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IOP Concise Physics

ElectromagnetismProblems and solutions

Carolina C Ilie and Zachariah S Schrecengost

Chapter 1

Mathematical techniques

There are a variety of mathematical techniques required to solve problems inelectromagnetism. The aim of this chapter is to provide problems that will buildconfidence in these techniques. Concepts from vector calculus and curvilinearcoordinate systems are the primary focus.

1.1 Theory

1.1.1 Dot and cross product

Given vectors = ˆ + ˆ + ˆA A x A y A zx y z and = ˆ + ˆ + ˆB B x B y B zx y z

θ ⋅ = + + =A B A B A B A B AB cosx x y y z z

θ × =ˆ ˆ ˆ

× =A B

x y zA A A

B B BA B ABwith sinx y z

x y z

where = ∣ ∣ = + +A A A A Ax y z2 2 2 , = ∣ ∣ = + +B B B B Bx y z

2 2 2 , and θ is the angle

between A and B .

1.1.2 Separation vector

This notation is outlined by David J Griffiths in his book Introduction toElectrodynamics (1999, 2013). Given a source point ′r and field point r , theseparation vector points from ′r to r and is given by

r = − ′ = − ′ ˆ + − ′ ˆ + − ′ ˆr r x x x y y y z z z( ) ( ) ( )

doi:10.1088/978-1-6817-4429-2ch1 1-1 ª Morgan & Claypool Publishers 2016

and the unit vector pointing from ′r to r is

r rr

ˆ =

= − ′ − ′

= − ′ ˆ + − ′ ˆ + − ′ ˆ− ′ + − ′ + − ′

r rr r

x x x y y y z z z

x x y y z z

( ) ( ) ( )

( ) ( ) ( ).

2 2 2

As explained by Griffiths, this notation greatly simplifies later equations.

1.1.3 Transformation matrix

Given vector = ˆ + ˆ + ˆA A x A y A zx y z in coordinate system K, the components of A incoordinate system ′K are determined by rotational matrix R given by

=

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟R

R R R

R R R

R R R

xx xy xz

yx yy yz

zx zy zz

with

′

′

′

=

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟

A

A

A

RAA

A.

x

y

z

x

y

z

1.1.4 Gradient

Given a scalar functionT , the gradients for various coordinate systems are given below.

Cartesian

∇ = ∂∂

ˆ + ∂∂

ˆ + ∂∂

ˆTTx

xTy

yTz

z

Cylindrical

ϕϕ∇ = ∂

∂ˆ + ∂

∂ˆ + ∂

∂ˆT

Ts

ss

T Tz

z1

Spherical

θθ

θ ϕϕ∇ = ∂

∂ˆ + ∂

∂ˆ + ∂

∂ˆT

Tr

rr

Tr

T1 1sin

1.1.5 Divergence

Given vector function v , the divergences for various coordinate systems are givenbelow.

Cartesian

∇ ⋅ = ∂∂

+∂∂

+ ∂∂

vvx

v

yvz

x y z

Electromagnetism

1-2

Cylindrical

ϕ∇ ⋅ = ∂

∂+

∂∂

+ ∂∂

ϕvs s

svs

v vz

1( )

1s

z

Spherical

θ θθ

θ ϕ∇ ⋅ = ∂

∂+ ∂

∂+

∂∂θ

ϕ( ) ( )vr r

r vr

vr

v1 1sin

sin1

sinr2

2

1.1.6 Curl

Given vector function v , the curls for various coordinate systems are given below.

Cartesian

∇ × = ∂∂

−∂∂

ˆ + ∂∂

− ∂∂

ˆ +∂∂

− ∂∂

ˆ⎜ ⎟⎛⎝⎜

⎞⎠⎟

⎛⎝

⎞⎠

⎛⎝⎜

⎞⎠⎟v

vy

v

zx

vz

vx

yv

xvy

zz y x z y x

Cylindrical

ϕϕ

ϕ∇ × = ∂

∂−

∂∂

ˆ + ∂∂

− ∂∂

ˆ + ∂∂

− ∂∂

ˆϕϕ⎜ ⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝

⎞⎠

⎡⎣⎢

⎤⎦⎥v

sv v

zs

vz

vs s s

svv

z1 1

( )z s z s

Spherical

θ θθ

ϕ θ ϕθ

θϕ

∇ × = ∂∂

− ∂∂

ˆ + ∂∂

− ∂∂

ˆ

+ ∂∂

− ∂∂

ˆ

ϕθ

ϕ

θ

⎡⎣⎢

⎤⎦⎥

⎡⎣⎢

⎤⎦⎥

⎡⎣⎢

⎤⎦⎥

( )vr

vv

rr

vr

rv

r rrv

v

1sin

sin1 1

sin( )

1( )

r

r

1.1.7 Laplacian

Given a scalar function T , the Laplacians for various coordinate systems are givenbelow.

Cartesian

∇ = ∂∂

+ ∂∂

+ ∂∂

TT

xT

yTz

22

2

2

2

2

2

Cylindrical

ϕ∇ = ∂

∂∂∂

+ ∂∂

+ ∂∂

⎜ ⎟⎛⎝

⎞⎠T

s ss

Ts s

T Tz

1 122

2

2

2

2

Spherical

θ θθ

θ θ ϕ∇ = ∂

∂∂∂

+ ∂∂

∂∂

+ ∂∂

⎜ ⎟ ⎜ ⎟⎛⎝

⎞⎠

⎛⎝

⎞⎠T

r rr

Tr r

Tr

T1 1sin

sin1

sin2

22

2 2 2

2

2

Electromagnetism

1-3

1.1.8 Line integral

Given vector function v and path P , a line integral is given by

Pl∫ ⋅

v d ,

a

b

where a and b are the end points, and ld is the infinitesimal displacement vectoralong P . In Cartesian coordinates l = ˆ + ˆ + ˆx x y y z zd d d d .

1.1.9 Surface integral

Given vector function v and surface S , a surface integral is given by

S∫ ⋅ v ad ,

where ad is the infinitesimal area vector that has direction normal to the surface.Note that ad always depends on the surface involved.

1.1.10 Volume integral

Given scalar function T and volume V , a volume integral is given by

V∫ τT d ,

where τd is the infinitesimal volume element. In Cartesian coordinatesτ = x y zd d d d .

1.1.11 Fundamental theorem for gradients

Pl∫ ∇ ⋅ = −

( ) ( )T T b T a( ) da

b

1.1.12 Fundamental theorem for divergences (Gauss’s theorem, Green’s theorem,divergence theorem)

V S∫ ∮τ∇ ⋅ = ⋅ ( )v v ad d

1.1.13 Fundamental theorem for curls (Stoke’s theorem, curl theorem)

S Pl∫ ∮∇ × ⋅ = ⋅ ( )v a vd d

1.1.14 Cylindrical polar coordinates

Here our infinitesimal quantities are

l ϕ ϕ = ˆ + ˆ + ˆs s s z zd d d d

andτ ϕ= s s zd d d d .

Electromagnetism

1-4

1.1.15 Spherical polar coordinates

Here our infinitesimal quantities are

l θ θ θ ϕ ϕ = ˆ + ˆ + ˆr r r rd d d sin d

and

τ θ θ ϕ= r rd sin d d d .2

1.1.16 One-dimensional Dirac delta function

The one-dimensional Dirac delta function is given by

δ − = ≠∞ =

⎧⎨⎩x ax ax a

( )0

and has the following properties

∫ δ − =−∞

∞

x a x( )d 1

∫ δ − =−∞

∞

f x x a x f a( ) ( )d ( )

δ δ=kxk

x( )1

( ).

1.1.17 Theory of vector fields

If the curl of a vector field F vanishes everywhere, then F can be written as thegradient of a scalar potentialV :

∇ × ↔ = −∇F F V .

If the divergence of a vector vanishes everywhere, then F can be expressed as the curlof a vector potential A:

∇ ⋅ = ↔ = ∇ × F F A0 .

1.2 Problems and solutionsProblem 1.1. Given vectors = ˆ + ˆ + ˆA x y z3 9 5 and = ˆ − ˆ + ˆB x y z7 4 , calculate

⋅ A B and × A B using vector components and find the angle between A and Busing both products.

Electromagnetism

1-5

Solution

⋅ = ˆ + ˆ + ˆ ⋅ ˆ − ˆ + ˆ

= + − + = − +

⋅ = −

× =ˆ ˆ ˆ

−

= − − ˆ + − ˆ + − − ˆˆ × ˆ = ˆ − ˆ − ˆ

A B x y z x y z

A B

A Bx y z

x y z

A B x y z

(3 9 5 ) ( 7 4 )

(3)(1) (9)( 7) (5)(4) 3 63 20

40

3 9 51 7 4

[(9)(4) ( 7)(5)] [(1)(5) (3)(4)] [(3)( 7) (1)(9)]

71 7 30

To find the angle θ between A and B we must first calculate A and B:

= + + =

= + − + =

A

B

3 9 5 115

1 ( 7) 4 66 .

2 2 2

2 2 2

Using the dot product, we have

θ θ

θ

⋅ = → = −

= °

−⎛⎝⎜

⎞⎠⎟A B AB cos cos

40

115 66117.3 .

1

Using the cross product, we have

θ θ

θ

× = → + − + − =

= °

A B AB sin 71 ( 7) ( 30) 115 66 sin

62.7 .

2 2 2

Note, however, that we can see that the angle between A and B is greater than °90 .For any argument γ , γ− ° ⩽ ⩽ °−90 sin ( ) 901 . Since the angle between A and B isgreater than °90 , we must adjust for this by subtracting our angle from °180 .Therefore, θ = ° − ° = °180 62.7 117.3 as expected.

Electromagnetism

1-6

Problem 1.2. The scalar triple product states ⋅ × = ⋅ × A B C B C A( ) ( ). Provethis by expressing each side in terms of its components.

Solution Starting with the left-hand side, the cross product is

× =ˆ ˆ ˆ

= − ˆ + − ˆ + − ˆ( )( ) ( )

B C

x y zB B B

C C C

B C B C x B C B C y B C B C z.

x y z

x y z

y z z y z x x z x y y x

Now, dotting A with × B C( )

⋅ × = − + − + −

= − + − + −

= − + − + −

⋅ × = ⋅ − ˆ + − ˆ + − ˆ⎡⎣ ⎤⎦

( )

( )

( )

( ) ( ) ( )

( ) ( )

( ) ( ) ( )

A B C A B C B C A B C B C A B C B C

A B C A B C A B C A B C A B C A B C

B C A C A B C A C A B C A C A

A B C B C A C A x C A C A y C A C A z .

x y z z y y z x x z z x y y x

x y z x z y y z x y x z z x y z y x

x y z z y y z x x z z x y y x

y z z y z x x z x y y x

Note the term in brackets is precisely × C A, therefore

⋅ × = ⋅ × ( ) ( )A B C B C A

as desired. This procedure can easily be applied again to prove the final part of thetriple product,

⋅ × = ⋅ × = ⋅ × ( ) ( ) ( )A B C B C A C A B .

Problem 1.3. Given source vector θ θ′ = ˆ + ˆr r x r ycos sin and field vector = ˆr zz,find the separation vector r and the unit vector rˆ.

Solution We have

r

r

θ θ

θ θ

= − ′ = ˆ − ˆ + ˆ

= − ˆ − ˆ + ˆ( )r r zz r x r y

r x r y zz

cos sin

cos sin .

To determine the unit vector rˆ, we must first find the magnitude of r ,

r θ θ θ θ= − + − + = + + = +( )r r z r z r z( cos ) ( sin ) cos sin .2 2 2 2 2 2 2 2 2

Electromagnetism

1-7

So

r rr

θ θˆ =

= − ˆ − ˆ + ˆ+

r x r y zz

r z

cos sin.

2 2

Problem 1.4. Given A in coordinate system K, find the rotational matrix to give thecomponents in system ′K .

Solution From the figures, we have

′ = ′ = ′ = −A A A A A A, , .x y y x z z

We want to find the rotational matrix R that satisfies

′

′

′

=

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟

A

A

A

RAA

A.

x

y

z

x

y

z

From our equations above

′

′

′

=−

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟

A

A

A

AA

A

0 1 0

1 0 0

0 0 1

.

x

y

z

x

y

z

Electromagnetism

1-8

Therefore,

=−

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟R

0 1 0

1 0 0

0 0 1

.

Problem 1.5. Find the gradient of the following functions:a) = + +T x y z4 2 3

b) =T x y zln2 3

c) = +T x y z2 3

Solutionsa) = + +T x y z4 2 3

∇ = ∂∂

ˆ + ∂∂

ˆ + ∂∂

ˆ = ˆ + ˆ + ˆTTx

xTy

yTz

z x x yy z z4 2 33 2

b) =T x y zln2 3

∇ = ∂∂

ˆ + ∂∂

ˆ + ∂∂

ˆ = ˆ + ˆ + ˆTTx

xTy

yTz

z xz y xx z

yy x z y z2 ln 3 ln3

2 32 2

c) = +T x y z2 3

∇ = ∂∂

ˆ + ∂∂

ˆ + ∂∂

ˆ = ˆ + ˆ + ˆTTx

xTy

yTz

z xyx x y z z2 32 2

Problem 1.6. Find the divergence of the following functions:a) = ˆ − ˆ + ˆv xyx y zy z z2 2 3

b) = + ˆ + + ˆ + + ˆv x y x y z y z x z( ) ( ) ( )

Solutionsa) = ˆ − ˆ + ˆv xyx y zy z z2 2 3

∇ ⋅ = ∂∂

+∂∂

+ ∂∂

= − +vvx

v

yvz

y yz z4 3x y z 2

b) = + ˆ + + ˆ + + ˆv x y x y z y z x z( ) ( ) ( )

∇ ⋅ = ∂∂

+∂∂

+ ∂∂

= + + =vvx

v

yvz

1 1 1 3x y z

Problem 1.7. Find the curl of the following functions:a) = ˆ − ˆ + ˆv xyx y zy z z2 2 3

b) = + ˆ + + ˆ + + ˆv x y x y z y z x z( ) ( ) ( )c) = ˆ + ˆv x x y ysin cos

Electromagnetism

1-9

Solutionsa) = ˆ − ˆ + ˆv xyx y zy z z2 2 3

∇ × = ∂∂

−∂∂

ˆ + ∂∂

− ∂∂

ˆ +∂∂

− ∂∂

ˆ

= ∂∂

−∂ −

∂ˆ +

∂∂

−∂

∂ˆ

+∂ −

∂−

∂∂

ˆ

= + ˆ + − ˆ + − ˆ

∇ × = ˆ − ˆ

⎜ ⎟⎛⎝⎜

⎞⎠⎟

⎛⎝

⎞⎠

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎡

⎣⎢⎢

⎤

⎦⎥⎥

( ) ( )

( )

( )

vvy

v

zx

vz

vx

yv

xvy

z

zy

y z

zx

xy

z

z

xy

y z

x

xy

yz

y x y x z

v y x xz

( ) 2 ( )

2 ( )

0 2 (0 0) (0 )

2

z y x z y x

3 2 3

2

2

2

b) = + ˆ + + ˆ + + ˆv x y x y z y z x z( ) ( ) ( )

∇ × = ∂∂

−∂∂

ˆ + ∂∂

− ∂∂

ˆ +∂∂

− ∂∂

ˆ

= ∂ +∂

− ∂ +∂

ˆ + ∂ +∂

− ∂ +∂

ˆ

+ ∂ +∂

− ∂ +∂

ˆ

∇ × = − ˆ − ˆ − ˆ

⎜ ⎟⎛⎝⎜

⎞⎠⎟

⎛⎝

⎞⎠

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

⎡⎣⎢

⎤⎦⎥

⎡⎣⎢

⎤⎦⎥

vvy

v

zx

vz

vx

yv

xvy

z

z xy

y zz

xx y

zz x

xy

y zx

x yy

z

v x y z

( ) ( ) ( ) ( )

( ) ( )

z y x z y x

c) = ˆ + ˆv x x y ysin cos

∇ × = ∂∂

−∂∂

ˆ + ∂∂

− ∂∂

ˆ +∂∂

− ∂∂

ˆ⎜ ⎟⎛⎝⎜

⎞⎠⎟

⎛⎝

⎞⎠

⎛⎝⎜

⎞⎠⎟v

vy

v

zx

vz

vx

yv

xvy

zz y x z y x

= ∂∂

− ∂∂

ˆ + ∂∂

− ∂∂

ˆ

+ ∂∂

− ∂∂

ˆ =

⎡⎣⎢

⎤⎦⎥

⎡⎣⎢

⎤⎦⎥

⎡⎣⎢

⎤⎦⎥

yy

zx

xz x

y

yx

xy

z

(0) (cos ) (sin ) (0)

(cos ) (sin )0

Electromagnetism

1-10

Problem 1.8. Prove ∇ × ∇ =T( ) 0.

Solution

∇ × ∇ =

ˆ ˆ ˆ

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

( )T

x y z

x y z

Tx

Ty

Tz

= ∂∂

∂∂

− ∂∂

∂∂

ˆ + ∂∂

∂∂

− ∂∂

∂∂

ˆ

+ ∂∂

∂∂

− ∂∂

∂∂

ˆ

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎜ ⎟

⎡⎣⎢

⎛⎝

⎞⎠

⎛⎝⎜

⎞⎠⎟⎤⎦⎥

⎡⎣⎢

⎛⎝

⎞⎠

⎛⎝

⎞⎠⎤⎦⎥

⎡⎣⎢

⎛⎝⎜

⎞⎠⎟

⎛⎝

⎞⎠⎤⎦⎥

yTz z

Ty

xz

Tx x

Tz

y

xTy y

Tx

z

∇ × ∇ =T( ) 0.

Problem 1.9. Find the Laplacian of the following functions:a) = + + +T x y xz 32

b) = +T y ze sin cos(2 )x

c) =T x ysin cosd) = ˆ + ˆ − ˆv xyx z y z22

Solutionsa) = + + +T x y xz 32

∇ = ∂∂

+ ∂∂

+ ∂∂

= + + =TT

xT

yTz

0 2 0 222

2

2

2

2

2

b) = +T y ze sin cos(2 )x

∇ = ∂∂

+ ∂∂

+ ∂∂

= − −

= −

TT

xT

yTz

y z y z

y z

e sin cos(2 ) 4 sin cos(2 )

e 5 sin cos(2 )

x

x

22

2

2

2

2

2

c) =T x ysin cos

∇ = ∂∂

+ ∂∂

+ ∂∂

= − − = −TT

xT

yTz

x y x y x ysin cos sin cos 2 sin cos22

2

2

2

2

2

Electromagnetism

1-11

d) = ˆ + ˆ − ˆv xyx z y z22

∇ = ∂∂

+ ∂∂

+ ∂∂

ˆ +∂∂

+∂∂

+∂∂

ˆ

+ ∂∂

+ ∂∂

+ ∂∂

ˆ

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

vv

xv

yvz

xv

x

v

y

v

zy

vx

vy

vz

z

x x x y y y

z z z

22

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

∇ = + + ˆ + + + ˆ+ + + ˆ = ˆv x y z y(0 0 0) (0 0 2) (0 0 0) 22

Problem 1.10. Test the divergence theorem with = ˆ + ˆ + − ˆv xyx y z y x z y z2 ( 2 )2 3 2

and the volume below.

Solution The divergence theorem states

V S∫ ∮τ∇ ⋅ = ⋅ v v ad d .

Starting with the left-hand side, we have the divergence

∇ ⋅ = + + = + +( )v y yz x y z x2 2 2 1 .3 2 3 2

We must split the volume into two pieces, (a) ⩽ ⩽y0 1 and (b) ⩽ ⩽y1 2.(a)

∫ ∫ ∫ ∫∇ ⋅ τ = + + =⎡⎣ ⎤⎦( )v y z x y x zd 2 1 d d d523

0

2

0

2

0

1

3 2

Electromagnetism

1-12

(b)

∫ ∫ ∫ ∫∇ ⋅ τ = + + =− ⎡⎣ ⎤⎦( )v y z x y x zd 2 1 d d d

17615

y

0

2

1

2

0

4 2

3 2

So,

V∫ ∇ ⋅ τ = + =v d

523

17615

43615

.

Now we solveS

∮ ⋅ v ad , which must be evaluated over the six sides.

(i) We must split this region into two sections (a) and (b), and = ˆa y z zd d d with=x 2.In (a), ⩽ ⩽y0 1,

∫ ∫ ∫ ⋅ = =v a y y zd 2(2) d d 4.0

2

0

1

In (b), ⩽ ⩽y1 2 and ⩽ ⩽ −z y0 4 2

∫ ∫ ∫ ⋅ = =−

v a y z yd 2(2) d d163

.

y

1

2

0

4 2

Electromagnetism

1-13

(ii) Here, = − ˆa y z xd d d and =x 0, so ⋅ = − ˆ =v a y y x xd 2(0) ( d d ) 0.(iii) Here, = ˆa x y zd d d and =z 2

∫ ∫ ∫ ⋅ = − =⎡⎣ ⎤⎦a x y xv d (2) 2y d d103

.0

2

0

1

2

(iv) Here, = − ˆa x y zd d d and =z 0

∫ ∫ ∫ ⋅ = − − =⎡⎣ ⎤⎦v a x y x yd (0) 2 ( d d ) 8.0

2

0

2

2

(v) Here = ˆa x z yd d d and =y 0, so ⋅ = − =v a z x zd 0 ( d d ) 02 3 .

(vi) Here, we have = ′ ˆa x z nd d d where ˆ = n nn. We can find n by crossing vectors

= − ˆ + ˆA y z2 and = ˆB x2 (the edges of the volume):

= × =ˆ ˆ ˆ

− = ˆ + ˆn A Bx y z

y z0 1 22 0 0

4 2 .

So

= + =n 4 2 2 52 2

and

ˆ = ˆ + ˆn y z2 5

55

5.

We can also obtain ′zd by considering

Electromagnetism

1-14

so ′ =z zd d52

. Now

= ˆ + ˆ = ˆ + ˆ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟a x z y z y z x zd

52

d d2 5

55

512

d d

and

= − → = −z y yz

4 2 22

.

So

∫ ∫ ∫ ⋅ = + −⎡⎣⎢

⎤⎦⎥( )v a y z x z y x zd

12

2 d d0

2

0

2

2 3 2

∫ ∫= − + − − =⎜ ⎟ ⎜ ⎟⎧⎨⎩

⎛⎝

⎞⎠

⎡⎣⎢

⎛⎝

⎞⎠⎤⎦⎥⎫⎬⎭

zz x z

zx z2

212

2 22

d d425

.0

2

0

22

3 2

Therefore

S∮ ⋅ = + + + + =v ad 4

163

103

8425

43615

as expected.

Problem 1.11. Test the curl theorem with = ˆ + ˆ + ˆv xy x yz y x zz5 42 2 2 and the surfacebelow.

Electromagnetism

1-15

Solution The curl theorem states

S Pl∫ ∮∇ × ⋅ = ⋅ v a v( ) d d .

Starting with the left-hand side, the curl is given by

∇ × =

ˆ ˆ ˆ∂

∂∂∂

∂∂ = − ˆ − ˆ − ˆv

x y z

x y z

xy yz x z

yzx xzy xyz

5 4

2 8 10

2 2 2

We also have = ˆa y z xd d d with ⩽ ⩽ − − +z y0 ( 2) 42 . So

S∫ ∫ ∫∇ × ⋅ = − = −

− − +

v a yz z y( ) d 2 d d1024

15.

y

0

4

0

( 2) 42

Now to solveP

l∮ ⋅ v d over the two paths (i) and (ii):

(i) Here we have =x 0, =z 0, and l = ˆy yd d . So l ⋅ = =v y yd (0 )d 02 .(ii) Here we have l = ˆ + ˆy y z zd d d , =x 0, and = − − +z y( 2) 42

P Pl∫ ∫ ∫ ⋅ = + = − − + = −⎡⎣ ⎤⎦v yz y z z y y yd d 4(0 ) d ( 2) 4 d

102415

.4

0

2 2 2 2

Electromagnetism

1-16

So,

Pl∮ ⋅ = + − = −

v d 0102415

102415

as expected.

Problem 1.12. Test the gradient theorem with = −T xz y z3 2 2 and path =z y2 and=z y3 from →(0, 0, 0) (0, 1, 1).

Solution The gradient theorem states

l∫ ∇ ⋅ = −

( ) ( )T T b T ad .a

b

Starting with the right side, we have

− = − − = −T T(0,1,1) (0,0,0) 3(0)(1 ) (1 )(1) 0 1.2 2

Now to solve l∫ ∇ ⋅

T da

b

, the gradient of T is given by

∇ = ∂∂

− ˆ + ∂∂

− ˆ + ∂∂

− ˆ( ) ( ) ( )Tx

xz y z xy

xz y z yz

xz y z z3 3 32 2 2 2 2 2

∇ = ˆ − ˆ + − ˆ( )T z x yz y xz y z3 2 6 .2 2

Electromagnetism

1-17

Here, l = ˆ + ˆy y z zd d d with =x 0. So

l∇ ⋅ = − + − = − −⎡⎣ ⎤⎦T yz y z y z yz y y zd 2 d 6(0)( ) d 2 d d .2 2

For path (i), we have = → =z y z y yd 2 d2 . So

l∇ ⋅ = − − = −( )T y y y y y y y yd 2 d (2 d ) 4 d2 2 3

and

l∫ ∫∇ ⋅ = − = −

T y yd 4 d 1a

b

0

1

3

as expected. For path (ii), we have = → =z y z y yd 3 d3 2 . So

l∇ ⋅ = − − = −( ) ( )T y y y y y y y yd 2 d 3 d 5 d3 2 2 4

and

l∫ ∫∇ ⋅ = − = −

T y yd 5 d 1a

b

0

1

4

also as expected.

Problem 1.13. Verify the following integration by parts given =f xy z2 and = ˆ + ˆ − ˆA z x xyy x zz42 2 and the surface below,

S S Pl∫ ∫ ∮∇ × ⋅ = × ∇ ⋅ + ⋅ ⎡⎣ ⎤⎦( ) ( )f A a A f a fAd d d .

Electromagnetism

1-18

Solution Starting with the left-hand side

∇ × =

ˆ ˆ ˆ∂

∂∂∂

∂∂

−

= − − ˆ + ˆ = + ˆ + ˆA

x y z

x y z

z xy x z

z xz y yz z x y yz

4

[2 ( 2 )] 4 2 ( 1) 4 .

2 2

Now

∇ × = + ˆ + ˆ( )f A xy z x y xy zz2 ( 1) 4 .2 2 3

Here we have = ′ ˆa x z nd d d where ˆ = ˆ + ˆn x y and =n 2 so ˆ = ˆ + ˆn x y22

22

. Alsofrom

we have

′ = = −x x y xd 2 d with 1 .

Now

= ˆ + ˆ = ˆ + ˆ⎛⎝⎜

⎞⎠⎟ ( )a x z x y x z x yd 2 d d

22

22

d d .

Therefore,

S∫ ∫ ∫∇ × ⋅ = + ˆ + ˆ ⋅ ˆ + ˆ⎡⎣ ⎤⎦( )f A a xy z x y xy zz x y x zd 2 ( 1) 4 ( )d d

0

1

0

1

2 2 3

∫ ∫= − + =x x z x x z2 (1 ) ( 1)d d790

.0

1

0

1

2 2

Next, we will solve theP

l∮ ⋅ fA d term for the four segments.

Electromagnetism

1-19

Segment (i)

= → = =z f xy0 (0) 0.2

Segment (ii)

= → = =x f y z0 (0) 0.2

Segment (iii)

l = ˆ + ˆ = = − → = −x x y y z y x y xd d d , 1, and 1 d d .

Segment (iv)

= → = =y f x z0 (0 ) 0.2

So

l ⋅ = + = − − −⎡⎣ ⎤⎦( ) ( )f A xy z x xy y x x x x xd d 4 d (1 ) 1 4 (1 ) d2 2 2

and

Pl∮ ∫ ⋅ = − − − =⎡⎣ ⎤⎦fA x x x x xd (1 ) 1 4 (1 ) d

160

.0

1

2

Now to solve theS

∫ × ∇ ⋅ A f a[ ] d term. First, we have

∇ = ˆ + ˆ + ˆf y zx xyzy xy z2 .2 2

Electromagnetism

1-20

So

× ∇ =ˆ ˆ ˆ

−A f

x y z

z xy x z

y z xyz xy

( ) 4

2

2 2

2 2

= + ˆ + − − ˆ + − ˆ( ) ( ) ( )x y x yz x x y z xy z y xyz xy z z4 2 2 4 .2 3 3 2 2 2 2 2 2 3 3

As before, = ˆ + ˆa x z x yd d d ( ). So

S∫ ∫ ∫× ∇ ⋅ = − + −

− − − −

⎡⎣

⎤⎦

A f a x x x x z

x x z x x z x z

[ ( )] d 4 (1 ) 2 (1 )

(1 ) (1 ) d d0

1

0

1

2 3 3 2

2 2 2 2 2

S∫ × ∇ ⋅ =A f a[ ( )] d

11180

.

So

S Pl∫ ∮× ∇ ⋅ + ⋅ = + =A f a fA[ ( )] d d

11180

160

790

as expected.

Problem 1.14. Find the divergence and curl of the following functions:a) θ ϕ θ θ ϕ ϕ = ˆ + ˆ + ˆv r r cos sin sin cos2

b) ϕ ϕ ϕ ϕ ϕ = ˆ + ˆ+ ˆv s s z zcos cos sin sin

Solutionsa) θ ϕ θ θ ϕ ϕ = ˆ + ˆ + ˆv r r cos sin sin cos2

θ θθ

θ ϕ∇ ⋅ = ∂

∂+ ∂

∂+

∂∂θ

ϕ( ) ( )vr r

r vr

vr

v1 1sin

sin1

sinr2

2

θ θθ θ ϕ

θ ϕθ ϕ= ∂

∂+ ∂

∂+ ∂

∂( ) ( )r r

rr r

1 1sin

sin cos sin1

sin(sin cos )

24

ϕθ

θ θ ϕ= + − + + −( ) ( )r

rr r

14

sinsin

sin cos1

( sin )2

3 2 2

Electromagnetism

1-21

ϕθ

θ ϕ= + − −( )rr r

4sinsin

1 2 sinsin2

ϕ θ θ∇ ⋅ = + − −( )v rr

4sin

csc 2 sin 1

θ θθ

ϕ

θ ϕθ

θϕ

∇ × = ∂∂

− ∂∂

ˆ

+ ∂∂

− ∂∂

ˆ + ∂∂

− ∂∂

ˆ

ϕθ

ϕ θ

⎡⎣⎢

⎤⎦⎥

⎡⎣⎢

⎤⎦⎥

⎡⎣⎢

⎤⎦⎥

( )vr

vv

r

rv

rrv

r rrv

v

1sin

sin

1 1sin

( )1

( )r r

θ θθ ϕ

ϕθ ϕ

θ ϕθ ϕ θ

θ ϕϕ

ϕ

= ∂∂

− ∂∂

ˆ

+ ∂∂

− ∂∂

ˆ

+ ∂∂

− ∂∂

ˆ

⎡⎣⎢

⎤⎦⎥

⎡⎣⎢

⎤⎦⎥

⎡⎣⎢

⎤⎦⎥

( )

( )

( )

rr

rr

rr

r rr r

1sin

sin cos (cos sin )

1 1sin

( sin cos )

1( cos sin )

2

2

2

θθ θ ϕ θ ϕ θ ϕ θ

θ ϕ ϕ

= − ˆ − ˆ

+ ˆr

rr

r

1sin

(2 sin cos cos cos cos )sin cos

cos sin

θ ϕ θ θ ϕ θ θ ϕ ϕ∇ × = − ˆ − ˆ + ˆvr

rr r

cos cos(2 csc )

sin cos cos sin

b) ϕ ϕ ϕ ϕ ϕ = ˆ + ˆ + ˆv s s z zcos cos sin sin

ϕ∇ ⋅ = ∂

∂+

∂∂

+ ∂∂

ϕvs s

svs

v vz

1( )

1s

z

ϕϕ

ϕ ϕ ϕ= ∂∂

+ ∂∂

+ ∂∂( )

s ss

s zz

1cos

1(cos sin ) ( sin )2

ϕ ϕ ϕ ϕ= + − + +( )s

2 cos1

sin cos sin2 2

ϕ ϕ ϕ ϕ∇ ⋅ = + + −v

s2 cos sin

cos sin2 2

Electromagnetism

1-22

ϕϕ

ϕ∇ × = ∂

∂−

∂∂

ˆ + ∂∂

− ∂∂

ˆ + ∂∂

− ∂∂

ˆϕϕ⎜ ⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝

⎞⎠

⎡⎣⎢

⎤⎦⎥v

sv v

zs

vz

vs s s

svv

z1 1

( )z s z s

ϕϕ ϕ ϕ ϕ ϕ ϕ

ϕ ϕϕ

ϕ

= ∂∂

− ∂∂

ˆ + ∂∂

− ∂∂

ˆ

+ ∂∂

− ∂∂

ˆ

⎡⎣⎢

⎤⎦⎥

⎡⎣⎢

⎤⎦⎥

⎡⎣⎢

⎤⎦⎥

sz

zs

zs

sz

s ss s z

1( sin ) (cos sin ) ( cos ) ( sin )

1( cos sin ) ( cos )

ϕ ϕ ϕ ϕ= ˆ + + ˆzs

ss

s zcos1

(cos sin sin )

ϕ ϕ ϕ∇ × = ˆ + + ˆvzs

ss

s zcossin

(cos )

Problem 1.15. Find the gradient and Laplacian of:a) θ ϕ θ ϕ= +T r (cos sin sin cos )2

b) ϕ ϕ= −T z ssin cos2 2

Solutionsa) θ ϕ θ ϕ= +T r (cos sin sin cos )2

θθ

θ ϕϕ∇ = ∂

∂ˆ + ∂

∂ˆ + ∂

∂ˆT

Tr

rr

Tr

T1 1sin

θ ϕ θ ϕ θ ϕ θ ϕ θ

θθ ϕ θ ϕ ϕ

= + ˆ + − + ˆ

+ − ˆ

r rr

r

rr

2 (cos sin sin cos )1

( sin sin cos cos )

1sin

(cos cos sin sin )

2

2

θ ϕ θ ϕ θ ϕ θ ϕ θ

θθ ϕ θ ϕ ϕ

= + ˆ + − ˆ

+ − ˆ

r r r

r

2 (cos sin sin cos ) (cos cos sin sin )

sin(cos cos sin sin )

θ ϕ θ ϕ θθ

θ ϕ ϕ∇ = + ˆ + + ˆ + + ˆT r r rr

2 sin( ) cos( )sin

cos( ) .

Note we could have written T as θ ϕ= +T r sin( )2 and then computed thegradient.

θ θθ

θ θ ϕ∇ = ∂

∂∂∂

+ ∂∂

∂∂

+ ∂∂

⎜ ⎟ ⎜ ⎟⎛⎝

⎞⎠

⎛⎝

⎞⎠

⎛⎝⎜

⎞⎠⎟T

r rr

Tr r

Tr

T1 1sin

sin1

sin2

22

2 2 2

2

2

Electromagnetism

1-23

θ ϕθ θ

θ θ ϕ

θ ϕθ ϕ

= ∂∂

+ + ∂∂

+

+ ∂∂

+

⎡⎣ ⎤⎦ ⎡⎣ ⎤⎦⎡⎣ ⎤⎦

r rr

rr

rr

12 sin( )

1sin

sin cos( )

1sin

cos( )

23

22

2 22

θ ϕθ

θ θ ϕ θ θ ϕ

θθ ϕ

= + + + − +

+ − +( )

[ ]6 sin( )1

sincos cos( ) sin sin( )

1sin

sin( )2

θ ϕ θθ

θ ϕ θ ϕθ

∇ = + + + − +T 5 sin( )

cossin

cos( )sin( )

sin.2

2

b) ϕ ϕ= −T z ssin cos2 2

ϕϕ∇ = ∂

∂ˆ + ∂

∂ˆ + ∂

∂ˆT

Ts

ss

T Tz

z1

ϕ ϕϕ

ϕ ϕ ϕ

ϕ ϕ

= ∂∂

− ˆ + ∂∂

− ˆ

+ ∂∂

− ˆ

( ) ( )

( )s

z s ss

z s

zz s z

sin cos1

sin cos

sin cos

2 2 2 2

2 2

ϕ ϕ ϕ ϕ ϕ ϕ= − ˆ + − − ˆ + ˆ⎡⎣ ⎤⎦ss

z s z zcos1

cos 2 cos ( sin ) 2 sin2 2

ϕ ϕ ϕ ϕ ϕ∇ = − ˆ + + ˆ + ˆ( )T ss

z s z zcoscos

2 sin 2 sin2 2

ϕ∇ = ∂

∂∂∂

+ ∂∂

+ ∂∂

⎜ ⎟⎛⎝

⎞⎠T

s ss

Ts s

T Tz

1 122

2

2

2

2

ϕ ϕ ϕ∂∂

= − → ∂∂

= − → ∂∂

∂∂

= −⎜ ⎟⎛⎝

⎞⎠

Ts

sTs

ss

sTs

cos cos cos2 2 2

ϕϕ ϕ ϕ

ϕϕ ϕ ϕ∂

∂= + → ∂

∂= − + − +( )T

z sT

z scos 2 cos sin sin 2 sin cos22

22 2 2

ϕ ϕ∂∂

= → ∂∂

=Tz

zTz

2 sin 2 sin2

ϕ ϕ ϕ ϕ ϕ∇ = − + − + − +( )Ts s

zs

cos 2sin cos sin 2 sin2

22 2

2

2

Electromagnetism

1-24

ϕ ϕ ϕ∇ = − + −⎛⎝⎜

⎞⎠⎟T

s szs

cos 2sin 2 sin2

22

2

2

Problem 1.16. Test the divergence theorem with ϕ θ θ θ = ˆ + ˆ +v r r rcos cos sinϕ ϕr sin and the volume below (the upper half of the sphere of radius R with a cone

of radius =a R

3cut out).

Solution The divergence theorem states

V S∫ ∮τ∇ ⋅ = ⋅ v v ad d .

Starting with the left-hand side, the divergence is

θ θθ

θ ϕ∇ ⋅ = ∂

∂+ ∂

∂+

∂∂θ

ϕ( ) ( )vr r

r vr

vr

v1 1sin

sin1

sinr2

2

ϕθ θ

θ θθ ϕ

ϕ= ∂∂

+ ∂∂

+ ∂∂( ) ( )

r rr

rr

rr

1cos

1sin

sin cos1

sin( sin )

23 2

ϕθ

θ θ θ ϕθ

= + − +( )3 cos1

sin2 sin cos sin

cossin

2 3

ϕ θ θ ϕθ

∇ ⋅ = + − +v 3 cos 2 cos sincossin

.2 2

For the volume,

π π θ π ϕ π⩽ ⩽ = = → ⩽ ⩽ ⩽ ⩽− −⎜ ⎟⎛⎝

⎞⎠

⎛⎝⎜

⎞⎠⎟r R

aR

0 , tan tan1

3 6 6 2, 0 2 .1 1

Electromagnetism

1-25

So

V∫ ∫ ∫ ∫τ ϕ θ θ ϕ

θθ ϕ θ∇ ⋅ = + − +

π

ππ

⎜ ⎟⎛⎝

⎞⎠( )v r rd 3 cos 2 cos sin

cossin

sin d d d

R

06

2

0

2

2 2 2

V∫ τ π∇ ⋅ = −v Rd

312

.3

Now for the right-hand side, we have three surfaces: the bottom (i), the outer shell(ii), and the inner part where the cone is cut out (iii). We have

ϕ ϕ ϕ θ θ θ = ˆ + ˆ + ˆv r r r rcos sin cos sin .

For (i), we have ϕ θ = ˆa r rd d d and θ = π2. So

π π ϕ ⋅ = =v a r rd cos2

sin2

d d 02

and

∫ ⋅ =v ad 0.i( )

For (ii), we have =r R and θ θ ϕ θ θ ϕ = ˆ = ˆa r r R rd sin d d sin d d2 2 . So

ϕ θ θ ϕ ⋅ =v a Rd cos sin d d3

and

∫ ∫ ∫ ϕ θ θ ϕ ⋅ = =π

π

π

v a Rd cos sin d d 0.ii( ) 0

2

6

23

For (iii), we have θ = π6and θ ϕ θ ϕ θ = − ˆ = − ˆa r r r rd sin d d d d1

2. So

π π ⋅ = − = −v a r rd12

cos6

sin6

38

2 2

and

∫ ∫ ∫ ϕ π ⋅ = − = −π

v a r r Rd3

8d d

312

.iii

R

( ) 0 0

2

2 3

Electromagnetism

1-26

Therefore,

S∮ π ⋅ = −v a Rd

312

3

as expected.

Problem 1.17. Test the curl theorem with ϕ ϕ ϕ ϕ = ˆ + ˆ + ˆv s z s zs zsin cos cos2

and half of a cylindrical shell with radius R and height h.

Solution The curl theorem states

S Pl∫ ∮∇ × ⋅ = ⋅ ( )v a vd d .

Starting with the left-handed side, we have

ϕ ϕ = ˆ = ˆa s z s R z sd d d d d .

Since we are dotting ad with ∇ × v , we only need the s component of the curl:

ϕ ϕϕ ϕ ϕ

ϕ

∇ × = ∂∂

−∂∂

ˆ = ∂∂

− ∂∂

ˆ

= − ˆ

ϕ⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥v

sv v

zs

szs

zs

z s

[ ]1 1

( cos ) (sin cos )

sin .

sz

So

ϕ ϕ∇ × ⋅ = −( )v a Rz zd sin d d .

Electromagnetism

1-27

We have

ϕ π⩽ ⩽ ⩽ ⩽z h0 and 0

so

S∫ ∫ ∫ ϕ ϕ∇ × ⋅ = − = −

π

( )v a Rz z h Rd sin d d .

h

0 0

2

For the left-hand side, we have four curves

with

ϕ ϕ ϕ ϕ = ˆ + ˆ + ˆv s z s z zsin cos cos .2

For curve (i), l ϕ ϕ = ˆd d , =z 0, and =s R. So

l ϕ ϕ ϕ ⋅ =v d sin cos d

and

∫ ϕ ϕ ϕ =π

sin cos d 0.0

For curve (ii), l = ˆz zd d , ϕ π= , and =s R. So

l ϕ π ⋅ = = = −v zs z zR z zR zd cos d cos d d

and

∫ − = −zR z h Rd12

.

h

0

2

Electromagnetism

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For curve (iii), l ϕ ϕ = ˆd d , =z h, and =s R. So

l ϕ ϕ ϕ ⋅ =v d sin cos d

and

∫ ϕ ϕ ϕ =π

sin cos d 0.

0

For curve (iv), l = ˆz zd d , ϕ = 0, and =s R. So

l ϕ ⋅ = = =v zs z zR z zR zd cos d cos(0)d d

and

∫ = −zR z h Rd12

.h

0

2 2

So,

P

l∮ ⋅ = − − = −v h R h R h Rd12

12

2 2 2

as expected.

Problem 1.18. Test the gradient theorem using ϕ=T sz sin2 and the half helix path(radius R, height h).

Electromagnetism

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Solution The gradient theorem states

Pl∫ ∇ ⋅ = − ( ) ( )T T b T ad .

Starting with the right-hand side

π π π π − = − − = − − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎛⎝

⎞⎠

⎛⎝

⎞⎠

⎛⎝

⎞⎠( ) ( )T b T a T R h T R Rh R h R,

2, ,

2, 0 sin

2(0) sin

2.2 2 2

Now, the gradient is

ϕϕ ϕ ϕ ϕ ϕ∇ = ∂

∂ˆ + ∂

∂ˆ + ∂

∂ˆ = ˆ + ˆ + ˆT

Ts

ss

T Tz

z z s z sz z1

sin cos 2 sin .2 2

We also have =s R and l ϕ ϕ ϕ ϕ = ˆ + ˆ = ˆ + ˆs z z R z zd d d d . So

l ϕ ϕ ϕ∇ ⋅ = +T Rz Rz zd cos d 2 sin d .2

We need a way to relate z and ϕ. Note that as ϕ increases, z increases linearly. So,using the equation of line

γ ϕ ϕ− = − )z z ( ,0 0

when =z 0 and ϕ = − π2,

γ ϕ π= +⎜ ⎟⎛⎝

⎞⎠z

2,

when =z h and ϕ = π2,

γ π π γπ

= + → =⎜ ⎟⎛⎝

⎞⎠h

h2 2

,

so

πϕ= −z

h h2

and

πϕ=z

hd d .

Using our expressions for z and zd , we have

lπ

ϕ ϕπ

ϕ ϕπ

ϕ∇ ⋅ = + + +⎡⎣⎢

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟⎤⎦⎥T R

h hR

h h hd

2cos 2

2sin d .

2

So

l∫ ∫ πϕ ϕ

πϕ ϕ

πϕ∇ ⋅ = + + + =

π

π

−

⎡⎣⎢

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟⎤⎦⎥T R

h hR

h h hh Rd

2cos 2

2sin d

a

b

2

2 22

as expected.

Electromagnetism

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Problem 1.19. Evaluate the following integrals:

a) ∫ δ− + −x x x x(2 4) ( 2)d1

3

2

b) ∫ δ+ −−

x x x( 4) ( 2)d1

1

2

c) ∫ δ π−x xsin( ) ( )dx32

2

6

d) ∫ δ+−

x x x(2 1) (4 )d2

2

3

e) ∫ δ +−∞

∞

x x x(2 1)d2

f) ∫ δ −x b x( )d

a

0

Solutionsa)

∫ δ− + −( )x x x x2 4 ( 2)d .1

3

2

Since ∈2 (1, 3) and = − +f x x x( ) 2 42 , we have

∫ δ− + − = = − + =( )x x x x f2 4 ( 2)d (2) 2(2) 2 4 10.1

3

2 2

b)

∫ δ+ −−

( )x x x4 ( 2)d .1

1

2

Since ∉ −2 ( 1, 1), we have

∫ δ+ − =−

( )x x x4 ( 2)d 0.1

1

2

c)

∫ δ π−⎛⎝⎜

⎞⎠⎟

xx xsin

32

( )d .2

6

Electromagnetism

1-31

Since π ∈ (2, 6) and =f x( ) sin( )x32

, we have

∫ δ π π π− = = = −⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

xx x fsin

32

( )d ( ) sin32

1.2

6

d)

∫ δ+−

( )x x x2 1 (4 )d .2

2

3

Since ∈ −0 ( 2, 2) and = +f x x( ) 2 13 , we have

∫ δ+ = = + =−

( ) ( )x x x f2 1 (4 )d14

(0)14

2(0) 114

.2

2

3 3

e)

∫ δ +−∞

∞

x x x(2 1)d .2

This can be rewritten as

∫ ∫ ∫δ δ δ+ = + = +−∞

∞

−∞

∞

−∞

∞⎡⎣⎢

⎛⎝⎜

⎞⎠⎟⎤⎦⎥

⎛⎝⎜

⎞⎠⎟x x x x x x x x x(2 1)d 2

12

d12

12

d .2 2 2

Since − ∈ −∞ ∞( , )12

and =f x x( ) 2, we have

∫ δ + = − = − =−∞

∞ ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟x x x f

12

12

d12

12

12

12

18

.22

f)

∫ δ −x b x( )d .

a

0

Here we have

∫ δ − = < <{x b x b a( )d 1 if 00 otherwise

.

a

0

Problem 1.20. Suppose we have two vector fields = ˆF y z12 and = ˆ + ˆ + ˆF xx yy zz2 .

Calculate the divergence and curl of each. Which can be written as the gradient of ascalar and which can be written as the curl of a vector? Find a scalar and a vectorpotential.

Electromagnetism

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Solution For F1, we have

∇ ⋅ = ∂∂

ˆ + ∂∂

ˆ + ∂∂

ˆ ⋅ ˆ =∂

∂=

⎛⎝⎜

⎞⎠⎟ ( )

( )F

xx

yy

zz y z

y

z01

22

and

∇ × =

ˆ ˆ ˆ∂

∂∂∂

∂∂ = ˆF

x y z

x y z

y

yx

0 0

2 .1

2

For F2, we have

∇ ⋅ = ∂∂

ˆ + ∂∂

ˆ + ∂∂

ˆ ⋅ ˆ + ˆ + ˆ = + + =⎛⎝⎜

⎞⎠⎟F

xx

yy

zz xx yy zz( ) 1 1 1 32

and

∇ × =

ˆ ˆ ˆ∂

∂∂∂

∂∂

= − ˆ + − ˆ + − ˆ =F

x y z

x y zx y z

x y z(0 0) (0 0) (0 0) 0.2

Since ∇ ⋅ =F 01 , F1 can be expressed as = ∇ × F A1 . We can find A by considering

∇ × =

ˆ ˆ ˆ∂

∂∂∂

∂∂

= ∂∂

−∂∂

ˆ + ∂∂

− ∂∂

ˆ +∂∂

− ∂∂

ˆ⎜ ⎟⎛⎝⎜

⎞⎠⎟

⎛⎝

⎞⎠

⎛⎝⎜

⎞⎠⎟

A

x y z

x y z

y

Ay

A

zx

Az

Ax

yA

xAy

z

0 0

.z y x z y x

2

By inspection:

∂∂

−∂∂

= ∂∂

− ∂∂

=∂∂

− ∂∂

=Ay

A

zAz

Ax

A

xAy

y0, 0, .z y x z y x 2

This is satisfied by

= ˆA y xy ,2

which is just one example. Since ∇ × =F 02 , F2 can be expressed as = −∇F V2 . Wecan findV by considering

Electromagnetism

1-33

= − ∂∂

ˆ + ∂∂

ˆ + ∂∂

ˆ⎛⎝⎜

⎞⎠⎟F

Vx

xVy

yVz

z .2

By inspection:

= −∂∂

= −∂∂

= −∂∂

xVx

yVy

zVz

, , .

This is satisfied by

= − + +⎛⎝⎜

⎞⎠⎟V

x y z2 2 2

2 2 2

which is again just one example.

BibliographyByron F W and Fuller R W 1992 Mathematics of Classical and Quantum Physics (New York:

Dover)Griffiths D J 1999 Introduction to Electrodynamics 3rd edn (Englewood Cliffs, NJ: Prentice Hall)Griffiths D J 2013 Introduction to Electrodynamics 4th edn (New York: Pearson)Halliday D, Resnick R and Walker J 2010 Fundamentals of Physics 9th edn (New York: Wiley)Halliday D, Resnick R and Walker J 2013 Fundamentals of Physics 10th edn (New York: Wiley)Purcell E M and Morin D J 2013 Electricity and Magnetism 3rd edn (Cambridge: Cambridge

University Press)Rogawski J 2011 Calculus: Early Transcendentals 2nd edn (San Francisco, CA: Freeman)

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