A Crash Course in Vector Calculus

SCALAR AND VECTOR FIELDS

This introductory chapter is a review of mathematical concepts required for thecourse. It is assumed that the reader is already familiar with elementary vectoranalysis.Physical quantities that we deal with in electromagnetism can be scalars or vec-tors.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 vectorsin physics are displacement, velocity, acceleration, force, electric field, magneticfield etc.A field is a quantity which can be specified everywhere in spaceas a function ofposition. 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, there-fore, be said to be a region of “temperature field” which is a scalar field becausethe temperatureT (x, y, z) is a scalar function of the position. An example of ascalar field in electromagnetism is theelectric potential.In a similar manner, a vector quantity which can be specified at every point in aregion of space is a vector field. For instance, every point onthe earth may beconsidered to be in the gravitational force field of the earth. we may specify thefield by the magnitude and the direction of acceleration due to gravity (i.e. forceper unit mass )g(x, y, z) at every point in space. As another example considerflow of water in a pipe. At each point in the pipe, the water molecule has a veloc-ity ~v(x, y, z). The water in the pipe may be said to be in a velocity field. Thereare several examples of vector field in electromagnetism, e.g., the electric field~E,the magnetic flux density~B etc.

1.1 Coordinate Systems :We are familiar with cartesian coordinate system. For systems exhibiting cylin-drical or spherical symmetry, it is convenient to use respectively the cylindricaland 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 originO andθ as the angle made by the radialvector with a reference line (usually chosen to coincide with the x-axis of thecartesian system). The radial unit vectorρ and the tangential (or angular) unitvector θ are taken respectively along the direction of increasing distanceρ andthat of increasing angleθ respectively, as shown in the figure.

θ

^θ

x

y

x

y

θθ

ρ

ρ

Relationship with the cartesian components are

x = ρ cos θ

y = ρ sin θ

so that the inverse relationships are

ρ =√

x2 + y2

θ = tan−1 y

x

By definition, the distanceρ > 0. we will take the range of anglesθ to be0 ≤ θ <2π (It is possible to define the range to be−π ≤ θ < +π). One has to be carefulin using the inverse tangent as the arc-tan function is defined in 0 ≤ θ < π. If yis negative, one has to addπ to the principal value ofθ calculated by the arc - tanfunction so that the point is in proper quadrant.

2

Example 1 :A vector ~A has cartesian componentsAx andAy. Write the vector in terms of itsradial and tangential components.Solution :Let us write

~A = Aρρ + Aθθ

Sinceρ andθ are orthonormal basis vectorsρ · ρ = θ · θ = 1 andρ · θ = 0. Thus

Ar = ~A · ρ, Aθ = ~A · θ

Note that (see figure)ρ makes an angleθ with the x-axis (ı) andπ/2 − θ with they-axis (). Similarly, the unit vectorθ makesπ/2 + θ with the x-axis andθ withthe y-axis. Thus

Aρ = ~A · ρ = Axı · ρ + Ay · ρ= Ax cos θ + Ay cos(π/2 − θ)

= Ax cos θ + Ay sin θ

Aθ = ~A · θ = Axıθ + Ay · θ= 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, thedifferential ele-ment also transform.

For instance, in 2 dimension the ele-ment of an area isdxdy but in polarcoordinates the element is notdθdρbut (ρdθ)dρ. This extra factorρ isimportant when we wish to integratea function using a different coordi-nate system.

θ

dθ ρ

d

dθ

ρ

ρ

If f(x, y) is a function ofx, y we may express the function in polar coordinatesand write it asg(ρ, θ). However, when we evaluate the integral

∫

f(x, y)dxdyin polar coordinates, the corresponding integral is

∫

g(r, θ)ρdθdρ. In general, ifx = f(u, v) andy = g(u, v), then, in going from(x, y) to (u, v), the differential

3

elementdxdy →| J | dudv whereJ is given by the determinant

J =

∣

∣

∣

∣

∣

∂x∂u

∂x∂v

∂y∂u

∂y∂v

∣

∣

∣

∣

∣

The differentiations are partial, i.e., while differentiating ∂x/∂u = ∂f(u, v)/∂u,the variablev is treated as constant. An useful fact is that the Jacobian oftheinverse transformation is1/J because the detrminant of the inverse of a matrix isequal to the inverse of the determinant of the original matrix.

Example 2 : Show that the Jacobian of the transformation from cartesianto polarcoordinates isρ.Solution :We have

J =

∣

∣

∣

∣

∣

∂x∂ρ

∂x∂θ

∂y∂ρ

∂y∂θ

∣

∣

∣

∣

∣

Usingx = ρ cos θ andy = ρ sin θ, we have

J =

∣

∣

∣

∣

∣

cos θ −ρ sin θsin θ ρ cos θ

∣

∣

∣

∣

∣

= ρ

Exercise :Show that the Jacobian of the inverse transformation from polar to cartesian is1/ρ = 1/

√x2 + y2.

Example 3 :find the area of a circle of radiusR.Solution :Take the origin to be at the centre of the circle and the plane of the circle to be theρ − θ plane. Since the area element in the polar coordinates isρdθdr, the area ofthe circle is

∫ 2π

0dθ∫ R

0ρdρ = 2π

[

ρ2

2

]R

0

= πR2

- a very well known result !Example 4 :

Find the integral∫

e−(x2+y2)dxdy where the region of integration is a unit circle

about the origin.Using polar coordinates the integrand becomese−ρ2

. The range of i integrationfor ρ is from 0 to 1 and forθ is from 0 to 2π. The integral is given by

I =∫ 2π

0dθ∫ 1

0e−ρ2

ρdρ = 2π∫ 1

0e−ρ2

ρdρ

4

The radial integral is evaluated by substitutionw = ρ2 so thatρdρ = dw/2. Thevalue of the integral is

I = 2π∫ 1

0e−wdw/2 = π[−e−w]10 = π(1 − 1

e)

Exercise :Evaluate

∫ ∫

xydxdy where the region of integration is the part of the area betweencircles of radii 1 and 2 that lies in the first quadrant. (Ans.15/8)Exercise :Evaluate the Gaussian integralI =

∫

∞

0 e−x2

dx[ Hint : The integration cannot be done using cartesian coordinates but is relativelyeasy using polar coordinates and properties of definite integrals. By changing thedummy variablex to y, one can writeI =

∫

∞

0 e−y2

dy, so that we can write

I2 =∫

∞

0

∫

∞

0e−(x2+y2)dxdy

Transform this to polar. Range of integration forρ is from 0 to ∞ and that ofθ isfrom 0 to π/2(why ?) Answer :

√π/2 ]

Differentiation of polar unit vectors with respect to time :It may be noted that the basis vectorsρ and θ, unlike ı and are not constantvectors but depend on the position of the point. The time derivative of the unitvectors are defined as follows

dρ

dt= lim

∆t→0

∆ρ

∆t= lim

∆t→0

ρ(t + ∆t) − ρ(t)

∆t

dθ

dt= lim

∆t→0

∆θ

∆t= lim

∆t→0

θ(t + ∆t) − θ(t)

∆t

One can evaluate the derivatives by laborious process of expressing the unit vec-tors ρ and θ in terms of constant unit vectors of cartesian system, differentiatingthe resulting expressions and finally transform back to the polar form. Alterna-tively, we can look at the problem geometrically, as shown inthe following figure.

5

θ

dθ

θ

θd^ (t)

(t+dt)^

θd

(t+dt)θ^

ρ(t)

ρ

In the figure, the positions of a particle are shown at timet andt + dt. The unitvectorsρ is shown in red while the unit vectorθ is shown in blue. It can be easilyseen by triangle law of addition of vectors that the magnitude of ∆ρ and∆θ is1.dθ = dθ. However, as the limitdt → 0, the direction of∆ρ is in the directionof θ while that of∆θ is in the direction of−ρ. Thus

dρ

dt= lim

∆t→0

∆ρ

∆t=

dθ

dtθ

dθ

dt= lim

∆t→0

∆θ

∆t= −dθ

dtρ

Now, dθ/dt is the angular velocity of the point, which is usually denoted by ω,Thus we have,

dρ

dt= ωθ

dθ

dt= −ωρ

Exercise :Obtain the above relationships directly by using the expression for the unit vectorsρ andθ in terms of the cartesian unit vectors, viz.,

ρ = ı cos θ + sin θ

θ = −ı sin θ + cos θ

6

1.1.2 Cylindrical coordinates :Cylnidrical coordinate system is obtained by extending thepolar coordinates byadding a z-axis along the height of a right circular cylinder. The z-axis of thecoordinate system is same as that in a cartesian system.In the figureρ is the distance of thefoot of the perpendicular drawn fromthe point to thex − y(ρ, θ) plane.Note thatρ here isnot the distanceof the point P from the origin, as isthe case in polar coordinate systems.(Some texts user to denote what weare calling asρ here. However, weuseρ to denote the distance from theorigin to the foot of the perpendicu-lar to avoid confusion.) In terms ofcartesian coordinates

x = ρ cos θ

y = ρ sin θ

z = z

so that the inverse relationships are

ρ =√

x2 + y2

θ = tan−1 y

xz = z

x

y

z

y

z

θ

P

θ

z

^

^θ

xO

P ’

ρ

ρ

Exercise :Find the cylindrical coordinate of the point3ı + 4 + k. [Hint : Determineρ andtan θ using above transformation] (Ans.5ρ + tan−1(4/3)θ + k)The line element in the system is given by

~dl = ρdρ + ρdθθ + dz

and the volume element isdV = ρdθdρdz

The Jacobian of transformation from cartesian to cylindrical is ρ as in the polarcoordinates sincez coordinate remains the same.

7

1.2.3 Spherical Polar Coordinates :Spherical coordinates are useful in dealing with problems which possess spheri-cal symmetry. The independent variables of the system are(r, θ, φ). Herer is thedistance of the pointP from the origin. Anglesθ andφ are similar to latitudesand longitudes.Two mutually perpendicular lines are chosen, taken to coincide with the x-axisand z-axis of the cartesian system. We take angleθ to be the angle made by theradius vector (i.e. the vector connecting the origin to P) with the z-axis (the angleθ is actually complementary to the latitude). The angleφ is the angle between thex-axis and the line joining the origin toP ′, the foot of the perpendicular fromPto the x-y plane.The unit vectorsr, θ andφ are respectively along the directions of increasingr, θandφ.

θ^

P

P ’

φ

^

φ

z

y

y

x

z

r

θ

r sinθ

r

^

O

r co

s

θ

The surface of constantr are spheres of radiusr about the centre. the surface ofconstantθ is a cone of semi-angleθ about the z-axis. The reference for measur-ing φ is the x-z plane of the cartesian system. A surface of constant φ is a planecontaining the z-axis which makes an angleφ with the reference plane.

Example 5 :

8

Express unit vectors of spherical coordinate system in terms of unit vectors ofcartesian system.Solution :From the point P drop a perpendicular on to the x-y plane. Denote ~OP ′ by ~ρ. Thefigure below shows the unit vectors in both the systems. By triangle law of vectoraddition,

~r = ~OP = ~OP ′ + ~P ′P = r sin θρ + r cos θk

However,ρ = cos φı + sin φ. Substituting this in the expression for~r, we get ondividing both sides by the magnitude ofr

r = sin θ cos φı + sin θ sin φ + r cos θk

The unit vectorθ is perpendicular toOP in the direction of increasingθ.The angle thatθ makes with the pos-itive z-axis isπ/2 + θ. Hence,

θ = cos(π

2+θ)k+sin(

π

2+θ)ρ = − sin θk+cos θρ

Substituting forρ, we get

θ = cos θ cos φı+cos θ sin φ−sin θk

O

r

θ^

x

y

z

ρ

r

z

k

j

^

^

^

iφ

θ

P θ

P ’

φφ

The azimuthal unit vector~φ is in the direction of increasing angleφ. It is perpen-dicular to~ρ and has no z- component. It can be easily seen that

φ = − sin φı + 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 =√

x2 + y2 + z2

9

θ = tan−1

√x2 + y2

z= cos−1 z

r

φ = tan−1 y

x

Range of the variables are as follows :

0 ≤ r < ∞ 0 ≤ θ ≤ π 0 ≤ φ ≤ 2π

Exercise :A particle moves along a spherical helix. its position coordinate at timet is givenby

x =cos t√1 + t2

, y =sin t√1 + t2

, z =t√

1 + t2

Express the equation of the path in spherical coordinates. (Ans.r = 1, cos θ = t/

√1 + t2 φ(t) = t)

The differential element of volumeis obtained by constructing a closedvolume by extendingr, θ andφ re-spectively bydr, dθ and dφ. Thelength elements in the direction ofr is dr, that alongθ rdθ while thatalongφ is r sin θdφ (see figure). Thevolume element, therefore, is

dV = dr·rdθ·r sin θdφ = r2 sin θdθdφdr

Thus the Jacobian of transformationis r2 sin θ.

x

z

θ

r

sinr θ dφφd

d

φ d

sin θ

θ

r

φ

dθ

y

rdr

Example 6:Find the volume of a solid region in the first octant that is bounded from above bythe spherex2 + y2 + z2 = 9 and from below by the conex2 + y2 = 3z2.Solution :Because of obvious spherical symmetry, the problem is best solved in sphericalpolar coordinates. The equation to sphere isr = 3 so that the range ofr vaiablefor our solid is from0 to 3.

10

The equation to the conex2 + y2 =3z2 becomesr2 sin2 θ = 3r2 cos2 θ.Solving, the semi-angle of cone istan 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 theazimuthal angleφ is from 0 to π/2.

The volume of the solid is∫

dV =∫

r2 sin θdθdφdr =∫ π/2

0dφ∫ π/3

0sin θdθ

∫ 3

0r2dr

=π

2. [− cos θ]π/3

0 .

[

r3

3

]3

0

=9π

4

Exercise :Using direct integration find the volume of the first octant bounded by a spherex2 + y2 + z2 = 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 scalarfunction it can be integrated and differentiated. However,as a vector field hasboth magnitude and direction it is necessary to define operations of calculus totake 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 lineintegral of the vector function may be defined as

11

∫

C

~F · ~dl

whereC is the curve along which theintegral is calculated. Like the inte-gral of a scalar function the integralabove is also interpreted as a limitof a sum. We first divide the curveC into a large number of infinitisi-mally small line segments such thatthe vector function is constant (inmagnitude and direction) over eachsuch line segment. The integrandis then equal to the product of thelength of the line segment and thecomponent of~F along the segment.The integration is defined as the limitof the sum of contributions from allsuch segments in the same manner asordinary integration is defined.

Fθdl

The concept of line integral is very useful in many branches of physics. In me-chanics, we define work done by a force~F in moving an object from an initialposition A to a final position B as the line integral of the force along the curvejoining the end points. Except in the case of conservative forces, the line integraldepends on the actual path along which the particle moves under the force.Example 7:A force ~F = zyı + x + z2xk acts on a particle that travels from the origin tothe point(1, 2, 3). Calculate the work done if the particle travels (i) along thepath(0, 0, 0) → (1, 0, 0) → (1, 2, 0) → (1, 2, 3) along straightline segments join-ing each pair of end points (ii) along the straightline joining the initial and finalpoints.Solution :

12

Along the path(0, 0, 0) → (1, 0, 0),~dl = ıdx andy = 0, z = 0. SinceFx = 0 along this segment, the in-tegral alongC1 is zero. Along thepathC2 joining (1, 0, 0) and(1, 2, 0),~dl = dyandx = 1. Along this pathFy = x = 1. The integral

∫ 2

0Fydy =

∫ 2

0dy = 2

Along the third path connecting(1, 2, 0) to (1, 2, 3), x = 1, y = 2.~dl = kdz andFz = z2x = z2. Theline integral is

∫ 3

0z2dz =

z3

3|30= 9

The work done is, therefore,0 + 2 +9 = 11.

(0,0,0) (1,0,0)

(1,2,0)

(1,2,3)

C

C

C

1

2

3

In order to calculate the work done when the particle moves along the straightlineconnecting the initial and final points, we need to write downthe equation to theline in a parametric form. If(x1, y1, z1) and (x2, y2, z2) are the end points, theequation is

x − x1

x2 − x1

=y − y1

y2 − y1

=z − z1

z2 − z1

= m

Substituting for the coordinates, we get the equation to theline as

x = m

y = 2m

z = 3m

Thus the differential elements aredx = dm, dy = 2dm anddz = 3dm. The lineintegral is given by

∫

~F · ~dl =∫

(Fxdx + Fydy + Fzdz)

=∫ 1

0(6m2.dm + m.2dm + 9m3.3dm)

13

= 2m3 + m2 + 27m4

4|10= 9.75

Example 8 :Find the line integral of~F = yı + over an anticlockwise circular loop of radius1 with the origin as the centre of the circle.Solution :

The length elementdl has a magni-tude1.dθ = dθ. Since the unit vectoralong~dl makes an angle of(π/2)+θwith the positivex− axis,

~dl = | dl | cos(π

2+ θ)ı+ | dl | sin(

π

2+ θ)

= − sin θı + cos θ

θ

dlθ

x

y

R=1

In polar coordinates,x = cos θ andy = sin θ (since the radius is 1). Thus∫

~F · ~dl =∫

Fxdx + Fydy

= −∫ 2π

0sin2 θdθ +

∫ 2π

0cos θdθ

= −π + 0 = −π

Exercise :A force ~F = xyı+(x2−z2)−xz2k acts on a particle. Calculate the work done ifthe particle is taken from the point(0, 0, 0) to the point(2, 1, 3) along straight linesegment connecting(0, 0, 0) → (0, 1, 0) → (2, 1, 0) → (2, 1, 3). What would bethe work done if the particle directly moved to the final pointalong the straightlineconnecting to origin. (Ans.−16, −13.8.)

Exercise :A vector field is given by

~F = (2x + 3y)ı + (3x + 2y)

14

Evaluate the line integral of the field around a circle of unitradius traversed inclockwise fashion. (Ans. 6π)

Exercise :Evaluate the line integral of ascalar functionxy along a parabolic pathy = x2

connecting the origin to the point(1, 1).

[ ]Hint : Remember that the arc length along a curve is given by√

(dx)2 + (dy)2.

The curve can be parametrized byx = t andy = 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 directionbeing defined as the outward normal to the surface. The concept of a surfaceintegral is related to flow. Suppose the vector field represents the rate at whichwater flows at a point in the region of flow. The flow may be measured in cubicmeter of water flowing per square meter of area per second. If an area is orientedperpendicular to the direction of flow, as shown in the figure to the left, maximumamount of water would flow through the surface. The amount of water passingthrough the area is theflux (measured in cubic meter per second).

n

V V

θ

^

n

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 areathat is perpendicular to the direction of flow will contribute to the flux.We defineflux through an area element~dS as the dot product of the vector field~V with the area vector~dS. When~V is parallel to ~dS, i.e. if the surface is orientedperpendicular to the direction of flow, the flux is maximum. Onthe other hand, asurface oriented parallel to the flow does not contribute to the flux.

15

Example 9 :A hemispherical bowl of radiusR is oriented such that the circular base is perpen-dicular to direction of flow. Calculate the flux through the curved surface of thebowl, assumuing the flow vector~V to be constant.Solution :Since the curved surface makes different angles at different positions, it is some-what difficult to calculate the flux through it. However, one can circumvent it bycalculating the flux through the circular base.

n

n

n

1

2

2

V

1n

^

^

^

Since the flow vector is constant all over the circular face which is oriented per-pendicular to the direction of flow, the flux through the base is −πR2V . Theminus sign is a result of the fact that the direction of the surface is opposite tothe direction of flow. Thus we may call the flux through the baseas inward flux.Since there is no source or sink of flow field (i.e. there is no accumulation ofwater) inside the hemisphere, whatever fluid enters throughthe base must leavethrough the curved face. Thus theoutward fluxfrom the curved face is+πR2V .

We may now generalize the above for a surface over which the field is not uniformby defining the flux through an area as the sum of contribution to the flux frominfinitisimal area elements which comprises the total area by treating the field tobe 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 normalni to which makes an an-gle θi with the vector field~Vi at theposition of the element, the total flux

Φ =∑

i

~Vi · ni∆Si =∑

i

ViδSi cos θi

In the limit of ∆Si → 0, the sumabove becomes a surface integral

Φ =∫

S

~V · dS

∆S i

θ i

ni

^

If the surface is closed, it encloses a volume and we define

Φ =∮

S

~V · dS

to be the net outward flux. In terms of cartesian components

Φ =∫

S(Vxdydz + Vydxdz + Vzdxdy)

Example 10 :A vector field is given by~B = xyı + yz + zxk. Evaluate the flux through eachface of a unit cube whose edges along the cartesian axes and one of the corners isat the origin.Solution :

Consider the base of the cube(OGCD), which is the x-y plane onwhich z = 0. On this face~B = xyı.The surface vectorn is along−k di-rection. Thus on this surface flux∫ ~B · ndxdy =

∫

xyı · kdxdy = 0sinceı · k = 0. For the top surface(ABFE), z = 1 andn = k. The fluxfrom this surface is∫

Vzdxdy =∫ 1

0xdx

∫ 1

0dy =

1

2x

y

z

BA

CD

E F

O G

In a similar way one can show that flux from left side (AEOD) is zero while thecontribution from the right side (BFGC) is 1/2. The back face(EFGO) contributes

17

zero while the front face (ABCD) contributes 1/2. The net flux, therefore, is 3/2.Exercise :

Find the flux of the vector field~V =Axı + By2 through a rectangularsurface in the x-y plane having di-mensionsa×b. The origin of the co-ordinate system is at one of the cor-ners of the rectangle and the x-axisalong its length. (Ans.Bab3/3)

x

y

z

O a

b

Example 11 :Calculate directly the flux through the curved surface of thehemispherical bowlof Example 10.Solution :

Use a spherical polar coordinates with thebase of the hemisphere being the x-y planeand the direction of the vector field as the z-axis. We have seen that an area element onthe surface is given bydS = R2 sin θdθdφ.Thus the fluxΦ =

∫ ~V · n is given by

Φ = | V | R2∫ 2

0πdφ

∫ π/2

0cos θ sin θdθ

= | V | R2.2π

[

sin2 θ

2

]2π

0

= π | V | R2

V

n

O θ

φ

x

y

z−axis

Exercise :Find the flux through a hemispherical bowl with its base on thex-y plane and theorigin at the centre of the base. The vector field, in spherical polar coordinates is~V = r sin θr + θ + φ. (Ans.π(1 + π/2))

18

Example 12 :A cylindrical object occupies a volume defined byx2 + y2 ≤ R2 and0 ≤ z ≤ h.Find the flux through each of the surfaces when the object is ina vector field~V = ıx + y + kz.Solution :Because of cylindrical symmetry, it is convenient to work ina cylindrical(ρ, θ, z)coordinates. The vector field is given by

~V = ρρ + zk

Bottom face hasz = 0 and the nor-mal to the face points in−k direc-tion. Thus the flux from this face∫ ~V · n =

∫

Vzρdθdρ = 0.The top face hasz = h and n = k.The flux is∫

Vzρdθdr = h∫ R

0ρdρ

∫ 2π

0dθ

= hR2

2.2π = πR2h

x

y

z

n

n

n

^

^

h

θ

ρ

The normal to the curved face is alongρ direction. An area element on the curvedface isRdθdz Thus the flux from this face is

∫

(Rρ + zk) · ρRdθdz = R2∫ 2π

0dθ∫ h

0dz = 2πR2h

The net flux from the faces of the object is3πR2h.Exercise :Find the flux of the vector field~V = 2ρ − 3rθ + zρk through surfaces of a rightcylinder of radius 1 and height 2. The base of the cylinder is in thez = 0 planewith 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 temperatureT (x, y, z) in some region of space. Thedistribution of temperature may be represented by drawing isothermal surfaces orcontours connecting points of identical temperatures,

T (x, y, z) = constant

One can draw such contours for different temperatures. If weare located at a point~r on one of these contours and move away along any direction other than alongthe contour, the temperature would change.

The change∆T in temperature as wemove away from a pointP (x, y, z) toa pointQ(x + ∆x, y + ∆y, z + ∆z)is given by

∆T =∂T

∂x∆x +

∂T

∂y∆y +

∂T

∂z∆z

where the derivatives in the aboveexpression are partial derivatives.

r + d r

r

T( )

θ

∆

T

Q

P

T( )

If the displacement from the initial position is infinitisimal, we get

dT =∂T

∂xdx +

∂T

∂ydy +

∂T

∂zdz

Note that the changedT involves a change in temperature with respect to each ofthe three directions. We define a vector called thegradient of T , denoted by∇Tor gradT as

∇T = ı∂T

∂x+

∂T

∂y+ k

∂T

∂z

using which, we get

dT = ∇T · (ıdx + dy + kdz) = ∇T · d~r

20

Note that∇T , the gradient of a scalarT is itself a vector. Ifθ is the angle betweenthe direction of∇T andd~r,

dT =| ∇T || d~r | cos θ = (∇T )r | d~r |

where(∇T )r is the component of the gradient in the direction ofd~r. If d~r lieson an isothermal surface thendT = 0. Thus,∇T is perpendicular to the surfacesof constantT . Whend~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 ofTand its direction is along the direction of greatest change.The above discussion is true for any scalar fieldV . If a vector field can be writtenas a gradient of some some scalar function, the latter is called thepotentialof thevector field. This fact is of importance in defining a conservative field of force inmechanics. 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 :

∫ f

i

~F · ~dl =∫ f

i∇V · d~l

=∫ f

idV = Vf − Vi

where the symbolsi andf represent the initial and thec final positions and in thelast step we have used an expression fordV similar to that derived fordT above.Thus the line integral of the force field is independent of thepath connecting theinitial and final points. If the initial and final points are the same, i.e., if the particleis taken through a closed loop under the force field, we have

∮

~F · ~dl = 0

Since the scalar product of force with displacement is equalto the work done bya force, the above is a statement of conservation of mechanical energy. Becauseof this reason, forces for which one can define a potential function are calledcon-servative forces.Example 13 :Find the gradient of the scalar functionV = x2y + y2z + z2x + 2xyz.

21

Solution :

∇V = ı∂V

∂x+

∂V

∂y+ k

∂V

∂z

= 2(xy + zx + yz)(ı + + k)

Exercise : Find the gradients of(i) xz − x2y + y2z2

(ii)x3 + y3 + z3

(iii) ln√

x2 + y2 + z2 (Ans. (xı + y + k)/(x2 + y2 + z2)

Gradient can be expressed in other coordinate systems by finding the length ele-ments in the direction of basis vectors. For example, in cylindrical coordinates thelength elements aredρ, ρdθ anddz alongρ, θ andk respectively. The expressionfor gradient is

∇V = ρ∂V

∂ρ+ θ

1

ρ

∂V

∂θ+ k

∂V

∂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 :Find the gradient ofV = e−(x2+y2) in cylindrical (polar) coordinates.Solution :In polar variables the function becomesV = e−ρ2

. Thus

∇V = ρ∂e−ρ2

∂ρ

= ρe−ρ2

.(−2)ρ = −2ρρV

Exercise :Find the gradient of the functionV of Example 15 in cartesian coordinates andthen transform into polar form to verify the answer.

22

Exercise :Find the gradient of the functionln

√ρ2 + z2 in cylindrical coordinates.(Ans.(ρρ+

kz)/(ρ2 + z2))

In spherical coordinates the length elements aredr, rdθ andr sin θdφ. Hence thegradient of a scalar functionU is given by

∇V = r∂V

∂r+ θ

1

r

∂V

∂θ+ k

1

r sin θ

∂V

∂φ

Exercise :Find the gradient ofV = r2 cos θ cos φ (Ans. 2r cos θ cos φr − r sin θ cos φθ −r cos θ sin φφ.)Exercise :A potential function is given in cylidrical coordinates ask/

√ρ2 + z2 Find the

force field it represents and express the field in spherical polar coordinates. (Ans.−k~r/r3)

1.4 Divergence of a Vector Field :Divergence of a vector field~F is a measure of net outward flux from a closedsurfaceS enclosing a volumeV , as the volume shrinks to zero.

div ~F ≡ ~∇ · ~F = lim∆V →0

∮

S~F · ~dS

∆V

where∆V is the volume (enclosed by the closed surfaceS) in which the point Pat which the divergence is being calculated is located. Since the volume shrinksto zero, the divergence is a point relationship and is a scalar.Consider a closed volumeV bounded byS. The volume may be mentally brokeninto a large number of elemental volumes closely packed together. It is easy tosee that the flux out of the boundaryS is equal to the sum of fluxes out of thesurfaces of the constituent volumes. This is because surfaces of boundaries of twoadjacent volumes have their outward normals pointing opposite to each other. Thefollowing figure illustrates it.

23

= +

We can generalize the above to closely packed volumes and conclude that the fluxout of the bounding surfaceS of a volumeV is equal to the sum of fluxes out ofthe elemental cubes. If∆V is the volume of an elemental cube with∆S as thesurface, then,

∫

S

~F · ~dS =∑

∫

∆S

~F · dS = lim∆V →0

(

1

∆V

∫

~F · ~dS)

∆V

The quantity in the bracket of the above expression was defined as the divergenceof ~F , giving

∫

S

~F · ~dS =∫

Vdiv ~FdV

This is known as theDivergence Theorem.We now calculate the divergence of~F from an infinitisimal volume over whichvariation of ~F is small so that one can retain only the first order term in a Taylorexpansion. Let the dimensions of the volume element be∆x × ∆y × ∆z and letthe 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 isfrom the y-component of~F only and is given by

24

[Fy(y + dy) − Fy(y)] dxdz

ExpandingFy(y+dy) in a Taylor se-ries and retaining only the fiirst orderterm

F(y + dy) = Fy +∂Fy

∂ydy

so that the flux from these two facesis

∂Fy

∂ydxdydz =

∂Fy

∂ydV

wheredV = dxdydz is the volumeof the cuboid.

x

y

z

(x+dx,y,z)

(x,y,z)

dx

dy

dz

n n

Combining the above with contributions from the two remaining pairs of faces,the total flux is

(

∂Fx

∂x+

∂Fy

∂y+

∂Fz

∂z

)

dV

Thus∫

S

~F · ~dS =∫

(

∂Fx

∂x+

∂Fy

∂y+

∂Fz

∂z

)

dV

Comparing with the statement of the divergence theorem, we have

div ~F =

(

∂Fx

∂x+

∂Fy

∂y+

∂Fz

∂z

)

Recalling that the operator∇ is given by

∇ = ı∂Fx

∂x+

∂Fy

∂y+ k

∂Fz

∂z

and using~F = ıFx + Fy + kFz, we can writediv ~F = ∇ · ~F .The following facts may be noted :

1. The divergence of a vector field is a scalar

2. ∇ · (~F + ~G) = ∇ · ~F + ∇ · ~G

25

3. ∇ · (φ~F ) = φ∇ · ~F + ∇φ · ~F

4. In cylindrical coordinates

∇ · ~F =1

ρ

∂

∂ρ(ρFρ) +

1

ρ

∂

∂θFθ +

∂

∂zFz

5. In spherical polar coordinates

∇ · ~F =1

r2

∂

∂r(r2Fr) +

1

r sin θ

∂

∂θ(Fθ sin θ) +

1

r sin θ

∂

∂φFφ

6. The divergence theorem is∮

S

~F · ~dS =∫

V∇ · ~FdV

Example 15 : Divergence of~r = ıx + y + kzDivergence of position vector~r is very useful to remember.

∇ · ~r =∂x

∂x+

∂y

∂y+

∂z

∂z= 1 + 1 + 1 = 3

One can also calculate easily in spherical coordinate since~r only has radial com-ponent

∇ · ~r =1

r2

∂

∂r(r2.r) =

1

r2.3r2 = 3

Exercise :Calculate the divergence of the vector field~r/r3 using all the three coordinate sys-tems. (Ans. 0)

Example 16 :A vector field is given by~F = 4xzı − y2 + yzk. Find the surface integral of thefield from the surfaces of a unit cube bounded by planesx = 0, x = 1, y = 0, y =1, z = 0 andz = 1. Verify that the result agrees with the divergence theorem.Solution :Divergence of~F is

∇ · ~F =∂Fx

∂x+

∂Fy

∂y+

∂Fz

∂z

=∂4xz

∂x+

∂(−y2)

∂y+

∂yz

∂z= 4z − 2y + y = 4z − y

26

The volume integral of above is

∫

∇ · ~FdV =∫ 1

0dx∫ 1

0dy∫ 1

0(4z − y)dxdydz =

3

2

Consider the surface integral fromthe six faces individually. For theface AEOD, the normal is along−.On this facey = 0 so that~F = 4xzı.Sinceı · = 0, the integrand is zero.For the surface BFGC, the normalis along and on this facey = 1.On this face the vector field is~F =4xzı − + zk. The surface integralis∫

~F · ~dS =∫

~F · dxdz

= −∫ 1

0dx∫ 1

0dz = −1

n

x

O

A

C

B

D

E F

z

y

(1,0,0)

= j= −jn ^

(0,0,1)

G (0,1,0)

Consider the top face (ABFE) for which the normal isk so that the surface integralis∫

Fzdydx. On this facez = 1 andFz = y. The contribution to the surfaceintegral from this face is

∫ 1

0dx∫ 1

0ydy =

1

2

For the bottom face (DOGC) the normal is along−k andz = 0. This givesFz = 0so that the integral vanishes.For the face EFGO the normal is along−ı so that the surface integral is− ∫ Fxdydz.On this facex = 0 giving Fx = 0. The surface integral is zero. For the front faceABCD, the normal is alongı and on this facex = 1 giving Fx = 4z. The surfaceintegral is

∫ 1

0dy∫ 1

04zdz = 2

Adding the six contributions above, the surface integral is3/2 consistent with thedivergence theorem.Exercise :Verify the divergence theorem by calculating the surface integral of the vectorfield ~F = ıx3 + y3 + kz3 for the cubical volume of Example 17. (Ans. Surface

27

integral has value 3)Example 17 :In Example 13 we found that the surface integral of a vector field xı + y + zkover a cylinder of radiusR and heighth is 3πR2h. Verify this result using thedivergence theorem.Solution :In Example 16 we have seen that the divergence of the field vector is 3. Since theintegrand is constant, the volume integral is3V = 3πR2h.Example 18 :A vector field is given by~F = x3 ı+y3. Verify Divergence theorem for a cylinderof radius 2 and height 5. The origin of the coordinate system is at the centre of thebase of the cylinderand z-axis along the axis.Solution :The problem is obviously simple in cylindrical coordinates. The divergence canbe easily seen to be3(x2 + y2) = 3ρ2. Recalling that the volume element isρdρdθdz, the integral is

∫

div ~FdV =∫

3ρ2dV = 3∫ 2

0ρ3dρ

∫ 2π

0dθ∫ 5

0dz = 120π

In order to calculate the surface integral, we first observe that the end faces havetheir normals along±k. Since the field does not have any z- component, thecontribution 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

ı = ρ cos θ − θ sin θ

= ρ sin θ + θ cos θ

k = k

Using these

~F = ρ3(cos4 θ + sin4 θ)ρ + ρ3(sin3 θ cos θ − cos3 θ sin θ)θ

The area element on the curved surface isRdθdzρ, whereR is the radius. Thusthe surface integral is

∫

FρdS = R4∫ 2π

0(cos4 θ + sin4 θ)dθ

∫ 5

0dz

28

= 16.3π

2.5 = 120π

where we have used∫ 2π0 sin4 θdθ =

∫ 2π0 cos4 θdθ = 3π/4.

Exercise :In the Exercise following Example 13, we had seen that surface integral of thevector field~V = 2ρ− 3ρθ + zρk through the surface of a cylinder of radius 1 andheight 2 is28π/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 originat the centre of the circular base. Calculate the surface integral of the vector field~F = x3 ı + y3 + z3k in the hemisphere and verify the divergence theorem.Solution :The divergence of~F is easily calculated

∇ · ~F =∂Fx

∂x+

∂Fy

∂y+

∂Fz

∂z

=∂x3

∂x+

∂y3

∂y+

∂z3

∂z

= 3(x2 + y2 + z2) = 3r2

wherer is the distance from origin. The volume integral over the hemisphere isconveniently calculated in spherical polar using the volume elementr2 sin θdθdφ.Since it is a hemisphere withz = 0 as the base, the range ofθ is 0 to π/2.

∇ · ~FdV = 3∫

r2dV = 3∫ 1

0r4dr

∫ π/2

0sin θdθ

∫ 2π

0dφ

= 3 × 1

5× 1 × 2π =

6π

5

The surface integrals are calculated conveniently in spherical polar. There is nocontribution 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 thebase of the hemi-sphere isz = 0.In order to calculate the flux from the curved face we need to express the forcefield and the unit vectors in spherical polar coordinates. Using the tranformationproperties given earlier and observing that we only requirethe radial componentof the vector field since the area element is radially directed. UsingR2 sin θdθdφas the area element, a bit of laborious algebra gives

29

~F · ~dS = R5∫ π/2

0sin5 θdθ

∫ 2π

0cos4 φdφ + R5

∫ π/2

0sin5 θdθ

∫ 2π

0sin4 φdφ

+ R5∫ π/2

0cos4 θ sin θdθ

∫ 2π

0dφ

Using∫ π/20 sin5 θdθ = 8/15 and

∫ 2π0 cos4 φdφ = 3π/4, the above integral can be

seen to give the correct result.

1.5 Curl of a Vector - Stoke’s TheoremWe have seen that the line integral of a vector field

∫ ~F · ~dl is essentially a sumof the component of~F along the curve. If the line integral is taken over a closedpath, we represent it as

∮ ~F · ~dl. If the vector field is consevative, i.e., if thereexists a scalar functionV such that one can write~F as∇V , the contour integralis zero. In other cases, it is, in general, non-zero.Consider a contourC enclosing a surfaceS. We may split the contour into a largenumber of elementary surface areas defined by a mesh of closedcontours.

Since adjacent contours are traversedin opposite directions, the only non-vanishing contribution to the integralcomes from the boundary of the con-tour C. If the surface area enclosedby thei − th cell is∆Si, then

∮

C

=∑

i

∫

Ci

~F · ~dl

=∑

i

∫

Ci

~F · ~dl

∆Si

∆Si

We define the quantity

lim∆Si→0

∫

Ci

~F · ~dl

∆Si

ni

as the curl of the vector~F at a point P which lies on the surface∆Si. Since thearea∆Si is infinitisimal it is a point relationship. The direction ofni is, as usual,

30

along the outward normal to the area element∆Si. For instance, the x-componentof the curl is given by

(curl)x = lim∆y,∆z→0

∫

Ci

~F · ~dl

∆y∆zni

Thus∮

C

~F · ~dl =∫

Scurl ~F · dS

This isStoke’s Theoremwhich relates the surface integral of a curl of a vectorto the line integral of the vector itself. The direction of~dl and ~dS are fixed by theright hand rule, i.e. when the fingers of the right hand are curled to point in thediurection of~dl, the thumb points in the direction of~dS.1.5.1 Curl in Cartesian Coordinates :We will obtain an expression for the curl in the cartesian coordinates. Let usconsider 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 thecorners 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 thelength (orthe bredth) so that we may retain only the first term in a Taylorexpansion incomputing the field variation.

A(y,z) B(y+ y,z)∆

D(y,z+ z)∆ C(y+ y,z+ z)∆ ∆

y

z

x

n

31

Contribution to the line integral from the two sides AB and CDare computed asfollows.On AB :

∫ ~F · ~dl =∫ ~F · dy =

∫

Fydy

ON CD :∫ ~F · ~dl =

∫ ~F · (−)dy =∫

FydyUsing Taylor expansion (retaining only the first order term), we can write

Fy |CD= Fy |AB +∂Fy

∂z∆z

Thus the line integral from the pair of sides AB and CD is

−∫

∂Fy

∂z∆zdy ≈ −∂Fy

∂z∆z∆y

In a similar way one can calculate the contributions from thesides BC and DAand show it to be

∫ ∂Fz

∂y∆ydz ≈ ∂Fz

∂y∆y∆z

Adding up we get

(curl F )x = lim∆y,∆z→0

(

∂Fz

∂y− ∂Fy

∂z

)

∆y∆z

∆y∆z=

∂Fz

∂y− ∂Fy

∂z

In a very similar way, one can obtain expressions for the y andz components

(curl F )y =∂Fx

∂z− ∂Fz

∂x

(curl F )z =∂Fy

∂x− ∂Fx

∂y

One can write the expression for the curl of~F by using the del operator as

curl ~F = ∇× ~F =

∣

∣

∣

∣

∣

∣

∣

ı k∂∂x

∂∂y

∂∂z

Fx Fy Fz

∣

∣

∣

∣

∣

∣

∣

Expression for curl in Cylidrical and Spherical Coordinates :In the cylindrical coordinates the curl is given by

∇× ~F =

(

1

ρ

∂Fz

∂θ− ∂Fθ

∂z

)

ρ

+

(

∂Fρ

∂z− ∂Fz

∂ρ

)

θ +

(

1

ρ

∂

∂ρ(ρFθ) −

1

ρ

∂Fρ

∂θ

)

k

32

In the spherical coordinates the corresponding expressionfor the curl is

∇× ~F =1

r sin θ

(

∂

∂θ(Fφ sin θ) − ∂Fθ

∂φ

)

r

+1

r

(

1

sin θ

∂Fr

∂φ− ∂

∂r(rFφ)

)

θ +1

r

(

∂

∂r(rFθ) −

∂Fr

∂θ

)

φ

Example 20 :

Verify Stoke’s theorem for ~F =x2 ı + 2x + z2k for the rectangle shownbelow, defined by sidesx = 0, y = 0, x = 1andy = 1.

O x

y

(1,0)A

B(1,1)C

The line integrals along the four sides are∮

~F · ~dl =∫ 1

0Fx |y=0 dx +

∫ 1

0Fy |x=0 dy +

∫ 0

1Fx |y=1 dx +

∫ 1

0Fy |x=1 dy

=∫ 1

0x2dx + 0 + +

∫ 0

1x2dx +

∫ 1

02dy

=1

3+ 0 − 1

3+ 2 = 2

Since the normal to the plane is alongk, we only needz− component of∇× ~Fto calculate the surface integral. It can be checked that

(∇× ~F )z =∂Fy

∂x− ∂Fx

∂y= 2 − 0 = 2

Thus∫

S(∇× ~F ) · kdS =

∫ 1

0

∫ 1

02dxdy = 2

which agrees with the line integral calculated.Exercise :

33

A vector field is given by~F = −yı + x + x2k. Calcu-late the line integral of the fieldalong the triangular path shownabove. Verify your result by Stoke’stheorem. (Ans. 1)

O A

B

(Hint : To calculate the line integral along a straightline,you need the equationto the line. For instance, the equation to the line BO isy = 2x. Check that∫

AB~F · ~dl = − ∫ ydx +

∫

xdy = 1. )Example 21 :A vector field is given by~F = −yı + z + x2k. Calculate the line integral ofthe field along a circular path of radiusR in the x-y plane with its centre at theorigin. Verify Stoke’s theorem by considering the circle todefine (i) the plane ofthe circle and (ii) a cylinder of heightz = h.Solution :The curl of ~F may be calculated as

∇× ~F = −ı + 2x + k

Because of symmetry, we use cylindrical (polar) coordinates. The transformationsarex = ρ cos θ, y = ρ sin θ, z = z. The unit vectors are

ı = ρ cos θ − θ sin θ

= ρ sin θ − θ cos θ

k = k

Substituting the above, the field~F and its curl are given by

~F = ρ(−ρ sin θ cos θ + z sin θ)

+ θ(ρ sin2 θ + z cos θ) + kρ2 cos2 θ

∇× ~F = ρ(− cos θ + 2ρ cos θ sin θ)

+ θ(sin θ + 2ρ cos2 θ) + k

34

The line integral of ~F around the circularloop :Since the line element is~dl = Rdθθ,∮

~F · ~dl =∫ 2π

0(R sin2 θ + z cos θ)Rdθ

On the circlez = 0. The integral oversin2 θgives 1/2. Hence

∮

~F · ~dl = πR2

x

y

z

−k

k

^

^

ρ

θ^

^

Let us calculate the surface integral of the curl of the field over two surfaces boundby the circular curve.(i) On the circular surface bound by the curve in the x-y plane, the outward normalis alongk (right hand rule). Thus

∫

S(∇× ~F ) · (kdS) =

∫

SdS = πR2

(ii) For the cylindrical cup, we have two surfaces : the curved face of the cylinderon whichn = ρ and the top circular face on whichn = k. The contribution fromthe top circular cap isπR2, as before because the two caps only differ in theirz values (the z-component of the curl is independent ofz). The surface integralfrom the curved surface is (the area element isRdθdzρ)

∫ 2π

0Rdθ

∫ h

0dz(−R sin θ cos θ + z cos θ)

For both the terms of the above integral, the angle integration gives zero. Thus thenet surface integral isπR2, as expected.Exercise :A vector field is given by~F = kρ2zθ. Check the validity of the Stoke’s theorm

35

by calculating the line integral aboutthe closed contour in the form of acircle at z = 4 and also calculat-ing the surface integral of the opensurface of the cylinder below it, asshown. (Hint : Express the curl incylindrical coordinates and take careof the signs of the surface elementsfrom the curved surface and the bot-tom cap. Ans.64kπ )

x

y

z

C

R=2

h=4

Exercise :Let C be a closed curve in the x-y plane in the shape of a quadrant of acircle ofradiusR.

If ~F = ıy+z+kx. calculate the lineintegral of the field along the contourshown in a direction which is anti-clockwise when looked from abovethe plane (z > 0). Take the sur-face of the quadrant enclosed by thecurve as the open surface boundedby the curve and verify Stoke’s theo-rem. (Ans.−πR2/4)

x

y

z

O (1,0,0)

(0,1,0)

Example 22 :A vector field is given by~F (r, θ, φ) = f(r)φ whereφ is the azimuthal anglevariable of a spherical coordinate system. Calculate the line integral over a circleof radiusR in the x-y plane centered at the origin. Consider an open surface in theform of a hemispherical bowl in the northern hemisphere bounded by the circle.

36

Solution :On the equatorial circle~dl = Rdφφ. Hence,

∮

~F · ~dl =∫ 2π

0f(R)Rdφ = 2πRf(R)

The expression for curl in spherical coordinates maybe used to calculate the curl of~F . Since the fieldonly has azimuthal component, the curl has radial andpolar (θ) components.

∇× ~F =1

r sin θ

∂

∂θ(f(r) sin θ)r − 1

r

∂

∂r(rf(r))θ

=f(r) cos θ

r sin θr −

(

f(r)

r+

∂f(r)

∂r

)

θ

The area element on the surface of the northern hemi-sphere isR2 sin θdθdφr.

r

y

z

xHence the surface integral is

∫ 2π

0dφ∫ π/2

0

f(r) cos θ

R sin θR2 sin θdθ = 2πRf(R)

∫ π/2

0cos θdθ

= 2πRf(R)

Exercise :Verify Stoke’s theorem for a vector field2zı + 3x + 5yk where the contour is anequatorial circle of radiusR and is anticlockwise when viewed from above andthe surface is the hemisphere shown in the preceding example.. (Ans.3R2π)1.6 Laplacian :Since gradient of a scalar field gives a vector field, we may compute the diver-gence of the resulting vector field to obtain yet another scalar field. The operatordiv(grad)= ∇ · ∇ is called theLaplacian and is written as∇2.If V is a scalar, then,

∇2V = ∇ · (∇V )

=

(

ı∂

∂x+

∂

∂y+ k

∂

∂z

)(

ı∂V

∂x+

∂V

∂y+ k

∂V

∂z

)

=∂2V

∂x2+

∂V

∂y2+

∂2V

∂z2

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Example 23 :Calculate the Laplacian of1/r = 1/

√x2 + y2 + z2.

Solution :

∂2

∂x2

1√x2 + y2 + z2

=∂

∂x

−x

(x2 + y62 + z2)3/2

=2x2 − y2 − z2

(x2 + y2 + z2)5/2

=3x2 − r2

r5

Adding similar contributions from∂2/∂y2 and∂2/∂z2, we get

∇2 1

r=

3(x2 + y2 + z2) − 3r2

r5=

3r2 − 3r2

r5= 0

Laplacian in cylindrical and spherical coordinates:In cylindrical :

∇2 =1

ρ

∂

∂ρ

(

ρ∂

∂ρ

)

+1

ρ2

∂2

∂θ2+

∂2

∂z2

In spherical :

∇2 =1

r2

∂

∂r

(

r2 ∂

∂r

)

+1

r2 sin2 θ

∂

∂θ

(

sin θ∂

∂θ

)

+1

r2 sin2 θ

∂2

∂φ2

Frequently the Laplacian of a vector field is used. It is simply a short hand notationfor the componentwise Laplacian

∇2 ~F = ı∇2Fx + ∇2Fy + k∇2Fz

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-δfunction. The peculiarity of the function is that though thevalue of the function

38

is zero everywhere, other than at one point, the integral of the function over anyregion which includes this singular point is finite. We define

δ(x − a) = 0 if x 6= a

with∫

f(x)δ(x − a)dx = f(a)

wheref(x) is any function that is continuous atx = a, provided that the range ofintegration includes the pointx = a. Strictly speaking,δ(x − a) is not a functionin 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 theδfunction as a limit of a sequence of functions. For instance,if we define a functionδǫ(x) such that

δǫ(x) =1

ǫfor − ǫ

2< x < +

ǫ

2

= 0 for | x |> ǫ

2

Thenδ(x) can be thought of as thelimit of δǫ(x) asǫ → 0. x

− ε /2 + ε/20

1/ε

δε(x)

One can easily extend the definition to three dimensions

δ(~r) = δ(x)δ(y)δ(z)

which has the property∫

f(~r)δ(~r − ~a) = f(~a)

provided, of course, the range of integration includes the point ~r = ~a.Example :Evaluate

∫ 50 cos xδ(x − π)dx.

Solution :The range of integration includes the pointx = π at which the argument of thedelta function vanishes. Thus, the value of the integral iscos π = −1.Exercise :Evaluate

∫

~r·(~a−~r)δ(~r−~b)dr, where~a = (1, 2, 3),~b = (3, 2, 1) and the integrationis 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 regionwhich contains apoint chargeq. The charge density is zero everywhere except at the point wherethe charge is located. However, the volume integral of the density in any regionwhich includes this point is equal toq itself. Thus ifq 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 + y2 + z2, we have

∂r

∂x=

x

r,

∂r

∂y=

y

r,

∂r

∂z=

z

r

using this it is easy to show that

∂2

∂x2(1

r) =

3x2 − r2

r5

Thus

∇2(

1

r

)

=

(

∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)

(1

r) =

3(x2 + y2 + z2) − 3r2

r5= 0

However, the above is not true at the origin as1/r diverges atr = 0 and is notdifferentiable at that point. Interestingly, however, theintegral of∇2(1/r) overany volume which includes the pointr = 0 is not zero. As the value of theintegrand is zero everywhere excepting at the origin, the point r = 0 has to betreated with care.Consider an infinitisimally small sphere of radiusr0 with the centre at the origin.Using divergence theorem, we have,

∫

V∇2

(

1

r

)

d3r =∫

V∇ · ∇

(

1

r

)

d3r =∫

S∇(

1

ri)

· dS

where the last integral is over the surfaceS of the sphere defined above. As thegradient is taken at points on the surface for whichr 6= 0, we may replace∇(1/r)with −1/r2

0 at all points on the surface. Thus the value of the integral is

− 1

r20

∫

SdS = − 1

r20

· 4πr20 = −4π

Hence

∇2(

1

r

)

= −4πδ(r)

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