Vector Analysis in Higher Dimensions · 2009-01-23 · 386 Chapter 8 Vector Analysis in Higher Dimensions 20. Here 1k dx1 G1 and k ,.-FORMS 3 3.,,,

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C H A P T E R 8

Vector Analysis in Higher Dimensions

8.1 AN INTRODUCTION TO DIFFERENTIAL FORMS

1. �dx1 − 3 dx2��7, 3� D dx1�7, 3� − 3 dx2�7, 3� D 7 − 3�3� D −2.2.

�2 dx C 6 dy − 5 dz��1,−1,−2�

D 2 dx�1,−1, 2� C 6 dy�1,−1,−2� − 5 dz�1,−1,−2�

D 2�1� C 6�−1� − 5�−2� D 6.

3. �3 dx1 G dx2��4,−1�, �2, 0�� D 3 det [ dx1�4,−1� dx1�2, 0�

dx2�4,−1� dx2�2, 0�] D 3 det [ 4 2

−1 0] D 3�2� D 6.

4.

�4 dx G dy − 7 dy G dz��0, 1,−1�, �1, 3, 2��

D 4 dx G dy��0, 1,−1�, �1, 3, 2�� − 7 dy G dz��0, 1,−1�, �1, 3, 2��

D 4 det [ dx�0, 1,−1� dx�1, 3, 2�

dy�0, 1,−1� dy�1, 3, 2�] − 7 det [ dy�0, 1,−1� dy�1, 3, 2�

dz�0, 1,−1� dz�1, 3, 2�]

D 4 det [ 0 1

1 3] − 7 det [ 1 3

−1 2] D 4�−1� − 7�5� D −39.

5.

�2 dx1 G dx3 G dx4 C dx2 G dx3 G dx5��a, b, c�

D 2 det

dx1(a) dx1(b) dx1(c)



C det




D 2 det

1 0 5

−1 9 0

4 1 0

C det

0 0 0

−1 9 0

2 −1 −2

D 2�−185� C 0 D −370.

6. ω�3,−1,4�(a) D �−9 dx C 4 dy C 192 dz��a1, a2, a3� D −9a1 C 4a2 C 192a3.

7.

ω�2,−1,−3,1��a, b� D �−6 dx1 G dx3 C dx2 G dx4��a, b�

D −6 det [ dx1�a� dx1�b�

dx3�a� dx3�b�] C det [ dx2�a� dx2�b�

dx4�a� dx4�b�]

D −6�a1b3 − a3b1� C a2b4 − a4b2.

382 2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist.No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Section 8.1 An Introduction to Differential Forms 383

8.

ω�0,−1,π/2��a,b� D �1 dx G dy − 1 dy G dz C 4 dx G dz��a,b�

D∣∣∣∣∣ dx�a� dx�b�

dy�a� dy�b�

∣∣∣∣∣ −∣∣∣∣∣ dy�a� dy�b�

dz�a� dz�b�

∣∣∣∣∣ C 4

∣∣∣∣∣ dx�a� dx�b�

dz�a� dz�b�

∣∣∣∣∣D∣∣∣∣∣ a1 b1

a2 b2

∣∣∣∣∣ −∣∣∣∣∣ a2 b2

a3 b3

∣∣∣∣∣ C 4

∣∣∣∣∣ a1 b1

a3 b3

∣∣∣∣∣D a1b2 − a2b1 − �a2b3 − a3b2� C 4�a1b3 − a3b1�

9.

ω�x,y,z��2, 0,−1�, �1, 7, 5�� D cos x

∣∣∣∣∣ 2 1

0 7

∣∣∣∣∣ − sin z

∣∣∣∣∣ 0 7

−1 5

∣∣∣∣∣ C �y2 C 3�

∣∣∣∣∣ 2 1

−1 5

∣∣∣∣∣D 14 cos x − 7 sin z C 11�y2 C 3�

10. From Definition 1.3 of exterior product,

�3 dx C 2 dy − x dz� G �x2 dx − cos y dy C 7 dz�

D 3x2 dx G dx C 2x2 dy G dx − x3 dz G dx − 3 cos y dx G dy − 2 cos y dy G dy C x cos y dz G dy

C 21 dx G dz C 14 dy G dz − 7x dz G dz

D 2x2 dy G dx − x3 dz G dx − 3 cos y dx G dy C x cos y dz G dy

C 21 dx G dz C 14 dy G dz using �4�,

D −�2x2 C 3 cos y� dx G dy C �x3 C 21� dx G dz

C �14 − x cos y� dy G dz using �3�.

11. Again from Definition 1.3 of exterior product,

�y dx − x dy� G �z dx G dy C y dx G dz C x dy G dz�

D yz dx G dx G dy − xz dy G dx G dy C y2 dx G dx G dz − xy dy G dx G dz

C xy dx G dy G dz − x2 dy G dy G dz

D 2xy dx G dy G dz using �3� and �4�.


�2 dx1 G dx2 − x3 dx2 G dx4� G �2x4 dx1 G dx3 C �x3 − x2� dx3 G dx4�

D 4x4 dx1 G dx2 G dx1 G dx3 − 2x3x4 dx2 G dx4 G dx1 G dx3

C 2�x3 − x2� dx1 G dx2 G dx3 G dx4 − x3�x3 − x2� dx2 G dx4 G dx3 G dx4

D −2x3x4 dx2 G dx4 G dx1 G dx3 C 2�x3 − x2� dx1 G dx2 G dx3 G dx4 using �4�,

D 2�x3x4 C x3 − x2� dx1 G dx2 G dx3 G dx4 using �3�.

2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist.No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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384 Chapter 8 Vector Analysis in Higher Dimensions


�x1 dx2 G dx3 − x2x3 dx1 G dx5� G �ex4x5 dx1 G dx4 G dx5 − x1 cos x5 dx2 G dx3 G dx4�

D x1ex4x5 dx2 G dx3 G dx1 G dx4 G dx5 − x2x3e

x4x5 dx1 G dx5 G dx1 G dx4 G dx5

− x21 cos x5 dx2 G dx3 G dx2 G dx3 G dx4 C x1x2x3 cos x5 dx1 G dx5 G dx2 G dx3 G dx4

D x1ex4x5 dx2 G dx3 G dx1 G dx4 G dx5 C x1x2x3 cos x5 dx1 G dx5 G dx2 G dx3 G dx4 using �4�,

D �x1ex4x5 − x1x2x3 cos x5�dx1 G dx2 G dx3 G dx4 G dx5 using �3�.

14. Using Definition 1.1,

dxi1 G dxi2 G · · · G dxij G · · · G dxil G · · · G dxik�a1, a2, ... , ak�

D det

dxi1�a1� dxi1�a2� ... dxi1�ak�

#

#

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dxij �a1� dxij �a2� ... dxij �ak�

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dxil�a1� dxil�a2� ... dxil�ak�

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dxik�a1� dxik�a2� ... dxik�ak�

D −det

dxi1�a1� dxi1�a2� ... dxi1�ak�

#

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dxil�a1� dxil�a2� ... dxil�ak�

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dxij �a1� dxij �a2� ... dxij �ak�

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dxik�a1� dxik�a2� ... dxik�ak�

(since switching rows l and j changes the sign of the determinant)

D −dxi1 G dxi2 G · · · G dxil G · · · G dxij G · · · G dxik�a1, a2, ... , ak�.

15. This is easier to show in person, but the point is that if you switch the two identical forms then, on the one hand,nothing has changed and, on the other hand, formula (3) says that you now have the negative of what you startedwith. So

dxi1 G dxi2 G · · · G dxij G · · · G dxij G · · · G dxik D −dxi1 G dxi2 G · · · G dxij G · · · G dxij G · · · G dxik

and thereforedxi1 G dxi2 G · · · G dxij G · · · G dxij G · · · G dxik D 0.


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Section 8.1 An Introduction to Differential Forms 385

16. A k-form ω on Rn may be written as ω D∑Fi1...ik dxi1 G · · · G dxik . For each summand, each of the k dxij ’s isone of dx1, dx2, ... dxn. If k > n, then, by the pigeon hole principle, there must be at least one repeated term dxlin dxi1 G · · · G dxik (i.e., it will look like dxi1 G · · · G dxl G · · · G dxl G · · · G dxik ). And so, by formula (4),we have that dxi1 G · · · G dxik D 0. Hence every term of ω is zero.

17. Let ω1 D ∑Fi1...ik dxi1 G · · · G dxik , ω2 D ∑Gi1...ik dxi1 G · · · G dxik , and η D ∑Hj1...jl dxj1 G · · · G

dxjl . Then

�ω1 C ω2� G η D

∑i1,...,ik

�Fi1...ik C Gi1...ik � dxi1 G · · · G dxik

G ∑j1,...,jl

Hj1...jl dxj1 G · · · G dxjl

D ∑i1,...,ikj1,...,jl

�Fi1...ik C Gi1...ik �Hj1...jl dxi1 G · · · G dxik G dxj1 G · · · G dxjl

D ∑i1,...,ikj1,...,jl

Fi1...ikHj1...jl dxi1 G · · · G dxik G dxj1 G · · · G dxjl

C ∑i1,...,ikj1,...,jl

Gi1...ikHj1...jl dxi1 G · · · G dxik G dxj1 G · · · G dxjl

D ω1 G η C ω2 G η.

18. Let ω D∑Fi1...ik dxi1 G · · · G dxik , and η D∑Gj1...jl dxj1 G · · · G dxjl . Then

ω G η D ∑i1,...,ikj1,...,jl

Fi1...ikGj1...jl dxi1 G · · · G dxik G dxj1 G · · · G dxjl .

Now move dxj1 to the front by switching, in reverse order, with each of the dxip ’s. There are k switches so, byformula (3), there are k sign changes and this last equation becomes

ω G η D ∑i1,...,ikj1,...,jl

Fi1...ikGj1...jl�−1�k dxj1 G dxi1 G · · · G dxik G dxj2 G · · · G dxjl .

Similarly, we use k more interchanges to move dxj2 into the second position. We repeat this for each of the ldxjq ’s. and our equation becomes

ω G η D ∑i1,...,ikj1,...,jl

Fi1...ikGj1...jl �−1�k�−1�k · · · �−1�k︸︷︷︸l times

dxj1 G · · · G dxjl G dxi1 G · · · G dxik D �−1�klη G ω.

19. Let ω D ∑Fi1...ik dxi1 G · · · G dxik , η D ∑Gj1...jl dxj1 G · · · G dxjl , and τ D ∑Hu1...um dxu1 G · · · G

dxum . Then

�ω G η� G τ D

∑i1,...,ikj1,...,jl

Fi1...ikGj1...jl dxi1 G · · · G dxik G dxj1 G · · · G dxjl

G ∑u1,...,um

Hu1...um dxu1 G · · · G dxum

D ∑i1,...,ikj1,...,jlu1,...,um

Fi1...ikGj1...jlHu1...um dxi1 G · · · G dxik G dxj1 G · · · G dxjl G dxu1 G · · · G dxum.

Similarly, calculate ω G �η G τ� and you will obtain the same result.


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20. Here ω D∑Fi1...ik dxi1 G · · · G dxik , and η D∑Gj1...jl dxj1 G · · · G dxjl and f is a function (or 0-form).First we note that

�fω� G η D ∑i1,...,ik

fFi1...ik dxi1 G · · · G dxik

G

∑j1,...,jl

Gj1...jl dxj1 G · · · G dxjl

D ∑

i1,...,ikj1,...,jl

fFi1...ikGj1...jl dxi1 G · · · G dxik G dxj1 G · · · G dxjl

D f ∑i1,...,ikj1,...,jl


D f�ω G η�.

We will use this result to establish the second equality,

f�ω G η� D �−1�klf �η G ω� by property 2 of Proposition 1.4,

D �−1�kl�fη� G ω by the result established above,

D �−1�kl�−1�klω G �fη� by property 2 of Proposition 1.4,

D ω G �fη�.

Therefore, �fω� G η D f�ω G η� D ω G �fη�.

8.2 MANIFOLDS AND INTEGRALS OF k-FORMS

1. Here the map is X�θ1, θ2, θ3� D �3 cos θ1, 3 sin θ1, 3 cos θ1 C 2 cos θ2, 3 sin θ1 C 2 sin θ2, 3 cos θ1 C 2 cos θ2 Ccos θ3, 3 sin θ1 C 2 sin θ2 C sin θ3�.Follow the lead of Example 2 from the text. Each component function is at least C1 so the mapping is at leastC1. To see one-one, consider the equation X�θ1, θ2, θ3� D X�

Bθ1,

Bθ2,

Bθ3�. The first two component equations

would then have cos θ1 D cosBθ1 and sin θ1 D sin

Bθ1. Since 0 … θ1,

Bθ1 < 2π we see that θ1 D

Bθ1. Using this

information in the next two component functions, we make the same conclusion for θ2 andBθ2. Finally, use all of

this information in the last set of equations to see that θ3 DBθ3. So X is one to one and C1. What is left to show

is that the tangent vectors Tθ1 , Tθ2 , and Tθ3 are linearly independent.

Tθ1 D �−3 sin θ1, 3 cos θ1,−3 sin θ1, 3 cos θ1,−3 sin θ1, 3 cos θ1�

Tθ2 D �0, 0,−2 sin θ2, 2 cos θ2,−2 sin θ2, 3 cos θ2�

Tθ3 D �0, 0, 0, 0, sin θ3, cos θ3�

Because of the leading pair of zeros in Tθ2 and Tθ3 we can see that if c1Tθ1 C c2Tθ2 C c3Tθ3 D 0, then c1 D 0.Looking at the second pair of zeros in Tθ3 we can then see that c2 D 0. This would then force c3 D 0. So Tθ1 ,Tθ2 , and Tθ3 are linearly independent. We have shown that the parametrized 3-manifold is a smooth parametrized3-manifold.

2. As in Example 3, let’s begin by describing the location of the point �x1, y1�. It is anywhere in the annular regiondescribed by �l1 cos θ1, l1 sin θ1� where 1 … l1 … 3 and 0 … θ1 < 2π. You can now describe �x2, y2� as beingthis same annular region centered at �x1, y1�. Together this means that the locus of �x2, y2� is the interior of adisk of radius 6. Using variables l2 and θ2 such that 1 … l2 … 3 and 0 … θ2 < 2π, the mapping is

X�l1, θ1, l2, θ2� D �l1 cos θ1, l1 sin θ1, l1 cos θ1 C l2 cos θ2, l1 sin θ1 C l2 sin θ2�.

As before, the component functions are at least C1 so the mapping is at least C1. As for one-one, considerX�l1, θ1, l2, θ2� D X�

Bl1,

Bθ1,

Bl2,

Bθ2�. From the first component functions we see that �x1, y1� lies on a circle

of radius l1 and � Bx1,By1� lies on a circle of radius

Bl1 so l1 D B

l1. Then, as in Exercise 1, cos θ1 D cosBθ1 and

sin θ1 D sinBθ1. As 0 … θ1,

Bθ1 < 2π, we see that θ1 D

Bθ1. Now the rest of the argument follows in exactly the


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Section 8.2 Manifolds and Integrals of k-forms 387

same way since �x2, y2� is related to �x1, y1� in the same way that �x1, y1� is related to the origin. We now needto show that the four tangent vectors are linearly independent.

Tl1 D �cos θ1, sin θ1, cos θ1, sin θ1�

Tθ1 D �−l1 sin θ1, l1 cos θ1,−l1 sin θ1, l1 cos θ1�

Tl2 D �0, 0, cos θ2, sin θ2�

Tθ2 D �0, 0,−l2 sin θ2, l2 cos θ2�

Look at the equation c1Tl1 C c2Tθ1 C c3Tl2 C c4Tθ2 D 0. Because of the leading pair of zeros in Tl2 andTθ2 we can see that c1 cos θ1 D c2l1 sin θ1 and c1 sin θ1 D −c2l1 cos θ1. Solve for c1 in the first equation andsubstitute into the second equation to get c2l1 sin2 θ1 D −c2l1 cos2 θ1. Because l1 cannot be zero, this impliesthat c2 D 0. This then implies that c1 D 0. Given that, we can make the same argument to show c3 D c4 D 0.Therefore the four tangent vectors are linearly independent and we have described the states of the robot arm asa smooth parametrized 4-manifold in R4.

3. This is a combination of Example 3 and Exercise 2. Let’s begin by describing the location of the point �x1, y1�. Itis anywhere on a circle of radius 3 centered at the origin. So �x1, y1� D �3 cos θ1, 3 sin θ1� where 0 … θ1 < 2π.We can then describe �x2, y2� as being this same annular region centered at �x1, y1�. Together this means thatthe locus of �x2, y2� is �3 cos θ1 C l2 cos θ2, 3 sin θ1 C l2 sin θ2� where 1 … l2 … 2 and 0 … θ2 < 2π. Similarlywe describe �x3, y3� in terms of �x2, y2� using variables l3 and θ3 such that 1 … l3 … 2 and 0 … θ2 < 2π. Themapping is

X�θ1, l2, θ2, l3, θ3� D �3 cos θ1, 3 sin θ1, 3 cos θ1 C l2 cos θ2, 3 sin θ1 C l2 sin θ2,

3 cos θ1 C l2 cos θ2 C l3 cos θ3, 3 sin θ1 C l2 sin θ2 C l3 sin θ3�.

As before, the component functions are at least C1 so the mapping is at least C1. As for one-one, considerX�θ1, l2, θ2, l3, θ3� D X�

Bθ1,

Bl2,

Bθ2,

Bl3,

Bθ3�. From the first two component functions we see that cos θ1 D cos

Bθ1

and sin θ1 D sinBθ1 and 0 … θ1,

Bθ1 < 2π so θ1 D

Bθ1. Now, �x2, y2� lies on a circle of radius l2 and � Bx2,

By2�

lies on a circle of radiusBl2 with each circle centered at the same point �x1, y1� D � Bx1,

By1�. So l2 D Bl2. Then,

as above, cos θ2 D cosBθ2 and sin θ2 D sin

Bθ2. As 0 … θ2,

Bθ2 < 2π, we see that θ2 D

Bθ2. Now the rest of the

argument follows in exactly the same way since �x3, y3� is related to �x2, y2� in the same way that �x2, y2� isrelated to �x1, y1�.

We now need to show that the five tangent vectors are linearly independent.

Tθ1 D �−3 sin θ1, 3 cos θ1,−3 sin θ1, 3 cos θ1,−3 sin θ1, 3 cos θ1�

Tl2 D �0, 0, cos θ2, sin θ2, cos θ2, sin θ2�

Tθ2 D �0, 0,−l2 sin θ2, l2 cos θ2,−l2 sin θ2, l2 cos θ2�

Tl3 D �0, 0, 0, 0, cos θ3, sin θ3�

Tθ3 D �0, 0, 0, 0,−l3 sin θ3, l3 cos θ3�

Look at the equation c1Tθ1 C c2Tl2 C c3Tθ2 C c4Tl3 C c5Tθ3 D 0. Because of the leading pair of zeros in all butthe vector Tθ1 we conclude that c1 D 0. The remainder of the argument is exactly as in Exercise 2. Because thefirst four components of Tl3 and Tθ3 are zero, we can see that c2 cos θ2 D c3l2 sin θ2 and c2 sin θ2 D −c3l2 cos θ2.Solve for c2 in the first equation and substitute into the second equation to get c3l2 sin2 θ2 D −c3l2 cos2 θ2.Because l2 cannot be zero, c3 D 0. This then implies that c2 D 0. Given that, we can make the same argument toshow c4 D c5 D 0. Therefore the five tangent vectors are linearly independent and we have described the statesof the robot arm as a smooth parametrized 5-manifold in R6.

4. We can use spherical coordinates to describe the parametrized space. The point �x1, y1, z1� can be written as�2 sinϕ1 cos θ1, 2 sinϕ1 sin θ1, 2 cosϕ1� where 0 … ϕ1 … π and 0 … θ1 < 2π. We can then write �x2, y2, z2�as �x1 C sinϕ2 cos θ2, y1 C sinϕ2 sin θ2, z1 C cosϕ2� where 0 … ϕ2 … π and 0 … θ2 < 2π. In other words,our mapping is

X�θ1, ϕ1, θ2, ϕ2� D �2 sinϕ1 cos θ1, 2 sinϕ1 sin θ1, 2 cosϕ1,

2 sinϕ1 cos θ1 C sinϕ2 cos θ2, 2 sinϕ1 sin θ1 C sinϕ2 sin θ2, 2 cosϕ1 C cosϕ2�.


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As in the previous exercises, the fact that the component functions are at least C1 tells us that the mapping is atleast C1. Checking one-one is a little more interesting than in the above exercises. Consider the implications of theequation X�θ1, ϕ1, θ2, ϕ2� D X�

Bθ1,

Bϕ1,Bθ2,

Bϕ2�. By the third component functions we see that cosϕ1 D cos Bϕ1.Because 0 … ϕ1,

Bϕ1 … π we see that ϕ1 D Bϕ1. Substituting this into the sixth component function implies thatϕ2 D Bϕ2. Now comparing the equations from the first two component functions we see that if ϕ1 D 0 or π thenθ1 need not be the same as

Bθ1. This is allowed—recall that the mapping might not be one-one on the boundary

of the domain. Other than on the boundary, cos θ1 D cosBθ1 and sin θ1 D sin

Bθ1 and so, as before θ1 D

Bθ1. Again,

substitute this into the equations that arise from the fourth and fifth component functions to conclude that, exceptwhen ϕ2 is 0 or π, we must have θ2 D

Bθ2.

We now need to show that the four tangent vectors are linearly independent.

Tθ1 D �−2 sinϕ1 sin θ1, 2 sinϕ1 cos θ1, 0,−2 sinϕ1 sin θ1, 2 sinϕ1 cos θ1, 0�

Tϕ1 D �2 cosϕ1 cos θ1, 2 cosϕ1 sin θ1,−2 sinϕ1, 2 cosϕ1 cos θ1, 2 cosϕ1 sin θ1,−2 sinϕ1�

Tθ2 D �0, 0, 0,− sinϕ2 sin θ2, sinϕ2 cos θ2, 0�

Tϕ2 D �0, 0, 0, cosϕ2 cos θ2, cosϕ2 sin θ2,− sinϕ2�

Look at the equation c1Tθ1 C c2Tϕ1 C c3Tθ2 C c4Tϕ2 D 0. There is a zero in the third component of all ofthe tangent vectors except for Tϕ1 . This tells us that c2 D 0. If that is the case, then there is a zero in the sixthcomponent of all of the remaining tangent vectors except for Tϕ2 so c4 D 0. But then the leading trio of zeros inTθ2 implies that c1 D 0 which in turn would mean that c3 D 0. Therefore the four tangent vectors are linearlyindependent and we have described the states of the robot arm as a smooth parametrized 4-manifold in R6.

5. This is just an exercise in linear algebra. If x ∈ Rn is orthogonal to vi for i D 1, ... , k, then x · vi D 0 fori D 1, ... , k. An arbitrary vector v in Span{v1, ... , vk} is of the form v D c1v1 C · · · C ckvk for scalarsc1, ... , ck ∈ R. The calculation is straightforward:

x · v D x · �c1v1 C · · · C ckvk� D c1�x · v1� C · · · C ck�x · vk� D c1�0� C · · · C ck�0� D 0.

In other words, x is orthogonal to v.

6. By Definition 2.1,

∫xω D

∫ π0ωx�t��x

′�t�� dt. We have, x′�t� D �−a sin t, b cos t, c� and also ω D b dx −a dy C xy dz so that∫

xω D

∫ π0

[b�−a sin t� − a�b cos t� C �ab cos t sin t�c] dt

D ab∫ π

0[− sin t − cos t C c sin t cos t] dt

D ab(cos t − sin t C c

2sin2 t) ∣∣∣∣π

0D −2ab.

7. Parametrize the unit circle C by x�t� D �cos t, sin t�, 0 … t … 2π. Then∫Cω D

∫ 2π

0ωx�t��− sin t, cos t� dt D

∫ 2π

0�sin t dx − cos t dy��− sin t, cos t� dt

D∫ 2π

0�− sin2 t − cos2 t� dt D

∫ 2π

0−1 dt D −2π.

8. Parametrize the segment as x�t� D �t, t, ... , t�, 0 … t … 3. Then x′�t� D �1, 1, ... , 1� and so

ωx�t��x′�t�� D �t dx1 C t2 dx2 C · · · C tn dxn��1, 1, ... , 1� D t C t2 C · · · C tn.

Hence,∫Cω D

∫ 3

0�t C t2 C · · · C tn� dt D (1

2t2 C 1

3t3 C · · · C 1

n C 1tnC1) ∣∣∣∣3

0DnC1∑kD2

3k

kD

n∑kD1

3kC1

k C 1.


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Section 8.2 Manifolds and Integrals of k-forms 389

9. By Definition 2.3,

∫Sω D

6DωX�s,t��Ts,Tt� ds dt. For X�s, t� D �s cos t, s sin t, t�, we have Ts D �cos t,

sin t, 0�, and Tt D �−s sin t, s cos t, 1�. Then

ωX�s,t��Ts,Tt� D �t dx G dy C 3 dz G dx − s cos t dy G dz��Ts,Tt�

D t∣∣∣∣∣ cos t −s sin t

sin t s cos t

∣∣∣∣∣ C 3

∣∣∣∣∣ 0 1

cos t −s sin t

∣∣∣∣∣ − s cos t

∣∣∣∣∣ sin t s cos t

0 1

∣∣∣∣∣D st − 3 cos t − s

2sin 2t.

Thus ∫Sω D

∫ 4π

0

∫ 1

0(st − 3 cos t − s

2sin 2t) ds dt D

∫ 4π

0(1

2t − 3 cos t − 1

4sin 2t) dt

D (14t2 − 3 sin t C 1

8cos 2t) ∣∣∣∣4π

0D 4π2.

10. (a) First calculate the two tangent vectors for this parametrization of the helicoid. We have Tu1 D �cos 3u2,sin 3u2, 0� and Tu2 D �−3u1 sin 3u2, 3u1 cos 3u2, 5�. Then

X�u1,u2��Tu1 ,Tu2� D det

−5 sin 3u2 cos 3u2 −3u1 sin 3u2

5 cos 3u2 sin 3u2 3u1 cos 3u2

−3u1 0 5

D −9u21 − 25 < 0

for all �u1, u2�. Therefore this particular parametrization is incompatible with .(b) There is more than one solution. One possible way to do this is to switch the ordering of the vari-

ables so that the resulting determinant is positive. Try the parametrization Y�u1, u2� D X�u2, u1� D�u2 cos 3u1, u2 sin 3u1, 5u1� for 0 … u1 … 2π and 0 … u2 … 5. Then the tangent vectors are Tu1 D�−3u2 sin 3u1, 3u2 cos 3u1, 5� and Tu2 D �cos 3u1, sin 3u1, 0�. Then

Y�u1,u2��Tu1 ,Tu2� D det

−5 sin 3u1 −3u2 sin 3u1 cos 3u1

5 cos 3u1 3u2 cos 3u1 sin 3u1

−3u2 5 0

D 9u21 C 25 > 0

for all �u1, u2�. Therefore this particular parametrization is now compatible with .(c) Since the goal is to change the sign of the resulting determinant, we can change to where

X�u1,u2��a, b� D −det

−5 sin 3u2 a1 b1

5 cos 3u2 a2 b2

−3u1 a3 b3

.(d) The discussion following Theorem 2.11 tells us what to do if the parametrization is compatible. Since the

parametrization X is incompatible with we make the following simple adjustment:∫S ω D −

∫X ω. We

pause to calculate

ωX�u1,u2��Tu1 ,Tu2� D 5u2

∣∣∣∣∣ cos 3u2 −3u1 sin 3u2

sin 3u2 3u1 cos 3u2

∣∣∣∣∣− �u2

1 cos2 3u2 C u21 sin2 3u2�

∣∣∣∣∣ sin 3u2 3u1 cos 3u2

0 5

∣∣∣∣∣D 15u1u2 − 5u2

1 sin 3u2.


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Hence,∫Sω D −

∫ 5

0

∫ 2π

0�15u1u2 − 5u2

1 sin 3u2� du2 du1 D 5

∫ 5

0

∫ 2π

0�u2

1 sin 3u2 − 3u1u2� du2 du1

D 5

∫ 5

0[u2

1 (− cos 3u2

3) − 3u1u

22

2] ∣∣∣∣u2D2π

u2D0du1 D 5

∫ 5

0�−6π2u1� du1

D −30π2∫ 5

0u1 du1 D −15π2u2

1

∣∣50 D −375π2.

11. (a) For the parametrization given, we calculate the tangent vectors as Tu1 D �cos u2, sin u2, 0�,Tu2 D �−u1 sinu2, u1 cos u2, 0�, and Tu3 D �0, 0, 1�. Then

X�u��Tu1 ,Tu2 ,Tu3� D det

cosu2 −u1 sinu2 0

sinu2 u1 cosu2 0

0 0 1

D u1.

As 0 … u1 …

√5, this is positive when u1 Z 0. Note that when u1 D 0 the parametrization is not one-one

and also that Tu2 D 0 so Tu1 ,Tu2 , and Tu3 are not linearly independent. In other words, the parametrizationis not smooth when u1 D 0. It is, however, smooth when u1 Z 0. You can easily see that the mappingis one-one and at least C1. To see that the tangent vectors are linearly independent, consider the equationc1Tu1 C c2Tu2 C c3Tu3 D 0. We see from the third components that c3 D 0. Look at the remainingequations and we see that

{ �cosu2�c1 − �u1 sin u2�c2 D 0

�sinu2�c1 C �u1 cos u2�c2 D 0.

Multiply the first equation by − sin u2 and the second by cosu2 and add to obtain u1c2 D 0. Because weare assuming that u1 Z 0, this implies that c2 D 0 and therefore c1 D 0. This shows that the tangent vectorsare linearly independent and hence the parametrization is smooth when u1 Z 0. The conclusion is then thatthe parametrization given is compatible with the orientation when it is smooth.

(b) We can read the boundary pieces right off of the original parametrization: they are paraboloids that intersectat z D −1 in a circle in the plane z D −1 of radius

√5 centered at �0, 0,−1�. The boundary is

�M D {�x, y, z�|z D x2 C y2 − 6, z … −1} ∪ {�x, y, z�|z D 4 − x2 − y2, z Ú −1}.

We can easily adapt the parametrization to each of these pieces. For the bottom, use

Y1 : [0,√

5] * [0, 2π�' R3I Y1�s1, s2� D �s1 cos s2, s1 sin s2, s21 − 6�.

For the top, use

Y2 : [0,√

5] * [0, 2π�' R3I Y2�s1, s2� D �s1 cos s2, s1 sin s2, 4 − s21�.

(c) On the bottom part of �M the outward-pointing unit vector

V1 D�2x, 2y,−1�√

4x2 C 4y2 C 1. In terms of Y1, this is V1 D

�2s1 cos s2, 2s1 sin s2,−1�√4s21 C 1

.

On the top part of �M the outward-pointing unit vector

V2 D�2x, 2y, 1�√

4x2 C 4y2 C 1. In terms of Y2, this is V2 D

�2s1 cos s2, 2s1 sin s2, 1�√4s21 C 1

.


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Section 8.3 The Generalized Stokes’s Theorem 391

12. The paraboloid can be parametrized as X�s, t� D �s, t, s2 C t2� where 0 … s2 C t2 … 4. Therefore, Ts D �1, 0, 2s�and Tt D �0, 1, 2t�. Note that this parametrization is compatible with the orientation derived from the normalN D �−2x,−2y, 1� as

X�s,t��Ts,Tt� D det[N Ts Tt] D

−2s 1 0

−2t 0 1

1 2s 2t

D 2s2 C 2t2 C 1 > 0.

Therefore, we may compute ∫S ω as ∫X ω. So we begin by calculating

ωX�s,t��Ts,Tt� D �es2Ct2 dx G dy C t dz G dx C s dy G dz��Ts,Tt�

D es2Ct2∣∣∣∣∣ 1 0

0 1

∣∣∣∣∣ C t∣∣∣∣∣ 2s 2t

1 0

∣∣∣∣∣ C s∣∣∣∣∣ 0 1

2s 2t

∣∣∣∣∣D es2Ct2 − 2t2 − 2s2.

Use this in the calculation:∫Sω D

60…s2Ct2…4[es

2Ct2 − 2�s2 C t2�] ds dt

D∫ 2π

0

∫ 2

0�er

2 − 2r2�r dr dθ using polar coordinates,

D∫ 2π

0(1

2er

2 − 12r4) ∣∣∣∣2

rD0dθ

D∫ 2π

0(1

2e4 − 1

2− 8) dθ D π�e4 − 17�.

13. We have, for the given parametrization, that Tu1 D �1, 0, 0, 4�2u1 − u3��,Tu2 D �0, 1, 0, 0�, and Tu3 D�0, 0, 1, 2�u3 − 2u1��. Thus,

ωX�u1,u2,u3��Tu1 ,Tu2 ,Tu3� D �u2 dx2 G dx3 G dx4 C 2u1u3 dx1 G dx2 G dx3��Tu1 ,Tu2 ,Tu3�

D u2

∣∣∣∣∣∣∣0 1 0

0 0 1

4�2u1 − u3� 0 2�u3 − 2u1�

∣∣∣∣∣∣∣ C 2u1u3

∣∣∣∣∣∣∣1 0 0

0 1 0

0 0 1

∣∣∣∣∣∣∣D u2�8u1 − 4u3� C 2u1u3 D 8u1u2 − 4u2u3 C 2u1u3.

Hence, ∫Mω D

∫ 1

0

∫ 1

0

∫ 1

0�8u1u2 − 4u2u3 C 2u1u3� du1 du2 du3

D∫ 1

0

∫ 1

0�4u2 − 4u2u3 C u3� du2 du3

D∫ 1

0�2 − 2u3 C u3� du3 D 2 − 1

2D 3

2.

8.3 THE GENERALIZED STOKES’S THEOREM


d�exyz� D �

�x�exyz� dx C �

�y�exyz� dy C �

�z�exyz� dz C exyz�yz dx C xz dy C xy dz�.


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d�x3y − 2xz2 C xy2z� D �3x2y − 2z2 C y2z� dx C �x3 C 2xyz� dy C �xy2 − 4xz� dz.

3. Again, using Definition 3.1,

d��x2 C y2� dx C xy dy� D d�x2 C y2� G dx C d�xy� G dy

D �2x dx C 2y dy� G dx C �y dx C x dy� G dy

D 2y dy G dx C y dx G dy using (4) from Section 8.1,

D −y dx G dy using (3) from Section 8.1.


d�x1 dx2 − x2 dx1 C x3x4 dx4 − x4x5 dx5�

D dx1 G dx2 − dx2 G dx1 C �x4 dx3 C x3 dx4� G dx4 − �x5 dx4 C x4 dx5� G dx5

D dx1 G dx2 − dx2 G dx1 C x4 dx3 G dx4 − x5 dx4 G dx5 using �4� from Section 8.1,

D 2 dx1 G dx2 C x4 dx3 G dx4 − x5 dx4 G dx5 using �3� from Section 8.1.


d�xz dx G dy − y2z dx G dz� D �z dx C x dz� G dx G dy − �2yz dy C y2 dz� G dx G dz

D x dz G dx G dy − 2yz dy G dx G dz using �4� from Section 8.1,

D �x C 2yz� dx G dy G dz using �3� from Section 8.1.


d�x1x2x3 dx2 G dx3 G dx4 C x2x3x4 dx1 G dx2 G dx3�

D �x2x3 dx1 C x1x3 dx2 C x1x2 dx3� G dx2 G dx3 G dx4

C �x3x4 dx2 C x2x4 dx3 C x2x3 dx4� G dx1 G dx2 G dx3

D x2x3 dx1 G dx2 G dx3 G dx4 C x2x3 dx4 G dx1 G dx2 G dx3 using �4� from Section 8.1,

D 0 using �3� from Section 8.1.

7. For this solution d̂xi means that the term dxi is omitted.

dω Dn∑iD1

d�xi�2

G dx1 G · · · G d̂xi G · · · G dxn

Dn∑iD1

2xi dxi G dx1 G · · · G d̂xi G · · · G dxn

Dn∑iD1�−1�i−12xi dx1 G · · · G dxn using equation �3� of Section 8.1 repeatedly

D 2�x1 − x2 C x3 − · · · C �−1�n−1xn� dx1 G · · · G dxn.

8. Let u D �u1, u2, ... , un�; then

dfx0�u� D �fx1�x0� dx1 C fx2�x0� dx2 C · · · C fxn�x0� dxn��u�

D fx1�x0�u1 C fx2�x0�u2 C · · · C fxn�x0�un

D �fx1�x0�, fx2�x0�, ... , fxn�x0�� · u

D §f�x0� · u

D Duf�x0� by Theorem 6.2 of Chapter 2.


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9. For ω D F�x, z� dy C G�x, y� dz, we have dω D �Fx dx C Fz dz� G dy C �Gx dx C Gy dy� G dz.Expanding, this gives dω D Fx dx G dy C Gx dx G dz C �Gy − Fz� dy G dz. But we are told thatdω D z dx G dy C y dx G dz so

�F

�xD z, �G

�xD y, and

�G

�y− �F

�zD 0.

The first equation implies that F�x, z� D xz C f�z� for some differentiable function f of z alone. Similarly,the second equation implies that G�x, y� D xy C g�y� for some differentiable function g of y alone. Usingthese results together with the third equation we see that x C g′�y� D x C f ′�z� or g′�y� D f ′�z�. This canonly be true if their common value is a constant C. So if g′�y� D f ′�z� D C, then f�z� D Cz C D1 andg�y� D Cy C D2 for arbitrary constants C, D1, and D2. We conclude that F�x, z� D xz C Cz C D1 andG�x, y� D xy C Cy C D2.

10. If ω D 2x dy G dz − z dx G dy, then dω D 2 dx G dy G dz − dz G dx G dy D dx G dy G dz. FromExercise 11 of Section 8.2, M is parametrized as X: D ' R3IX�u1, u2, u3� D �u1 cosu2, u1 sin u2, u3� whereD D {�u1, u2, u3�|u2

1 − 6 … u3 … 4 − u21, 0 … u1 …

√5, 0 … u2 < 2π}. If we orient M by the 3-form

D dx G dy G dz, then

X(u)�Tu1 ,Tu2 ,Tu3� D det

cos u2 −u1 sin u2 0

sin u2 u1 cos u2 0

0 0 1

D u1 Ú 0.

As before, this is strictly positive when the parametrization is smooth so the parametrization is compatible withthe orientation.

Therefore, using this orientation,

∫M

dω D∫

Xdω D

∫ 2π

0

∫ √5

0

∫ 4−u21

u21−6

u1 du3 du1 du2 D 2π

∫ √5

0u1�10 − 2u2

1� du1

D 4π(52u2

1 − 14u4

1) ∣∣∣∣√

5

0D 4π(25

2− 25

4) D 25π.

On the other hand, �M is parametrized on the bottom surface as

Y1 : [0,√

5] * [0, 2π�' R3I Y1�s1, s2� D �s1 cos s2, s1 sin s2, s21 − 6�

with tangent vector normal to �M

V1 D�2s1 cos s2, 2s1 sin s2,−1�√

4s21 C 1.

The boundary �M is parametrized on the top surface as

Y2 : [0,√

5] * [0, 2π�' R3I Y2�s1, s2� D �s1 cos s2, s1 sin s2, 4 − s21�

with tangent vector normal to �M

V2 D�2s1 cos s2, 2s1 sin s2, 1�√

4s21 C 1.


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Then we have that the induced orientation on �M is given by �M�a1, a2� D �V, a1, a2�. Therefore we seethat on the bottom part of �M

�MY1�s��Ts1 ,Ts2� D X�s1,s2,s21−6��V1,Ts1 ,Ts2�

D det

2s1 cos s2√4s21 C 1

cos s2 −s1 sin s2

2s1 sin s2√4s21 C 1

sin s2 s1 cos s2

−1√4s21 C 1

2s1 0

D − 4s31 C s1√

4s21 C 1… 0.

The parametrization Y1 is incompatible with the induced orientation on �M. Along the top part of �M

�MY2�s��Ts1 ,Ts2� D X�s1,s2,4−s21�

�V2,Ts1 ,Ts2�

D det

2s1 cos s2√4s21 C 1

cos s2 −s1 sin s2

2s1 sin s2√4s21 C 1

sin s2 s1 cos s2

1√4s21 C 1

−2s1 0

D 4s31 C s1√

4s21 C 1Ú 0.

This parametrization is compatible with the induced orientation on �M.Therefore we set up our integral (changing signs in the first integrand because of the incompatibility of the

parametrization) to obtain the following.∫�Mω D −

∫Y1

ω C∫

Y2

ω

D −∫ 2π

0

∫ √5

0

2s1 cos s2

∣∣∣∣∣ sin s2 s1 cos s22s1 0

∣∣∣∣∣ − �s21 − 6�

∣∣∣∣∣ cos s2 −s1 sin s2sin s2 s1 cos s2

∣∣∣∣∣ ds1 ds2

C∫ 2π

0

∫ √5

0

2s1 cos s2

∣∣∣∣∣ sin s2 s1 cos s2−2s1 0

∣∣∣∣∣ − �4 − s21�∣∣∣∣∣ cos s2 −s1 sin s2

sin s2 s1 cos s2

∣∣∣∣∣ ds1 ds2

D∫ 2π

0

∫ √5

0[8s31 cos2 s2 C �2s21 − 10�s1] ds1 ds2

D∫ 2π

0[s41 cos 2s2 C 3

2s41 − 5s21]∣∣∣∣s1D

√5

s1D0ds2

D∫ 2π

0[25 cos 2s2 C 25/2] ds2 D [�25/2� sin 2s2 C 25s2/2]

∣∣∣2π0

D 25π.

11. One integral is easy. Since ω D xy dz G dw and �M D {�x, y, z,w�|x D 0, 8 − 2y2 − 2z2 − 2w2 D 0}, wesee that x D 0 along �M so

∫�M ω D

∫�M 0 D 0.


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Now dω D d�xy� G dz G dw D x dy G dz G dw C y dx G dz G dw. We can orient M any way we wish,so we won’t worry about this—we’ll choose the orientation to be compatible with the parametrization.

X : D' R4, X�u1, u2, u3� D �8 − 2u21 − 2u2

2 − 2u23, u1, u2, u3�

where D D {�u1, u2, u3�|u21 C u2

2 C u23 … 4} (i.e., the solid ball of radius 2). Then∫

Mdω D

∫X

dω DlB

dωX�u��Tu1 ,Tu2 ,Tu3� du1 du2 du3

DlB

�8 − 2u21 − 2u2

2 − 2u23�

∣∣∣∣∣∣∣1 0 0

0 1 0

0 0 1

∣∣∣∣∣∣∣ C u1

∣∣∣∣∣∣∣−4u1 −4u2 −4u3

0 1 0

0 0 1

∣∣∣∣∣∣∣ du1 du2 du3

DlB

�8 − 2�u21 C u2

2 C u23� − 4u2

1� du1 du2 du3.

At this point it is helpful to switch to spherical coordinates. The previous quantity is then

D∫ 2π

0

∫ π0

∫ 2

0�8 − 2ρ2 − 4ρ2 sin2 ϕ cos2 θ�ρ2 sinϕ dρ dϕ dθ

D 8 · �volume of B� − 2

∫ 2π

0

∫ π0

∫ 2

0ρ4�sinϕ C 2 sin3 ϕ cos2 θ� dρ dϕ dθ

D 8 · (43π23) − 2

∫ 2π

0

∫ π0

325�sinϕ C 2 sinϕ�1 − cos2 ϕ� cos2 θ� dϕ dθ

D 256π3

− 645

∫ 2π

0{�− cosϕ�

∣∣∣∣π0C 2 cos2 θ�− cosϕ C �cos3 ϕ�/3�

∣∣∣∣πϕD0

} dθ

D 256π3

− 645

∫ 2π

0{2 C 2 cos2 θ · (2 − 2

3)} dθ D 256π

3− 128

5

∫ 2π

0(1 C 4

3cos2 θ) dθ

D 256π3

− 1285

∫ 2π

0(5

3C 2

3cos 2θ) dθ �using the half angle formula�

D 256π3

− 1285(5

3�2π� C 1

3sin 2θ

∣∣∣∣2π0) D 256π

3− 256π

3D 0.

12. (a) Using the generalized version of Stokes’s theorem (Theorem 3.2), we have

13

∫�Mx dy G dz − y dx G dz C z dx G dy D 1

3

∫M

d�x dy G dz − y dx G dz C z dx G dy�

D 13

∫M

dx G dy G dz − dy G dx G dz C dz G dx G dy

D 13

∫M

3 dx G dy G dz using formula (3) of Section 8.1,

D∫M

dx G dy G dz DlM

dx dy dz D volume of M.

(See Definition 2.6 and Example 6 of Section 8.2.)(b) This generalizes the result demonstrated in part (a). Notice that the kth summand is �−1�k−1xk multiplied

by the (n − 1)-form which is the wedge product of the dxi’s in order with dxk missing. In other words, thekth summand is

�−1�k−1xk dx1 G · · · G d̂xk G · · · G dxn


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where d̂xk means that dxk is omitted. (Make the obvious adjustments to the expression if it is the first or lastterm that is omitted.) Then

d�of the kth summand� D �−1�k−1 dxk G dx1 G · · · G d̂xk G · · · G dxn.

Let ω denote the (n − 1)-form in the integrand. Then, using the generalized Stokes’s theorem,

1n

∫�Mω D 1

n

∫M

dω

D 1n

∫M( n∑kD1�−1�k−1 dxk G dx1 G · · · G d̂xk G · · · G dxn) .

Use formula (3) of Section 8.1 to “move” each dxk back into the slot from which it has been omitted andcollect terms to obtain

1n

∫�Mω D 1

n

∫Mn dx1 G · · · G dxn D

∫· · ·∫M

dx1 · · · dxn.

It is entirely reasonable to take this last n-dimensional integral to represent the n-dimensional volume of M.

8.4 TRUE/FALSE EXERCISES FOR CHAPTER 8

1. True.2. False. (There is a negative sign missing.)3. True.4. False.5. True.6. False. (A negative sign is missing.)7. True.8. False. (There should be no negative sign.)9. True.

10. True.11. False. �X�1, 1,−1� D X�1, 1, 1�, so X is not one-one on D.)12. True. (Both manifolds are the same helicoid.)13. False. (The agreement is only up to sign.)14. True.15. False. (This is only true if n is even.)16. False. (A negative sign is missing.)17. True.18. False. �dω D 0.�19. True. (dω would be an �n C 1�-form, and there are no nonzero ones on Rn.)20. True. (This is the generalized Stokes’s theorem, since �M D ∅.)

8.5 MISCELLANEOUS EXERCISES FOR CHAPTER 8

1. (a) First, by definition of the exterior product and derivative

d�f G g� Dn∑iD1

�

�xi�fg� dxi D

n∑iD1( �f

�xig C f �g

�xi) dxi by the product rule,

D gn∑iD1

�f

�xidxi C f

n∑iD1

�g

�xidxi

D g G df C f G dg

D df G g C �−1�0f G dg.


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Section 8.5 Miscellaneous Exercises for Chapter 8 397

(b) If k D 0, then write ω D f so that

d�ω G η� D d�f G η� D d (∑ fGj1...jl dxj1 G · · · G dxjl) D∑ d�fGj1...jl� G dxj1 G · · · G dxjl

D∑�df G Gj1...jl C f G dGj1...jl� G dxj1 G · · · G dxjl from (a),

D df G ∑Gj1...jl dxj1 G · · · G dxjl C f G ∑ dGj1...jl dxj1 G · · · G dxjl

D df G η C �−1�0f G dη.

(c) If l D 0, then write η D g so that

d�ω G η� D d�ω G g� D d (∑ gFi1...ik dxi1 G · · · G dxik )D∑ d�gFi1...ik � G dxi1 G · · · G dxik

D∑�dg G Fi1...ik C g G dFi1...ik � G dxi1 G · · · G dxik

D dg G ω C g G dω

D �−1�kω G dg C dω G g

by part 2 of Proposition 1.4 (recall dg is a 1-form).(d) In general,

d�ω G η� D d

∑1…i1<···<ik…n

Fi1...ik dxi1 G · · · G dxik G ∑1…j1<···<jl…n

Gj1...jl dxj1 G · · · G dxjl

D d

∑1…i1<···<ik…n1…j1<···<jl…n


D∑ d�Fi1...ikGj1...jl� G dxi1 G · · · G dxik G dxj1 G · · · G dxjl so by part (a),

D∑�dFi1...ik G Gj1...jl C Fi1...ik G dGj1...jl � G dxi1 G · · · G dxik G dxj1 G · · · G dxjl

D∑ dFi1...ik G Gj1...jl G dxi1 G · · · G dxik G dxj1 G · · · G dxjl

C ∑Fi1...ik G dGj1...jl G dxi1 G · · · G dxik G dxj1 G · · · G dxjl

D∑ dFi1...ik G dxi1 G · · · G dxik G Gj1...jl G dxj1 G · · · G dxjl

C ∑Fi1...ik G �−1�k dxi1 G · · · G dxik G dGj1...jl G dxj1 G · · · G dxjl

since Gj1...jl is a 0-form and dGj1...jl is a 1-form,

D dω G η C �−1�kω G dη.

2. (a) Define X : [0, 1] * [0, 1] * [0, 1] * [0, 1] ' R5,X�u1, u2, u3, u4, u5� D �u1, u2, u3, u4, u1u2u3u4�.Then Tu1 D �1, 0, 0, 0, u2u3u4�,Tu2 D �0, 1, 0, 0, u1u3u4�,Tu3 D �0, 0, 1, 0, u1u2u4�, and Tu4 D �0,0, 0, 1, u1u2u3�. From this we see that

X�u��Tu1 ,Tu2 ,Tu3 ,Tu4� D det

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

D 1.


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(b) We can now calculate∫M

dx1 G dx2 G dx3 G dx4

D∫ 1

0

∫ 1

0

∫ 1

0

∫ 1

0det

1 0 0 0

0 1 0 0

0 0 1 0

u2u3u4 u1u3u4 u1u2u4 u1u2u3

du1 du2 du3 du4

D∫ 1

0

∫ 1

0

∫ 1

0

∫ 1

0u1u2u3 du1 du2 du3 du4 D 1

8.

3. (a) The curve C may be parametrized as x�t� D �t, f �t��, a … t … b. Then x′�t� D �1, f ′�t�� and this iscompatible with the orientation of C. By Definition 2.1, we have∫

Cω D

∫xω D

∫ baωx�t��x

′�t�� dt.

For ω D y dx this is ∫ baf�t� · 1 dt D

∫ baf�t� dt D area under the graph.

(b) Parametrize S by

X : [a, b] * [c, d] ' R3I X�u1, u2� D �u1, u2, f �u1, u2��.

The upward unit normal N is given by

N D �−fx,−fy, 1�√f 2x C f 2

y C 1.

The parametrization is compatible with the orientation since

Tu1 * Tu2 D

∣∣∣∣∣∣∣i j k

1 0 fu1

0 1 fu2

∣∣∣∣∣∣∣ D �−fu1 ,−fu2 , 1�

is parallel to N (when N is expressed in terms of the parametrization). Thus,∫Sω D

∫Xω D

∫ dc

∫ baωX�u1,u2��Tu1 ,Tu2� du1 du2.

For ω D z dx G dy, this is

∫ dc

∫ baf�u1, u2�

∣∣∣∣∣ 1 0

0 1

∣∣∣∣∣ du1 du2 D∫ dc

∫ baf�u1, u2� du1 du2 D area under the graph.

(c) Parametrize M using

X : D' Rn, X�u1, ... , un−1� D �u1, ... , un−1, f �u1, ... , un−1��.


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Then, depending on how M is oriented,∫Mω D ;

∫Xω D ;

∫· · ·∫DωX�u��Tu1 , ... ,Tun−1� du1 · · · dun−1

D ;

∫· · ·∫Df�u1, ... , un−1�det

1. . .

1

du1 · · · dun−1

D ;

∫· · ·∫Df�u1, ... , un−1� du1 · · · dun−1 D ;�n-dimensional volume under the graph�.

If you orient M with the unit normal

N D �−1�n�fx1 , ... , fxn−1 ,−1�√

�fx1�2 C · · · C �fxn−1�

2 C 1

we can guarantee a C sign above.4. (a) Define a parametrization

X : [0, 3] * [0, 2π�' R3I X�u1, u2� D �cosu2, u1, sin u2�.

Then we may define X�u��a, b� D det[N a b]. Note that X is compatible with this orientation as Tu1 D�0, 1, 0� and Tu2 D �− sin u2, 0, cos u2� so that

X�u��Tu1 ,Tu2� D det

cos u2 0 − sin u2

0 1 0

sinu2 0 cosu2

D 1 > 0.

(Note that the first column is the normal N in terms of the parametrization.)(b) The boundary �M consists of two disjoint pieces. The left piece is {�x, 0, z�|x2 C z2 D 1}, parametrized

by Y1 : [0, 2π� ' R3,Y1�t� D �cos t, 0, sin t�. The right piece is {�x, 3, z�|x2 C z2 D 1}, parametrizedby Y2 : [0, 2π�' R3,Y2�t� D �cos t, 3, sin t�.

(c) We must first determine V, a unit vector tangent toM, normal to �M, and pointing away fromM. If you thinkabout the boundary pieces we looked at in part (b), a vector corresponding to the left side is V1 D �0,−1, 0�and corresponding to the right side is V2 D �0, 1, 0�. Then, along the left circle of �M,

�MY1�t��a� D X�0,t��V1, a�

and along the right circle of �M,�M

Y2�t��a� D X�3,t��V2, a�.

Note that

�MY1�t��Tt� D det

cos t 0 − sin t

0 −1 0

sin t 0 cos t

D −1.

So the parametrization Y1 is incompatible with �M . However,

�MY2�t��Tt� D det

cos t 0 − sin t

0 1 0

sin t 0 cos t

D 1.

So the parametrization Y2 is compatible with �M .


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(d) If ω D z dx C �x C y C z� dy − x dz, we have

dω D dz G dx C �dx C dy C dz� G dy − dx G dz D dx G dy − dy G dz − 2 dx G dz.

Then using the orientation and the parametrization X from part (a), we have∫M

dω D∫

Xdω D

∫ 2π

0

∫ 3

0

∣∣∣∣∣ 0 − sin u2

1 0

∣∣∣∣∣ −∣∣∣∣∣ 1 0

0 cos u2

∣∣∣∣∣ − 2

∣∣∣∣∣ 0 − sin u2

0 cos u2

∣∣∣∣∣ du1 du2

D∫ 2π

0

∫ 3

0�sinu2 − cosu2� du1 du2 D 3�− cos u2 − sinu2�|2π0 D 0.

On the other hand, using the parametrizations Y1 and Y2 for �M in parts (b) and (c), we have (after reversingthe sign for the left piece because of the incompatibility with �M)∫

�Mω D −

∫Y1

ω C∫

Y2

ω

D −∫ 2π

0[sin t�− sin t� C �cos t C sin t� · 0 − cos t�cos t�] dt

C∫ 2π

0[sin t�− sin t� C �cos t C 3 C sin t� · 0 − cos t�cos t�] dt D 0.

5. If S4 is the unit 4-sphere in R5, then let B denote the 5-dimensional unit ball

B D {x1, x2, x3, x4, x5�|x21 C x2

2 C x23 C x2

4 C x25 … 1}.

Note that �B D S4. Then using the generalized Stokes’s theorem, we have∫S4ω D

∫B

dω.

For ω D x3 dx1 G dx2 G dx4 G dx5 C x4 dx1 G dx2 G dx3 G dx5 we have dω D dx3 G dx1 G dx2 G dx4 G

dx5 C dx4 G dx1 G dx2 G dx3 G dx5 D dx1 G · · · G dx5 − dx1 G · · · G dx5 D 0. Hence∫S4 ω D

∫B 0 D 0.

6. (a) Let ω D f . Then df D∑i

�f

�xidxi and

d�df� D∑i

d ( �f

�xi) G dxi D∑

i

∑j

�2f

�xj�xiG dxj

G dxi

D∑i<j

�2f

�xj�xidxj G dxi C ∑

i>j

�2f

�xj�xidxj G dxi,

since the terms where i D j contain dxi G dxi D 0. By exchanging the roles of i and j in the second sum,we find

d�df� D∑i<j

�2f

�xj�xidxj G dxi C ∑

i<j

�2f

�xi�xjdxi G dxj

D∑i<j(− �2f

�xj�xiC �2f

�xj�xi) dxi G dxj D 0

since the mixed partials are equal because f is of class C2.


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(b) Now

d�dω� D d

d

∑1…i1<···<ik…n

Fi1...ik dxi1 G · · · G dxik

D d

∑1…i1<···<ik…n

dFi1...ik G dxi1 G · · · G dxik

D ∑

1…i1<···<ik…n[d�dFi1...ik � G dxi1 G · · · G dxik

C �−1�1 dFi1...ik G d�dxi1 G · · · G dxik�] from Exercise 1,

D − ∑1…i1<···<ik…n

dFi1...ik G d�dxi1 G · · · G dxik�

since d�dFi1...ik � D 0 from part (a). But

d�dxi1 G · · · G dxik� D d�1 dxi1 G · · · G dxik� D d�1� G dxi1 G · · · G dxik D 0.

Hence d�dω� D 0, as desired.7. (a) If ω is a 0-form, write ω D f . Then, using the first row of the chart, the 1-form dω corresponds to the vector

field §f . Hence, from the second row of the chart, d�dω� is the 2-form that corresponds to § * §f . Thusd�dω� D 0 “translates” to the statement § * �§f� D 0.

(b) If ω is a 1-form, it corresponds to the vector field F and, using the second row of the chart, dω is the 2-formthat corresponds to § * F, another vector field. Then, using the third row of the chart, d�dω� is the 3-formthat corresponds to § · �§ * F�. Hence, d�dω� D 0 “translates” to the statement that § · �§ * F� D 0.

8. (a) The outward unit normal N D �x, y, z� gives orientation form X�u��a1, a2� D det[N a1 a2] where X is aparametrization of S. For a specific parametrization we can use

X : [0, π] * [0, 2π�' R3I X�u1, u2� D �sinu1 cos u2, sin u1 sinu2, cos u1�.

Then Tu1 D �cosu1 cos u2, cos u1 sinu2,− sin u1� and Tu2 D �− sinu1 sin u2, sin u1 cosu2, 0�, so that

X�u��Tu1 ,Tu2� D det

sinu1 cosu2 cosu1 cos u2 − sin u1 sinu2

sinu1 sinu2 cos u1 sin u2 sinu1 cosu2

cosu1 − sin u1 0

D sinu1 Ú 0.

In fact, this quantity is strictly greater than 0 when the parametrization is smooth and so the parametrizationis compatible with the orientation.

Next we note that on S we have ω D x dy G dz C y dz G dx C z dx G dy as the denominators in ωare all 1 on S. Therefore,∫

Sω D

∫Xω

D∫ 2π

0

∫ π0

sin u1 cos u2

∣∣∣∣∣ cosu1 sin u2 sin u1 cos u2

− sin u1 0

∣∣∣∣∣C sin u1 sinu2

∣∣∣∣∣ − sin u1 0

cos u1 cos u2 − sin u1 sin u2

∣∣∣∣∣C cos u1

∣∣∣∣∣ cosu1 cos u2 − sin u1 sin u2

cos u1 sin u2 sinu1 cosu2

∣∣∣∣∣ du1 du2


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D∫ 2π

0

∫ π0�sin3 u1 C cos2 u1 sin u1� du1 du2 D

∫ 2π

0

∫ π0

sinu1 du1 du2

D 2π�− cos u1�|π0 D 4π.

(b) For ω as given we calculate

d [ x

�x2 C y2 C z2�3/2 ] D �y2 C z2 − 2x2� dx − 3xy dy − 3xz dz

�x2 C y2 C z2�5/2

d [ y

�x2 C y2 C z2�3/2 ] D −3xy dx − �x2 C z2 − 2y2� dy − 3yz dz

�x2 C y2 C z2�5/2

d [ z

�x2 C y2 C z2�3/2 ] D −3xz dx − 3yz dy − �x2 C y2 − 2z2�dz

�x2 C y2 C z2�5/2

Hence,

dω D 1�x2 C y2 C z2�5/2 [�y2 C z2 − 2x2� dx G dy G dz

C �x2 − 2y2 C z2� dy G dz G dx

C�x2 C y2 − 2z2� dz G dx G dy]This is identically equal to 0 wherever it is defined.

(c) Since M does not include the origin, we have∫M dω D

∫M 0 D 0 from part (b).

�M consists of two pieces. The outer piece S1 is the unit sphere x2 C y2 C z2 D 1, oriented by theoutward unit normal n1 D �x, y, z�. The inner piece is the sphere x2 C y2 C z2 D a2 of radius a, orientedby inward unit normal n2 D �−x,−y,−z�/a. Then, using Proposition 2.4, we have∫

�Mω D

6�M

F · dS where F D xi C yj C zk�x2 C y2 C z2�3/2

.

In the following calculation we will use the fact that x2 C y2 C z2 is 1 on S1 and is a2 on S2.∫�Mω D

6S1

F · n1 dS C6S2

F · n2 dS

D6S1

1 dS C6S2

− 1a2

dS

D �1��surface area of S1� − 1a2�surface area of S2�

D 4π − 1a2�4πa2� D 0.

This verifies Theorem 3.2.(d) No—since ω is not defined at the origin, Theorem 3.2 does not apply.(e) Let M be the 3-manifold bounded on the outside by S, oriented with the outward normal, and on the inside

by Sε, oriented by the inward normal. Then 0 /∈ M, so we have

0 D∫M

dω D∫

�Mω D

∫Sω C

∫Sε

ω D∫Sω − 4π.

The last equality follows from part (c). The conclusion is that ∫S ω D 4π.


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9. Because �M D ∅, the note following Theorem 3.2 advises us to take ∫�M ω G η to be 0 in the equation∫�M ω G η D ∫M d�ω G η�. Now substitute the results of Exercise 1 to get

0 D∫M

d�ω G η� D∫M

dω G η C �−1�kω G dη D∫M

dω G η C �−1�k∫Mω G dη.

Pull this last piece to the other side to obtain the result

�−1�kC1∫Mω G dη D

∫M

dω G η.

10. By the generalized Stokes’s theorem,∫�Mfω D

∫M

d�fω�

D∫M�df G ω C f G dω� by the result of Exercise 1,

D∫M�df G ω C f dω�.

Hence ∫Mf dω D

∫�Mfω −

∫M

df G ω.


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Vector Analysis in Higher Dimensions · 2009-01-23 · 386 Chapter 8 Vector Analysis in Higher Dimensions 20. Here 1k dx1 G1 and k ,.-FORMS 3 3.,,,