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1: Double Integrals over Rectangles

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97817_16_ch16_p1128-1129.qk_97817_16_ch16_p1128-1129 11/9/10 9:22 AM Page 1128

1128

CHAPTER 16

VECTOR CALCULUS

Similarly, if we keep v constant by putting v ෇ v0 , we get a grid curve C2 given by

r͑u, v0 ͒ that lies on S, and its tangent vector at P0 is

ru ෇

5

Ѩx

Ѩy

Ѩz

͑u0 , v0 ͒ i ϩ

͑u0 , v0 ͒ j ϩ

͑u0 , v0 ͒ k

Ѩu

Ѩu

Ѩu

If ru ϫ rv is not 0, then the surface S is called smooth (it has no “corners”). For a smooth

surface, the tangent plane is the plane that contains the tangent vectors ru and rv , and the

vector ru ϫ rv is a normal vector to the tangent plane.

Figure 13 shows the self-intersecting

surface in Example 9 and its tangent plane

at ͑1, 1, 3͒.

EXAMPLE 9 Find the tangent plane to the surface with parametric equations x ෇ u 2,

y ෇ v 2, z ෇ u ϩ 2v at the point ͑1, 1, 3͒.

v

SOLUTION We first compute the tangent vectors:

z

(1, 1, 3)

ru ෇

Ѩy

Ѩz

Ѩx

k ෇ 2u i ϩ k

Ѩu

Ѩu

Ѩu

rv ෇

Ѩx

Ѩy

Ѩz

k ෇ 2v j ϩ 2 k

Ѩv

Ѩv

Ѩv

y

x

Thus a normal vector to the tangent plane is

Խ Խ

i

ru ϫ rv ෇ 2u

0

FIGURE 13

j k

0 1 ෇ Ϫ2v i Ϫ 4u j ϩ 4uv k

2v 2

Notice that the point ͑1, 1, 3͒ corresponds to the parameter values u ෇ 1 and v ෇ 1, so

the normal vector there is

Ϫ2 i Ϫ 4 j ϩ 4 k

Therefore an equation of the tangent plane at ͑1, 1, 3͒ is

Ϫ2͑x Ϫ 1͒ Ϫ 4͑y Ϫ 1͒ ϩ 4͑z Ϫ 3͒ ෇ 0

x ϩ 2y Ϫ 2z ϩ 3 ෇ 0

or

Surface Area

Now we define the surface area of a general parametric surface given by Equation 1. For

simplicity we start by considering a surface whose parameter domain D is a rectangle, and

we divide it into subrectangles Rij . Let’s choose ͑u i*, vj*͒ to be the lower left corner of Rij.

(See Figure 14.)

z

R ij

r

Ỵ√

Pij

Sij

Ỵu

(u *i , √ *j )

FIGURE 14

The image of the

subrectangle Rij is the patch Sij .

0

0

u

x

y

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

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97817_16_ch16_p1128-1129.qk_97817_16_ch16_p1128-1129 11/9/10 9:22 AM Page 1129

PARAMETRIC SURFACES AND THEIR AREAS

SECTION 16.6

1129

The part Sij of the surface S that corresponds to Rij is called a patch and has the point Pij

with position vector r͑u i*, vj*͒ as one of its corners. Let

ru* ෇ ru͑u i*, vj*͒

Sij

Pij

(a)

rv* ෇ rv͑u i*, vj*͒

and

be the tangent vectors at Pij as given by Equations 5 and 4.

Figure 15(a) shows how the two edges of the patch that meet at Pij can be approximated

by vectors. These vectors, in turn, can be approximated by the vectors ⌬u r*u and ⌬v r*v

because partial derivatives can be approximated by difference quotients. So we approximate Sij by the parallelogram determined by the vectors ⌬u r*u and ⌬v r*v . This parallelogram

is shown in Figure 15(b) and lies in the tangent plane to S at Pij . The area of this parallelogram is

͑⌬u ru*͒ ϫ ͑⌬v r*͒

෇ ru* ϫ r*v ⌬u ⌬v

v

Խ

Խ Խ

Խ

and so an approximation to the area of S is

m

n

͚ ͚ Խ r* ϫ r* Խ ⌬u ⌬v

u

v

i෇1 j෇1

Ỵ√ r √*

Ỵu r u*

(b)

Our intuition tells us that this approximation gets better as we increase the number of subrectangles, and we recognize the double sum as a Riemann sum for the double integral

xxD ru ϫ rv du dv. This motivates the following definition.

Խ

Խ

6

Definition If a smooth parametric surface S is given by the equation

r͑u, v͒ ෇ x͑u, v͒ i ϩ y͑u, v͒ j ϩ z͑u, v͒ k

FIGURE 15

Approximating a patch

by a parallelogram

͑u, v͒ ʦ D

and S is covered just once as ͑u, v͒ ranges throughout the parameter domain D,

then the surface area of S is

Խ

Խ

A͑S͒ ෇ yy ru ϫ rv dA

D

where

ru ෇

Ѩx

Ѩy

Ѩz

k

Ѩu

Ѩu

Ѩu

rv ෇

Ѩx

Ѩy

Ѩz

k

Ѩv

Ѩv

Ѩv

EXAMPLE 10 Find the surface area of a sphere of radius a.

SOLUTION In Example 4 we found the parametric representation

x ෇ a sin ␾ cos ␪

y ෇ a sin ␾ sin ␪

z ෇ a cos ␾

where the parameter domain is

D ෇ ͕͑␾, ␪͒

Խ 0 ഛ ␾ ഛ ␲, 0 ഛ ␪ ഛ 2␲ ͖

We first compute the cross product of the tangent vectors:

Խ

i

Ѩx

r␾ ϫ r␪ ෇ Ѩ␾

Ѩx

Ѩ␪

j

Ѩy

Ѩ␾

Ѩy

Ѩ␪

ԽԽ

k

Ѩz

i

Ѩ␾ ෇ Ϫa cos ␾ cos ␪

Ѩz

Ϫa sin ␾ sin ␪

Ѩ␪

j

a cos ␾ sin ␪

a sin ␾ cos ␪

k

Ϫa sin ␾

0

Խ

෇ a 2 sin 2␾ cos ␪ i ϩ a 2 sin2␾ sin ␪ j ϩ a 2 sin ␾ cos ␾ k

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

97817_16_ch16_p1130-1137.qk_97817_16_ch16_p1130-1137 11/9/10 9:23 AM Page 1130

1130

CHAPTER 16

VECTOR CALCULUS

Thus

Խr

Խ

ϫ r␪ ෇ sa 4 sin 4␾ cos 2␪ ϩ a 4 sin 4␾ sin 2␪ ϩ a 4 sin 2␾ cos 2␾

෇ sa 4 sin 4␾ ϩ a 4 sin 2␾ cos 2␾ ෇ a 2 ssin 2␾ ෇ a 2 sin ␾

since sin ␾ ജ 0 for 0 ഛ ␾ ഛ ␲. Therefore, by Definition 6, the area of the sphere is

Խ

Խ

A ෇ yy r␾ ϫ r␪ dA ෇ y

D

෇ a2 y

2␲

0

2␲

0

y

0

a 2 sin ␾ d␾ d␪

d␪ y sin ␾ d␾ ෇ a 2͑2␲͒2 ෇ 4␲a 2

0

Surface Area of the Graph of a Function

For the special case of a surface S with equation z ෇ f ͑x, y͒, where ͑x, y͒ lies in D and f has

continuous partial derivatives, we take x and y as parameters. The parametric equations are

x෇x

rx ෇ i ϩ

so

and

y෇y

ͩ ͪ

Ѩf

Ѩx

z ෇ f ͑x, y͒

ry ෇ j ϩ

k

ͩ ͪ

Ѩf

Ѩy

k

Խ Խ

7

i

j

rx ϫ ry ෇ 1

0

0

1

k

Ѩf

Ѩf

Ѩf

෇Ϫ

jϩk

Ѩx

Ѩx

Ѩy

Ѩf

Ѩy

Thus we have

8

Խr

x

Խ

ϫ ry ෇

ͱͩ ͪ ͩ ͪ

Ѩf

Ѩx

2

ϩ

Ѩf

Ѩy

2

ϩ1෇

ͱ ͩ ͪ ͩ ͪ

Ѩz

Ѩx

2

ϩ

Ѩz

Ѩy

2

and the surface area formula in Definition 6 becomes

Notice the similarity between the surface area

formula in Equation 9 and the arc length formula

L෇

y

b

a

from Section 8.1.

ͱ ͩ ͪ

dy

dx

2

9

dx

A͑S͒ ෇

yy

D

v

ͱ ͩ ͪ ͩ ͪ

Ѩz

Ѩx

2

ϩ

Ѩz

Ѩy

2

dA

EXAMPLE 11 Find the area of the part of the paraboloid z ෇ x 2 ϩ y 2 that lies under

the plane z ෇ 9.

SOLUTION The plane intersects the paraboloid in the circle x 2 ϩ y 2 ෇ 9, z ෇ 9. There-

fore the given surface lies above the disk D with center the origin and radius 3. (See

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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

97817_16_ch16_p1130-1137.qk_97817_16_ch16_p1130-1137 11/9/10 9:23 AM Page 1131

PARAMETRIC SURFACES AND THEIR AREAS

SECTION 16.6

z

1131

Figure 16.) Using Formula 9, we have

9

A෇

yy

D

ͱ ͩ ͪ ͩ ͪ

Ѩz

Ѩx

Ѩz

Ѩy

2

ϩ

2

dA

෇ yy s1 ϩ ͑2x͒ 2 ϩ ͑2y͒ 2 dA

D

෇ yy s1 ϩ 4͑x 2 ϩ y 2 ͒ dA

D

x

3

D

y

Converting to polar coordinates, we obtain

FIGURE 16

A෇y

2␲

0

y

3

0

s1 ϩ 4r 2 r dr d␪ ෇ y

2␲

0

෇ 2␲ ( 18 ) 23 ͑1 ϩ 4r 2 ͒3͞2

]

3

0

3

d␪ y rs1 ϩ 4r 2 dr

0

(37s37 Ϫ 1)

6

The question remains whether our definition of surface area 6 is consistent with the

surface area formula from single-variable calculus (8.2.4).

We consider the surface S obtained by rotating the curve y ෇ f ͑x͒, a ഛ x ഛ b, about

the x-axis, where f ͑x͒ ജ 0 and f Ј is continuous. From Equations 3 we know that parametric equations of S are

y ෇ f ͑x͒ cos ␪

x෇x

z ෇ f ͑x͒ sin ␪

aഛxഛb

0 ഛ ␪ ഛ 2␲

To compute the surface area of S we need the tangent vectors

rx ෇ i ϩ f Ј͑x͒ cos ␪ j ϩ f Ј͑x͒ sin ␪ k

r␪ ෇ Ϫf ͑x͒ sin ␪ j ϩ f ͑x͒ cos ␪ k

Խ

Thus

i

j

rx ϫ r␪ ෇ 1 f Ј͑x͒ cos ␪

0 Ϫf ͑x͒ sin ␪

k

f Ј͑x͒ sin ␪

f ͑x͒ cos ␪

Խ

෇ f ͑x͒ f Ј͑x͒ i Ϫ f ͑x͒ cos ␪ j Ϫ f ͑x͒ sin ␪ k

and so

Խr

x

Խ

ϫ r␪ ෇ s͓ f ͑x͔͒ 2 ͓ f Ј͑x͔͒ 2 ϩ ͓ f ͑x͔͒ 2 cos 2␪ ϩ ͓ f ͑x͔͒ 2 sin 2␪

෇ s͓ f ͑x͔͒ 2 ͓1 ϩ ͓ f Ј͑x͔͒ 2 ͔ ෇ f ͑x͒s1 ϩ ͓ f Ј͑x͔͒ 2

because f ͑x͒ ജ 0. Therefore the area of S is

Խ

Խ

A ෇ yy rx ϫ r␪ dA

D

෇y

2␲

0

y

b

a

f ͑x͒s1 ϩ ͓ f Ј͑x͔͒ 2 dx d␪

b

෇ 2␲ y f ͑x͒s1 ϩ ͓ f Ј͑x͔͒ 2 dx

a

This is precisely the formula that was used to define the area of a surface of revolution in

single-variable calculus (8.2.4).

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

97817_16_ch16_p1130-1137.qk_97817_16_ch16_p1130-1137 11/9/10 9:23 AM Page 1132

1132

VECTOR CALCULUS

CHAPTER 16

16.6

Exercises

1–2 Determine whether the points P and Q lie on the given

surface.

z

I

II

z

1. r͑u, v͒ ෇ ͗2u ϩ 3v, 1 ϩ 5u Ϫ v, 2 ϩ u ϩ v ͘

P͑7, 10, 4͒, Q͑5, 22, 5͒

2. r͑u, v͒ ෇ ͗u ϩ v, u 2 Ϫ v, u ϩ v 2 ͘

x

P͑3, Ϫ1, 5͒, Q͑Ϫ1, 3, 4͒

x

y

y

3–6 Identify the surface with the given vector equation.

3. r͑u, v͒ ෇ ͑u ϩ v͒ i ϩ ͑3 Ϫ v͒ j ϩ ͑1 ϩ 4u ϩ 5v͒ k

4. r͑u, v͒ ෇ 2 sin u i ϩ 3 cos u j ϩ v k,

z

III

IV

z

0ഛvഛ2

5. r͑s, t͒ ෇ ͗s, t, t Ϫ s ͘

2

2

6. r͑s, t͒ ෇ ͗ s sin 2t, s 2, s cos 2t ͘

x

; 7–12 Use a computer to graph the parametric surface. Get a

printout and indicate on it which grid curves have u constant and

which have v constant.

z

V

7. r͑u, v͒ ෇ ͗u 2, v 2, u ϩ v ͘,

Ϫ1 ഛ u ഛ 1, Ϫ1 ഛ v ഛ 1

y

x

y

VI

z

8. r͑u, v͒ ෇ ͗u, v 3, Ϫv ͘ ,

Ϫ2 ഛ u ഛ 2, Ϫ2 ഛ v ഛ 2

9. r͑u, v͒ ෇ ͗u cos v, u sin v, u 5 ͘ ,

Ϫ1 ഛ u ഛ 1, 0 ഛ v ഛ 2␲

y

x

x

10. r͑u, v͒ ෇ ͗u, sin͑u ϩ v͒, sin v ͘ ,

Ϫ␲ ഛ u ഛ ␲, Ϫ␲ ഛ v ഛ ␲

y

11. x ෇ sin v,

y ෇ cos u sin 4 v, z ෇ sin 2u sin 4 v,

0 ഛ u ഛ 2␲, Ϫ␲͞2 ഛ v ഛ ␲͞2

y ෇ cos u sin v, z ෇ sin v,

0 ഛ u ഛ 2␲, 0 ഛ v ഛ 2␲

12. x ෇ sin u,

19–26 Find a parametric representation for the surface.

19. The plane through the origin that contains the vectors i Ϫ j

and j Ϫ k

13–18 Match the equations with the graphs labeled I–VI and

curves have u constant and which have v constant.

13. r͑u, v͒ ෇ u cos v i ϩ u sin v j ϩ v k

Ϫ␲ ഛ u ഛ ␲

15. r͑u, v͒ ෇ sin v i ϩ cos u sin 2v j ϩ sin u sin 2v k

16. x ෇ ͑1 Ϫ u͒͑3 ϩ cos v͒ cos 4␲ u,

y ෇ ͑1 Ϫ u͒͑3 ϩ cos v͒ sin 4␲ u,

z ෇ 3u ϩ ͑1 Ϫ u͒ sin v

Խ Խ

y ෇ sin 3 u cos 3 v,

18. x ෇ (1 Ϫ u ) cos v,

;

Խ Խ

21. The part of the hyperboloid 4x 2 Ϫ 4y 2 Ϫ z 2 ෇ 4 that lies in

22. The part of the ellipsoid x 2 ϩ 2y 2 ϩ 3z 2 ෇ 1 that lies to the

left of the xz-plane

23. The part of the sphere x 2 ϩ y 2 ϩ z 2 ෇ 4 that lies above the

cone z ෇ sx 2 ϩ y 2

24. The part of the sphere x 2 ϩ y 2 ϩ z 2 ෇ 16 that lies between

z ෇ sin 3 v

y ෇ (1 Ϫ u ) sin v,

Graphing calculator or computer required

contains the vectors ͗2, 1, 4 ͘ and ͗Ϫ3, 2, 5͘

front of the yz-plane

14. r͑u, v͒ ෇ u cos v i ϩ u sin v j ϩ sin u k,

17. x ෇ cos 3 u cos 3 v,

20. The plane that passes through the point ͑0, Ϫ1, 5͒ and

z෇u

the planes z ෇ Ϫ2 and z ෇ 2

25. The part of the cylinder y 2 ϩ z 2 ෇ 16 that lies between the

planes x ෇ 0 and x ෇ 5

CAS Computer algebra system required

1. Homework Hints available at stewartcalculus.com

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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SECTION 16.6

26. The part of the plane z ෇ x ϩ 3 that lies inside the cylinder

38. r͑u, v͒ ෇ ͑1 Ϫ u 2 Ϫ v 2 ͒ i Ϫ v j Ϫ u k;

x ϩy ෇1

2

PARAMETRIC SURFACES AND THEIR AREAS

2

1133

͑Ϫ1, Ϫ1, Ϫ1͒

39–50 Find the area of the surface.

CAS

27–28 Use a computer algebra system to produce a graph that

39. The part of the plane 3x ϩ 2y ϩ z ෇ 6 that lies in the

looks like the given one.

first octant

27.

28.

40. The part of the plane with vector equation

r͑u, v͒ ෇ ͗u ϩ v, 2 Ϫ 3u, 1 ϩ u Ϫ v ͘ that is given by

0 ഛ u ഛ 2, Ϫ1 ഛ v ഛ 1

3

z

z 0

_3

_3

y

0 5

x

41. The part of the plane x ϩ 2y ϩ 3z ෇ 1 that lies inside the

0

cylinder x 2 ϩ y 2 ෇ 3

42. The part of the cone z ෇ sx 2 ϩ y 2 that lies between the

_1

_1

0

y

0

1 1

0

plane y ෇ x and the cylinder y ෇ x 2

_1

x

43. The surface z ෇ 3 ͑x 3͞2 ϩ y 3͞2 ͒, 0 ഛ x ഛ 1, 0 ഛ y ഛ 1

2

44. The part of the surface z ෇ 1 ϩ 3x ϩ 2y 2 that lies above the

triangle with vertices ͑0, 0͒, ͑0, 1͒, and ͑2, 1͒

; 29. Find parametric equations for the surface obtained by

Ϫx

rotating the curve y ෇ e , 0 ഛ x ഛ 3, about the x-axis and

use them to graph the surface.

45. The part of the surface z ෇ xy that lies within the

cylinder x 2 ϩ y 2 ෇ 1

; 30. Find parametric equations for the surface obtained by

46. The part of the paraboloid x ෇ y 2 ϩ z 2 that lies inside the

rotating the curve x ෇ 4y 2 Ϫ y 4, Ϫ2 ഛ y ഛ 2, about the

y-axis and use them to graph the surface.

cylinder y 2 ϩ z 2 ෇ 9

47. The part of the surface y ෇ 4x ϩ z 2 that lies between the

; 31. (a) What happens to the spiral tube in Example 2 (see Fig-

planes x ෇ 0, x ෇ 1, z ෇ 0, and z ෇ 1

ure 5) if we replace cos u by sin u and sin u by cos u ?

(b) What happens if we replace cos u by cos 2u and sin u

by sin 2u?

48. The helicoid (or spiral ramp) with vector equation

r͑u, v͒ ෇ u cos v i ϩ u sin v j ϩ v k, 0 ഛ u ഛ 1, 0 ഛ v ഛ ␲

49. The surface with parametric equations x ෇ u 2 , y ෇ u v,

z ෇ 12 v 2, 0 ഛ u ഛ 1, 0 ഛ v ഛ 2

; 32. The surface with parametric equations

x ෇ 2 cos ␪ ϩ r cos͑␪͞2͒

50. The part of the sphere x 2 ϩ y 2 ϩ z 2 ෇ b 2 that lies inside the

y ෇ 2 sin ␪ ϩ r cos͑␪͞2͒

cylinder x 2 ϩ y 2 ෇ a 2, where 0 Ͻ a Ͻ b

z ෇ r sin͑␪͞2͒

51. If the equation of a surface S is z ෇ f ͑x, y͒, where

where Ϫ 12 ഛ r ഛ 12 and 0 ഛ ␪ ഛ 2␲, is called a Möbius

strip. Graph this surface with several viewpoints. What is

Խ Խ

52–53 Find the area of the surface correct to four decimal places

33–36 Find an equation of the tangent plane to the given

by expressing the area in terms of a single integral and using

your calculator to estimate the integral.

parametric surface at the specified point.

33. x ෇ u ϩ v,

34. x ෇ u 2 ϩ 1,

y ෇ 3u 2,

z ෇ u Ϫ v ; ͑2, 3, 0͒

y ෇ v 3 ϩ 1,

cylinder x 2 ϩ y 2 ෇ 1

37–38 Find an equation of the tangent plane to the given

parametric surface at the specified point. Graph the surface and

the tangent plane.

37. r͑u, v͒ ෇ u 2 i ϩ 2u sin v j ϩ u cos v k;

2

u ෇ 1, v ෇ 0

2

53. The part of the surface z ෇ eϪx Ϫy that lies above the

u ෇ 1, v ෇ ␲͞3

36. r͑u, v͒ ෇ sin u i ϩ cos u sin v j ϩ sin v k;

u ෇ ␲͞6 , v ෇ ␲͞6

CAS

52. The part of the surface z ෇ cos͑x 2 ϩ y 2 ͒ that lies inside the

z ෇ u ϩ v ; ͑5, 2, 3͒

35. r͑u, v͒ ෇ u cos v i ϩ u sin v j ϩ v k;

Խ Խ

x 2 ϩ y 2 ഛ R 2, and you know that fx ഛ 1 and fy ഛ 1,

what can you say about A͑S͒?

disk x ϩ y ഛ 4

2

CAS

2

54. Find, to four decimal places, the area of the part of the sur-

face z ෇ ͑1 ϩ x 2 ͒͑͞1 ϩ y 2 ͒ that lies above the square

x ϩ y ഛ 1. Illustrate by graphing this part of the

surface.

Խ Խ Խ Խ

55. (a) Use the Midpoint Rule for double integrals (see Sec-

tion 15.1) with six squares to estimate the area of the

surface z ෇ 1͑͞1 ϩ x 2 ϩ y 2 ͒, 0 ഛ x ഛ 6, 0 ഛ y ഛ 4.

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1134

CAS

CAS

CHAPTER 16

VECTOR CALCULUS

61. Find the area of the part of the sphere x 2 ϩ y 2 ϩ z 2 ෇ 4z

(b) Use a computer algebra system to approximate the surface area in part (a) to four decimal places. Compare

with the answer to part (a).

that lies inside the paraboloid z ෇ x 2 ϩ y 2.

62. The figure shows the surface created when the cylinder

y 2 ϩ z 2 ෇ 1 intersects the cylinder x 2 ϩ z 2 ෇ 1. Find the

area of this surface.

56. Find the area of the surface with vector equation

r͑u, v͒ ෇ ͗ cos 3u cos 3v, sin 3u cos 3v, sin 3v ͘ , 0 ഛ u ഛ ␲,

0 ഛ v ഛ 2␲. State your answer correct to four decimal

z

places.

CAS

57. Find the exact area of the surface z ෇ 1 ϩ 2x ϩ 3y ϩ 4y 2,

1 ഛ x ഛ 4, 0 ഛ y ഛ 1.

x

58. (a) Set up, but do not evaluate, a double integral for the area

of the surface with parametric equations x ෇ au cos v,

y ෇ bu sin v, z ෇ u 2, 0 ഛ u ഛ 2, 0 ഛ v ഛ 2␲.

(b) Eliminate the parameters to show that the surface is an

elliptic paraboloid and set up another double integral for

the surface area.

(c) Use the parametric equations in part (a) with a ෇ 2 and

b ෇ 3 to graph the surface.

(d) For the case a ෇ 2, b ෇ 3, use a computer algebra system

to find the surface area correct to four decimal places.

;

CAS

59. (a) Show that the parametric equations x ෇ a sin u cos v,

y ෇ b sin u sin v, z ෇ c cos u, 0 ഛ u ഛ ␲, 0 ഛ v ഛ 2␲,

63. Find the area of the part of the sphere x 2 ϩ y 2 ϩ z 2 ෇ a 2

that lies inside the cylinder x 2 ϩ y 2 ෇ ax.

64. (a) Find a parametric representation for the torus obtained

;

represent an ellipsoid.

(b) Use the parametric equations in part (a) to graph the

ellipsoid for the case a ෇ 1, b ෇ 2, c ෇ 3.

(c) Set up, but do not evaluate, a double integral for the surface area of the ellipsoid in part (b).

;

y

by rotating about the z-axis the circle in the xz-plane

with center ͑b, 0, 0͒ and radius a Ͻ b. [Hint: Take as

parameters the angles ␪ and ␣ shown in the figure.]

(b) Use the parametric equations found in part (a) to graph

the torus for several values of a and b.

(c) Use the parametric representation from part (a) to find

the surface area of the torus.

z

(x, y, z)

60. (a) Show that the parametric equations x ෇ a cosh u cos v,

y ෇ b cosh u sin v, z ෇ c sinh u, represent a hyperboloid

;

16.7

of one sheet.

(b) Use the parametric equations in part (a) to graph the

hyperboloid for the case a ෇ 1, b ෇ 2, c ෇ 3.

(c) Set up, but do not evaluate, a double integral for the surface area of the part of the hyperboloid in part (b) that

lies between the planes z 3 and z 3.

0

ă

x

y

(b, 0, 0)

Surface Integrals

The relationship between surface integrals and surface area is much the same as the relationship between line integrals and arc length. Suppose f is a function of three variables

whose domain includes a surface S. We will define the surface integral of f over S in such

a way that, in the case where f ͑x, y, z͒ ෇ 1, the value of the surface integral is equal to the

surface area of S. We start with parametric surfaces and then deal with the special case

where S is the graph of a function of two variables.

Parametric Surfaces

Suppose that a surface S has a vector equation

r͑u, v͒ ෇ x͑u, v͒ i ϩ y͑u, v͒ j ϩ z͑u, v͒ k

͑u, v͒ ʦ D

We first assume that the parameter domain D is a rectangle and we divide it into subrect-

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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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SECTION 16.7

1135

angles Rij with dimensions ⌬u and ⌬v. Then the surface S is divided into corresponding

patches Sij as in Figure 1. We evaluate f at a point Pij* in each patch, multiply by the area

⌬Sij of the patch, and form the Riemann sum

R ij

Ỵ√

D

SURFACE INTEGRALS

Ỵu

m

n

͚ ͚ f ͑P*͒ ⌬S

ij

0

ij

i෇1 j෇1

u

Then we take the limit as the number of patches increases and define the surface integral

of f over the surface S as

r

m

z

S

P *ij

S

Sij

0

x

yy f ͑x, y, z͒ dS ෇

1

y

lim

n

͚ ͚ f ͑P*͒ ⌬S

ij

m, n l ϱ i෇1 j෇1

ij

Notice the analogy with the definition of a line integral (16.2.2) and also the analogy with

the definition of a double integral (15.1.5).

To evaluate the surface integral in Equation 1 we approximate the patch area ⌬Sij by the

area of an approximating parallelogram in the tangent plane. In our discussion of surface

area in Section 16.6 we made the approximation

Խ

Խ

⌬Sij Ϸ ru ϫ rv ⌬u ⌬v

FIGURE 1

where

ru ෇

Ѩx

Ѩy

Ѩz

k

Ѩu

Ѩu

Ѩu

rv ෇

Ѩx

Ѩy

Ѩz

k

Ѩv

Ѩv

Ѩv

are the tangent vectors at a corner of Sij . If the components are continuous and ru and rv

are nonzero and nonparallel in the interior of D, it can be shown from Definition 1, even

when D is not a rectangle, that

We assume that the surface is covered only

once as ͑u, v͒ ranges throughout D. The value

of the surface integral does not depend on the

parametrization that is used.

2

yy f ͑x, y, z͒ dS ෇ yy f ͑r͑u, v͒͒ Խ r

u

S

Խ

ϫ rv dA

D

This should be compared with the formula for a line integral:

y

C

Խ

b

Խ

f ͑x, y, z͒ ds ෇ y f ͑r͑t͒͒ rЈ͑t͒ dt

a

Observe also that

yy 1 dS ෇ yy Խ r

u

S

D

Խ

ϫ rv dA ෇ A͑S͒

Formula 2 allows us to compute a surface integral by converting it into a double integral over the parameter domain D. When using this formula, remember that f ͑r͑u, v͒͒ is

evaluated by writing x ෇ x͑u, v͒, y ෇ y͑u, v͒, and z ෇ z͑u, v͒ in the formula for f ͑x, y, z͒.

EXAMPLE 1 Compute the surface integral xxS x 2 dS, where S is the unit sphere

x 2 ϩ y 2 ϩ z 2 ෇ 1.

SOLUTION As in Example 4 in Section 16.6, we use the parametric representation

x ෇ sin ␾ cos ␪

y ෇ sin ␾ sin ␪

z ෇ cos ␾

0ഛ ␾ഛ␲

0 ഛ ␪ ഛ 2␲

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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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1136

CHAPTER 16

VECTOR CALCULUS

r͑␾, ␪ ͒ ෇ sin ␾ cos ␪ i ϩ sin ␾ sin ␪ j ϩ cos ␾ k

that is,

As in Example 10 in Section 16.6, we can compute that

Խr

Խ

ϫ r␪ ෇ sin ␾

Therefore, by Formula 2,

yy x

2

S

D

෇y

Here we use the identities

2␲

0

cos2␪ ෇ 12 ͑1 ϩ cos 2␪ ͒

෇y

sin ␾ ෇ 1 Ϫ cos ␾

2

Խ

Խ

dS ෇ yy ͑sin ␾ cos ␪ ͒2 r␾ ϫ r␪ dA

Instead, we could use Formulas 64 and 67 in

the Table of Integrals.

sin 2␾ cos 2␪ sin ␾ d␾ d␪ ෇ y

0

2␲ 1

2

0

2

y

2␲

cos 2␪ d␪

0

͑1 ϩ cos 2␪ ͒ d␪

[

y

0

0

sin 3␾ d␾

͑sin ␾ Ϫ sin ␾ cos 2␾͒ d␾

2␲

0

] [Ϫcos ␾ ϩ

෇ 12 ␪ ϩ 12 sin 2␪

y

1

3

]

cos 3␾ 0 ෇

4␲

3

Surface integrals have applications similar to those for the integrals we have previously

considered. For example, if a thin sheet (say, of aluminum foil) has the shape of a surface

S and the density (mass per unit area) at the point ͑x, y, z͒ is ␳ ͑x, y, z͒, then the total mass

of the sheet is

m ෇ yy ␳ ͑x, y, z͒ dS

S

and the center of mass is ͑x, y, z͒, where

x෇

1

m

yy x ␳ ͑x, y, z͒ dS

y෇

S

1

m

yy y ␳ ͑x, y, z͒ dS

z෇

S

1

m

yy z ␳ ͑x, y, z͒ dS

S

Moments of inertia can also be defined as before (see Exercise 41).

Graphs

Any surface S with equation z ෇ t͑x, y͒ can be regarded as a parametric surface with parametric equations

x෇x

and so we have

rx ෇ i ϩ

y෇y

ͩ ͪ

Ѩt

Ѩx

z ෇ t͑x, y͒

ry ෇ j ϩ

k

ͩ ͪ

Ѩt

Ѩy

k

Thus

rx ϫ ry ෇ Ϫ

3

and

Խr

x

Խ

ϫ ry ෇

Ѩt

Ѩt

jϩk

Ѩx

Ѩy

ͱͩ ͪ ͩ ͪ

Ѩz

Ѩx

2

ϩ

Ѩz

Ѩy

2

ϩ1

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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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SECTION 16.7

SURFACE INTEGRALS

1137

Therefore, in this case, Formula 2 becomes

4

ͱͩ ͪ ͩ ͪ

Ѩz

Ѩx

yy f ͑x, y, z͒ dS ෇ yy f ( x, y, t͑x, y͒)

S

D

Ѩz

Ѩy

2

ϩ

2

ϩ 1 dA

Similar formulas apply when it is more convenient to project S onto the yz-plane or

xz-plane. For instance, if S is a surface with equation y ෇ h͑x, z͒ and D is its projection

onto the xz-plane, then

ͱͩ ͪ ͩ ͪ

Ѩy

Ѩx

yy f ͑x, y, z͒ dS ෇ yy f ( x, h͑x, z͒, z)

S

z

D

Ѩy

Ѩz

2

ϩ

2

ϩ 1 dA

EXAMPLE 2 Evaluate xxS y dS, where S is the surface z ෇ x ϩ y 2, 0 ഛ x ഛ 1, 0 ഛ y ഛ 2.

(See Figure 2.)

SOLUTION Since

Ѩz

෇1

Ѩx

y

Formula 4 gives

ͱ ͩ ͪ ͩ ͪ

x

yy y dS ෇ yy y

FIGURE 2

S

D

෇y

Ѩz

෇ 2y

Ѩy

and

1

0

y

2

0

1

Ѩz

Ѩy

2

ϩ

2

dA

ys1 ϩ 1 ϩ 4y 2 dy dx

෇ y dx s2

0

Ѩz

Ѩx

y

2

0

ys1 ϩ 2y 2 dy

]

2

෇ s2 ( 14 ) 23 ͑1 ϩ 2y 2 ͒3͞2 0 ෇

13s2

3

If S is a piecewise-smooth surface, that is, a finite union of smooth surfaces S1 , S2, . . . ,

Sn that intersect only along their boundaries, then the surface integral of f over S is defined

by

yy f ͑x, y, z͒ dS ෇ yy f ͑x, y, z͒ dS ϩ и и и ϩ yy f ͑x, y, z͒ dS

S

z

S£ (z=1+x )

S¡ (≈+¥=1)

x

SOLUTION The surface S is shown in Figure 3. (We have changed the usual position of

the axes to get a better look at S.) For S1 we use ␪ and z as parameters (see Example 5

in Section 16.6) and write its parametric equations as

0

FIGURE 3

Sn

v EXAMPLE 3 Evaluate xxS z dS, where S is the surface whose sides S1 are given by the

cylinder x 2 ϩ y 2 ෇ 1, whose bottom S2 is the disk x 2 ϩ y 2 ഛ 1 in the plane z ෇ 0, and

whose top S3 is the part of the plane z ෇ 1 ϩ x that lies above S2 .

y

S™

S1

x ෇ cos ␪

y ෇ sin ␪

z෇z

where

0 ഛ ␪ ഛ 2␲

and

0 ഛ z ഛ 1 ϩ x ෇ 1 ϩ cos ␪

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1138

CHAPTER 16

VECTOR CALCULUS

Therefore

Խ

Խ

i

r␪ ϫ rz ෇ Ϫsin ␪

0

Խr

and

j

k

cos ␪ 0 ෇ cos ␪ i ϩ sin ␪ j

0

1

Խ

ϫ rz ෇ scos 2␪ ϩ sin 2␪ ෇ 1

Thus the surface integral over S1 is

yy z dS ෇ yy z Խ r␪ ϫ r Խ dA

z

S1

D

෇y

2␲

y

0

1ϩcos ␪

0

෇ 12 y

2␲

0

z dz d␪ ෇ y

2␲ 1

2

͑1 ϩ cos ␪ ͒2 d␪

0

͓1 ϩ 2 cos ␪ ϩ 12 ͑1 ϩ cos 2␪ ͔͒ d␪

[

2␲

0

]

෇ 12 32 ␪ ϩ 2 sin ␪ ϩ 14 sin 2␪

3␲

2

Since S2 lies in the plane z ෇ 0, we have

yy z dS ෇ yy 0 dS ෇ 0

S2

S2

The top surface S3 lies above the unit disk D and is part of the plane z ෇ 1 ϩ x. So,

taking t͑x, y͒ ෇ 1 ϩ x in Formula 4 and converting to polar coordinates, we have

yy

ͱ ͩ ͪ ͩ ͪ

z dS ෇ yy ͑1 ϩ x͒

S3

D

෇y

2␲

0

y

1

0

2␲

y y

෇ s2

y (

෇ s2

0

0

ͫ

Ѩz

Ѩx

2

ϩ

Ѩz

Ѩy

2

dA

͑1 ϩ r cos ␪ ͒s1 ϩ 1 ϩ 0 r dr d␪

෇ s2

1

0

2

1

2

͑r ϩ r 2 cos ␪ ͒ dr d␪

ϩ 13 cos ␪) d␪

sin ␪

ϩ

2

3

ͬ

2␲

෇ s2 ␲

0

Therefore

yy z dS ෇ yy z dS ϩ yy z dS ϩ yy z dS

S

S1

S2

S3

3␲

ϩ 0 ϩ s2 ␲ ෇ ( 32 ϩ s2 )␲

2

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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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