Problem 3.37 For each of the following scalar fields, obtain an analytical solutionfor ∇T and generate a corresponding arrow representation.

(a) T = 10+ x, for −10≤ x ≤ 10

(b) T = x2, for −10≤ x ≤ 10

(c) T = 100+ xy, for −10≤ x ≤ 10

(d) T = x2y2, for −10≤ x,y ≤ 10

(e) T = 20+ x+ y, for −10≤ x,y ≤ 10

(f) T = 1+sin(πx/3), for −10≤ x ≤ 10

(g) T = 1+cos(πx/3), for −10≤ x ≤ 10

(h) T = 15+ r cosφ , for

{

0≤ r ≤ 100≤ φ ≤ 2π.

(i) T = 15+ r cos2 φ , for

{

0≤ r ≤ 100≤ φ ≤ 2π.

Solution:

(a)

T = 10+ x

∇T = x∂T∂x

+ y∂T∂y

+ z∂T∂ z

= x.

The direction of∇T displays the fact thatT increases linearly withx only.

(b)

T = x2

∇T = x∂T∂x

+ y∂T∂y

+ z∂T∂ z

= x2x.

(c)

T = 100+ xy

∇T = x∂T∂x

+ y∂T∂y

+ z∂T∂ z

= xy+ yx.

The magnitude of the gradient increases monotonically withx.

(d)

T = x2y2

∇T = x∂T∂x

+ y∂T∂y

+ z∂T∂ z

= x2xy2 + y2x2y.

(e)

T = 20+ x+ y

∇T = x∂T∂x

+ y∂T∂y

+ z∂T∂ z

= x+ y.

(f)

T = 1+sin(πx/3)

∇T = x∂T∂x

+ y∂T∂y

+ z∂T∂ z

= x2π6

cos(πx/3).

(g)

T = 1+cos(πx/3)

∇T = x∂T∂x

+ y∂T∂y

+ z∂T∂ z

= −x2π6

sin(πx/3).

(h)

T = 15+ r cosθ

∇T = r∂T∂ r

+ φφφ1r

∂T∂φ

= r cosφ − φφφsinφ= x.

(i)

T = 15+ r cos2 θ

∇T = r∂T∂ r

+ φφφ1r

∂T∂φ

= r cos2 φ − φφφ2sinφ cosφ .

Problem 3.44 Each of the following vector fields is displayed in Fig. P3.44 in theform of a vector representation. Determine∇ ·A analytically and then compare theresult with your expectations on the basis of the displayed pattern.

(a) A = −xcosxsiny+ ysinxcosy, for −π ≤ x,y ≤ π

Figure P3.44(a)

Solution:

A = −xcosxsiny+ ysinxcosy

∇ ·A =∂Ax

∂x+

∂Ay

∂y

=∂∂x

(−cosxsiny)+∂∂y

(−sinxcosy)

= sinxsiny−sinxsiny = 0

Yes,A is divergenceless everywhere.

(b) A = −xsin2y+ ycos2x, for −π ≤ x,y ≤ π

Figure P3.44(b)

Solution:

A = −xsin2y+ ycos2x

∇ ·A =∂Ax

∂x+

∂Ay

∂y

=∂∂x

(−sin2y)+∂∂y

(cos2x) = 0

Yes,A is divergenceless everywhere.

(c) A = −xxy+ yy2, for −10≤ x,y ≤ 10

Figure P3.44(c)

Solution:

A = −xxy+ yy2

∇ ·A =∂Ax

∂x+

∂Ay

∂y

=∂∂x

(−xy)+∂∂y

(y2) = −y+2y = y

NO, A is not divergenceless everywhere. It is divergenceless only aty = 0.

(d) A = −xcosx+ ysiny, for −π ≤ x,y ≤ π

Figure P3.44(d)

Solution:

A = −xcosx+ ysiny

∇ ·A =∂Ax

∂x+

∂Ay

∂y

=∂∂x

(−cosx)+∂∂y

(siny) = sinx+cosy

NO, A is not divergenceless everywhere.

(e) A = xx, for −10≤ x ≤ 10

Figure P3.44(e)

Solution:

A = xx

∇ ·A =∂Ax

∂x+

∂Ay

∂y+

∂Az

∂ z

= 1

This indicates that the divergence ofA is the same at all points in the defined space.In other words, every small volume is a source of flux (more flux leaving thevolumethan entering it), and the net generated flux is the same at all locations.

(f) A = xxy2, for −10≤ x,y ≤ 10

Figure P3.44(f)

Solution:

A = xxy2

∇ ·A =∂Ax

∂x+

∂Ay

∂y+

∂Az

∂ z

= y2

(g) A = xxy2 + yx2y, for −10≤ x,y ≤ 10

Figure P3.44(g)

Solution:

A = xxy2 + yx2y

∇ ·A =∂Ax

∂x+

∂Ay

∂y+

∂Az

∂ z

= y2 + x2

(h) A = xsin(πx

10

)

+ ysin(πy

10

)

, for −10≤ x,y ≤ 10

Figure P3.44(h)

Solution:

A = xsin(πx/10)+ ysin(πy/10)

∇ ·A =∂Ax

∂x+

∂Ay

∂y+

∂Az

∂ z

=π10

[cos(πx/10)+cos(πy/10)]

(i) A = r r + φφφr cosφ , for

{

0≤ r ≤ 100≤ φ ≤ 2π.

Figure P3.44(i)

Solution:

A = r r + φφφr cosφ

∇ ·A =1r

∂∂ r

(rAr)+1r

∂Aφ

∂φ+

∂Az

∂ z

= 2−sinφ

(j) A = r r2 + φφφr2sinφ , for

{

0≤ r ≤ 100≤ φ ≤ 2π.

Figure P3.44(j)

Solution:

A = r r2 + φφφr2sinφ

∇ ·A =1r

∂∂ r

(rAr)+1r

∂Aφ

∂φ+

∂Az

∂ z

= 3r + r cosφ

Problem 3.46 For the vector fieldE = xxz − yyz2 − zxy, verify the divergencetheorem by computing:

(a) the total outward flux flowing through the surface of a cube centered at theorigin and with sides equal to 2 units each and parallel to the Cartesian axes,and

(b) the integral of∇ ·E over the cube’s volume.

Solution:(a) For a cube, the closed surface integral has 6 sides:

n

∫

E ·ds= Ftop+Fbottom+Fright +Fleft +Ffront +Fback,

Ftop =∫ 1

x=−1

∫ 1

y=−1

(

xxz− yyz2− zxy)∣

∣

z=1 · (zdy dx)

= −∫ 1

x=−1

∫ 1

y=−1xy dy dx =

(

(

x2y2

4

)∣

∣

∣

∣

1

y=−1

)∣

∣

∣

∣

∣

1

x=−1

= 0,

Fbottom=∫ 1

x=−1

∫ 1

y=−1

(

xxz− yyz2− zxy)∣

∣

z=−1 · (−zdy dx)

=∫ 1

x=−1

∫ 1

y=−1xy dy dx =

(

(

x2y2

4

)∣

∣

∣

∣

1

y=−1

)∣

∣

∣

∣

∣

1

x=−1

= 0,

Fright =∫ 1

x=−1

∫ 1

z=−1

(

xxz− yyz2− zxy)∣

∣

y=1 · (y dz dx)

= −∫ 1

x=−1

∫ 1

z=−1z2 dz dx = −

(

(

xz3

3

)∣

∣

∣

∣

1

z=−1

)∣

∣

∣

∣

∣

1

x=−1

=−43

,

Fleft =∫ 1

x=−1

∫ 1

z=−1

(

xxz− yyz2− zxy)∣

∣

y=−1 · (−y dz dx)

= −∫ 1

x=−1

∫ 1

z=−1z2 dz dx = −

(

(

xz3

3

)∣

∣

∣

∣

1

z=−1

)∣

∣

∣

∣

∣

1

x=−1

=−43

,

Ffront =∫ 1

y=−1

∫ 1

z=−1

(

xxz− yyz2− zxy)∣

∣

x=1 · (x dz dy)

=∫ 1

y=−1

∫ 1

z=−1z dz dy =

(

(

yz2

2

)∣

∣

∣

∣

1

z=−1

)∣

∣

∣

∣

∣

1

y=−1

= 0,

Fback=∫ 1

y=−1

∫ 1

z=−1

(

xxz− yyz2− zxy)∣

∣

x=−1 · (−x dz dy)

=∫ 1

y=−1

∫ 1

z=−1z dz dy =

(

(

yz2

2

)∣

∣

∣

∣

1

z=−1

)∣

∣

∣

∣

∣

1

y=−1

= 0,

n

∫

E ·ds= 0+0+−43

+−43

+0+0 =−83

.

(b)

∫∫∫

∇·E dv =∫ 1

x=−1

∫ 1

y=−1

∫ 1

z=−1∇·(xxz− yyz2− zxy)dz dy dx

=∫ 1

x=−1

∫ 1

y=−1

∫ 1

z=−1(z− z2)dz dy dx

=

(

(

xy

(

z2

2− z3

3

))∣

∣

∣

∣

1

z=−1

)∣

∣

∣

∣

∣

1

y=−1

∣

∣

∣

∣

∣

∣

1

x=−1

=−83

.

Problem 3.55 Verify Stokes’s theorem for the vector fieldB = (r cosφ + φφφsinφ)by evaluating:

(a) n

∫

CB · dℓℓℓ over the path comprising a quarter section of a circle, as shown in

Fig. P3.55, and

(b)∫

S(∇×××B) ·ds over the surface of the quarter section.

x

y

(0, 3)

L1

L3(−3, 0)

Figure P3.55: Problem 3.55.

Solution:(a)

n

∫

CB ·dℓℓℓ =

∫

L1

B ·dℓℓℓ+∫

L2

B ·dℓℓℓ+∫

L3

B ·dℓℓℓ

Given the shape of the path, it is best to use cylindrical coordinates.B is alreadyexpressed in cylindrical coordinates, and we need to choosedℓℓℓ in cylindricalcoordinates:

dℓℓℓ = r dr + φφφr dφ + z dz.

Along pathL1, dφ = 0 anddz = 0. Hence,dℓℓℓ = r dr and

∫

L1

B ·dℓℓℓ =∫ r=3

r=0(r cosφ + φφφsinφ) · r dr

∣

∣

∣

∣

φ=90◦

=∫ 3

r=0cosφ dr

∣

∣

∣

∣

φ=90◦= r cosφ |3r=0

∣

∣

∣

φ=90◦= 0.

Along L2, dr = dz = 0. Hence,dℓℓℓ = φφφr dφ and

∫

L2

B ·dℓℓℓ =∫ 180◦

φ=90◦(r cosφ + φφφsinφ) · φφφr dφ

∣

∣

∣

r=3

= −3cosφ |180◦90◦ = 3.

Along L3, dz = 0 anddφ = 0. Hence,dℓℓℓ = r dr and

∫

L3

B ·dℓℓℓ =∫ 0

r=3(r cosφ + φφφsinφ) · r dr

∣

∣

∣

φ=180◦

=∫ 0

r=3cosφ dr|φ=180◦ = −r|03 = 3.

Hence,n

∫

CB ·dℓℓℓ = 0+3+3 = 6.

(b)

∇×B = z1r

(

∂∂ r

(

rBφ −∂Br

∂φ

))

= z1r

(

∂∂ r

(r sinφ)− ∂∂φ

(cosφ)

)

= z1r(sinφ +sinφ) = z

2r

sinφ .

∫

S(∇×B) ·ds=

∫ 3

r=0

∫ 180◦

φ=90◦

(

z2r

sinφ)

· zr dr dφ

= −2r|3r=0cosφ∣

∣

∣

180◦

φ=90◦= 6.

Hence, Stokes’s theorem is verified.