Section 15.8 Triple Integrals in Spherical Coordinates
596 ¤ CHAPTER 15 MULTIPLE INTEGRALS
10. (a) = sin cos , = sin sin , and = cos , so the equation = 2+ 2becomes
cos = ( sin cos )2+ ( sin sin )2 or cos = 2sin2. If 6= 0, this becomes cos = sin2 or = cos csc2 or = cot csc . ( = 0 corresponds to the origin which is included in the surface.) (b) The equation = 2− 2becomes cos = ( sin cos )2− ( sin sin )2
or cos = 2(sin2)(cos2 − sin2) ⇔ cos = 2sin2 cos 2. If 6= 0, this becomes cos = sin2 cos 2. ( = 0 corresponds to the origin which is included in the surface.)
11. ≤ 1 represents the (solid) unit ball. 0 ≤ ≤6 restricts the solid to that portion on or above the cone = 6, and 0 ≤ ≤ further restricts the solid to that portion on or to the right of the -plane.
12. 1 ≤ ≤ 2 represents the solid region between and including the spheres of radii 1 and 2, centered at the origin. 2 ≤ ≤ restricts the solid to that portion on or below the -plane.
13. 2 ≤ ≤ 4 represents the solid region between and including the spheres of radii 2 and 4, centered at the origin. 0 ≤ ≤ 3 restricts the solid to that portion on or above the cone = 3, and 0 ≤ ≤ further restricts the solid to that portion on or to the right of the -plane.
14. ≤ 2 represents the solid sphere of radius 2 centered at the origin. Notice that 2+ 2= ( sin cos )2+ ( sin sin )2= 2sin2. Then
= csc ⇒ sin = 1 ⇒ 2sin2 = 2+ 2 = 1, so ≤ csc restricts the solid to that portion on or inside the circular cylinder
2+ 2= 1.
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SECTION 15.8 TRIPLE INTEGRALS IN SPHERICAL COORDINATES ¤ 597 15. ≥
2+ 2 because the solid lies above the cone. Squaring both sides of this inequality gives 2≥ 2+ 2 ⇒ 22≥ 2+ 2+ 2= 2 ⇒ 2= 2cos2 ≥122 ⇒ cos2 ≥12. The cone opens upward so that the inequality is cos ≥√12, or equivalently 0 ≤ ≤ 4. In spherical coordinates the sphere = 2+ 2+ 2is cos = 2 ⇒
= cos . 0 ≤ ≤ cos because the solid lies below the sphere. The solid can therefore be described as the region in spherical coordinates satisfying 0 ≤ ≤ cos , 0 ≤ ≤4.
16. (a) The hollow ball is a spherical shell with outer radius 15 cm and inner radius 14.5 cm. If we center the ball at the origin of the coordinate system and use centimeters as the unit of measurement, then spherical coordinates conveniently describe the hollow ball as 145 ≤ ≤ 15, 0 ≤ ≤ 2, 0 ≤ ≤ .
(b) If we position the ball as in part (a), one possibility is to take the half of the ball that is above the -plane which is described by 145 ≤ ≤ 15, 0 ≤ ≤ 2, 0 ≤ ≤ 2.
17. The region of integration is given in spherical coordinates by
= {( ) | 0 ≤ ≤ 3 0 ≤ ≤ 2 0 ≤ ≤ 6}. This represents the solid region in the first octant bounded above by the sphere = 3 and below by the cone
= 6.
6 0
2 0
3
0 2sin =6
0 sin 2 0 3
0 2
=
− cos 6 0
2 0
1
333 0
=
1 −
√3 2
2
(9) = 9
4
2 −√ 3
18. The region of integration is given in spherical coordinates by
= {( ) | 0 ≤ ≤ sec 0 ≤ ≤ 2 0 ≤ ≤ 4}.
= sec ⇔ cos = 1 ⇔ = 1, so is the solid region above the cone = 4 and below the plane = 1.
4 0
2
0
sec
0 2sin =4 0
2
0
1
33sin =sec
=0
=4 0
2
0 1
3sec3 sin =134
0 sec3 sin 2
0
=134
0 tan sec2 2
0 = 131
2tan24 0
2
0
=131
2 − 0
(2) =3
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 ) .
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598 ¤ CHAPTER 15 MULTIPLE INTEGRALS
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. In spherical coordinates, is represented by {( ) | 0 ≤ ≤ 5 0 ≤ ≤ 2 0 ≤ ≤ }. Thus
(2+ 2+ 2)2 = 0
2
0
5
0(2)22sin =
0 sin 2
0 5 0 6
=
− cos 0
2
0
1 775
0= (2)(2)78,125 7
= 312,5007 ≈ 140,2497
22. In spherical coordinates, is represented by
( )
0 ≤ ≤ 1 0 ≤ ≤ 2 0 ≤ ≤3
. Thus
22 =3 0
2
0
1
0 ( sin sin )2( cos )22sin
=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 −13cos33 0
1
2 −14sin 22
0
1 771
0
=
1 5
1
2
5
−13
1
2
3
−15+13
( − 0)1
7 − 0
=48047 · ·17 = 336047
23. In spherical coordinates, is represented by {( ) | 2 ≤ ≤ 3 0 ≤ ≤ 2 0 ≤ ≤ } and
2+ 2= 2sin2 cos2 + 2sin2 sin2 = 2sin2
cos2 + sin2
= 2sin2. Thus
(2+ 2) = 0
2
0
3
2(2sin2) 2sin =
0 sin3 2
0 3 2 4
=
0(1 − cos2) sin
2
0
1 553
2=
− cos +13cos3
0 (2) ·15(243 − 32)
=
1 −13 + 1 −13
(2)211 5
=168815
24. In spherical coordinates, is represented by
( )
0 ≤ ≤ 3 0 ≤ ≤ 2 0 ≤ ≤ 2
. Thus
(9 − 2− 2) =2 0
2
0
3 0
9 − (2sin2 cos2 + 2sin2 sin2)
2sin
=2 0
2
0
3
0(9 − 2sin2) 2sin
=2 0
2
0
33−155sin2=3
=0sin
=2 0
2
0
81 sin −2435 sin3
=2
0 2 0
81 sin −2435 (1 − cos2) sin
= 2
−81 cos −2435
1
3cos3 − cos 2 0
= 2
0 + 81 +2435
−23
= 4865
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SECTION 15.8 TRIPLE INTEGRALS IN SPHERICAL COORDINATES ¤ 599 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
442 1=
−√22+ 1
(2) ·14(16 − 1) =152 1 −√22
27. The solid region is given by =
( ) | 0 ≤ ≤ 0 ≤ ≤ 26 ≤ ≤ 3
and its volume is
=
=3
6
2
0
0 2sin =3
6 sin 2
0 0 2
= [− cos ]36[]20 1
33 0 =
−12+√23 (2)1
33
= √33−13
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 33
0
2
0
0 · 2sin = 3 43
0
sin
2
0
0
3
= 3
43
−cos 0
2
0
1 44
0 = 3
43(2)(2)1 44
=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
33=4 cos
=0 sin =2
0
3 0
64
3 cos3
sin
=2
0
−163 cos4=3
=0 =2
0 −163
1 16− 1
= 52
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 21).
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
=2
4
2
0
2
0 2sin =2
4 sin 2
0 2 0 2
=
− cos 2
4
2
0
1 332
0=√ 2 2
(2)8
3
= 8√32
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SECTION 15.8 TRIPLE INTEGRALS IN SPHERICAL COORDINATES ¤ 599 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
1 442
1=
−√22+ 1
(2) ·14(16 − 1) =152 1 −√22
27. The solid region is given by =
( ) | 0 ≤ ≤ 0 ≤ ≤ 26 ≤ ≤ 3
and its volume is
=
=3
6
2
0
0 2sin =3
6 sin 2
0 0 2
= [− cos ]36[]20 1 33
0 =
−12+√23 (2)1
33
= √33−13
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 33
0
2
0
0
· 2sin = 3 43
0
sin
2
0
0
3
= 3
43
−cos 0
2
0
1
44
0 = 3
43(2)(2)1
44
=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
33=4 cos
=0 sin =2
0
3 0
64 3 cos3
sin
=2
0
−163 cos4=3
=0 =2
0 −163
1
16− 1
= 52
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 21).
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
=2
4
2
0
2
0 2sin =2
4 sin 2
0 2 0 2
=
− cos 2
4
2
0
1
332 0=√
2 2
(2)8
3
= 8√32
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1
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 = 22
0 cos sin ·144 = 124
−14cos 22
0 =44. By the symmetry of the problem = = 0, and
=2
0
2 0
0 4cos2 sin = 2552
0 cos2 sin = 255
−13cos32
0 = 1525. Hence ( ) =
0 0158. 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+ 11 3
=13 2 −√
2,
=2
0
4 0
1
0 3sin cos = 2
−14cos 24 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
33
= 193.
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 44
0
0 = (2)1 44
() = 124
(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 22
0
1
44 0
0+
2
0
1
22 0
1
33 0
= ()1
44
() + (2)1
22 1
33
=1212(32+ 42)
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= 22so 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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SECTION 15.8 TRIPLE INTEGRALS IN SPHERICAL COORDINATES ¤ 603 and 0 ≤ ≤√2. Then the integral becomes
4 0
2 0
√2
0 ( sin cos ) ( sin sin ) 2sin
=4
0 sin3 2
0 sin cos √2
0 4 =4 0
1 − cos2
sin 1
2sin22 0
1
55√2 0
=1
3cos3 − cos 4 0 ·12·15
√25
=√ 2
12 −√22 −1 3− 1
·2√52 =4√152−5
42. The region of integration is the solid sphere 2+ 2+ 2≤ 2, so 0 ≤ ≤ 2, 0 ≤ ≤ , and 0 ≤ ≤ . Also
2 + 2 + 3= (2+ 2+ 2) = 2 = 3cos , so the integral becomes
0
2
0
0
3cos
2sin =
0 sin cos 2
0
0 5 =1
2sin2 0
2
0
1
66 0 = 0
43. The region of integration is the solid sphere 2+ 2+ ( − 2)2 ≤ 4 or equivalently
2sin2 + ( cos − 2)2= 2− 4 cos + 4 ≤ 4 ⇒ ≤ 4 cos , so 0 ≤ ≤ 2, 0 ≤ ≤2, and 0 ≤ ≤ 4 cos . Also (2+ 2+ 2)32= (2)32= 3, so the integral becomes
2 0
2
0
4 cos 0
3
2sin =2 0
2
0 sin 1
66=4 cos
=0
=162 0
2
0 sin
4096 cos6
=16(4096)2
0 cos6 sin 2
0 =20483
−17cos72 0
2
0
=20483 1
7
(2) =409621
44. The solid region between the ground and an altitude of 5 km (5000 m) is given by
=
( ) | 6370 × 106≤ ≤ 6375 × 106 0 ≤ ≤ 2 0 ≤ ≤ . Then the mass of the atmosphere in this region is
=
=2
0
0
6375×106
6370×106 (61909 − 0000097) 2sin
=2
0
0 sin 6375×106 6370×106
619092− 00000973
=
2
0 [− cos ]0
61909
3 3−00000974 46375×106 6370×106
= (2)(2)61909
3
(6375 × 106)3− (6370 × 106)3
−00000974
(6375 × 106)4− (6370 × 106)4
≈ 4
1944 × 1017
≈ 244 × 1018kg
45. In cylindrical coordinates, the equation of the cylinder is = 3, 0 ≤ ≤ 10.
The hemisphere is the upper part of the sphere radius 3, center (0 0 10), equation
2+ ( − 10)2= 32, ≥ 10. In Maple, we can use the coords=cylindrical option in a regular plot3d command. In Mathematica, we can use ParametricPlot3D.
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SECTION 15.8 TRIPLE INTEGRALS IN SPHERICAL COORDINATES ¤ 603
and 0 ≤ ≤√2. Then the integral becomes
4 0
2 0
√2
0 ( sin cos ) ( sin sin ) 2sin
=4
0 sin3 2
0 sin cos √2
0 4 =4 0
1 − cos2
sin 1
2sin22 0
1
55√2 0
=1
3cos3 − cos 4 0 ·12·15
√25
=√ 2
12 −√22 −1 3− 1
·2√52 =4√152−5
42. The region of integration is the solid sphere 2+ 2+ 2≤ 2, so 0 ≤ ≤ 2, 0 ≤ ≤ , and 0 ≤ ≤ . Also
2 + 2 + 3= (2+ 2+ 2) = 2 = 3cos , so the integral becomes
0
2
0
0
3cos
2sin =
0 sin cos 2
0
0 5 =1
2sin2 0
2
0
1
66 0 = 0
43. The region of integration is the solid sphere 2+ 2+ ( − 2)2 ≤ 4 or equivalently
2sin2 + ( cos − 2)2= 2− 4 cos + 4 ≤ 4 ⇒ ≤ 4 cos , so 0 ≤ ≤ 2, 0 ≤ ≤2, and 0 ≤ ≤ 4 cos . Also (2+ 2+ 2)32= (2)32= 3, so the integral becomes
2 0
2
0
4 cos 0
3
2sin =2 0
2
0 sin 1
66=4 cos
=0
=162 0
2
0 sin
4096 cos6
=16(4096)2
0 cos6 sin 2
0 =20483
−17cos72 0
2
0
=20483 1
7
(2) =409621
44. The solid region between the ground and an altitude of 5 km (5000 m) is given by
=
( ) | 6370 × 106≤ ≤ 6375 × 106 0 ≤ ≤ 2 0 ≤ ≤ . Then the mass of the atmosphere in this region is
=
=2
0
0
6375×106
6370×106 (61909 − 0000097) 2sin
=2
0
0 sin 6375×106 6370×106
619092− 00000973
=
2
0 [− cos ]0
61909
3 3−00000974 46375×106 6370×106
= (2)(2)61909 3
(6375 × 106)3− (6370 × 106)3
−00000974
(6375 × 106)4− (6370 × 106)4
≈ 4
1944 × 1017
≈ 244 × 1018kg
45. In cylindrical coordinates, the equation of the cylinder is = 3, 0 ≤ ≤ 10.
The hemisphere is the upper part of the sphere radius 3, center (0 0 10), equation
2+ ( − 10)2= 32, ≥ 10. In Maple, we can use the coords=cylindrical option in a regular plot3d command. In Mathematica, we can use ParametricPlot3D.
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604 ¤ CHAPTER 15 MULTIPLE INTEGRALS
46. We begin by finding the positions of Los Angeles and Montréal in spherical coordinates, using the method described in the exercise:
Montréal Los Angeles
= 3960mi = 3960mi
= 360◦− 7360◦= 28640◦ = 360◦− 11825◦= 24175◦
= 90◦− 4550◦= 4450◦ = 90◦− 3406◦= 5594◦
Now we change the above to Cartesian coordinates using = cos sin , = sin sin and = cos to get two position vectors of length 3960 mi (since both cities must lie on the surface of the earth). In particular:
Montréal: h78367 −266267 282447i Los Angeles: h−155280 −288991 221784i
To find the angle between these two vectors we use the dot product:
h78367 −266267 282447i · h−155280 −288991 221784i = (3960)2cos ⇒ cos ≈ 08126 ⇒
≈ 06223 rad. The great circle distance between the cities is = ≈ 3960(06223) ≈ 2464 mi.
47. If is the solid enclosed by the surface = 1 +15sin 6 sin 5, it can be described in spherical coordinates as
=
( ) | 0 ≤ ≤ 1 + 15sin 6 sin 5 0 ≤ ≤ 2 0 ≤ ≤ . Its volume is given by
() =
= 0
2
0
1 + (sin 6 sin 5)5
0 2sin = 13699 [using a CAS].
48. The given integral is equal to lim
→∞
2
0
0
0 −22sin = lim
→∞
2
0
0 sin
0 3−2. Now use integration by parts with = 2, = −2to get
lim→∞2(2)
2
−12
−2 0 −
0 2
−12
−2
= lim
→∞4
−122−2+
−12−2 0
= 4 lim
→∞
−122−2−12−2+12
= 41
2
= 2
(Note that 2−2→ 0 as → ∞ by l’Hospital’s Rule.) 49. (a) From the diagram, = cot 0to =√
2− 2, = 0 to = sin 0(or use 2− 2= 2cot20). Thus
=2
0
sin 0 0
√2−2
cot 0
= 2 sin 0 0
√
2− 2− 2cot 0
=23
−(2− 2)32− 3cot 0
sin 0 0
=23
−
2− 2sin2032
− 3sin30cot 0+ 3
=233 1 −
cos30+ sin20cos 0
=233(1 − cos 0)
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