
0
2π
sin4 φdφ
dφ
cos4 φdφ = 3π/4, the above integral can be
1.5 Curl of a Vector - Stoke’s Theorem
R
~ is essentially a sum
We have seen that the line integral of a vector field F~ · dl
of the component of F~ along
the curve. If the line integral is taken over a closed
H
~ If the vector field is consevative, i.e., if there
path, we represent it as F~ · dl.
exists a scalar function V such that one can write F~ as ∇V , the contour integral
is zero. In other cases, it is, in general, non-zero.
Consider a contour C enclosing a surface S. We may split the contour into a large
number of elementary surface areas defined by a mesh of closed contours.
Since adjacent contours are traversed
in opposite directions, the only nonvanishing contribution to the integral
comes from the boundary of the contour C. If the surface area enclosed
by the i − th cell is ∆Si , then
I
=
C
XZ
i
=
X
i
Ci
R
~
F~ · dl
~
F~ · dl
∆Si
∆Si
Ci
We define the quantity
lim
∆Si →0
R
~
F~ · dl
n̂i
∆Si
Ci
as the curl of the vector F~ at a point P which lies on the surface ∆Si . Since the
area ∆Si is infinitisimal it is a point relationship. The direction of n̂i is, as usual,
30
along the outward normal to the area element ∆Si . For instance, the x-component
of the curl is given by
(curl)x =
lim
∆y,∆z→0
R
~
F~ · dl
n̂i
∆y∆z
Ci
Thus
I
C
~ =
F~ · dl
Z
S
curlF~ · dS
This is Stoke’s Theorem which relates the surface integral of a curl of a vector
~ and dS
~ are fixed by the
to the line integral of the vector itself. The direction of dl
right hand rule, i.e. when the fingers of the right hand are curled to point in the
~ the thumb points in the direction of dS.
~
diurection of dl,
1.5.1 Curl in Cartesian Coordinates :
We will obtain an expression for the curl in the cartesian coordinates. Let us
consider a rectangular contour ABCD in the y-z plane having dimensions ∆y ×
∆z. The rectangle is oriented with its edges parallel to the axes and one of the
corners is located at (y, z). We will calculate the line integral of a vector field
F~ along this contour. We assume the field to vary slowly over the length (or
the bredth) so that we may retain only the first term in a Taylor expansion in
computing the field variation.
z
D(y,z+ ∆z)
C(y+ ∆ y,z+ ∆ z)
^n
A(y,z)
B(y+ ∆ y,z)
y
x
31
Contribution to the line integral from the two sides AB and CD are computed as
follows.
R
~ = R F~ · ̂dy = R Fy dy
On AB : F~ · dl
R
~ = R F~ · (−̂)dy = R Fy dy
ON CD : F~ · dl
Using Taylor expansion (retaining only the first order term), we can write
∂Fy
∆z
∂z
Thus the line integral from the pair of sides AB and CD is
Fy |CD = Fy |AB +
∂Fy
∂Fy
∆zdy ≈ −
∆z∆y
∂z
∂z
In a similar way one can calculate the contributions from the sides BC and DA
and show it to be
Z
∂Fz
∂Fz
∆ydz ≈
∆y∆z
∂y
∂y
Adding up we get
−
(curl F )x =
Z
lim
∆y,∆z→0
∂Fz
∂y
−
∂Fy
∂z
∆y∆z
∆y∆z
=
∂Fz
∂Fy
−
∂y
∂z
In a very similar way, one can obtain expressions for the y and z components
∂Fx ∂Fz
−
∂z
∂x
∂Fy ∂Fx
−
=
∂x
∂y
(curl F )y =
(curl F )z
One can write the expression for the curl of F~ by using the del operator as
curl F~ = ∇ × F~ =
ı̂
̂
∂
∂x
∂
∂y
k̂ ∂
∂z
Fx Fy Fz
Expression for curl in Cylidrical and Spherical Coordinates :
In the cylindrical coordinates the curl is given by
∇ × F~ =
+
!
1 ∂Fz
∂Fθ
ρ̂
−
ρ ∂θ
∂z
!
!
1 ∂Fρ
1 ∂
∂Fρ ∂Fz
−
θ̂ +
(ρFθ ) −
k̂
∂z
∂ρ
ρ ∂ρ
ρ ∂θ
32
In the spherical coordinates the corresponding expression for the curl is
!
1
∂
∂Fθ
∇ × F~ =
r̂
(Fφ sin θ) −
r sin θ ∂θ
∂φ
!
1 ∂Fr
∂
1
1
− (rFφ ) θ̂ +
+
r sin θ ∂φ
∂r
r
∂
∂Fr
(rFθ ) −
∂r
∂θ
!
φ̂
Example 20 :
y
Verify Stoke’s theorem for F~
=
x2 ı̂ + 2x̂ + z 2 k̂ for the rectangle shown
below, defined by sides x = 0, y = 0, x = 1
and y = 1.
B(1,1)
C
x
O
A (1,0)
The line integrals along the four sides are
I
~ =
F~ · dl
=
=
Z
1
0
Z
0
1
Fx |y=0 dx +
Z
x2 dx + 0 + +
1
0
Z
Fy |x=0 dy +
0
1
x2 dx +
1
1
+0− +2=2
3
3
Z
1
0
Z
0
1
Fx |y=1 dx +
Z
0
1
Fy |x=1 dy
2dy
Since the normal to the plane is along k̂, we only need z− component of ∇ × F~
to calculate the surface integral. It can be checked that
(∇ × F~ )z =
∂Fy ∂Fx
−
=2−0=2
∂x
∂y
Thus
Z
S
(∇ × F~ ) · k̂dS =
Z
0
1
Z
which agrees with the line integral calculated.
Exercise :
33
0
1
2dxdy = 2
B
A vector field is given by
F~ = −yı̂ + x̂ + x2 k̂. Calculate the line integral of the field
along the triangular path shown
above. Verify your result by Stoke’s
theorem.
(Ans. 1)
O
A
(Hint : To calculate the line integral along a straightline, you need the equation
to the line. ForR instance,
the equation to the line BO is y = 2x. Check that
R
~ = − ydx + R xdy = 1. )
~ · dl
F
AB
Example 21 :
A vector field is given by F~ = −yı̂ + z̂ + x2 k̂. Calculate the line integral of
the field along a circular path of radius R in the x-y plane with its centre at the
origin. Verify Stoke’s theorem by considering the circle to define (i) the plane of
the circle and (ii) a cylinder of height z = h.
Solution :
The curl of F~ may be calculated as
∇ × F~ = −ı̂ + 2x̂ + k̂
Because of symmetry, we use cylindrical (polar) coordinates. The transformations
are x = ρ cos θ, y = ρ sin θ, z = z. The unit vectors are
ı̂ = ρ̂ cos θ − θ̂ sin θ
̂ = ρ̂ sin θ − θ̂ cos θ
k̂ = k̂
Substituting the above, the field F~ and its curl are given by
F~ = ρ̂(−ρ sin θ cos θ + z sin θ)
+ θ̂(ρ sin2 θ + z cos θ) + k̂ρ2 cos2 θ
∇ × F~ = ρ̂(− cos θ + 2ρ cos θ sin θ)
+ θ̂(sin θ + 2ρ cos2 θ) + k̂
34
z
^
k
The line integral of F~ around the circular
loop :
~ = Rdθθ̂,
Since the line element is dl
I
~ =
F~ · dl
Z
0
2π
(R sin2 θ + z cos θ)Rdθ
On the circle z = 0. The integral over sin2 θ
gives 1/2. Hence
I
y
^
θ
^ρ
~ = πR2
F~ · dl
^
−k
x
Let us calculate the surface integral of the curl of the field over two surfaces bound
by the circular curve.
(i) On the circular surface bound by the curve in the x-y plane, the outward normal
is along k̂ (right hand rule). Thus
Z
S
(∇ × F~ ) · (k̂dS) =
Z
S
dS = πR2
(ii) For the cylindrical cup, we have two surfaces : the curved face of the cylinder
on which n̂ = ρ̂ and the top circular face on which n̂ = k̂. The contribution from
the top circular cap is πR2 , as before because the two caps only differ in their
z values (the z-component of the curl is independent of z). The surface integral
from the curved surface is (the area element is Rdθdz ρ̂)
Z
0
2π
Rdθ
Z
0
h
dz(−R sin θ cos θ + z cos θ)
For both the terms of the above integral, the angle integration gives zero. Thus the
net surface integral is πR2 , as expected.
Exercise :
A vector field is given by F~ = kρ2 z θ̂. Check the validity of the Stoke’s theorm
35
z
by calculating the line integral about
the closed contour in the form of a
circle at z = 4 and also calculating the surface integral of the open
surface of the cylinder below it, as
shown. (Hint : Express the curl in
cylindrical coordinates and take care
of the signs of the surface elements
from the curved surface and the bottom cap. Ans. 64kπ )
h=4
C
y
R=2
x
Exercise :
Let C be a closed curve in the x-y plane in the shape of a quadrant of a circle of
radius R.
y
If F~ = ı̂y+̂z+ k̂x. calculate the line
integral of the field along the contour
shown in a direction which is anticlockwise when looked from above
the plane (z > 0). Take the surface of the quadrant enclosed by the
curve as the open surface bounded
by the curve and verify Stoke’s theorem. (Ans. −πR2 /4)
(0,1,0)
O
z
x
(1,0,0)
Example 22 :
A vector field is given by F~ (r, θ, φ) = f (r)φ̂ where φ is the azimuthal angle
variable of a spherical coordinate system. Calculate the line integral over a circle
of radius R in the x-y plane centered at the origin. Consider an open surface in the
form of a hemispherical bowl in the northern hemisphere bounded by the circle.
36
Solution :
~ = Rdφφ̂. Hence,
On the equatorial circle dl
I
~ =
F~ · dl
Z
0
2π
f (R)Rdφ = 2πRf (R)
z
The expression for curl in spherical coordinates may
be used to calculate the curl of F~ . Since the field
only has azimuthal component, the curl has radial and
polar (θ) components.
^r
1 ∂
1 ∂
(f (r) sin θ)r̂ −
(rf (r))θ̂
r sin θ ∂θ
r ∂r !
f (r) cos θ
f (r) ∂f (r)
=
θ̂
r̂ −
+
r sin θ
r
∂r
∇ × F~ =
y
The area element on the surface of the northern hemisphere is R2 sin θdθdφr̂.
x
Hence the surface integral is
Z
0
2π
dφ
Z
π/2
0
f (r) cos θ 2
R sin θdθ = 2πRf (R)
R sin θ
= 2πRf (R)
Z
0
π/2
cos θdθ
Exercise :
Verify Stoke’s theorem for a vector field 2zı̂ + 3x̂ + 5y k̂ where the contour is an
equatorial circle of radius R and is anticlockwise when viewed from above and
the surface is the hemisphere shown in the preceding example.. (Ans. 3R2 π)
1.6 Laplacian :
Since gradient of a scalar field gives a vector field, we may compute the divergence of the resulting vector field to obtain yet another scalar field. The operator
div(grad) = ∇ · ∇ is called the Laplacian and is written as ∇2 .
If V is a scalar, then,
∇2 V
= ∇ · (∇V )
!
!
∂
∂V
∂
∂
∂V
∂V
= ı̂
ı̂
+ ̂
+ k̂
+ ̂
+ k̂
∂x
∂y
∂z
∂x
∂y
∂z
∂2V
∂V
∂2V
=
+
+
∂x2
∂y 2
∂z 2
37
Example 23 :
√
Calculate the Laplacian of 1/r = 1/ x2 + y 2 + z 2 .
Solution :
∂2
−x
1
∂
√ 2
=
2
2
2
2
∂x x + y + z
∂x (x + y62 + z 2 )3/2
2x2 − y 2 − z 2
=
(x2 + y 2 + z 2 )5/2
3x2 − r 2
=
r5
Adding similar contributions from ∂ 2 /∂y 2 and ∂ 2 /∂z 2 , we get
∇2
3(x2 + y 2 + z 2 ) − 3r 2
3r 2 − 3r 2
1
=
=
=0
r
r5
r5
Laplacian in cylindrical and spherical coordinates:
In cylindrical :
1 ∂
∂
1 ∂2
∂2
∇ =
ρ
+ 2 2+ 2
ρ ∂ρ
∂ρ
ρ ∂θ
∂z
!
2
In spherical :
∂
1 ∂
r2
∇2 = 2
r ∂r
∂r
!
1
∂
∂
+ 2 2
sin θ
∂θ
r sin θ ∂θ
!
+
1
∂2
r 2 sin2 θ ∂φ2
Frequently the Laplacian of a vector field is used. It is simply a short hand notation
for the componentwise Laplacian
∇2 F~ = ı̂∇2 Fx + ̂∇2 Fy + k̂∇2 Fz
Exercise :
Show that
∇ × (∇ × F~ ) = ∇(∇ · F~ ) − ∇2 F~
1.7 Dirac- Delta Function :
In electromagnetism, we often come across use of a function known as Dirac- δ
function. The peculiarity of the function is that though the value of the function
38
is zero everywhere, other than at one point, the integral of the function over any
region which includes this singular point is finite. We define
δ(x − a) = 0 if x 6= a
with
Z
f (x)δ(x − a)dx = f (a)
where f (x) is any function that is continuous at x = a, provided that the range of
integration includes the point x = a. Strictly speaking, δ(x − a) is not a function
in the usual sense as Riemann integral of any function which is zero everywhere,
excepting at discrete set of points should be zero. However, one can look at the δ
function as a limit of a sequence of functions. For instance, if we define a function
δǫ (x) such that
δ ε(x)
1/ε
ǫ
ǫ
1
for − < x < +
ǫ
2
2
ǫ
= 0 for | x |>
2
δǫ (x) =
Then δ(x) can be thought of as the
limit of δǫ (x) as ǫ → 0.
− ε /2
0 + ε/2
x
One can easily extend the definition to three dimensions
δ(~r) = δ(x)δ(y)δ(z)
which has the property
Z
f (~r)δ(~r − ~a) = f (~a)
provided, of course, the range of integration includes the point ~r = ~a.
Example :
R
Evaluate 05 cos xδ(x − π)dx.
Solution :
The range of integration includes the point x = π at which the argument of the
delta function vanishes. Thus, the value of the integral is cos π = −1.
Exercise :
R
Evaluate ~r ·(~a−~r)δ(~r −~b)dr , where ~a = (1, 2, 3), ~b = (3, 2, 1) and the integration
is over a sphere of radius 1.5 centered at (2,2,2)
(Ans. -4)
39
A physical example is the volume density of charge in a region which contains a
point charge q. The charge density is zero everywhere except at the point where
the charge is located. However, the volume integral of the density in any region
which includes this point is equal to q itself. Thus if q is located at the point ~r = ~a,
we can write
ρ(~r) = qδ(~r − ~a)
Example :
Show that ∇2 (1/r) is a delta function.
Solution√:
As r = x2 + y 2 + z 2 , we have
x
∂r
y
∂r
= ,
= ,
∂x
r
∂y
r
using this it is easy to show that
∂r
z
=
∂z
r
∂2 1
3x2 − r 2
( )=
∂x2 r
r5
Thus
∇
2
1
=
r
∂2
1
∂2
∂2
3(x2 + y 2 + z 2 ) − 3r 2
( )=
+
+
=0
∂x2 ∂y 2 ∂z 2 r
r5
!
However, the above is not true at the origin as 1/r diverges at r = 0 and is not
differentiable at that point. Interestingly, however, the integral of ∇2 (1/r) over
any volume which includes the point r = 0 is not zero. As the value of the
integrand is zero everywhere excepting at the origin, the point r = 0 has to be
treated with care.
Consider an infinitisimally small sphere of radius r0 with the centre at the origin.
Using divergence theorem, we have,
Z
Z
Z
1 3
1
3
2 1
dr=
d r = ∇ i · dS
∇
∇·∇
r
r
r
V
V
S
where the last integral is over the surface S of the sphere defined above. As the
gradient is taken at points on the surface for which r 6= 0, we may replace ∇(1/r)
with −1/r02 at all points on the surface. Thus the value of the integral is
Z
1
1
− 2 dS = − 2 · 4πr02 = −4π
r0 S
r0
Hence
2 1
= −4πδ(r)
∇
r
40