Section 15.6 Triple Integrals
568 ¤ CHAPTER 15 MULTIPLE INTEGRALS
2. There are six different possible orders of integration.
( + 2) =2 0
1 0
3
0 ( + 2) =2 0
1 0
+133=3
=0 =2 0
1
0 (3 + 9)
=2 0
3
22+ 9=1
=0 =2 0
3 2 + 9
=3
42+ 92 0= 21
( + 2) =1 0
2 0
3
0 ( + 2) =1 0
2 0
+133=3
=0 =1 0
2
0 (3 + 9)
=1 0
3
22 + 9=2
=0 =1
0 (6 + 18) =
32+ 181 0= 21
( + 2) =2 0
3 0
1
0 ( + 2) =2 0
3 0
1
22+ 2=1
=0 =2 0
3 0
1 2 + 2
=2 0
1
2 +133=3
=0 =2 0
3 2 + 9
=3
42+ 92 0 = 21
( + 2) =3 0
2 0
1
0 ( + 2) =3 0
2 0
1
22+ 2=1
=0 =3 0
2 0
1
2 + 2
=3 0
1
42+ 2=2
=0 =3 0
1 + 22
=
+2333 0= 21
( + 2) =1 0
3 0
2
0 ( + 2) =1 0
3 0
1
22 + 2=2
=0 =1 0
3 0
2 + 22
=1 0
2 +233=3
=0 =1
0 (6 + 18) =
32+ 181 0= 21
( + 2) =3 0
1 0
2
0 ( + 2) =3 0
1 0
1
22 + 2=2
=0 =3 0
1 0
2 + 22
=3 0
2+ 22=1
=0 =3 0
1 + 22
=
+2333 0 = 21 3. 2
0
2 0
−
0 (2 − ) =2 0
2 0
2− =−
=0 =2 0
2 0
( − )2− ( − )
=2 0
2 0
2−
=2 0
2−122=2
=0 =2 0
4−125
=1
55−12162
0=325 −6412 =1615 4. 1
0
2
+
0 6 =1 0
2
6=+
=0 =1 0
2
6( + ) =1 0
2
(62 + 62)
=1 0
23 + 322=2
= =1
0 234 = 23551 0= 235 5. 2
1
2
0
ln
0 − =2 1
2
0
−−=ln
=0 =2 1
2
0
−− ln + 0
=2 1
2
0 (−1 + ) =2 1
− +122=2
=0
=2 1
−2 + 22
=
−2+2332
1= −4 +163 + 1 −23 = 53
6.
1 0
1 0
√1−2 0
+ 1 =
1 0
1 0
+ 1·
=√
1−2
=0
=
1 0
1 0
√ 1 − 2
+ 1
=
1 0
−13(1 − 2)32
+ 1
=1
=0
=1 3
1 0
1
+ 1 = 1
3ln( + 1)
1 0
= 13(ln 2 − ln 1) =13ln 2
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SECTION 15.6 TRIPLE INTEGRALS ¤ 569 7.
0
1 0
√1−2
0 sin = 0
1 0
sin =√
1−2
=0 = 0
1 0 √
1 − 2 sin
= 0 sin
−13(1 − 2)32=1
=0 = 0
1
3sin = −13cos
0 = −13(−1 − 1) =23
8. 1 0
1 0
2−2−2
0 =1 0
1 0
=2−2−2
=0 =1 0
1
0(2−2−2− )
=1 0
−122−2−2−122=1
=0 =1 0
−121−2−12 +122−2
=
1
41−2−142−142−21
0=14 −14 −14 −14 + 0 +142=142−12
9.
=3 0
0
+
− =3 0
0
=+
=− =3 0
0 22
=3 0
2
33=
=0 =3 0
2
33 = 1643
0= 816 = 272
10.
=1 0
1
0 =1 0
1
=
=0
=1 0
1
(− ) =1 0
− =1
= =1 0
− − + 2
=1
22−122− ( − 1)+1331
0 [integrate by parts]
=12 −12 +13 − 1 = 12 −76
11.
2+ 2 =
4 1
4
0
2+ 2 =
4 1
4
·1
tan−1
=
=0
=4 1
4
tan−1(1) − tan−1(0)
=4 1
4
4 − 0
=44 1
=4
=
= 44
1(4 − ) = 4
4 −1224 1= 4
16 − 8 − 4 +12
=98
12. Here = {( ) | 0 ≤ ≤ 0 ≤ ≤ − 0 ≤ ≤ }, so
sin = 0
−
0
0 sin = 0
−
0
sin =
=0 = 0
−
0 sin
= 0
− cos =−
=0 =
0 [− cos( − ) + ]
=
sin( − ) − cos( − ) +122
0 [integrate by parts]
= 0 − 1 +122− 0 − 1 − 0 =122− 2
13. Here = {( ) | 0 ≤ ≤ 1 0 ≤ ≤√
0 ≤ ≤ 1 + + }, so
6 =1 0
√ 0
1++
0 6 =1 0
√ 0
6=1++
=0
=1 0
√
0 6(1 + + ) =1 0
32+ 322+ 23=√
=0
=1
0 (32+ 33+ 252) =
3+344+47721 0= 6528
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SECTION 15.6 TRIPLE INTEGRALS ¤ 569 7.
0
1 0
√1−2
0 sin = 0
1 0
sin =√
1−2
=0 = 0
1 0 √
1 − 2 sin
= 0 sin
−13(1 − 2)32=1
=0 = 0
1
3sin = −13cos
0 = −13(−1 − 1) =23
8. 1 0
1 0
2−2−2
0 =1 0
1 0
=2−2−2
=0 =1 0
1
0(2−2−2− )
=1 0
−122−2−2−122=1
=0 =1 0
−121−2−12 +122−2
=
1
41−2−142−142−21
0=14 −14 −14 −14 + 0 +142=142−12
9.
=3 0
0
+
− =3 0
0
=+
=− =3 0
0 22
=3 0
2 33=
=0 =3 0
2
33 = 1643
0= 816 = 272
10.
=1 0
1
0 =1 0
1
=
=0
=1 0
1
(− ) =1 0
− =1
= =1 0
− − + 2
=1
22−122− ( − 1)+1331
0 [integrate by parts]
=12 −12 +13 − 1 = 12 −76
11.
2+ 2 =
4 1
4
0
2+ 2 =
4 1
4
·1
tan−1
=
=0
=4 1
4
tan−1(1) − tan−1(0)
=4 1
4
4 − 0
=44 1
=4
=
= 44
1(4 − ) = 4
4 −1224 1= 4
16 − 8 − 4 +12
=98
12. Here = {( ) | 0 ≤ ≤ 0 ≤ ≤ − 0 ≤ ≤ }, so
sin = 0
−
0
0 sin = 0
−
0
sin =
=0 = 0
−
0 sin
= 0
− cos =−
=0 =
0 [− cos( − ) + ]
=
sin( − ) − cos( − ) +122
0 [integrate by parts]
= 0 − 1 +122− 0 − 1 − 0 =122− 2
13. Here = {( ) | 0 ≤ ≤ 1 0 ≤ ≤√
0 ≤ ≤ 1 + + }, so
6 =1 0
√ 0
1++
0 6 =1 0
√ 0
6=1++
=0
=1 0
√
0 6(1 + + ) =1 0
32+ 322+ 23=√
=0
=1
0 (32+ 33+ 252) =
3+344+47721 0= 6528
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570 ¤ CHAPTER 15 MULTIPLE INTEGRALS
14. Here =
( ) | −1 ≤ ≤ 1 0 ≤ ≤ 2 2− 1 ≤ ≤ 1 − 2. Thus,
( − ) =1
−1
2 0
1−2
2−1( − )
=1
−1
2
0 ( − )(1 − 2− (2− 1))
=1
−1
2
0 (2 − 23− 2 + 22)
=1
−1
2 − 23 − 2+ 22=2
=0
=1
−1(4 − 43− 4 + 42)
=
22− 4− 4 +4331
−1= −53 −113 = −163
15. Here = {( ) | 0 ≤ ≤ 2 0 ≤ ≤ 2 − 0 ≤ ≤ 2 − − }.
Thus,
2 =2 0
2−
0
2−−
0 2
=2 0
2−
0 2(2 − − )
=2 0
2−
0 [(2 − )2− 3]
=2 0
(2 − )1
33
−144=2−
=0
=2 0
1
3(2 − )4−14(2 − )4
=2 0
1
12(2 − )4
=1 12
−15
(2 − )52
0= −601(0 − 32) =158
16. The projection of onto the -plane is the triangle bounded by the lines
= , = 0, and = 1. Then
= {( ) | 0 ≤ ≤ 1 ≤ ≤ 1 0 ≤ ≤ − }, and
=1 0
1
−
0 =1 0
1
( − )
=1 0
1
(2− 2) =1 0
1
33−1222=1
=
=1 0
1
3 −122−134+124
=1
62−163+30151
0=16−16 +301 =301
17. The projection of onto the -plane is the disk 2+ 2≤ 1. Using polar coordinates = cos and = sin , we get
=
4
42+ 42
= 12
42− (42+ 42)2
= 82
0
1
0(1 − 4) = 82
0 1
0( − 5)
= 8(2)1
22−1661 0=163
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572 ¤ CHAPTER 15 MULTIPLE INTEGRALS 22. Here =
( ) | −1 ≤ ≤ 4 − 2+ 2≤ 4, so
=
2
−2
√4−2
−√
4−2
4−
−1
=
2
−2
√4−2
−√
4−2(4 − + 1)
=
2
−2
5 −122=√
4−2
=−√
4−2 =
2
−2
10
4 − 2
= 10
2
√4 − 2+ 2 sin−1
2
2
−2
using trigonometric substitution or Formula 30 in the Table of Integrals
= 10
2 sin−1(1) − 2 sin−1(−1)
= 20 2 −
−2
= 20
Alternatively, use polar coordinates to evaluate the double integral:
2
−2
√4−2
−√
4−2(5 − ) =
2
0
2
0 (5 − sin )
=2
0
5
22−133sin =2
=0
=2
0
10 −83sin
= 10 +83cos 2
0 = 20
23. (a) The wedge can be described as the region
=
( ) | 2+ 2≤ 1, 0 ≤ ≤ 1, 0 ≤ ≤
=
( ) | 0 ≤ ≤ 1, 0 ≤ ≤ , 0 ≤ ≤ 1 − 2 So the integral expressing the volume of the wedge is
=1 0
0
√1− 2
0 .
(b) A CAS gives1 0
0
√1− 2
0 = 4 −13. (Or use Formulas 30 and 87 from the Table of Integrals.)
24. (a) Divide into 8 cubes of size ∆ = 8. With ( ) =
2+ 2+ 2, the Midpoint Rule gives
2+ 2+ 2 ≈
2
= 1
2
= 1
2
= 1
∆
= 8[ (1 1 1) + (1 1 3) + (1 3 1) + (1 3 3) + (3 1 1) + (3 1 3) + (3 3 1) + (3 3 3)]
≈ 23964 (b) Using a CAS we have
2+ 2+ 2 =4 0
4 0
4 0
2+ 2+ 2 ≈ 24591. This differs from the
estimate in part (a) by about 2.5%.
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1
SECTION 15.6 TRIPLE INTEGRALS ¤ 573
25. Here ( ) = cos() and ∆ =12 ·12 ·12 =18, so the Midpoint Rule gives
( ) ≈
=1
=1
=1
∆
= 18
1
41414 + 1
41434 + 1
43414 + 1
43434 + 3
41414 + 3
41434 + 3
43414 + 3
43434
= 18
cos641 + cos643 + cos643 + cos649 + cos643 + cos649 + cos649 + cos2764
≈ 0985 26. Here ( ) = √ and ∆ = 2 ·12· 1 = 1, so the Midpoint Rule gives
( ) ≈
=1
=1
=1
∆
= 1
11412
+ 11432
+ 13412
+ 13432 +
31412 +
31432 +
33412 +
33432
= 18+ 38+ 38+ 98+√
338+√
398+√
398+√
3278≈ 70932 27. = {( ) | 0 ≤ ≤ 1, 0 ≤ ≤ 1 − , 0 ≤ ≤ 2 − 2},
the solid bounded by the three coordinate planes and the planes
= 1 − , = 2 − 2.
28. =
( ) | 0 ≤ ≤ 2 0 ≤ ≤ 2 − 0 ≤ ≤ 4 − 2
the solid bounded by the three coordinate planes, the plane = 2 − , and the cylindrical surface = 4 − 2.
29.
[continued]
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SECTION 15.6 TRIPLE INTEGRALS ¤ 577 Then
( ) =2 0
2 2−
(+−2)2
0 ( ) =2 0
2 2−
(+−2)2
0 ( )
=2 0
2 0
2
2−+2 ( ) =1 0
2 2
2
2−+2 ( )
=2 0
2 0
2
2−+2 ( ) =1 0
2 2
2
2−+2 ( )
33.
The diagrams show the projections of onto the -, -, and -planes.
Therefore
1 0
1
√
1−
0 ( ) =1 0
2 0
1−
0 ( ) =1 0
1−
0
2
0 ( )
=1 0
1−
0
2
0 ( ) =1 0
1−√ 0
1−
√ ( )
=1 0
(1−)2 0
√1−
( )
34.
The projections of onto the
- and -planes are as in the first two diagrams and so
1 0
1−2 0
1−
0 ( ) =1 0
√1−
0
1−
0 ( )
=1 0
1−
0
1−2
0 ( ) =1 0
1−
0
1−2
0 ( )
Now the surface = 1 − 2intersects the plane = 1 − in a curve whose projection in the -plane is = 1 − (1 − )2 or = 2 − 2. So we must split up the projection of on the -plane into two regions as in the third diagram. For ( ) in 1, 0 ≤ ≤ 1 − and for ( ) in 2, 0 ≤ ≤√
1 − , and so the given integral is also equal to
1 0
1−√ 1−
0
√1−
0 ( ) +1 0
1 1−√
1−
1−
0 ( )
=1 0
2−2 0
1−
0 ( ) +1 0
1 2−2
√1−
0 ( )
35.
1 0
1
0 ( ) =
( ) where = {( ) | 0 ≤ ≤ , ≤ ≤ 1, 0 ≤ ≤ 1}.
[continued]
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2
578 ¤ CHAPTER 15 MULTIPLE INTEGRALS
If 1, 2, and 3are the projections of onto the -, - and -planes then
1= {( ) | 0 ≤ ≤ 1, ≤ ≤ 1} = {( ) | 0 ≤ ≤ 1, 0 ≤ ≤ },
2= {( ) | 0 ≤ ≤ 1, 0 ≤ ≤ } = {( ) | 0 ≤ ≤ 1, ≤ ≤ 1}, and
3= {( ) | 0 ≤ ≤ 1, 0 ≤ ≤ } = {( ) | 0 ≤ ≤ 1, ≤ ≤ 1}.
Thus we also have
= {( ) | 0 ≤ ≤ 1, 0 ≤ ≤ , 0 ≤ ≤ } = {( ) | 0 ≤ ≤ 1, 0 ≤ ≤ , ≤ ≤ 1}
= {( ) | 0 ≤ ≤ 1, ≤ ≤ 1, ≤ ≤ 1} = {( ) | 0 ≤ ≤ 1, 0 ≤ ≤ , ≤ ≤ }
= {( ) | 0 ≤ ≤ 1, ≤ ≤ 1, ≤ ≤ } .
Then 1
0
1
0 ( ) =1 0
0
0 ( ) =1 0
0
1
( )
=1 0
1
1
( ) =1 0
0
( )
=1 0
1
( )
36.
1 0
1
0 ( ) =
( ) where = {( ) | 0 ≤ ≤ , ≤ ≤ 1, 0 ≤ ≤ 1}.
Notice that is bounded below by two different surfaces, so we must split the projection of onto the -plane into two regions as in the second diagram. If 1, 2, and 3are the projections of on the -, - and -planes then
1 = 1∪ 2= {( ) | 0 ≤ ≤ 1, 0 ≤ ≤ } ∪ {( ) | 0 ≤ ≤ 1, ≤ ≤ 1}
= {( ) | 0 ≤ ≤ 1, ≤ ≤ 1} ∪ {( ) | 0 ≤ ≤ 1, 0 ≤ ≤ },
2= {( ) | 0 ≤ ≤ 1, ≤ ≤ 1} = {( ) | 0 ≤ ≤ 1, 0 ≤ ≤ }, and
3= {( ) | 0 ≤ ≤ 1, ≤ ≤ 1} = {( ) | 0 ≤ ≤ 1, 0 ≤ ≤ }.
Thus we also have
= {( ) | 0 ≤ ≤ 1, 0 ≤ ≤ , ≤ ≤ 1} ∪ {( ) | 0 ≤ ≤ 1, ≤ ≤ 1, ≤ ≤ 1}
= {( ) | 0 ≤ ≤ 1, ≤ ≤ 1, ≤ ≤ 1} ∪ {( ) | 0 ≤ ≤ 1, 0 ≤ ≤ , ≤ ≤ 1}
= {( ) | 0 ≤ ≤ 1, 0 ≤ ≤ , 0 ≤ ≤ } = {( ) | 0 ≤ ≤ 1, ≤ ≤ 1, 0 ≤ ≤ }
= {( ) | 0 ≤ ≤ 1, 0 ≤ ≤ , 0 ≤ ≤ } .
Then 1
0
1
0 ( ) =1 0
0
1
( ) +1 0
1
1
( )
=1 0
1
1
( ) +1 0
0
1
( )
=1 0
0
0 ( ) =1 0
1
0 ( )
=1 0
0
0 ( )
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582 ¤ CHAPTER 15 MULTIPLE INTEGRALS (b) ( ) where = −11
0
3 3
√9−2
0 (2+ 2) ≈ 0375,
= −11 0
3 3
√9−2
0 (2+ 2) = 4564 ≈ 2209,
= −11 0
3 3
√9−2
0 (2+ 2) =1516 = 09375.
(c) =1 0
3 3
√9−2
0 (2+ 2)2 =10,464175 ≈ 5979 51. (a) ( ) is a joint density function, so we know
R3 ( ) = 1. Here we have
R3 ( ) =∞
−∞
∞
−∞
∞
−∞ ( ) =2 0
2 0
2
0
= 2 0 2
0 2
0 = 1
222 0
1
222 0
1
222 0= 8
Then we must have 8 = 1 ⇒ = 18. (b) ( ≤ 1 ≤ 1 ≤ 1) =1
−∞
1
−∞
1
−∞ ( ) =1 0
1 0
1 0 1
8
= 181 0 1
0 1
0 = 181 221
0
1 221
0
1 221
0=181 2
3
= 641
(c) ( + + ≤ 1) = (( ) ∈ ) where is the solid region in the first octant bounded by the coordinate planes and the plane + + = 1. The plane + + = 1 meets the -plane in the line + = 1, so we have
( + + ≤ 1) =
( ) =1 0
1−
0
1−−
0 1
8
=181 0
1−
0 1
22=1−−
=0 = 161 1 0
1−
0 (1 − − )2
=161 1 0
1−
0 [(3− 22+ ) + (22− 2)2+ 3]
=161 1 0
(3− 22+ )122+ (22− 2)133+ 1
44=1−
=0
=1921 1
0( − 42+ 63− 44+ 5) = 1921 1 30
=57601
52. (a) ( ) is a joint density function, so we know
R3 ( ) = 1. Here we have
R3 ( ) =∞
−∞
∞
−∞
∞
−∞ ( ) =∞
0
∞
0
∞
0 −(05+02+01)
= ∞
0 −05∞
0 −02∞
0 −01
= lim
→∞
0−05 lim
→∞
0−02 lim
→∞
0−01
= lim
→∞
−2−05 0 lim
→∞
−5−02 0 lim
→∞
−10−01 0
= lim
→∞
−2(−05− 1)
lim→∞
−5(−02− 1)
lim→∞
−10(−01− 1)
= · (−2)(0 − 1) · (−5)(0 − 1) · (−10)(0 − 1) = 100
So we must have 100 = 1 ⇒ =1001 . (b) We have no restriction on , so
( ≤ 1 ≤ 1) =1
−∞
1
−∞
∞
−∞ ( ) =1 0
1 0
∞ 0
1
100−(05+02+01)
=1001 1
0 −051
0 −02∞
0 −01
=1001
−2−051 0
−5−021 0 lim
→∞
−10−01
0 [by part (a)]
=1001 (2 − 2−05)(5 − 5−02)(10) = (1 − −05)(1 − −02) ≈ 007132
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