Section 15.8 Triple Integrals in Spherical Coordinates

Section 15.8 Triple Integrals in Spherical Coordinates

9. Write the equation in spherical coordinates. (a) x²+ y²+ z²= 9 (b) x²− y²− z²= 1.

Solution:

SECTION 15.8 TRIPLE INTEGRALS IN SPHERICAL COORDINATES ¤ 595 3. (a) From Equations 1 and 2,  =

²+ ²+ ²=

1²+ 0²+ (−1)²=√2, cos  = 

 = −1

√2 = −

√2

2 ⇒  = 3

4 [since   0], and cos  = 

 sin  = 1

√2 sin^3₄ = 1

√2

 1

√2

 = 1 ⇒  = 0. Thus, spherical coordinates are

√ 2 03

4

 .

(b)  = (√

3 )²+ (√

3 )²+ (√

2 )² =√

8 = 2√2, cos  = 

 =

√2 2√

2 = 1

2 ⇒  = 

3, and

cos  = 

 sin =

√3 2√

2 sin 3

=

√3

2√ 2 ·

√3 2

=

√2

2 ⇒  = 

4. Thus, spherical coordinates are 2√

2 4

3

.

4. (a)  =

²+ ²+ ²=√1 + 0 + 3 = 2, cos  = 

 =

√3

2 ⇒  = 

6, and cos  = 

 sin  = 1

2 sin(6) = 1 ⇒

 = 0. Thus spherical coordinates are 2 0

6

.

(b)  =√3 + 1 + 12 = 4, cos  = 

 = 2√ 3

4 =

√3

2 ⇒  = 

6, and cos  = 

 sin  =

√3 4 sin(6) =

√3

2 ⇒

 = 11

6 [since   0]. Thus spherical coordinates are

 411

6  6

 .

5. Since  = ^₃ but  and  can vary, the surface is the top half of a right circular cone with vertex at the origin and axis the positive -axis. (See Figure 4.)

6. ²− 3 + 2 = 0 ⇒ ( − 1)( − 2) = 0 ⇒  = 1 or  = 2. Thus the equation represents two surfaces. In the case

 = 1, the distance from any point to the origin is 1. Because  and  can vary, the surface is a sphere centered at the origin with radius 1. (See Figure 2.) Similarly,  = 2 is a sphere centered at the origin with radius 2.

Also,  = 1 ⇒ ²= 1 ⇒ ²+ ²+ ²= 1which we recognize as the equation of the unit sphere, and similarly,

 = 2 ⇒ ²= 4 ⇒ ²+ ²+ ²= 4.

7. From Equations 1 we have  =  cos , so  cos  = 1 ⇔  = 1, and the surface is the horizontal plane  = 1.

8.  = cos  ⇒ ² =  cos  ⇔ ²+ ²+ ²=  ⇔ ²+ ²+ ²−  +¹4 = ¹₄ ⇔ ²+ ²+ ( −¹2)²= ¹₄. Therefore, the surface is a sphere of radius ¹₂centered at

0 0¹₂ .

9. (a) From Equation 2 we have ²= ²+ ²+ ², so ²+ ²+ ²= 9 ⇔ ²= 9 ⇒  = 3 (since  ≥ 0).

(b) From Equations 1 we have  =  sin  cos ,  =  sin  sin , and  =  cos , so the equation ²− ²− ² = 1 becomes ( sin  cos )²− ( sin  sin )²− ( cos )²= 1 ⇔ (²sin²)(cos² − sin²) − ²cos² = 1 ⇔

²(sin² cos 2 − cos²) = 1.

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21. (a) Express the triple integralRRR

Ef (x, y, z) dV as an iterated integral in spherical coordinates for the given function f and solid region E.

(b) Evaluate the iterated integral.

1106 CHAPTER 15 Multiple Integrals

15.8 Exercises

1–2 Plot the point whose spherical coordinates are given. Then find the rectangular coordinates of the point.

1. (a) s2, 3y4, y2d (b) s4, 2y3, y4d 2. (a) s5, y2, y3d (b) s6, 0, 5y6d

3–4 Change from rectangular to spherical coordinates.

3. (a) s3, 3, 0d (b)

(

1, 2s3, 2s3

)

4. (a) s0, 4, 24d (b)

(

22, 2, 2s6

)

5–6 Describe in words the surface whose equation is given.

5.  − 3y4 6. ²23 1 2 − 0

7–8 Identify the surface whose equation is given.

7.  cos  − 1 8.  − cos 

9–10 Write the equation in spherical coordinates.

9. (a) x²1y²1z²− 9 (b) x²2y²2z²− 1 10. (a) z − x²1y² (b) z − x²2y²

11–14 Sketch the solid described by the given inequalities.

11.  < 1, 0 <  < y6, 0 <  <  12. 1 <  < 2, y2 <  < 

13. 1 <  < 3, 0 <  < y2,  <  < 3y2 14.  < 2,  < csc 

15. A solid lies inside the sphere x²1y²1z²− 4z and outside the cone z −sx²1y² . Write a description of the solid in terms of inequalities involving spherical coordinates.

16. (a) Find inequalities that describe a hollow ball with diam- eter 30 cm and thickness 0.5 cm. Explain how you have positioned the coordinate system that you have chosen.

(b) Suppose the ball is cut in half. Write inequalities that describe one of the halves.

17–18 Sketch the solid whose volume is given by the integral and evaluate the integral.

17.

y

0^y6

y

0^y2

y

0³ ²sin  d d d

18.

y

0^y4

y

0²

y

0^sec ² sin  d d d

19–20 Set up the triple integral of an arbitrary continuous function fsx, y, zd in cylindrical or spherical coordinates over the solid shown.

19. z 20.

x y

z

x 2 y

1 3

2

21–22

(a) Express the triple integral yyyE fsx, y, zd dV as an iterated integral in spherical coordinates for the given function f and solid region E.

(b) Evaluate the iterated integral.

21. fsx, y, zd −sx²1y²1z² 22. fsx, y, zd − xy

≈+¥+z@=4

≈+¥+z@=9 E x

z

y z=œ„„„„„„≈+¥

≈+¥+z@=8

x 0 z

y E

23–36 Use spherical coordinates.

23. Evaluate yyyB sx²1y²1z²d² dV, where B is the ball with center the origin and radius 5.

24. Evaluate yyyE y²z² dV, where E lies above the cone  − y3 and below the sphere  − 1.

25. Evaluate yyyE sx²1y²d dV, where E lies between the spheres x²1y²1z²− 4 and x²1y²1z²− 9.

26. Evaluate yyyE y² dV, where E is the solid hemisphere x²1y²1z²<9, y > 0.

27. Evaluate yyyE xe^x21y21z2 dV, where E is the portion of the unit ball x²1y²1z²<1 that lies in the first octant.

28. Evaluate yyyEsx²1y²1z² dV, where E lies above the cone z −sx²1y² and between the spheres x²1y²1z²− 1

and x²1y²1z²− 4.

29. Find the volume of the part of the ball  < a that lies between the cones  − y6 and  − y3.

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Solution:

1574 ¤ CHAPTER 15 MULTIPLE INTEGRALS

19. The solid  is most conveniently described if we use cylindrical coordinates:

 =

(  ) | 0 ≤  ≤ ^2 0 ≤  ≤ 3 0 ≤  ≤ 2 . Then



 (  )  =2 0

3 0

2

0  ( cos   sin  )    .

20. The solid  is most conveniently described if we use spherical coordinates:

 = 

(  ) | 1 ≤  ≤ 2 ^2 ≤  ≤ 2 0 ≤  ≤ ^2

. Then



 (  )  =2 0

2

2

2

1  ( sin  cos   sin  sin   cos ) ²sin    .

21. (a) The solid can be described in spherical coordinates by  = {(  ) | 2 ≤  ≤ 3^₂ ≤  ≤ ^32 ^₂ ≤  ≤ }.

Thus,



²+ ²+ ² =

2

32

2

3

2  · ²sin    .

(b) 

2

32

2

3

2 ³sin     =

2sin  32

2 3 2 ³

=

− cos =

=2

=32

=2

⁴ 4

=3

=2

= (1)() · 1

4(81 − 16) = 65

4 22. (a) The solid can be described in spherical coordinates by  = {(  ) | 0 ≤  ≤ 2√

2, 0 ≤  ≤ 2, 0 ≤  ≤ ^4}.

Thus,

  =4 0

2

0

2√ 2

0  sin  cos  ·  sin  sin  · ²sin    .

(b) 4 0

2

0

2√ 2

0 ⁴sin³ cos  sin     =4

0 sin³ 2

0 cos  sin  2√ 2

0 ⁴. Since

2

0 cos  sin   = ¹₂2

0 sin 2  = ¹₂

−¹2cos 22

0 = −¹4(1 − 1) = 0, the original iterated integral equals 0.

23. In spherical coordinates,  is represented by {(  ) | 0 ≤  ≤ 5 0 ≤  ≤ 2 0 ≤  ≤  }. Thus,



(²+ ²+ ²)² = 0

2

0

5

0(²)²²sin     =

0 sin  2

0  5 0 ⁶

=

− cos  0

2

0

1 7⁷5

0= (2)(2)_78,125

7



= ^312,500₇  ≈ 140,2497 24. In spherical coordinates,  is represented by

(  )

 0 ≤  ≤ 1 0 ≤  ≤ 2 0 ≤  ≤ ^₃. Thus,



 ²² =3 0

2

0

1

0 ( sin  sin )²( cos )²²sin    

=3

0 sin³ cos² 2

0 sin² 1 0 ⁶

=3

0 (1 − cos²) cos² sin   2

0 1

2(1 − cos 2) 1 0 ⁶

=₁

5cos⁵ −¹3cos³3 0

₁

2 −¹4sin 22

0

₁

7⁷1 0

=

1 5

₁

2

5

−¹3

₁

2

3

−¹5+ ¹₃

( − 0)₁

7− 0

= ₄₈₀⁴⁷ ·  · ¹7 = ₃₃₆₀⁴⁷ 

25. In spherical coordinates,  is represented by {(  ) | 2 ≤  ≤ 3 0 ≤  ≤ 2 0 ≤  ≤  } and

²+ ²= ²sin² cos² + ²sin² sin² = ²sin²

cos² + sin²

= ²sin². Thus,

[continued]

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29. Use spherical coordinates. Find the volume of the part of the ball ρ ≤ a that lies between the cones φ = ^π₆ and φ =^π₃.

Solution:

SECTION 15.8 TRIPLE INTEGRALS IN SPHERICAL COORDINATES ¤ 599 25. In spherical coordinates,  is represented by

(  )

 0 ≤  ≤ 1 0 ≤  ≤ ^2 0 ≤  ≤^2

. Thus



^{2+2+2} =2 0

2 0

1

0( sin  cos )^2²sin     =2

0 sin² 2

0 cos  1

0 ³^2

=2 0

1

2(1 − cos 2) 2

0 cos  



1

2²^21 0−1

0 ^2



integrate by parts with  = ²,  = ^2

=1

2 − ¹4sin 22

0 [sin ]^2₀ ₁

2²^2−¹2^21 0=

4 − 0

(1 − 0) 0 +¹₂

= ^₈

26. In spherical coordinates, the cone  =

²+ ²is equivalent to  = 4 (as in Example 4) and  is represented by {(  ) | 1 ≤  ≤ 2 0 ≤  ≤ 2 0 ≤  ≤ 4 }. Also

²+ ²+ ²=

²= , so





²+ ²+ ² =4 0

2

0

2

1  · ²sin     =4

0 sin  2

0 2 1 ³

= [− cos ]^40

^2

0

1 4⁴²

1=

−^√2²+ 1

(2) · ¹4(16 − 1) = ¹⁵2 1 −^√2²



27. The solid region is given by  =

(  ) | 0 ≤  ≤  0 ≤  ≤ 2^6 ≤  ≤ ^3

and its volume is

 =

 =3

6

2

0



0 ²sin     =3

6 sin  2

0  0 ²

= [− cos ]^36 []^2₀ 1 3³

0 =

−¹2+^√₂³ (2)1

3³

= ^√3₃⁻¹³ 28. If we center the ball at the origin, then the ball is given by

 = {(  ) | 0 ≤  ≤  0 ≤  ≤ 2 0 ≤  ≤ } and the distance from any point (  ) in the ball to the center (0 0 0) is

²+ ²+ ²= . Thus the average distance is 1

 ()





  = 1

4 3³

  0

 2

0

  0

 · ²sin     = 3 4³

  0

sin  

 2

0



  0

³

= 3

4³

−cos  0

2

0

1 4⁴

0 = 3

4³(2)(2)1 4⁴

= ³₄

29. (a) Since  = 4 cos  implies ²= 4 cos  ⇔ ²+ ²+ ²= 4 ⇔ ²+ ²+ ( − 2)²= 4, the equation is that of a sphere of radius 2 with center at (0 0 2). Thus

 =2

0

3 0

4 cos 

0 ²sin     =2

0

3 0

1

3³=4 cos 

=0 sin    =2

0

3 0

64 3 cos³

sin   

=2

0

−¹⁶3 cos⁴=3

=0  =2

0 −¹⁶3

₁

16 − 1

 = 52

0 = 10

(b) By the symmetry of the problem  = = 0. Then

=2

0

3 0

4 cos 

0 ³cos  sin     =2

0

3

0 cos  sin 

64 cos⁴

 

=2

0 64

−¹6cos⁶=3

=0  =2

0 21

2  = 21

Hence (  ) = (0 0 21(10)) = (0 0 21).

30. In spherical coordinates, the sphere ²+ ²+ ²= 4is equivalent to  = 2 and the cone  =

²+ ²is represented by  = ^₄ (as in Example 4). Thus, the solid is given by

(  )

 0 ≤  ≤ 2 0 ≤  ≤ 2 ^4 ≤  ≤ ^2

and

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30. Use spherical coordinates. Find the average distance from a point in a ball of radius a to its center.

Solution:

1

SECTION 15.8 TRIPLE INTEGRALS IN SPHERICAL COORDINATES ¤ 599 25. In spherical coordinates,  is represented by

(  )

 0 ≤  ≤ 1 0 ≤  ≤^2 0 ≤  ≤ ^2

. Thus



^{2+2+2} =2 0

2 0

1

0( sin  cos )^2²sin     =2

0 sin² 2

0 cos   1

0 ³^2

=2 0

1

2(1 − cos 2) 2

0 cos  



1 2²^21

0−1 0 ^2



integrate by parts with  = ²,  = ^2

=1

2 − ¹4sin 22

0 [sin ]^2₀ 

1

2²^2−¹2^21 0=

4 − 0

(1 − 0) 0 + ¹₂

= ^₈

26. In spherical coordinates, the cone  =

²+ ²is equivalent to  = 4 (as in Example 4) and  is represented by {(  ) | 1 ≤  ≤ 2 0 ≤  ≤ 2 0 ≤  ≤ 4 }. Also

²+ ²+ ²=

²= , so





²+ ²+ ² =4 0

2

0

2

1  · ²sin     =4

0 sin  2

0  2 1 ³

= [− cos ]^40

2

0

₁

4⁴2 1=

−^√2² + 1

(2) ·¹4(16 − 1) = ¹⁵2 1 − ^√2²



27. The solid region is given by  =

(  ) | 0 ≤  ≤  0 ≤  ≤ 2^6 ≤  ≤ ^3

and its volume is

 =

 =^3

6

^2

0

^

0 ²sin     =^3

6 sin  ^2

0 ^

0 ²

= [− cos ]^36 []^2₀ 1 3³^

0=

−¹2 +^√₂³ (2)1

3³

= ^√3₃⁻¹³ 28. If we center the ball at the origin, then the ball is given by

 = {(  ) | 0 ≤  ≤  0 ≤  ≤ 2 0 ≤  ≤ } and the distance from any point (  ) in the ball to the center (0 0 0) is

²+ ²+ ²= . Thus the average distance is 1

 ()





  = 1

4 3³

  0

 2

0

 

0  · ²sin     = 3 4³

  0

sin  

 2

0



  0

³

= 3

4³

−cos  0

2

0

₁

4⁴

0= 3

4³(2)(2)₁

4⁴

= ³₄

29. (a) Since  = 4 cos  implies ²= 4 cos  ⇔ ²+ ²+ ²= 4 ⇔ ²+ ²+ ( − 2)²= 4, the equation is that of a sphere of radius 2 with center at (0 0 2). Thus

 =2

0

3 0

4 cos 

0 ²sin     =2

0

3 0

₁

3³=4 cos 

=0 sin    =2

0

3 0

₆₄

3 cos³

sin   

=2

0

−¹⁶3 cos⁴=3

=0  =2

0 −¹⁶3

1 16− 1

 = 52

0 = 10

(b) By the symmetry of the problem = = 0. Then

=2

0

3 0

4 cos 

0 ³cos  sin     =2

0

3

0 cos  sin 

64 cos⁴

 

=2

0 64

−¹6cos⁶=3

=0  =2

0 21

2  = 21

Hence (  ) = (0 0 21(10)) = (0 0 21).

30. In spherical coordinates, the sphere ²+ ²+ ²= 4is equivalent to  = 2 and the cone  =

²+ ²is represented by  = ^₄ (as in Example 4). Thus, the solid is given by

(  )

 0 ≤  ≤ 2 0 ≤  ≤ 2 ^₄ ≤  ≤ ^2

and

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37. Use cylindrical or spherical coordinates, whichever seems more appropriate. Find the volume and centroid of the solid E that lies above the cone z =p

x²+ y² and below the sphere x²+ y²+ z²= 1.

Solution:

SECTION 15.8 TRIPLE INTEGRALS IN SPHERICAL COORDINATES ¤ 601

34. Place the center of the base at (0 0 0), then the density is (  ) = ,  a constant. Then

 =2

0

2 0



0( cos ) ²sin     = 22

0 cos  sin  · ¹4⁴ = ¹₂⁴

−¹4cos 22

0 = ^₄⁴. By the symmetry of the problem  =  = 0, and

 =2

0

2 0



0 ⁴cos² sin     = ²₅⁵2

0 cos² sin   = ²₅⁵

−¹3cos³2

0 = ₁₅²⁵. Hence (  ) =

0 0₁₅⁸. 35. In spherical coordinates  =

²+ ²becomes  = ^₄ (as in Example 4). Then

 =2

0

4 0

1

0 ²sin     =2

0 4

0 sin  1

0 ² = 2

−^√2²+ 1₁

3

= ¹₃ 2 −√

2,

 =2

0

4 0

1

0 ³sin  cos     = 2

−¹4cos 24 0

₁

4

= ^₈ and by symmetry =  = 0.

Hence (  ) =



0 0 3 8

2 −√ 2

 .

36. Place the center of the sphere at (0 0 0), let the diameter of intersection be along the -axis, one of the planes be the -plane and the other be the plane whose angle with the -plane is  = ^₆. Then in spherical coordinates the volume is given by

 =6 0

 0



0 ²sin     =6 0  

0 sin  

0 ² = ^₆(2)1 3³

= ¹₉³.

37. (a) If we orient the cylinder so that its axis is the -axis and its base lies in the -plane, then the cylinder is described, in cylindrical coordinates, by  = {(  ) | 0 ≤  ≤  0 ≤  ≤ 2 0 ≤  ≤ }. Assuming constant density , the moment of inertia about its axis (the -axis) is

=

(²+ ²) (  )  =2

0

 0



0 (²)     = 2

0 

0 ³ 0 

= 

2

0

₁

4⁴ 0



0 =  (2)₁

4⁴

() = ¹₂⁴

(b) By symmetry, the moments of inertia about any two diameters of the base will be equal, and one of the diameters lies on the -axis, so we compute:

=

(²+ ²) (  )  =2

0

 0



0 (²sin² + ²)    

= 2

0

 0



0 ³sin²    + 2

0

 0



0 ²  

= 2

0 sin²  

0 ³ 

0  + 2

0 

0   0 ²

= ₁

2 −¹4sin 22

0

₁

4⁴ 0

 0+ 

2

0

₁

2² 0

₁

3³ 0

=  ()1 4⁴

() +  (2)1 2² 1

3³

= ₁₂¹²(3²+ 4²)

38. Orient the cone so that its axis is the -axis and its base lies in the -plane, as shown in the figure. (Then the -axis is the axis of the cone and the -axis contains a diameter of the base.) A right circular cone with axis the -axis and vertex at the origin has equation ²= ²(²+ ²). Here we have the bottom frustum, shifted upward  units, and with ²= ²²so that the cone includes the point ( 0 0). Thus an equation of the cone in rectangular coordinates is  =  − ^

²+ ², 0 ≤  ≤ . In cylindrical

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