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### Transcript of (b) · PDF file (0, 3) L1 (−3, 0) L3 Figure P3.55: Problem 3.55. Solution: (a) ♥...

• Problem 3.37 For each of the following scalar fields, obtain an analytical solution for ∇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 ≤ 10 0≤ φ ≤ 2π.

(i) T = 15+ r cos2 φ , for {

0≤ r ≤ 10 0≤ φ ≤ 2π.

Solution:

(a)

T = 10+ x

∇T = x̂ ∂T ∂x

+ ŷ ∂T ∂y

+ ẑ ∂T ∂ z

= x̂.

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

• (b)

T = x2

∇T = x̂ ∂T ∂x

+ ŷ ∂T ∂y

+ ẑ ∂T ∂ z

= x̂2x.

• (c)

T = 100+ xy

∇T = x̂ ∂T ∂x

+ ŷ ∂T ∂y

+ ẑ ∂T ∂ z

= x̂y+ ŷx.

The magnitude of the gradient increases monotonically withx.

• (d)

T = x2y2

∇T = x̂ ∂T ∂x

+ ŷ ∂T ∂y

+ ẑ ∂T ∂ z

= x̂2xy2 + ŷ2x2y.

• (e)

T = 20+ x+ y

∇T = x̂ ∂T ∂x

+ ŷ ∂T ∂y

+ ẑ ∂T ∂ z

= x̂+ ŷ.

• (f)

T = 1+sin(πx/3)

∇T = x̂ ∂T ∂x

+ ŷ ∂T ∂y

+ ẑ ∂T ∂ z

= x̂ 2π 6

cos(πx/3).

• (g)

T = 1+cos(πx/3)

∇T = x̂ ∂T ∂x

+ ŷ ∂T ∂y

+ ẑ ∂T ∂ z

= −x̂ 2π 6

sin(πx/3).

• (h)

T = 15+ r cosθ

∇T = r̂ ∂T ∂ r

+ φ̂φφ 1 r

∂T ∂φ

= r̂ cosφ − φ̂φφsinφ = x̂.

• (i)

T = 15+ r cos2 θ

∇T = r̂ ∂T ∂ r

+ φ̂φφ 1 r

∂T ∂φ

= r̂ cos2 φ − φ̂φφ2sinφ cosφ .

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

(a) A = −x̂cosxsiny+ ŷsinxcosy, for −π ≤ x,y ≤ π

Figure P3.44(a)

Solution:

A = −x̂cosxsiny+ ŷsinxcosy

∇ ·A = ∂Ax ∂x

+ ∂Ay ∂y

= ∂ ∂x

(−cosxsiny)+ ∂ ∂y

(−sinxcosy)

= sinxsiny−sinxsiny = 0

Yes,A is divergenceless everywhere.

• (b) A = −x̂sin2y+ ŷcos2x, for −π ≤ x,y ≤ π

Figure P3.44(b)

Solution:

A = −x̂sin2y+ ŷcos2x

∇ ·A = ∂Ax ∂x

+ ∂Ay ∂y

= ∂ ∂x

(−sin2y)+ ∂ ∂y

(cos2x) = 0

Yes,A is divergenceless everywhere.

• (c) A = −x̂xy+ ŷy2, for −10≤ x,y ≤ 10

Figure P3.44(c)

Solution:

A = −x̂xy+ ŷy2

∇ ·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 = −x̂cosx+ ŷsiny, for −π ≤ x,y ≤ π

Figure P3.44(d)

Solution:

A = −x̂cosx+ ŷsiny

∇ ·A = ∂Ax ∂x

+ ∂Ay ∂y

= ∂ ∂x

(−cosx)+ ∂ ∂y

(siny) = sinx+cosy

NO, A is not divergenceless everywhere.

• (e) A = x̂x, for −10≤ x ≤ 10

Figure P3.44(e)

Solution:

A = x̂x

∇ ·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 thevolume than entering it), and the net generated flux is the same at all locations.

• (f) A = x̂xy2, for −10≤ x,y ≤ 10

Figure P3.44(f)

Solution:

A = x̂xy2

∇ ·A = ∂Ax ∂x

+ ∂Ay ∂y

+ ∂Az ∂ z

= y2

• (g) A = x̂xy2 + ŷx2y, for −10≤ x,y ≤ 10

Figure P3.44(g)

Solution:

A = x̂xy2 + ŷx2y

∇ ·A = ∂Ax ∂x

+ ∂Ay ∂y

+ ∂Az ∂ z

= y2 + x2

• (h) A = x̂sin (πx

10

)

+ ŷsin (πy

10

)

, for −10≤ x,y ≤ 10

Figure P3.44(h)

Solution:

A = x̂sin(πx/10)+ ŷsin(πy/10)

∇ ·A = ∂Ax ∂x

+ ∂Ay ∂y

+ ∂Az ∂ z

= π 10

[cos(πx/10)+cos(πy/10)]

• (i) A = r̂ r + φ̂φφr cosφ , for {

0≤ r ≤ 10 0≤ φ ≤ 2π.

Figure P3.44(i)

Solution:

A = r̂ r + φ̂φφr cosφ

∇ ·A = 1 r

∂ ∂ r

(rAr)+ 1 r

∂Aφ ∂φ

+ ∂Az ∂ z

= 2−sinφ

• (j) A = r̂ r2 + φ̂φφr2sinφ , for {

0≤ r ≤ 10 0≤ φ ≤ 2π.

Figure P3.44(j)

Solution:

A = r̂ r2 + φ̂φφr2sinφ

∇ ·A = 1 r

∂ ∂ r

(rAr)+ 1 r

∂Aφ ∂φ

+ ∂Az ∂ z

= 3r + r cosφ

• Problem 3.46 For the vector fieldE = x̂xz − ŷyz2 − ẑxy, verify the divergence theorem by computing:

(a) the total outward flux flowing through the surface of a cube centered at the origin 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

(

x̂xz− ŷyz2− ẑxy )∣

z=1 · (ẑdy dx)

= − ∫ 1

x=−1

∫ 1

y=−1 xy dy dx =

(

(

x2y2

4

)∣

1

y=−1

)∣

1

x=−1 = 0,

Fbottom= ∫ 1

x=−1

∫ 1

y=−1

(

x̂xz− ŷyz2− ẑxy )∣

z=−1 · (−ẑdy dx)

= ∫ 1

x=−1

∫ 1

y=−1 xy dy dx =

(

(

x2y2

4

)∣

1

y=−1

)∣

1

x=−1 = 0,

Fright = ∫ 1

x=−1

∫ 1

z=−1

(

x̂xz− ŷyz2− ẑxy )∣

y=1 · (ŷ dz dx)

= − ∫ 1

x=−1

∫ 1

z=−1 z2 dz dx = −

(

(

xz3

3

)∣

1

z=−1

)∣

1

x=−1 =

−4 3

,

Fleft = ∫ 1

x=−1

∫ 1

z=−1

(

x̂xz− ŷyz2− ẑxy )∣

y=−1 · (−ŷ dz dx)

= − ∫ 1

x=−1

∫ 1

z=−1 z2 dz dx = −

(

(

xz3

3

)∣

1

z=−1

)∣

1

x=−1 =

−4 3

,

Ffront = ∫ 1

y=−1

∫ 1

z=−1

(

x̂xz− ŷyz2− ẑxy )∣

x=1 · (x̂ dz dy)

= ∫ 1

y=−1

∫ 1

z=−1 z dz dy =

(

(

yz2

2

)∣

1

z=−1

)∣

1

y=−1 = 0,

• Fback= ∫ 1

y=−1

∫ 1

z=−1

(

x̂xz− ŷyz2− ẑxy )∣

x=−1 · (−x̂ dz dy)

= ∫ 1

y=−1

∫ 1

z=−1 z dz dy =

(

(

yz2

2

)∣

1

z=−1

)∣

1

y=−1 = 0,

n

E ·ds= 0+0+ −4 3

+ −4 3

+0+0 = −8 3

.

(b)

∫∫∫

∇·E dv = ∫ 1

x=−1

∫ 1

y=−1

∫ 1

z=−1 ∇·(x̂xz− ŷyz2− ẑxy)dz dy dx

= ∫ 1

x=−1

∫ 1

y=−1

∫ 1

z=−1 (z− z2)dz dy dx

=

(

(

xy

(

z2

2 − z

3

3

))∣

1

z=−1

)∣

1

y=−1

1

x=−1

= −8 3

.

• Problem 3.55 Verify Stokes’s theorem for the vector fieldB = (r̂ cosφ + φ̂φφsinφ) by evaluating:

(a) n ∫

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

Fig. P3.55, and

(b) ∫

S (∇×××B) ·ds over the s