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2 Green’s Theorem on Rectangular Domains

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

=−





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



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