2 Green’s Theorem on Rectangular Domains
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Advanced Calculus Demystified
250
parallel to the y-axis. The magnitude of ∇ × W can be estimated by
|∇ × W( p)| ≈
1
Area(D)
W · ds
C
where D is the disk whose boundary is the loop C. Since the radius of C is .1, the
area of D is π(.1)2 . This gives us
1
Area(D)
W · ds =
C
1
(.5) ≈ 15.92
π(.1)2
We conclude that at the origin
∇ × W ≈ 0, 15.92, 0
Problem 133
The circulation of the vector ﬁeld around any loop in a plane parallel to the x z-plane
would be zero, since the vector ﬁeld is constant on such a plane. A vector V that is
perpendicular to such a loop points in the y-direction. The circulation around any
horizontal loop would also have to be zero, since the vector ﬁeld is perpendicular to
such a loop. A vector W that is perpendicular to such a loop points in the z-direction.
The curl must then be perpendicular to both V and W , and so must point in the
x-direction.
One can also see this algebraically. Such a vector ﬁeld must look like 0, 0, f (y) .
The curl of this is
i
∂
∂x
0
j
∂
∂y
0
k
∂
= f (y), 0, 0
∂z
f (y)
Problem 134
Let V be the unit cube. Then Gauss’ Theorem tells us
∇ · W dx dy dz
W · dS =
∂V
V
1
1
1
=
∇ · W dx dy dz
0
0
0
Answers to Problems
251
1
1
1
=
2xyz + 2xyz + 2xyz dx dy dz
0
0
1
0
1
1
=
6xyz dx dy dz
0
0
1
0
1
=
3yz dy dz
0
0
1
3
z dz
2
=
0
=
3
4
Problem 135
By Gauss’ Theorem the integral of ∇ · W over the ball B is equal to the integral of
W over the unit sphere S, with outward-pointing normal vector. The unit sphere is
parameterized in the usual way by
(θ, φ) = (sin φ cos θ, sin φ sin θ, cos φ)
0 ≤ θ ≤ 2π,
0≤φ≤π
As in Example 11-8,
∂
∂
×
= − sin2 φ cos θ, − sin2 φ sin θ, − sin φ cos φ
∂θ
∂φ
which agrees with the orientation on S.
We now compute
W · dS
∇ · W dx dy dz =
B
S
π 2π
0, 0, ecos φ · − sin2 φ cos θ, − sin2 φ sin θ,
=
0
0
− sin φ cos φ dθ dφ
Advanced Calculus Demystified
252
π 2π
− sin φ cos φecos φ dθ dφ
=
0
0
π
− sin φ cos φecos φ dφ
= 2π
0
−1
= 2π
ueu du
1
= 2π(ueu − eu )|−1
1
=−
4π
e
Problem 136
The region V is parameterized by
(r, θ, z) = (r cos θ, r sin θ, z)
π
1 ≤ r ≤ 2, 0 ≤ θ ≤ , 0 ≤ z ≤ 2
2
We use Gauss’ Theorem to transform the integral:
∇ · W dx dy dz
W · dS =
∂V
V
=
3(x 2 + y 2 ) dx dy dz
V
We now use the parameterization to change variables:
3(x 2 + y 2 ) dx dy dz
V
π
2
2
2
=
0
0
1
cos θ sin θ 0
3((r cos θ )2 + (r sin θ)2 ) −r sin θ r cos θ 0 dr dθ dz
0
0 1
Answers to Problems
π
2
2
2
=
3r 2 (r ) dr dθ dz
0
0
1
π
2
2
2
=
3r 3 dr dθ dz
0
0
1
π
2
2
=
0
0
π
2
2
=
0
=
253
3 4
r
4
2
dθ dz
1
45
dθ dz
4
0
45π
4
Problem 137
First, we observe that the polar equation r = cos θ is a circle of radius 12 , centered
on the point ( 12 , 0). Hence, the surface C is a cylinder of height 2. The surface D
is a disk, which caps off the top of the cylinder C, like an upside-down can. Let E
denote the “bottom” of the can. That is, E is the set of points in the xy-plane which
are within 12 of a unit away from the point ( 12 , 0, 0). We assume an orientation is
given on E so that C, D, and E together form the (oriented) boundary of V , the
points inside the “can.” Applying Gauss’ Theorem gives us
∇ · W dx dy dz =
V
W · dS
C+D+E
W · dS +
=
C+D
W · dS
E
Now notice that on E the vector ﬁeld W = 0, 0, 0 , so
W · dS = 0
E
Advanced Calculus Demystified
254
Putting these results together gives us
∇ · W dx dy dz =
V
W · dS
C+D
We may thus obtain the desired answer by evaluating the integral on the left-hand
side of this last equation. The ﬁrst thing we will need to evaluate this is the
divergence of W:
∇ ·W=
∂
xyz = x z
∂y
Next, we will need to parameterize V . Notice that V is a solid cylinder of radius 12 ,
whose central axis has been translated 12 of a unit away from the z-axis, in the
positive x-direction. A parameterization is thus given by
1
(r, θ, z) = r cos θ + , r sin θ, z
2
1
0≤r ≤ ,
2
0 ≤ θ ≤ 2π,
0≤z≤1
The partials of this parameterization are
∂
= cos θ, sin θ, 0
∂r
∂
= −r sin θ, r cos θ, 0
∂θ
∂
= 0, 0, 1
∂z
The determinant of the matrix which consists of these vectors is
cos θ
−r sin θ
0
sin θ
r cos θ
0
0
0 =r
1
Answers to Problems
We now use
255
to evaluate the integral:
∇ · W dx dy dz
x z dx dy dz
V
V
1
2
1 2π
r cos θ +
=
0
0
0
1
2
1 2π
r 2 z cos θ +
=
0
0
1
3
0
rz
dr dθ dz
2
0
1 2π
=
1
(z)(r ) dr dθ dz
2
1
2
3
z cos θ +
1
4
1
2
2
z dθ dz
0
1 2π
1
1
z cos θ + z dθ dz
24
16
=
0
0
1
=
π
z dz
8
0
=
π
16
Problem 138
S1 , and S2 with the opposite orientation, together bound a volume V of R3 .
Oriented surfaces with
the same oriented boundary
S1 and S2, with opposite
orientation, bound a volume V
Advanced Calculus Demystified
256
By Gauss’ Theorem
W · dS
∇ · W dx dy dz =
∂V
V
=
W · dS
S1 −S2
W · dS −
=
S1
W · dS
S2
∇ · W dx dy dz = 0. We conclude
But ∇ · W = 0 implies
V
W · dS =
S1
W · dS
S2
Problem 139
∇ · W( p) ≈
=
1
Volume(B)
1
4
π(.1)3
3
W · dS
∂B
(.5)
≈ 119.36662
Chapter 11 Quiz
Problem 140
1. The key to this problem is to notice that W = ∇ f (x, y, z), where f (x, y, z) =
xy2 z 2 . So,
W · ds =
C
(∇ f ) · ds
C
= f (1, 1, 1) − f (0, 0, 0)
=1−0
=1
Answers to Problems
257
2. Using Green’s Theorem
∂
∂
(− ln x) − (1) dx dy
∂x
∂y
1, ln x · ds =
σ
∂σ
=
−
σ
1
dx dy
x
To evaluate this integral we will need the partials of the parameterization:
∂φ
= v 2 , 3u 2 v
∂u
∂φ
= 2uv, u 3
∂v
The determinant of the matrix of partials is thus
v2
2uv
3u 2 v
= u 3 v 2 − 6u 3 v 2 = −5u 3 v 2
u3
We now integrate:
2
σ
2
1
− dx dy =
x
1
−1
(−5u 3 v 2 ) du dv
uv 2
1
2
2
=
5u 2 du dv
1
1
2
5 3
u
3
=
1
2
dv
1
2
35
dv
3
=
1
=
35
3