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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.
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