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2 Green’s Theorem on Rectangular Domains

# 2 Green’s Theorem on Rectangular Domains

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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

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

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

=−

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

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

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

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

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

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

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