Problem 3.43 For the scalar function U = R1 sin2 θ , determine its directional
derivative along the range direction R̂ and then evaluate it at P = (5, π /4, π /2).
Solution:
1 2
sin θ ,
R
1 ∂U
sin2 θ
∂U
1 ∂U
2 sin θ cos θ
∇U = R̂
+ φ̂φ
= −R̂ 2 − θ̂θ
+ θ̂θ
,
∂R
R ∂θ
R sin θ ∂ φ
R
R
sin2 θ
dU
= ∇U · R̂ = − 2 ,
dl
R
¯
2
¯
dU ¯
sin (π /4)
= −0.02 .
=−
¯
dl (5,π /4,π /2)
25
U=
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 = −x̂ cos x sin y + ŷ sin x cos y, for −π ≤ x, y ≤ π
Figure P3.44(a)
Solution:
A = −x̂ cos x sin y + ŷ sin x cos y
∂ Ax ∂ Ay
+
∇·A =
∂x
∂y
∂
∂
(− cos x sin y) + (− sin x cos y)
=
∂x
∂y
= sin x sin y − sin x sin y = 0
Yes, A is divergenceless everywhere.
(b) A = −x̂ sin 2y + ŷ cos 2x, for −π ≤ x, y ≤ π
Figure P3.44(b)
Solution:
A = −x̂ sin 2y + ŷ cos 2x
∂ Ax ∂ Ay
+
∇·A =
∂x
∂y
∂
∂
(− sin 2y) + (cos 2x) = 0
=
∂x
∂y
Yes, A is divergenceless everywhere.
(c)
A = −x̂ xy + ŷ y2 , for −10 ≤ x, y ≤ 10
Figure P3.44(c)
Solution:
A = −x̂ xy + ŷ y2
∂ Ax ∂ Ay
+
∇·A =
∂x
∂y
∂
∂
(−xy) + (y2 ) = −y + 2y = y
=
∂x
∂y
NO, A is not divergenceless everywhere. It is divergenceless only at y = 0.
(d) A = −x̂ cos x + ŷ sin y, for −π ≤ x, y ≤ π
Figure P3.44(d)
Solution:
A = −x̂ cos x + ŷ sin y
∂ Ax ∂ Ay
+
∇·A =
∂x
∂y
∂
∂
(− cos x) + (sin y) = sin x + cos y
=
∂x
∂y
NO, A is not divergenceless everywhere.
(e)
A = x̂ x, for −10 ≤ x ≤ 10
Figure P3.44(e)
Solution:
A = x̂ x
∂ Ax ∂ Ay ∂ Az
+
+
∂x
∂y
∂z
=1
∇·A =
This indicates that the divergence of A is the same at all points in the defined space.
In other words, every small volume is a source of flux (more flux leaving the volume
than entering it), and the net generated flux is the same at all locations.
(f) A = x̂ xy2 , for −10 ≤ x, y ≤ 10
Figure P3.44(f)
Solution:
A = x̂ xy2
∂ Ax ∂ Ay ∂ Az
+
+
∂x
∂y
∂z
2
=y
∇·A =
(g) A = x̂ xy2 + ŷ x2 y, for −10 ≤ x, y ≤ 10
Figure P3.44(g)
Solution:
A = x̂ xy2 + ŷ x2 y
∂ Ax ∂ Ay ∂ Az
+
+
∇·A =
∂x
∂y
∂z
2
2
= y +x
(h) A = x̂ sin
¡ πx ¢
10
+ ŷ sin
¡ πy ¢
10 , for −10 ≤ x, y ≤ 10
Figure P3.44(h)
Solution:
A = x̂ sin(π x/10) + ŷ sin(π y/10)
∂ Ax ∂ Ay ∂ Az
∇·A =
+
+
∂x
∂y
∂z
π
= [cos(π x/10) + cos(π y/10)]
10
(i) A = r̂ r + φ̂φ r cos φ , for
½
0 ≤ r ≤ 10
0 ≤ φ ≤ 2π .
Figure P3.44(i)
Solution:
A = r̂ r + φ̂φ r cos φ
1 ∂
1 ∂ Aφ ∂ Az
+
(rAr ) +
r ∂r
r ∂φ
∂z
= 2 − sin φ
∇·A =
(j) A =
r̂ r2 + φ̂φ r2 sin φ ,
for
½
0 ≤ r ≤ 10
0 ≤ φ ≤ 2π .
Figure P3.44(j)
Solution:
A = r̂ r2 + φ̂φ r2 sin φ
1 ∂
1 ∂ Aφ ∂ Az
+
(rAr ) +
r ∂r
r ∂φ
∂z
= 3r + r cos φ
∇·A =
Problem 3.52 Verify Stokes’s theorem for the vector field B = (r̂r cos φ + φ̂φ sin φ )
by evaluating:
Z
(a)
B · dl over the semicircular contour shown in Fig. P3.52(a), and
n
ZC
(b)
S
× B) · ds over the surface of the semicircle.
(∇×
y
2
y
2
L2
1
-2 L3
0 L1
(a)
x
2
0
L2
L3
L4
L1
1
2
x
(b)
Figure P3.52: Contour paths for (a) Problem 3.52 and
(b) Problem 3.53.
Solution:
(a)
Z
B · dl =
n
Z
L1
B · dl +
Z
L2
B · dl +
Z
L3
B · dl,
B · dl = (r̂r cos φ + φ̂φ sin φ ) · (r̂ dr + φ̂φr d φ + ẑ dz) = r cos φ dr + r sin φ d φ ,
µZ 2
µZ 0
¶¯
¶¯
¯
¯
r cos φ dr ¯¯
B · dl =
+
r sin φ d φ ¯¯
r=0
L1
φ =0
φ =0, z=0
z=0
¡ 1 2 ¢¯2
= 2 r ¯r=0 + 0 = 2,
¶¯
¶¯
µZ π
µZ 2
Z
¯
¯
r sin φ d φ ¯¯
B · dl =
r cos φ dr ¯¯ +
L
φ =0
r=2
Z
2
Z
L3
Z
= 0+
µZ
B · dl =
z=0
π
(−2 cos φ )|φ =0 = 4,
¶¯
0
¯
+
r cos φ dr ¯¯
r=2
φ =π ,z=0
¡
¢¯0
= − 21 r2 ¯r=2 + 0 = 2,
B · dl = 2 + 4 + 2 = 8.
n
r=2, z=0
π
¶¯
¯
r sin φ d φ ¯¯
φ =π
µZ
z=0
(b)
∇×B = ∇×(r̂r cos φ + φ̂φ sin φ )
µ
¶
µ
¶
1 ∂
∂
∂
∂
= r̂
0 − (sin φ ) + φ̂φ
(r cos φ ) − 0
r ∂φ
∂z
∂z
∂r
¶
µ
∂
1 ∂
(r cos φ )
(r(sin φ )) −
+ ẑ
r ∂r
∂φ
¶
µ
1
1
= r̂0 + φ̂φ0 + ẑ (sin φ + (r sin φ )) = ẑ sin φ 1 +
,
r
r
µ
¶¶
ZZ
Z π Z 2 µ
1
∇×B · ds =
· (ẑr dr d φ )
ẑ sin φ 1 +
r
φ =0 r=0
Z π Z 2
³¡
¢¯2 ´¯¯π
sin φ (r + 1) dr d φ = − cos φ ( 12 r2 + r) ¯r=0 ¯
=
φ =0 r=0
φ =0
= 8.