Section 6.2 Volume
SECTION 6.2 VOLUMES ¤ 571 17. =
1
−1
(1 − 2)2 = 2
1 0
(1 − 2)2
= 2
1 0
(1 − 22+ 4)
= 2
−233+1551 0
= 2 ·158 =1615
18. For 0 ≤ 2, a cross-section is an annulus with inner radius 2 − 1 and outer radius 4 − 1, the area of which is
1() = (4 − 1)2− (2 − 1)2. For 2 ≤ ≤ 4, a cross-section is an annulus with inner radius − 1 and outer radius 4 − 1, the area of which is 2() = (4 − 1)2− ( − 1)2.
=4
0 () = 2 0
(4 − 1)2− (2 − 1)2
+ 4 2
(4 − 1)2− ( − 1)2
= 82
0+ 4
2(8 + 2 − 2)
= 16 +
8 + 2−1334 2
= 16 +
32 + 16 − 643
−
16 + 4 −83
= 763
19. R1about OA (the line = 0):
=
1 0
() =
1 0
()2 =
1 331
0= 13
20. R1about OC (the line = 0):
=
1 0
() =
1 0
(12− 2) =
−1331 0=
1 −13
=23
21. R1about AB (the line = 1):
=
1 0
() =
1
0 (1 − )2 =
1
0 (1 − 2 + 2) =
− 2+1331
0= 13
22. R1about BC (the line = 1):
=
1 0
() =
1 0
[(1 − 0)2− (1 − )2] =
1 0
[1 − (1 − 2 + 2)]
=
1 0
(−2+ 2) =
−133+ 21 0=
−13 + 1
= 23
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
572 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION 23. R2about OA (the line = 0):
=
1 0
() =
1 0
12−√4
2
=
1
0 (1 − 12) =
−23321 0=
1 −23
= 13
24. R2about OC (the line = 0):
=
1 0
() =
1 0
[(4)2] =
1 0
8 = 1
991 0=19
25. R2about AB (the line = 1):
=
1 0
() =
1 0
[12− (1 − 4)2] =
1
0 [1 − (1 − 24+ 8)]
=
1 0
(24− 8) = 2
55−1991 0= 2
5−19
= 1345
26. R2about BC (the line = 1):
=
1 0
() =
1 0
(1 −√4
)2 =
1 0
(1 − 214+ 12)
=
−8554+23321 0=
1 −85+23
= 151
27. R3about OA (the line = 0):
=
1 0
() =
1 0
√4
2
− 2
=
1 0
(12− 2) =
2
332−1331 0= 2
3 −13
=13
Note: Let R = R1∪ R2∪ R3. If we rotate R about any of the segments , , , or , we obtain a right circular cylinder of height 1 and radius 1. Its volume is 2 = (1)2· 1 = . As a check for Exercises 19, 23, and 27, we can add the answers, and that sum must equal . Thus,13 +13 +13 = .
28. R3about OC (the line = 0):
=
1 0
() =
1 0
[2− (4)2] =
1 0
(2− 8) = 1
33−1991 0= 1
3−19
= 29
Note: See the note in the solution to Exercise 27. For Exercises 20, 24, and 28, we have23 +19 +29 = .
29. R3about AB (the line = 1):
=
1 0
() =
1
0 [(1 − 4)2− (1 − )2] =
1
0 [(1 − 24+ 8) − (1 − 2 + 2)]
=
1 0
(8− 24− 2+ 2) = 1
99−255−133+ 21
0= 1
9 −25−13+ 1
= 1745
Note: See the note in the solution to Exercise 27. For Exercises 21, 25, and 29, we have13 +1345 +1745 = .
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
572 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION 23. R2about OA (the line = 0):
=
1 0
() =
1 0
12−√4
2
=
1 0
(1 − 12) =
−23321 0=
1 −23
= 13
24. R2about OC (the line = 0):
=
1 0
() =
1 0
[(4)2] =
1 0
8 = 1 991
0=19
25. R2about AB (the line = 1):
=
1 0
() =
1 0
[12− (1 − 4)2] =
1 0
[1 − (1 − 24+ 8)]
=
1 0
(24− 8) = 2
55−1991
0= 2 5−19
= 1345
26. R2about BC (the line = 1):
=
1 0
() =
1
0 (1 −√4
)2 =
1
0 (1 − 214+ 12)
=
−8554+23321 0=
1 −85+23
= 151
27. R3about OA (the line = 0):
=
1 0
() =
1 0
√4
2
− 2
=
1 0
(12− 2) =
2
332−1331 0= 2
3 −13
=13
Note: Let R = R1∪ R2∪ R3. If we rotate R about any of the segments , , , or , we obtain a right circular cylinder of height 1 and radius 1. Its volume is 2 = (1)2· 1 = . As a check for Exercises 19, 23, and 27, we can add the answers, and that sum must equal . Thus,13 +13 +13 = .
28. R3about OC (the line = 0):
=
1 0
() =
1 0
[2− (4)2] =
1 0
(2− 8) = 1
33−1991
0= 1 3−19
= 29
Note: See the note in the solution to Exercise 27. For Exercises 20, 24, and 28, we have23 +19 +29 = .
29. R3about AB (the line = 1):
=
1 0
() =
1 0
[(1 − 4)2− (1 − )2] =
1 0
[(1 − 24+ 8) − (1 − 2 + 2)]
=
1 0
(8− 24− 2+ 2) = 1
99−255−133+ 21
0= 1
9 −25−13+ 1
= 1745
Note: See the note in the solution to Exercise 27. For Exercises 21, 25, and 29, we have13 +1345 +1745 = .
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
572 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION
23. R2about OA (the line = 0):
=
1 0
() =
1 0
12−√4
2
=
1 0
(1 − 12) =
−23321 0=
1 −23
= 13
24. R2about OC (the line = 0):
=
1 0
() =
1 0
[(4)2] =
1 0
8 = 1
991 0=19
25. R2about AB (the line = 1):
=
1 0
() =
1 0
[12− (1 − 4)2] =
1
0 [1 − (1 − 24+ 8)]
=
1 0
(24− 8) = 2
55−1991 0= 2
5−19
= 1345
26. R2about BC (the line = 1):
=
1 0
() =
1
0 (1 −√4
)2 =
1
0 (1 − 214+ 12)
=
−8554+23321 0=
1 −85+23
= 151
27. R3about OA (the line = 0):
=
1 0
() =
1 0
√4
2
− 2
=
1 0
(12− 2) =
2
332−1331 0= 2
3 −13
=13
Note: Let R = R1∪ R2∪ R3. If we rotate R about any of the segments , , , or , we obtain a right circular cylinder of height 1 and radius 1. Its volume is 2 = (1)2· 1 = . As a check for Exercises 19, 23, and 27, we can add the answers, and that sum must equal . Thus,13 +13 +13 = .
28. R3about OC (the line = 0):
=
1 0
() =
1 0
[2− (4)2] =
1 0
(2− 8) = 1
33−1991 0= 1
3−19
= 29
Note: See the note in the solution to Exercise 27. For Exercises 20, 24, and 28, we have23 +19 +29 = .
29. R3about AB (the line = 1):
=
1 0
() =
1 0
[(1 − 4)2− (1 − )2] =
1 0
[(1 − 24+ 8) − (1 − 2 + 2)]
=
1 0
(8− 24− 2+ 2) = 1
99−255−133+ 21
0= 1
9 −25−13+ 1
= 1745
Note: See the note in the solution to Exercise 27. For Exercises 21, 25, and 29, we have13 +1345 +1745 = .
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
572 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION 23. R2about OA (the line = 0):
=
1 0
() =
1 0
12−√4
2
=
1
0 (1 − 12) =
−23321 0=
1 −23
= 13
24. R2about OC (the line = 0):
=
1 0
() =
1 0
[(4)2] =
1 0
8 = 1 991
0=19
25. R2about AB (the line = 1):
=
1 0
() =
1 0
[12− (1 − 4)2] =
1 0
[1 − (1 − 24+ 8)]
=
1 0
(24− 8) = 2
55−1991 0= 2
5−19
= 1345
26. R2about BC (the line = 1):
=
1 0
() =
1 0
(1 −√4
)2 =
1 0
(1 − 214+ 12)
=
−8554+23321 0=
1 −85+23
= 151
27. R3about OA (the line = 0):
=
1 0
() =
1 0
√4
2
− 2
=
1 0
(12− 2) =
2
332−1331 0= 2
3 −13
=13
Note: Let R = R1∪ R2∪ R3. If we rotate R about any of the segments , , , or , we obtain a right circular cylinder of height 1 and radius 1. Its volume is 2 = (1)2· 1 = . As a check for Exercises 19, 23, and 27, we can add the answers, and that sum must equal . Thus,13 +13 +13 = .
28. R3about OC (the line = 0):
=
1 0
() =
1 0
[2− (4)2] =
1 0
(2− 8) = 1
33−1991
0= 1 3−19
= 29
Note: See the note in the solution to Exercise 27. For Exercises 20, 24, and 28, we have23 +19 +29 = .
29. R3about AB (the line = 1):
=
1 0
() =
1
0 [(1 − 4)2− (1 − )2] =
1
0 [(1 − 24+ 8) − (1 − 2 + 2)]
=
1 0
(8− 24− 2+ 2) = 1
99−255−133+ 21 0= 1
9 −25−13+ 1
= 1745
Note: See the note in the solution to Exercise 27. For Exercises 21, 25, and 29, we have13 +1345 +1745 = .
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
572 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION 23. R2about OA (the line = 0):
=
1 0
() =
1 0
12−√4
2
=
1 0
(1 − 12) =
−23321 0=
1 −23
= 13
24. R2about OC (the line = 0):
=
1 0
() =
1 0
[(4)2] =
1 0
8 = 1
991 0=19
25. R2about AB (the line = 1):
=
1 0
() =
1 0
[12− (1 − 4)2] =
1
0 [1 − (1 − 24+ 8)]
=
1 0
(24− 8) = 2
55−1991
0= 2 5−19
= 1345
26. R2about BC (the line = 1):
=
1 0
() =
1
0 (1 −√4
)2 =
1
0 (1 − 214+ 12)
=
−8554+23321 0=
1 −85+23
= 151
27. R3about OA (the line = 0):
=
1 0
() =
1 0
√4
2
− 2
=
1 0
(12− 2) =
2
332−1331 0= 2
3 −13
=13
Note: Let R = R1∪ R2∪ R3. If we rotate R about any of the segments , , , or , we obtain a right circular cylinder of height 1 and radius 1. Its volume is 2 = (1)2· 1 = . As a check for Exercises 19, 23, and 27, we can add the answers, and that sum must equal . Thus,13 +13 +13 = .
28. R3about OC (the line = 0):
=
1 0
() =
1 0
[2− (4)2] =
1 0
(2− 8) = 1
33−1991 0= 1
3−19
= 29
Note: See the note in the solution to Exercise 27. For Exercises 20, 24, and 28, we have23 +19 +29 = .
29. R3about AB (the line = 1):
=
1 0
() =
1 0
[(1 − 4)2− (1 − )2] =
1 0
[(1 − 24+ 8) − (1 − 2 + 2)]
=
1 0
(8− 24− 2+ 2) = 1
99−255−133+ 21
0= 1
9 −25−13+ 1
= 1745
Note: See the note in the solution to Exercise 27. For Exercises 21, 25, and 29, we have13 +1345 +1745 = .
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
