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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
288 SECTION 6.1
CHAPTER 6
SECTION 6.1
1. (a)∫ 2
−1
[(x + 2) − x2] dx
(b)∫ 1
0
[(√y) − (−√
y )] dy +∫ 4
1
[(√y ) − (y − 2)] dy
2. (a)∫ 0
−4
[(−4x) − x2
]dx
(b)∫ 16
0
[(−1
4y
)− (−√
y)]dy
3. (a)∫ 2
0
[(2x2
)−
(x3
)]dx
(b)∫ 8
0
[(y1/3
)−
(12y
)1/2]dy
4. (a)∫ 1
0
[√x− x3
]dx
(b)∫ 1
0
[y1/3 − y2
]dy
5. (a)∫ 4
0
[(0) −
(−√x)]
dx +∫ 6
4
[(0) − (x− 6)] dx
(b)∫ 0
−2
[(y + 6) −
(y2
)]dy
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
SECTION 6.1 289
6. (a)∫ 8
−1
[x1/3 −
(x− 2
3
)]dx
(b)∫ 2
−1
[(3y + 2) − y3
]dy
7. (a)∫ 0
−2
[(8 + x
3
)− (−x)
]dx +
∫ 4
0
[(8 + x
3
)− (x)
]dx
(b)∫ 2
0
[(y) − (−y)] dy +∫ 4
2
[(y) − (3y − 8)] dy
8. (a)∫ 3/2
0
[2x− x] dx +∫ 3
3/2
[3 − x] dx
(b)∫ 3
0
[y − 1
2y
]dy
9. (a)∫ 5
−4
[(√4 + x
)−
(−√
4 + x)]
dx
(b)∫ 3
−3
[(5) −
(y2 − 4
)]dy
10. (a)∫ 2
0
[x− (−x)] dx
(b)∫ 0
−2
[2 − (−y)] dy +∫ 2
0
[2 − y] dy
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
290 SECTION 6.1
11. (a)∫ 3
−1
[(2x) − (x− 1)] dx +∫ 5
3
[(9 − x) − (x− 1)] dx
(b)∫ 4
−2
[(y + 1) −
(12y
)]dy +
∫ 6
4
[(9 − y) −
(12y
)]dy
12. (a)∫ 1
−1
[x3 − (x2 + x− 1)
]dx
(b)∫ −1
−5/4
[(−1
2+
12
√4y + 5
)−
(−1
2− 1
2
√4y + 5
)]dy
+∫ 1
−1
[(−1
2+
12
√4y + 5
)− y1/3
]dy
13. (a)∫ 1
−1
[(x1/3
)−
(x2 + x− 1
)]dx
(b)∫ −1
−5/4
[(−1
2+
12
√4y + 5
)−
(−1
2− 1
2
√4y + 5
)]dy
+∫ 1
−1
[(−1
2+
12
√4y + 5
)−
(y3
)]dy
14. (a)∫ 3
−1
[(x + 1) −
(−x− 13
)]dx +
∫ 5
3
[(13 − 3x) −
(−x− 13
)]dx
(b)∫ 0
−2
[(13 − y
3
)− (−3y − 1)
]dy +
∫ 4
0
[(13 − y
3
)− (y − 1)
]dy
15. A =∫ 3
0
[(4y − y2
4
)−
(y4
)]dy
=∫ 3
0
(34y − 1
4y2
)dy
=[38y
2 − 112y
3]30
= 98
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
SECTION 6.1 291
16. A =∫ 2
−1
[(4 − y2) − (2 − y)
]dy
=∫ 2
−1
(2 + y − y2) dy
=[2y + y2
2 − y3
3
]2
−1=
92
17. A = 2∫ 2
0
[(12 − 2y2
)−
(y2
)]dy
= 2∫ 2
0
(12 − 3y2
)dy
= 2[12y − y3
]20
= 2 (16) = 32
2 4 6 8 10 12x
-2-1
12
y
18. A =∫ 1
0
[y1/3 − (2y2 − y)
]dy
=∫ 1
0
(y1/3 + y − 2y2) dy
=[34y4/3 +
y2
2− 2
3y3
]1
0
=712
19. A =∫ 0
−2
[(y3 − y
)−
(y − y2
)]dy +
∫ 1
0
[(y − y2
)−
(y3 − y
)]dy
=∫ 0
−2
(y3 + y2 − 2y
)dy +
∫ 1
0
(2y − y2 − y3
)dy
=[
14y
4 + 13y
3 − y2]0
−2+
[y2 − 1
3y3 − 1
4y4]1
0= 8
3 + 512 = 37
12
-6 -4 -2 1x
-2
2
y
20. A =∫ 0
−2
[18(2y3 + y2 − 2y) − 1
8y3
]dy
+∫ 1
0
[18y3 − 1
8(2y3 + y2 − 2y)
]dy
=18
[y4
4+
y3
3− y2
]0
−2
+18
[−y4
4− y3
3+ y2
]1
0
=3796
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
292 SECTION 6.1
21. A =∫ π/4
−π/4
[sec2 x− cosx
]dx
= 2∫ π/4
0
[sec2 x− cosx
]dx
= 2 [tanx + sinx]π/40 = 2[1 +
√2/2
]= 2 +
√2
22. A =∫ π/4
−π/4
(tan2 x− sin2 x) dx =∫ π/4
−π/4
(sec2 x− 3
2+
cos 2x2
)dx
=[tanx− 3
2x +
sin 2x4
]π/4−π/4
=52− 3π
4
23. A =∫ −π/2
−π
[sin 2x− 2 cosx] dx +∫ π/2
−π/2
[2 cosx− sin 2x] dx
+∫ π
π/2
[sin 2x− 2 cosx] dx
=[− 1
2 cos 2x− 2 sinx]−π/2
−π+
[2 sinx + 1
2 cos 2x]π/2−π/2
+[− 1
2 cos 2x− 2 sinx]ππ/2
= 8
24. A =∫ π/3
0
(sin 2x− sinx) dx +∫ π/2
π/3
(sinx− sin 2x) dx
=[−cos 2x
2+ cosx
]π/30
+[− cosx +
cos 2x2
]π/2π/3
=12
25. A =∫ π/2
0
(sin4 x cosx) dx
=∫ 1
0
u4 du, (u = sinx)
=[u5
5
]1
0
=15
pÇÇÇÇ2
x
0.1
0.2
y
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
SECTION 6.1 293
26. A =∫ π/8
0
(cos 2x− sin 2x) dx +∫ π/4
π/8
(sin 2x− cos 2x) dx
=[sin 2x
2+
cos 2x2
]π/80
+[−cos 2x
2− sin 2x
2
]π/4π/8
= 2√
2 − 1
8 4
x
1y
π π
27. A =∫ 1
0
[3x− 1
3x
]dx +
∫ 3
1
[−x + 4 − 1
3x
]dx =
[43x2
]1
0
+[−2
3x2 + 4x
]3
1
= 4
28. A =∫ 2
0
[(x + 1) −
(1 − x
2
)]dx +
∫ 3
2
[(x + 1) − (4x− 8)] dx =[x2
4
]2
0
+[−3x2
2+ 9x
]3
2
=52
29. A =∫ 1
−2
[x− (−2)] dx +∫ 5
1
[1 − (−2)] dx +∫ 7
5
[−3
2x +
172
− (−2)]dx
=[
12 x
2 + 2x]1
−2+
[3x
]5
1+
[− 3
4 x2 + 21
2 x]7
5=
392
30. A =∫ 1
0
[y1/3 − (−y)
]dy
=∫ 1
0
(y1/3 + y) dy
=[34y4/3 +
y2
2
]1
0
=54
31. A =∫ 0
−3
[6 − x2 − x
]dx +
∫ 3
0
[6 − x2 − (−x)
]dx
=[6x− 1
3x3 − 1
2x2
]0
−3
+[6x− 1
3x3 +
12x2
]3
0
= 27
32. A =∫ 4p
0
(√4px− x2
4p
)dx
=[√
4p23x3/2 − x3
12p
]4p
0
=163p2
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
294 SECTION 6.1
33.∫ √
c
0
[c− x2
]dx =
12
∫ 2
0
[4 − x2
]dx
[cx− 1
3 x3]√c
0= 1
2
[4x− 1
3 x3]20
23 c
3/2 = 83 and c = 42/3
34. We want∫ c
0
cosx dx =12
∫ π/2
0
cosx dx =⇒ [sinx]c0 =12
[sinx]π/20 =12
=⇒ sin c =12
=⇒ c =π
6
35. A =∫ 1
0
√3 dx +
∫ 2
1
√4 − x2 dx;
A =∫ √
3
0
(√4 − y2 − y√
3
)dy
36. A =∫ 1
0
(√4 − x2 −
√3x
)dx
37. A =∫ 2
0
[√4 − x2 − (2 −
√4x− x2)
]dx 38. A =
∫ √3
0
(√
4 − x2 − x2
3) dx
39. The area under the curve is Ac =∫ a
0
bxn dx =ban+1
n + 1.
For the rectangle, Ar = ban+1. Thus the ratio is1
n + 1.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
SECTION 6.1 295
40. (a) A =∫ √
1+aa
0
(1 + a− ax2) dx
=[x + ax− ax3
3
]√ 1+aa
0
=2(1 + a)3/2
2a1/2.
(b) A′ =23
[3a(1 + a)1/2 − (1 + a)3/2
2a3/2
]= 0
=⇒ a =12.
41. A ∼=∫ 1.79
−1.49
[x + 2 − (x4 − 2x2)
]dx
=[12 x
2 − 2x− 15 x
5 + 23 x
3]1.78−1.49
∼= 7.93
42. A ∼= 0.67
43. V = 8 · 12∫ 3
−3
[4 − 4
9x2
]dx
= 96 · 2∫ 3
0
[4 − 4
9x2
]dx
= 192[4x− 4
27 x3]30
= 1536 cu. in.∼= 0.89 cu. ft.
44. (a) A =∫ b
1
1x2
dx =[− 1x
]b1
= 1 − 1b.
(b) 1 − 1b→ 1 as b → ∞.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
296 SECTION 6.2
45. (a) A =∫ b
1
1√xdx =
[2√x]b1
= 2√b− 2.
(b) 2√b− 2 → ∞ as b → ∞.
46. (a) r > 1 : A =∫ b
1
1xr
dx =[ x1−r
1 − r
]b1
=1
r − 1[1 − b1−r
];
A → 1r − 1
as b → ∞
(b) 0 < r < 1 : A =b1−r
1 − r− 1
1 − r; A → ∞ as b → ∞
SECTION 6.2
1. V =∫ 1
0
π[(x)2 − (0)2
]dx = π
[x3
3
]1
0
=π
3
2. V =∫ 3
0
π (3 − x)2 dx =[−π(3 − x)3
3
]3
0
= 9π
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
SECTION 6.2 297
3. V =∫ 3
−3
π[(9)2 −
(x2
)2]dx = 2
∫ 3
0
π(81 − x4
)dx
= 2π[81x− x5
5
]3
0
=1944π
5
4. V =∫ 2
0
π[82 − x6
]dx = π
[64x− x7
7
]2
0
=7687
π
5. V =∫ 1
0
π[(√
x)2 −
(x3
)2]dx
=∫ 1
0
π(x− x6
)dx
= π
[12x2 − 1
7x7
]1
0
=5π14
6. V =∫ 1
0
π[x2/3 − x4
]dx = π
[35x5/3 − x5
5
]1
0
=2π5
7. V =∫ 2
1
π[(x3
)2 − (1)2]dx +
∫ 9
2
π[(10 − x)2 − (1)2
]dx
=∫ 2
1
π(x6 − 1
)dx +
∫ 9
2
π(99 − 20x + x2
)dx
= π
[17x7 − x
]2
1
+ π
[99x− 10x2 +
13x3
]9
2
=3790π
21
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
298 SECTION 6.2
8. V =∫ 4
1
π[x− 1] dx +∫ 5
4
π[(6 − x)2 − 1
]dx
= π
[x2
2− x
]4
1
+ π
[− (6 − x)3
3− x
]5
4
=356π
9. V =∫ 2
−1
π[(x + 2)2 −
(x2
)2]dx
=∫ 2
−1
π(x2 + 4x + 4 − x4
)dx
= π[13x
3 + 2x2 + 4x− 15x
5]2−1
=725π
10. V =∫ 1
−2
π[(2 − x)2 − x4
]dx = π
[− (2 − x)3
3− x5
5
]1
−2
=725π
11. V =∫ 2
−2
π[√
4 − x2]2
dx = 2∫ 2
0
π(4 − x2
)dx
= 2π[4x− x3
3
]2
0
=323π
12. V = 2 (Volume of cone of radius 1, height 1) = 2 · 13π =
23π
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
SECTION 6.2 299
13. V =∫ π/4
0
π sec2 x dx = π [tanx]π/40 = π
14. V =∫ 3π/4
π/4
π[csc2 x− 0
]dx = π [− cotx]3π/4π/4 = 2π
15. V =∫ π/2
0
π[(x + 1)2 − (cosx)2
]dx
=∫ π/2
0
π
[(x + 1)2 −
(12
+12
cos 2x)]
dx
= π
[13
(x + 1)3 − 12x− 1
4sin 2x
]π/20
= π2
24
(π2 + 6π + 6
)
16. V =∫ π/2
π/4
π sin2 x dx = π
[x
2− 1
4sin 2x
]π/2π/4
=18π(π + 2)
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
300 SECTION 6.2
17. V =∫ 4
0
π(y
2
)2
dy =π
12[y3
]40
=16π3
18. V = (Volume of cone of radius 6, height 2)
=13π 62 · 2 = 24π
19. V =∫ 2
0
π[(8)2 −
(y3
)2]dy
=∫ 2
0
π(64 − y6
)dy
= π[64y − 1
7y7]20
= 7687 π
20. V =∫ 2
−2
π[42 − y4
]dy = π
[16y − y5
5
]2
−2
=2565
π
21. V =∫ 1
0
π
[(y1/3
)2
−(y2
)2]dy
=∫ 1
0
π[y2/3 − y4
]dy
= π[35y
5/3 − 15y
5]10
= 25π
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
SECTION 6.2 301
22. V =∫ 1
0
π[(√y)2 − (y3)2
]dy =
∫ 1
0
π(y − y6) dy
= π
[y2
2− y7
7
]1
0
=514
π
23. V =∫ 4
0
π
[y2 −
(y2
)2]dy +
∫ 8
4
π
[42 −
(y2
)2]dy
=∫ 4
0
π
[34y2
]dy +
∫ 8
4
π
[16 − 1
4y2
]dy
= π[14y
3]40
+ π[16y − 1
12y3]84
= 1283 π
24. V =∫ 3
0
π
[(3 − y
2
)2
− (3 − y)2]dy +
∫ 6
3
π(3 − y
2
)2
dy
= π
[−2
3
(3 − y
2
)3
+(3 − y)3
3
]3
0
+ π
[−2
3
(3 − y
2
)3]6
3
= 9π
25. V =∫ 1
−1
π[(
2 − y2)2 −
(y2
)2]dy
= 2∫ 1
0
π[4 − 4y2
]dy = 2π
[4y − 4
3y3
]1
0
=163π
26. V =∫ 3
0
π(9 − y2) dy = π
[9y − y3
3
]3
0
= 18π (half sphere of radius 3.)
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
302 SECTION 6.2
27. (a) V =∫ r
−r
(2√r2 − x2
)2
dx = 8∫ r
0
(r2 − x2
)dx = 8
[r2x− 1
3x3
]r0
=163r3
(b) V =∫ r
−r
√3
4
(2√r2 − x2
)2
dx = 2√
3∫ r
0
(r2 − x2
)dx =
4√
33
r3
28. For each x ∈ [−3, 3], the length of the base of the cross-section at x is 2y =43
√9 − x2.
(a) The area of each triangle is√
34
s2.
Thus V =∫ 3
−3
√3
4· 16
9(9 − x2) dx =
4√
39
∫ 3
−3
(9 − x2) dx
=4√
39
[9x− x3
3
]3
−3
= 16√
3.
(b) The area of each square is s2
Thus V =∫ 3
−3
169
(9 − x2) dx =169
∫ 3
−3
(9 − x2) dx
=169
[9x− x3
3
]3
−3
= 64.
29. (a) V =∫ 2
−2
(4 − x2
)2dx = 2
∫ 2
0
(16 − 8x2 + x4
)dx = 2
[16x− 8
3x3 +
15x5
]2
0
=51215
(b) V =∫ 2
−2
π
2
(4 − x2
2
)2
dx =π
4
∫ 2
0
(4 − x2
)2dx =
π
4
(25615
)=
6415
π
(c) V =∫ 2
−2
√3
4(4 − x2
)2dx =
√3
2
∫ 2
0
(4 − x2
)2dx =
√3
2
(25615
)=
12815
√3
30. (a) V =∫ 1
0
2√xh dx +
∫ 3
1
2 · 1√2
√3 − xh dx =
4h3
[x3/2
]1
0+
[−2
√2h
3(3 − x)3/2
]3
1
= 4h
(b) V =∫ 1
0
(12· 2√x ·
√3√x
)dx +
∫ 3
1
(12·√
2√
3 − x ·√
3√2·√
3 − x
)dx
=√
3∫ 1
0
x dx +√
32
∫ 3
1
(3 − x) dx =3√
32
(c) V =∫ 1
0
(12· 2√x ·
√x
)dx +
∫ 3
1
(12·√
2√
3 − x ·√
22
·√
3 − x
)dx
=∫ 1
0
x dx +12
∫ 3
1
(3 − x) dx =32
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
SECTION 6.2 303
31. (a) V =∫ 4
0
[(√y ) − (−√
y )]2 dy =∫ 4
0
4y dy =[2y2
]40
= 32
(b) V =∫ 4
0
π
2(√y)2 dy =
π
2
∫ 4
0
y dy =π
2
[12y2
]4
0
= 4π
(c) V =∫ 4
0
√3
4[(√y ) − (−√
y )]2 dy =√
3∫ 4
0
y dy = 8√
3
32. (a) V =∫ 1
−1
(3 − 3y2)h dy = h[3y − y3
]1−1
= 4h
(b) V =∫ 1
−1
12(3 − 3y2)
√3
2(3 − 3y2) dy =
9√
34
∫ 1
−1
(1 − 2y2 + y4) dy
=9√
34
[y − 2
3y3 +
y5
5
]1
−1
=125
√3
(c) V =∫ 1
−1
12(3 − 3y2)
12(3 − 3y2) dy =
1√3·[Volume in (b)] =
125
33. (a) V =∫ 4
0
(4 − x)2 dx =[16x− 4x2 +
x3
3
]4
0
=643.
(b) V =14
∫ 4
0
(4 − x)2 dx =14
[16x− 4x2 +
x3
3
]4
0
=163.
34. (a) Area of each triangle = y2, thus V = 2∫ a
0
(b2 − b2
a2x2
)dx = 2
[b2x− b2
3a2x3
]a0
=43ab2.
(b) Area of each square = 4y2, thus V = 4· the answer to part (a) =163ab2.
(c) Area of each triangle = 2y, thus V = 2∫ a
0
2
√b2 − b2
a2x2 dx
= 4ba
∫ a
0
√a2 − x2 dx = 4b
a
[x2
√a2 − x2 + a2
2 sin−1 xa
]a0
= abπ.
35. (a) V =∫ π/2
0
√3 sinx dx = −
√3[cosx
]π/20
=√
3
(b) V =∫ π/2
0
4 sinx dx = −4[cosx
]π/20
= 4
36. (a) V =∫ π/4
0
π[12 (secx− tanx)2
]dx =
π
4
∫ π/4
0
(2 sec2 x− 2 secx tanx− 1
)dx
=π
4
[4 − 2
√2 − π
4
]
(b) V =∫ π/4
0
π (secx− tanx)2 dx =∫ π/4
0
(2 sec2 x− 2 secx tanx− 1
)dx = 4 − 2
√2 − π
4
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
304 SECTION 6.2
37. V =∫ a
−a
π
(b
√1 − x2
a2
)2
dx =2πb2
a2
∫ a
0
π(a2 − x2
)dx =
2πb2
a2π
[a2x− 1
3x3
]a0
=2πb2
a2
(23a3
)=
43πab2
38. V =∫ b
−b
π(ab
√b2 − y2
)2
dy =πa2
b2
∫ b
−b
(b2 − y2) dy
=πa2
b2
[b2y − y3
3
]b−b
=43πa2b
39. The specified frustum is generated by revolving
the region Ω about the y-axis.
V =∫ h
0
π
[r −R
hy + R
]2
dy
= π
[h
3 (r −R)
(r −R
hy + R
)3]h
0
=πh
3 (r −R)(r3 −R3
)=
πh
3(r2 + rR + R2
)
40. V = 2∫ a/2
0
π(√
3x)2
dx = 6π[x3
3
]a/20
=14πa3 (twice the volume of cone of radius
√3
2 a, height a2 )
41. Capacity of basin = 12
(43πr
3)
= 23πr
3.
(a) Volume of water =∫ r
r/2
π[√
r2 − x2]2
dx
= π
∫ r
r/2
(r2 − x2
)dx = π
[r2x− 1
3x3
]rr/2
=524
πr3.
The basin is(
524πr
3)(100) /
(23πr
3)
= 31 14% full.
(b) Volume of water =∫ r
2r/3
π[√
r2 − x2]2
dx = π
∫ r
2r/3
(r2 − x2
)dx =
881
πr3.
The basin is(
881πr
3)(100)
/ (23πr
3)
= 14 2227% full.
42. V =∫ b
a
π(√
r2 − x2)2
dx = π
∫ b
a
(r2 − x2) dx = π
[r2x− x3
3
]ba
= πr2(b− a) − 13π(b3 − a3).
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
SECTION 6.2 305
43. V =∫ r
h
π(r2 − y2) dy = π
[r2y − y3
3
]rh
=π
3[2r3 − 3r2h + h3
].
44. Imagine the punchbowl upside down on the x, y-plane centered over the origin.
(a) V = π
∫ 12
1
(144 − y2) dy = π
[144y − y3
3
]12
1
= π(1584 − 17273
) in3 or about 13.7 gallons.
(b) V = π
∫ 10
1
(144 − y2) dy = π
[144y − y3
3
]10
1
= 963π in3 or about 13.1 gallons.
45. (a)(b) A(b) =
∫ b
1
x− 23 dx = 3(b
13 − 1)
(c) V (b) =∫ b
1
π(x− 23 )2 dx = 3π(1 − b−
13 )
(d) As b → ∞, A(b) → ∞ and V (b) → 3π.
46. (a) A(c) =∫ 1
c
x− 23 dx =
[3x
13
]1
c= 3(1 − c
13 ).
(c) V (c) = π
∫ 1
c
(x− 23 )2 dx = π
[−3x− 1
3
]1
c= 3π(c−
13 − 1).
(d) As c → 0, c13 → 0 and c−
13 → ∞. Thus A(c) → 3, and V (c) → ∞.
47. If the depth of the liquid in the container is h feet, then the volume of the liquid is:
V (h) =∫ h
0
π(√
y + 1)2
dy =∫ h
0
π [y + 1] dy.
Differentiation with respect to t gives
dV
dt=
dV
dh· dhdt
= π(h + 1)dh
dt.
Now, sincedV
dt= 2, it follows that
dh
dt=
2π(h + 1)
. Thus
dh
dt
∣∣∣∣h=1
=22π
=1π
ft/min anddh
dt
∣∣∣∣h=2
=23π
ft/min.
48. At time t the water in the container is at height h. The volume V of water is given by
V =∫ h
0
π[f(y)]2 dy.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
306 SECTION 6.2
Differentiating with respect to t givesdV
dt=
dV
dh
dh
dt= π[f(h)]2
dh
dt.
The surface area of the water at time t is given by S = π[f(h)]2 and dV/dt = kS = kπ[f(h)]2, k < 0
the constant of proportionality. Thus, dh/dt = k
49. (b) The x-coordinates of the points of intersection are: x = 0, x = 21/4 ∼= 1.1892.
(c) A ∼= 2∫ 1.1892
0
(2x− x5
)dx ∼= 2(0.9428) = 1.8856
(d) V =∫ 1.1892
0
π(4x2 − x10
)dx ∼= 5.1234
50. (b) x-coordinates of the points of
intersection: x = 1, 3.5747
(c) A ∼= 3.1148
(d) V ∼= 22.2025
51. V =∫ 4
0
π[4 −
(2 −
√x)2
]dx =
∫ 4
0
π[4√x− x
]dx = π
[83x3/2 − 1
2x2
]4
0=
40π3
52. V =∫ 3
0
π([(x + 1) + 1]2 −
[(x− 1)2 + 1
]2)dx =
∫ 3
0
π[(x + 2)2 − (x2 − 2x + 2)2
]dx
=∫ 3
0
π(12x− 7x2 + 4x3 − x4) dx = π
[6x2 − 7
3x3 + x4 − x5
5
]3
0
=1175
π
53. V =∫ π
0
π[2 sinx− sin2 x
]dx = π
[− 2 cosx
]π0− π
2
∫ π
0
(1 − cos 2x) dx
= 4π − π
2
[x− 1
2sin 2x
]π0
= 4π − 12π2
54. V =∫ π
π/4
π[(1 − cosx)2 − (1 − sinx)2
]dx = π
∫ π
π/4
(2 sinx− 2 cosx + cos 2x) dx
= π
[−2 cosx− 2 sinx +
12
sin 2x]ππ/4
=32
+ 2√
2
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
SECTION 6.2 307
55. V =∫ 5
0
π([3x− (−1)]2 −
[x2 − 2x− (−1)
]2)dx
= π
∫ 5
0
[3x + 1]2 dx−∫ 5
0
[x− 1]4 dx
= π[19 (3x + 1)3
]50−
[(x− 1)5
5
]5
0
= 250π
56. (a) V =∫ 1
0
π[(2 +
√y)2 − (2 + y2)2
]dy = π
∫ 1
0
(4y1/2 + y − 4y2 − y4
)dy =
4930
π
(b) V =∫ 1
0
π[(3 − y2)2 − (3 −√
y)2]dy = π
∫ 1
0
(6√y + y4 − 6y2 − y
)dy =
1710
π
57. (a) V =∫ 4
0
π
[(√4x
)2
− x2
]dx
= π
∫ 4
0
[4x− x2
]dx
= π[2x2 − 1
3 x3]40
=32π3
(b) V =∫ 4
0
π
[(14y2 − 4
)2
− (y − 4)2]dy
= π
∫ 4
0
[116
y4 − 3y2 + 8y]dy
= π[
180 y
5 − y3 + 4y2]40
=64π5
58. (a) V =∫ 16
0
π
[(5 − y
4
)4
− (5 −√y)2
]dy = π
∫ 16
0
(10√y − 7
2y +
y2
16
)dy
= π
[203y3/2 − 7
4y2 +
y3
48
]16
0
= 64π
(b) V =∫ 16
0
π
[(√y + 1)2 −
(y4
+ 1)2
]dy = π
∫ 16
0
(y
2+ 2
√y − y2
16
)dy
= π
[y2
4+
43y3/2 − y3
48
]16
0
= 64π
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
308 SECTION 6.3
59. (a) V =∫ 4
0
π(x3/2
)2
dx = π
∫ 4
0
x3 dx = π
[14x4
]4
0
= 64π
(b) V =∫ 8
0
π(4 − y2/3
)2
dy = π
∫ 8
0
(16 − 8y2/3 + y4/3
)dy
= π[16y − 24
5 y5/3 + 37y
7/3]80
= 102435 π
(c) V =∫ 4
0
π
[(8)2 −
(8 − x3/2
)2]dx = π
∫ 4
0
(16x3/2 − x3
)dx
= π[325 x5/2 − 1
4x4]40
= 7045 π
(d) V =∫ 8
0
π
[(4)2 −
(y2/3
)2]dy = π
∫ 8
0
(16 − y4/3
)dy = π
[16y − 3
7y7/3
]8
0
=5127
π
60. (a) V =∫ 8
0
πy4/3 dy = π
[37y7/3
]8
0
=3847
π
(b) V =∫ 4
0
π(8 − x3/2)2 dx = π
∫ 4
0
(64 − 16x3/2 + x3
)dx = π
[64x− 32
5x5/2 +
x4
4
]4
0
=5765
π
(c) V =∫ 8
0
π[42 − (4 − y2/3)2
]dy = π
∫ 8
0
(8y2/3 − y4/3) dy = π
[245y5/3 − 3
7y7/3
]8
0
=345635
π
(d) V =∫ 4
0
π[82 − x3
]dx = π
[64x− x4
4
]4
0
= 192π
SECTION 6.3
1. V =∫ 1
0
2πx [x− 0] dx = 2π∫ 1
0
x2 dx
= 2π[13x3
]1
0
=2π3
2. V =∫ 3
0
2πx(3 − x) dx = 2π∫ 3
0
(3x− x2) dx
= 2π[3x2
2− x3
3
]3
0
= 9π
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
SECTION 6.3 309
3. V =∫ 4
0
2πx[√
x − 0]dx = 2π
∫ 4
0
x3/2 dx
= 2π[25x
5/2]40
=1285
π
4. V =∫ 2
0
2πxx3 dx = 2π[x5
5
]2
0
=645π
5. V =∫ 1
0
2πx[√
x − x3]dx
= 2π∫ 1
0
(x3/2 − x4
)dx
= 2π[25x5/2 − 1
5x5
]1
0
=2π5
6. V =∫ 1
0
2πx[x1/3 − x2
]dx = 2π
∫ 1
0
(x4/3 − x3
)dx
= 2π[37x7/3 − x4
4
]1
0
=5π14
7. V =∫ 2
0
2πx [2x− x] dx +∫ 4
2
2πx [4 − x] dx
= 2π∫ 2
0
x2 dx + 2π∫ 4
2
(4x− x2
)dx
= 2π[13x
3]20
+ 2π[2x2 − 1
3x3]42
= 16π
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
310 SECTION 6.3
8. V =∫ 3
1
2πx(x− 1) dx +∫ 5
3
2πx(6 − x− 1) dx
= 2π∫ 3
1
(x2 − x) dx + 2π∫ 5
3
(5x− x2) dx
= 2π[x3
3− x2
2
]3
1
+ 2π[5x2
2− x3
3
]5
3
= 24π
9. V =∫ 1
0
2πx[(√
x)−
(−√x
)]dx +
∫ 4
1
2πx[(√
x)− (x− 2)
]dx
= 4π∫ 1
0
x3/2 dx + 2π∫ 4
1
(x3/2 − x2 + 2x
)dx
= 4π[25x
5/2]10
+ 2π[25x
5/2 − 13x
3 + x2]41
= 725 π
10. V =∫ 1
0
2πx · 2√x dx +
∫ 4
1
2πx(2 − x +√x) dx
= 4π∫ 1
0
x3/2 dx + 2π∫ 4
1
(2x− x2 + x3/2) dx
= 4π[25x5/2
]1
0
+ 2π[x2 − x3
3+
25x5/2
]4
1
=725π
11. V =∫ 3
0
2πx[√
9 − x2 −(−√
9 − x2)]
dx
= 4π∫ 3
0
x(9 − x2
)1/2dx
= 4π[− 1
3
(9 − x2
)3/2]3
0= 36π
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
SECTION 6.3 311
12. V =∫ 1
0
2πx · 2x dx +∫ 2
1
2πx · 2√
2 − x dx
= 4π∫ 1
0
x2 dx− 4π∫ 0
1
(2 − u)√u du (u = 2 − x)
= 4π[x3
3
]1
0
− 4π[43u3/2 − 2
5u5/2
]0
1
=7615
π
13. V =∫ 2
0
2πy [6 − 3y] dy
= 6π∫ 2
0
(2y − y2
)dy
= 6π[y2 − 1
3y3]20
= 8π
14. V =∫ 5
0
2πy · y dy = 2π[y3
3
]5
0
=2503
π
15. V =∫ 9
0
2πy [(√y ) − (−√
y )] dy
= 4π∫ 9
0
y3/2 dy
= 4π[25y
5/2]90
= 19445 π
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
312 SECTION 6.3
16. V =∫ 8
0
2πyy1/3 dy = 2π∫ 8
0
y4/3 dy
= 2π[37y7/3
]8
0
=7687
π
17. V =∫ 1
0
2πy[y1/3 − y2
]dy
= 2π∫ 1
0
(y4/3 − y3
)dy
= 2π[37y
7/3 − 14y
4]10
= 514π
18. V =∫ 1
0
2πy(√y − y3) dy = 2π
∫ 1
0
(y3/2 − y4) dy
= 2π[25y5/2 − y5
5
]1
0
=25π
19. V =∫ 1
0
2πy [(√y ) − (−√
y )] dy +∫ 4
1
2πy [(√y ) − (y − 2)] dy
= 4π∫ 1
0
y3/2 dy + 2π∫ 4
1
(y3/2 − y2 + 2y
)dy
= 4π[25y
5/2]10
+ 2π[25y
5/2 − 13y
3 + y2]41
= 725 π
20. V =∫ 1
0
2πy2√y dy +
∫ 4
1
2πy(2 − y +√y) dy
= 4π∫ 1
0
y3/2 dy + 2π∫ 4
1
(2y − y2 + y3/2) dy
= 4π[25y5/2
]1
0
+ 2π[y2 − y3
3+
25y5/2
]4
1
=725π
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
SECTION 6.3 313
21. V =∫ 4
0
2πy[y − y
2
]dy +
∫ 8
4
2πy[4 − y
2
]dy
= π
∫ 4
0
y2 dy + π
∫ 8
4
(8y − y2
)dy
= π[13y
3]40
+ π[4y2 − 1
3y3]84
= 64π
22. V =∫ 1
0
2πy(y) dy +∫ 7
1
2πy dy +∫ 8
7
2πy(8 − y) dy
= 2π∫ 1
0
y2dy + 2π∫ 7
1
y dy + 2π∫ 8
7
(8y − y2) dy
= 2π[y3
3
]1
0
+ 2π[y2
2
]7
1
+ 2π[4y2 − y3
3
]8
7
= 56π
23. V =∫ 1
0
2πy[√
1 − y2 − (1 − y)]dy
= 2π∫ 1
0
[y(1 − y2
)1/2 − y + y2]dy
= 2π[−1
3(1 − y2
)3/2 − 12y2 +
13y3
]1
0
=π
3
24. V =∫ 1
0
2πy2√y dy +
∫ 2
1
2πy2(2 − y) dy
= 4π∫ 1
0
y3/2 dy + 4π∫ 2
1
(2y − y2) dy
= 4π[25y5/2
]1
0
+ 4π[y2 − y3
3
]2
1
=6415
π
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
314 SECTION 6.3
25. (a) V =∫ 1
0
2πx[1 −
√x]dx (b) V =
∫ 1
0
π y4 dy
= π[15 y
5]10
= 15π
26. (a) V =∫ 1
0
π[(2 −
√x)2 − 12
]dx
(b) V =∫ 1
0
2π(2 − y)y2 dy = 2π∫ 1
0
(2y2 − y3) dy = 2π[23y3 − y4
4
]1
0
=56π
27. (a) V =∫ 1
0
π(x− x4
)dx
= π[12 x
2 − 15 x
5]10
=3π10
(b) V =∫ 1
0
2πy(√
y − y2)dy
28. (a) V =∫ 1
0
2π(x + 3)(√x− x2) dx = 2π
∫ 1
0
(x3/2 + 3
√x− 3x2 − x3
)dx
= 2π[25x5/2 + 2x3/2 − x3 − x4
4
]1
0
=2310
π
(b) V =∫ 1
0
π[(√y + 3)2 − (y2 + 3)2
]dy
29. (a) V =∫ 1
0
2πx · x2 dx = 2π∫ 1
0
x3 dx
= 2π[14x
4]10
=π
2
(b) V =∫ 1
0
π(1 − y) dy
30. (a) V =∫ 1
0
π[(x2 + 1)2 − 1
]dx = π
∫ 1
0
(x4 + 2x2) dx =1315
π
(b) V =∫ 1
0
2π(y + 1)(1 −√y) dy
31. V =∫ a
0
2πx
[2b
√1 − x2
a2
]dx =
4πba
∫ a
0
x(a2 − x2
)1/2dx =
4πba
[−1
3(a2 − x2
)3/2]a0
=43πa2b
32. V =∫ b
0
2πy2ab
√b2 − y2 dy =
4πab
∫ b
0
y(b2 − y2)1/2 dy
=4πab
[−1
3(b2 − y2)3/2
]b0
=43πab2
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
SECTION 6.3 315
33. By the shell method
V =∫ a/2
0
2πx(√
3x)dx +
∫ a
a/2
2πx[√
3 (a− x)]dx
= 2π√
3∫ a/2
0
x2 dx + 2π√
3∫ a
a/2
(ax− x2
)dx
= 2π√
3[13x3
]a/20
+ 2π√
3[a
2x2 − 1
3x3
]aa/2
=√
34
a3π
34. V =∫ √
r2−a2
0
2πx(√
r2 − x2 − a)dx = 2π
∫ √r2−a2
0
[x(r2 − x2)1/2 − ax
]dx
= 2π[−1
3(r2 − x2)3/2 − a
2x2
]√r2−a2
0
=13π(2r3 + a3 − 3ar2)
35. (a) V =∫ 8
0
2πy[4 − y2/3
]dy = 2π
∫ 8
0
(4y − y5/3
)dy = 2π
[2y2 − 3
8y8/3
]8
0
= 64π
(b) V =∫ 4
0
2π (4 − x)[x3/2
]dx = 2π
∫ 4
0
(4x3/2 − x5/2
)dx
= 2π[85x
5/2 − 27x
7/2]40
= 102435 π
(c) V =∫ 8
0
2π (8 − y)[4 − y2/3
]dy = 2π
∫ 8
0
(32 − 4y − 8y2/3 + y5/3
)dy
= 2π[32y − 2y2 − 24
5 y5/3 + 38y
8/3]80
= 7045 π
(d) V =∫ 4
0
2πx[x3/2
]dx = 2π
∫ 4
0
x5/2 dx = 2π[27x7/2
]4
0
=5127
π
36. (a) V =∫ 4
0
2πx(8 − x3/2) dx = 2π∫ 4
0
(8x− x5/2) dx = 2π[4x2 − 2
7x7/2
]4
0
=3847
π
(b) V =∫ 8
0
2π(8 − y)y2/3 dy = 2π∫ 8
0
(8y2/3 − y5/3) dy = 2π[245y5/3 − 3
8y8/3
]8
0
=5765
π
(c) V =∫ 4
0
2π(4 − x)(8 − x3/2) dx = 2π∫ 4
0
(32 − 8x− 4x3/2 + x5/2) dx
= 2π[32x− 4x2 − 8
5x5/2 +
27x7/2
]4
0
=345635
π
(d) V =∫ 8
0
2πy y2/3 dy = 2π∫ 8
0
y5/3 dy = 2π[38y8/3
]8
0
= 192π
37. (a) F ′(x) = sinx + x cosx− sinx = x cosx = f(x).
(b) V =∫ π/2
0
2πx · cosx dx = 2π [x sinx + cosx]π/20 = π2 − 2π
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
316 SECTION 6.3
38. (a) (b) By the shell method
V =∫ 2
0
2πx(x2 − 2x + 2) dx +∫ 4
2
2πx (x + 2 − 2x + 2) dx
= 2π∫ 2
0
(x3 − 2x2 + 2x) dx + 2π∫ 4
2
(4x− x2) dx
= 2π[x4
4− 2
3x3 + x2
]2
0
+ 2π[2x2 − x3
3
]4
2
= 16π
39. (a) V =∫ 1
0
2√
3πx2 dx +∫ 2
1
2πx√
4 − x2 dx (b) V =∫ √
3
0
π
[4 − 4
3y2
]dy
(c) V =∫ √
3
0
π
[4 − 4
3y2
]dy = π
[4y − 4
9y3
]√3
0
=8π
√3
3
40. (a) V =∫ 1
0
π(√
3x)2 dx +∫ 2
1
π(√
4 − x2)2
dx
(b) V =∫ √
3
0
2πy[√
4 − y2 − y√3
]dy
(c) use (a): V = 3π∫ 1
0
x2 dx + π
∫ 2
1
(4 − x2) dx = π[x3
]10
+ π
[4x− x3
3
]2
1
=83π
41. (a) V =∫ 1
0
2√
3πx(2 − x) dx +∫ 2
1
2π(2 − x)√
4 − x2 dx
(b) V =∫ √
3
0
π
[(2 − y√
3
)2
−(2 −
√4 − y2
)2]dy
42. (a) V =∫ 1
0
π[(√
3x + 1)2 − 12]dx +
∫ 2
1
π
[(√4 − x2 + 1
)2
− 12
]dx
(b) V =∫ √
3
0
2π(y + 1)(√
4 − y2 − y√3
)dy
43. (a) V = 2∫ b+a
b−a
2πx√
a2 − (x− b)2 dx
(b) V =∫ a
−a
π
[(b +
√a2 − y2
)2
−(b−
√a2 − y2
)2]dy
44. V =∫ a
−a
2π(a− x)2√a2 − x2 dx = 4πa
∫ a
−a
√a2 − x2 dx− 4π
∫ a
−a
x√a2 − x2 dx
= 4πa( Area of half circle ) − 4π[−1
3(a2 − x2)3/2
]a−a
= 4πa · πa2 − 0 = 4π2a3
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
SECTION 6.3 317
45. V =∫ r
0
2πx(h− h
rx) dx = 2πh
[x2
2− x3
3r
]r0
=πr2h
3.
46. V = 2∫ r
√r2−h2/4
2πx√r2 − x2 dx = −2π
∫ 0
h2/4
u12 du = 2π
[23u
32
]h2/4
0=
πh3
6.
∧u = r2 − x2, du = −2x dx
47. (a) V =∫ r
a
2πx(r2 − x2) dx = 2π∫ r
a
(r2x− x3) dx = 2π[
12 r
2x2 − 14x
4]ra
= 12π
(r2 − a2
)2
(b) V =∫ r2−a2
0
π
[(√r2 − y
)2
− a2
]dy = π
∫ r2−a2
0
(r2 − y − a2) dy = π[(r2 − a2)y − 1
2y2]r2−a2
0
= 12π(r2 − a2)2
48. (a)(b) V =
∫ 1
0
π sin2 πx2 dx ∼= 1.1873
(c) V =∫ 1
0
2πx sin πx2 dx =[− cos πx2
]1
0= 2−3 −2 −1 1 2 3
x
−1
1y
49. (a) (b) points of intersection: x = 0, x ∼= 1.8955
(c) A ∼=∫ 1.8955
0
(sinx− 1
2x)dx ∼= 0.4208
(d) V ∼=∫ 1.8955
0
2π(sinx− 1
2x)dx ∼= 2.6226
1 2 3x
1
y
50. (a)
1 2 3x
1
2
y
(b) first quadrant points of intersection: x ∼= 0.1939, x ∼= 2.7093
(c) A ∼=∫ 2.7093
0.1939
(32 − 1
2x− 2(x + 1)2
)dx ∼= 0.8114
(d) V ∼=∫ 2.7093
0.1939
2πx(
32 − 1
2x− 2(x + 1)2
)dx ∼= 6.4873
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
318 SECTION 6.4
SECTION 6.4
1. A =∫ 4
0
√x dx =
163
xA =∫ 4
0
x√x dx =
645, x =
125
yA =∫ 4
0
12(√
x)2
dx = 4, y =34
Vx = 2πyA = 8π, Vy = 2πxA =1285
π
2. A =∫ 2
0
x3 dx = 4
xA =∫ 2
0
xx3 dx =325, x =
85
yA =∫ 2
0
12(x3)2 dx =
647, y =
167
Vx = 2πyA =1287
π, Vy = 2πxA =645π
3. A =∫ 1
0
(x1/3 − x2
)dx =
512
xA =∫ 1
0
x(x1/3 − x2
)dx =
528
, x =37
yA =∫ 1
0
12
[(x1/3
)2
−(x2
)2]dx =
15, y =
1225
Vx = 2πyA = 25π, Vy = 2πxA = 5
14π
4. A =∫ 1
0
(√x− x3) dx =
512
xA =∫ 1
0
x(√x− x3) dx =
15, x =
1225
yA =∫ 1
0
12(x− x6) dx =
528
, y =37
Vx = 2πyA =514
π, Vy = 2πxA =25π
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
SECTION 6.4 319
5. A =∫ 3
1
(2x− 2) dx = 4
xA =∫ 3
1
x (2x− 2) dx =283, x =
73
yA =∫ 3
1
12
[(2x)2 − (2)2
]dx =
403, y =
103
Vx = 2πyA = 803 π, Vy = 2πxA = 56
3 π
6. A =12· 3 · 1 =
32
xA =∫ 2
1
x(6 − 3x) dx = 2, x =43
yA =∫ 2
1
12(62 − (3x)2
)dx =
152, y = 5
Vx = 2πyA = 15π, Vy = 2πxA = 4π
7. A =∫ 2
0
[6 −
(x2 + 2x
)]dx =
163
xA =∫ 2
0
x[6 −
(x2 + 2
)]dx = 4, x =
34
yA =∫ 2
0
12
[(6)2 −
(x2 + 2
)2]dx =
35215
, y =225
Vx = 2πyA = 70415 π, Vy = 2πxA = 8π
8. A =∫ 3
0
[(x2 + 1) − 1
]dx = 9
xA =∫ 3
0
x[(x2 + 1) − 1
]dx =
814, x =
94
yA =∫ 3
0
12[(x2 + 1)2 − 12
]dx =
99930
, y =11130
Vx = 2πyA =3335
π, Vy = 2πxA =812π
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
320 SECTION 6.4
9. A =∫ 1
0
[(1 − x) −
(1 −
√x)2
]dx =
13
xA =∫ 1
0
x[(1 − x) −
(1 −
√x
)2]dx =
215
, x =25
y = 25 by symmetry
Vx = 2πyA = 415π, Vy = 4
15π by symmetry
10. A =π
4− 1
2=
π − 24
xA =∫ 1
0
x(√
1 − x2 + x− 1)dx =
16, x =
23(π − 2)
y =2
3(π − 2)by symmetry
Vx = 2πyA =π
3, Vy = 2πxA =
π
3
11. A =∫ 2
1
x2 dx =73
xA =∫ 2
1
x(x2
)dx =
154, x =
4528
yA =∫ 2
1
12(x2
)2dx =
3110
, y =9370
Vx = 2πyA = 315 π, Vy = 2πxA = 15
2 π
12. A =∫ 8
1
(x1/3 − 1) dx =174
xA =∫ 8
1
x(x1/3 − 1) dx =32114
, x =642119
yA =∫ 8
1
12(x2/3 − 12) dx =
5810
, y =11685
Vx = 2πyA =585π, Vy = 2πxA =
3217
π
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
SECTION 6.4 321
13. A = 12bh = 4; by symmetry, x = 3
yA =∫ 3
1
y [(6 − y) − y] dy =203, y =
53
Vx = 2πyA = 403 π, Vy = 2πxA = 24π
14. A =92
xA =∫ 3
0
x(2x− x) dx = 9, x = 2
yA =∫ 3
0
12(4x2 − x2
)dx =
272, y = 3
Vx = 2πyA = 27π, Vy = 2πxA = 18π
15.(
52 , 5
)16. A =
∫ 3
−1
(4x− x2 − 2x + 3) dx =323
xA =∫ 3
−1
x(2x− x2 + 3) dx =323
=⇒ x = 1
yA =∫ 3
−1
12[(4x− x2)2 − (2x− 3)2
]dx =
325
=⇒ y =35
17.(1, 8
5
)18. A =
∫ 3
−1
(2x + 3 − x2) dx =323
xA =∫ 3
−1
x(2x + 3 − x2) dx =323
=⇒ x = 1
yA =∫ 3
−1
12[(2x + 3)2 − x4
]dx =
54415
=⇒ y =175
19.(
103 , 40
21
)
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
322 SECTION 6.4
20. A =∫ 2
0
(x− x2 +√
2x) dx = 2
xA =∫ 2
0
x(x− x2 +√
2x) dx =2815
=⇒ x =1415
yA =∫ 2
0
12[(x− x2)2 − 2x
]dx = −22
15=⇒ y = −11
15
21. (2, 4)
22. A =∫ 6
1
(6x− x2 − 6 + x) dx; xA =∫ 6
1
x(6x− x2 − 6 + x) dx;
yA =∫ 6
1
12[(6x− x2)2 − (6 − x)2
]dx. =⇒ x =
72, y = 5
23.(− 3
5 , 0)
24. A =∫ a
0
(√a−
√x)2 dx; xA =
∫ a
0
x(√a−
√x)2 dx;
yA =∫ a
0
12(√a−
√x)4 dx. =⇒ x =
a
5, y =
a
5
25. (a) (0, 0) by symmetry
(b) Ω1 smaller quarter disc, Ω2 the larger quarter disc
A1 =116
π, A2 = π; x1 = y1 =23π
, x2 = y2 =83π
(Example 1)
xA =(
83π
)(π) − 2
3π
(116
π
)=
6324
, A =1516
π
x =(
6324
)/(15π16
)=
145π
, y = x =145π
(symmetry)
(c) x = 0, y =145π
26. A =12πab; x = 0 by symmetry.
yA =∫ a
−a
12
(b
a
√a2 − x2
)2
dx =23ab2 =⇒ y =
4b3π
.
27. Use theorem of Pappus. Centroid of rectangle is located
c +
√(a
2
)2
+(b
2
)2
units
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
SECTION 6.4 323
from line l. The area of the rectangle is ab. Thus,
volume = 2π
⎡⎣c +
√(a
2
)2
+(b
2
)2⎤⎦ (ab) = πab
(2c +
√a2 + b2
).
28. (a) A1 = 12h · b
3 = 16hb; A2 = 1
2b · h3 = 1
6hb.
A3 = 12bh− (A1 + A2) = 1
2bh− 13bh = 1
6bh.
(b) Hypotenuse has equation hx + by − bh = 0, so distance from ( b3 ,
h3 ) to hypotenuse is
d =|hb3 + bh
3 − bh|√h2 + b2
=bh
3√h2 + b2
(c) V = 2πdA = 2πbh
3√h2 + b2
· 12bh =
πb2h2
3√h2 + b2
29. (a)(
23a,
13h
)(b)
(23a + 1
3b,13h
)(c)
(13a + 1
3b,13h
)
30. (a) V = 2πyA = 2π13h
12bh =
13πbh2 [using Exercise 29(c)]
(b) V = 2πxA = 2π13(a + b)
12bh =
13π(a + b)bh.
31. (a) V = 23πR
3 sin3 θ + 13πR
3 sin2 θ cos θ = 13πR
3 sin2 θ (2 sin θ + cos θ)
(b) x =V
2πA=
13πR
3 sin2 θ (2 sin θ + cos θ)2π
(12R
2 sin θ cos θ + 14πR
2 sin2 θ) =
2R sin θ (2 sin θ + cos θ)3 (π sin θ + 2 cos θ)
32. (a) x = 0, y = 0 (by symmetry).
(b) x = 0 (by symmetry about y-axis), y = r +4r3π
by Example 6.
(c) x = 0 (by symmetry about y-axis).
y =1A
∫ r
−r
12
[(r +
√r2 − x2
)2
− (−r)2]dx =
1A
[r
∫ r
−r
√r2 − x2 dx +
12
∫ r
−r
(r2 − x2) dx]
=1
πr2
2 + 4r2
[rπr2
2+
23r3
]=
r
3
(3π + 4π + 8
)
(d) as in (c): x = 0, (by symmetry about y-axis), y = − r3
(3π+4π+8
)(e) x = 0, y = 0 (by symmetry).
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
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324 SECTION 6.4
(f) A = πr2 + 4r2
xA =∫ r
−r
x(2r +
√r2 − x2
)dx +
∫ 2r
r
x 2√r2 − (x− r)2 dx = 0 + xΩ2AΩ2
=⇒ x = xΩ2
AΩ2
A=
(r +
4r3π
)πr2/2
πr2 + 4r2=
r
6
(3π + 4π + 4
)
By symmetry about the line y = x, y = x =r
6
(3π + 4π + 4
)
(g) A =32πr2 + 4r2
xA = xS∪Ω1∪Ω2 AS∪Ω1∪Ω2 =⇒ x =r
6
(3π + 4π + 4
)πr2 + 4r2
32πr
2 + 4r2=
r
3
(3π + 4π + 8
)
y = 0, (by symmetry about x-axis).
33. An annular region; see Exercise 25(a).
34. Extend the figure as indicated in the diagram below. Denote the area of Ω by A and the centroid
by (x, y). The centroid of the parallelogram is the center ([b + 2a]/2, h/2) and the area of the
parallelogram is 2A.
The upper triangle has centroid (b + x, h + y) and, by congruence, area A.
The union of the configurations is a triangle with centroid (2x, 2y) and area 4A. (The linear dimensions
have been doubled.) Therefore, by Principle 2,
xA +(b + 2a
2
)2A + (b + x)A = (2x)4A
yA +(h
2
)2A + (b + y)AT = (2y)4A
Solve these equations for x and y
and you will see that x = a+b3 , y = h
3 .
Now verify that these values of x and y satisfy the equations of the medians.
35. (a) A = 12 (b)
(1635 ,
1635
)(c) V = 16
35 π (d) V = 1635 π
36. (a) A = 43 (b)
(2, 23
5
)(c) V = 184
15 π (d) V = 163 π
37. (a) A = 2503 (b)
(− 9
8 ,29021
) ∼= (−1.125, 13.8095)
38. (a) A = 929 (b)
(− 6
115 ,12769
) ∼= (−0.0522, 1.8406)
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
SECTION 6.4 325
PROJECT 6.4
1. Let P = {x0, x1, . . . , xn} be a partition of [a, b]. P breaks up [a, b] into n subintervals [xi−1, xi]. Choose
x∗i as the midpoint of [xi−1, xi]. By revolving the ith midpoint rectangle about x-axis, we obtain a
solid cylinder of volume Vi = π [f (x∗i )]
2 Δxi and centroid (center) on the x-axis at x = x∗i . The union
of all these cylinders has centroid at x = xP where
xpVp = πx∗1 [f (x∗
1)]2 Δx1 + · · · + πx∗
n [f (x∗n)]2 Δxn
(Here VP represents the union of the n cylinders.) As ‖P‖ → 0, the union of the cylinders tends to the
shape of S and the equation just derived tends to the given formula.
2. Let P = {x0, x1, . . . , xn} be a partition of [a, b]. P breaks up [a, b] into n subintervals [xi−1, xi].
Choose x∗i as the midpoint of [xi−1, xi]. By revolving the ith midpoint rectangle about y-axis,
we obtain a solid cylinder of volume Vi = 2πx∗if (x∗
i ) Δxi and centroid (center) on the y-axis
at y = 12f (x∗
i ). The union of all these cylindrical shells has centroid at y = yP where
ypVp = πx∗1 [f (x∗
1)]2 Δx1 + · · · + πx∗
n [f (x∗n)]2 Δxn
(Here VP represents volume of the union of the n cylindrical shells.) As ‖P‖ → 0, the union
of the cylinders tends to the shape of S and the equation just derived tends to given formula.
3. (a) x
(13πr2h
)=
∫ h
0
πx( r
hx)2
dx =14πr2h2
x =(
14πr
2h2) /(
13πr
2h)
= 34h.
The centroid of the cone lies on the axis of the cone at a distance 34h from the vertex.
(b) The ball is obtained by rotating f(x) =√r2 − x2, x ∈ [−r, r], around the x-axis.
Vx =23πr3; xVx =
∫ r
−r
πx(r2 − x2) dx = π
[r2x
2
2− x4
4
]r−r
= 0
=⇒ x = 0; no surprise here.
(c) Vx =∫ a
0
πb2
a2
(a2 − x2
)dx =
23πab2, xVx =
∫ a
0
πxb2
a2
(a2 − x2
)dx =
14πa2b2
x =(
14πa
2b2) /(
23πab
2)
= 38a; centroid
(38a, 0
)(d) (i) Vx =
∫ 1
0
π(√
x)2
dx =12π, xVx =
∫ 1
0
πx(√
x)2
dx =13π
x =(
13π
) /(12π
)= 2
3 ; centroid(
23 , 0
)(ii) Vy =
∫ 1
0
2πx√x dx =
45π, yVy =
∫ 1
0
πx(√
x)2
dx =13π
y =(
13π
) /(45π
)= 5
12 ; centroid(0, 5
12
)
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
326 SECTION 6.5
(e) (i) Vx =∫ 2
0
π(4 − x2)2 dx =25615
π
xVx =∫ 2
0
πx(4 − x2)2 dx = π
[− (4 − x2)3
6
]2
0
=32π3
=⇒ x =58; y = 0
(ii) Vy =∫ 2
0
2πx(4 − x2) dx = 8π
yVy =∫ 2
0
πx(4 − x2)2 dx =323π =⇒ y =
43; x = 0
SECTION 6.5
1. W =∫ 4
1
x(x2 + 1
)2dx =
16
[(x2 + 1
)3]4
1= 817.5 ft-lb
2. W =∫ 8
3
2x√x + 1 dx =
∫ 9
4
2(u− 1)√u du = 2
[25u5/2 − 2
3u3/2
]9
4
=215215
ft-lb
3. W =∫ 3
1
x√x2 + 7 dx =
13
[(x2 + 7)
32
]3
0=
13(64 − 7
32 ) newton-meters
4. W =∫ π
4
0
(x2 + cos 2x) dx =[x3
3+
sin 2x2
]π4
0
= (π3
192+
12) newton-meters
5. W =∫ π
π/6
(x + sin 2x) dx =[12x2 − 1
2cos 2x
]ππ/6
=3572
π2 − 14
newton-meters
6. W =∫ π/2
0
cos 2x√2 + sin 2x
dx =[(2 + sin 2x)1/2
]π/20
= 0 newton-meters
7. By Hooke’s law, we have 600 = −k(−1). Therefore k = 600.
The work required to compress the spring to 5 inches is given by
W =∫ 5
10
600(x− 10) dx = 600[12x2 − 10x
]5
10
= 7500 in-lb, or 625 ft-lb
8. Work done by spring = −5 =∫ 3
1
−kx dx = −k
[x2
2
]3
1
= −4k =⇒ k =54.
We want s such that −6 =∫ s
0
−54x dx = −5
4s2
2=⇒ s =
4√
3√5
feet.
9. To counteract the restoring force of the spring we must apply a force F (x) = kx.
Since F (4) = 200, we see that k = 50 and therefore F (x) = 50x.
(a) W =∫ 1
0
50x dx = 25 ft-lb (b) W =∫ 3/2
0
50x dx =2254
ft-lb
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
SECTION 6.5 327
10. (a) W =∫ a
0
−kx dx = −k
2a2 =⇒
∫ 2a
0
−kx dx = −k
2(2a)2 = 4
(−k
2a2
)= 4W
(b)∫ na
0
−kx dx = −k
2n2a2 = n2W
(c)∫ 2a
a
−kx dx = −k
2(4a2 − a2) = 3
(−k
2a2
)= 3W
(d)∫ na
a
−kx dx =∫ na
0
−kx dx−∫ a
0
−kx dx = n2W −W = (n2 − 1)W
11. Let L be the natural length of the spring.∫ 2.1−L
2−L
kx dx =12
∫ 2.2−L
2.1−L
kx dx
[12kx
2]2.1−L
2−L= 1
2
[12kx
2]2.2−L
2.1−L
(2.1 − L)2 − (2 − L)2 = 12
[(2.2 − L)2 − (2.1 − L)2
].
Solve this equation for L and you will find that L = 1.95. Answer: 1.95 ft
12. (a) W =∫ b
a
σs(x)A(x) dx =∫ 6
0
62.5x 4π dx = 4, 500π ft-lb
(b) W =∫ 6
0
62.5 (x + 5) · 4π dx = 12, 000π ft-lb.
13. (a) W =∫ 3
0
(x + 3) (60) (8)(2√
9 − x2)dx
= 960∫ 3
0
x(9 − x2
)1/2dx
+2880∫ 3
0
√9 − x2 dx︸ ︷︷ ︸
area of quartercircle of radius 3
= 960[− 1
3
(9 − x2
)3/2]3
0+ 2880
[94π
]= (8640 + 6480π) ft-lb
(b) W =∫ 3
0
(x + 7)(60)(8)(2√
9 − x2)dx= 960
∫ 3
0
x(9 − x2
)1/2dx + 6720
∫ 3
0
√9 − x2 dx
= (8640 + 15120π) ft-lb
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
328 SECTION 6.5
14. (a) W =∫ 2
0
62.5(4 − x)(6)(2√
4 − (2 − x)2)dx +
∫ 4
2
62.5(4 − x)(6)(2√
4 − (x− 2)2)dx
750∫ 2
0
(4 − x)√
4 − (2 − x)2 dx = 750∫ 2
0
(2 + u)√
4 − u2 du = 750[83 + 2π
](u = 2 − x, du = −dx, 4 − x = 2 + (2 − x); u(0) = 2, u(2) = 0)
750∫ 4
2
(4 − x)√
4 − (x− 2)2 dx = 750∫ 4
2
(2 − u)√
4 − u2 du = 750[− 8
3 + 2π]
(u = x− 2, du = dx, 4 − x = 2 − (x− 2); u(2) = 0, u(4) = 2)
W = 750(4π) = 3000π ft-lb
(b) W =∫ 2
0
62.5(9 − x)(6)(2√
4 − (2 − x)2)dx +
∫ 4
2
62.5(9 − x)(6)(2√
4 − (x− 2)2)dx
= 10, 500π ft-lb
15. Set w(x) = the weight of the chain from height x to the ground. (Weight measured in pounds, height
measured in feet.) Then w(x) = 1.5x and
W =∫ 50
0
1.5x dx = 1.5[x2
2
]50
0
= 1875 ft-lb
16. The lifting force also acts in the negative direction.
17. By similar triangles
h
r=
h− x
yso that y =
r
h(h− x).
Thus, the area of a cross section of the
fluid at a depth of x feet is
πy2 = πr2
h2(h− x)2.
(a) W =∫ h/2
0
xσ
[πr2
h2(h− x)2
]dx =
σπr2
h2
∫ h/2
0
(h2x− 2hx2 + x3
)dx =
11192
σπr2h2 ft-lb
(b) W =∫ h/2
0
(x + k)σ[πr2
h2(h− x)2
]dx =
11192
πr2h2σ +724
πr2hkσ ft-lb
18. W =∫ h
0
σ(h− x)πr2
h2(h− x)2 dx = σπ
r2
h2
[− (h− x)4
4
]h0
=14σπr2h2 ft-lb.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
SECTION 6.5 329
19. y =34x2, 0 ≤ x ≤ 4
(a) W =∫ 12
0
σ(12 − y)πx2 dy =43πσ
∫ 12
0
(12y − y2) dy
=43πσ
[6y2 − y3
3
]12
0
=43πσ(288) = 384πσ newton-meters.
(b) W =∫ 12
0
σ(13 − y)πx2 dy =43πσ
∫ 12
0
(13y − y2) dy
=43πσ
[13y2
2− y3
3
]12
0
=43πσ(360) = 480πσ newton-meters.
20. W =∫ r2
r1
F dr =∫ r2
r1
−GmM
r2dr =
[GmM
r
]r2r1
= GmM
[1r2
− 1r1
]
21. W =∫ 80
0
(80 − x)15 dx = 15[80x− 1
2x2
]80
0
= 48, 000 ft-lb
22. (a) W = wd ft-lb
(b) component of force along the inclined plane=w sin θ lb
distance traveled=d
sin θft; W = (w sin θ)
d
sin θ= wd ft-lb.
23. (a) W = 200 · 100 = 20, 000 ft-lb (b) W =∫ 100
0
[(100 − x)2 + 200] dx
=∫ 100
0
(400 − 2x) dx
=[400x− x2
]1000
= 30, 000 ft-lb
24. (a) Loses 50 pounds over 100 feet, so weight x feet from top is
W (x) = 200 − 50100
(100 − x) = 150 +x
2
Thus, work =∫ 100
0
(150 +
x
2
)dx = 17, 500 ft-lb
(b) Just add the work done lifting the chain, namely∫ 100
0
2x dx = 10, 000 ft-lb.
Total work =27, 500 ft-lb
25. The bag is raised 8 feet and loses a total of 1 pound at a constant rate. Thus, the bag loses sand at
the rate of 1/8 lb/ft. After the bag has been raised x feet it weighs 100 − x
8pounds.
W =∫ 8
0
(100 − x
8
)dx =
[100x− x2
16
]8
0
= 796 ft-lb.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
330 SECTION 6.5
26. Weight at depth x is 40 − 120
(8.3)(40 − x) = 0.415x + 23.4.
Thus W =∫ 40
0
(0.415x + 23.4) dx =[0.415
x2
2+ 23.4x
]40
0
= 1268 ft-lb.
27. (a) W =∫ l
0
xσ dx =12σl2 ft-lb
(b) W =∫ l
0
(x + l)σ dx =32σl2 ft-lb
28. Work = wh +∫ h
0
σx dx =(wh +
12σh2
)ft-lb.
29. Thirty feet of cable and the steel beam weighing a total of: 800 + 30(6) = 980 lbs. are raised 20 feet.
The work required is: (20)(980) = 19, 600 ft-lb.
Next, the remaining 20 feet of cable is raised a varying distance and wound onto the steel drum. Thus
the total work is given by
W = 19, 600 +∫ 20
0
6x dx = 19, 600 + 1, 200 = 20, 800 ft-lb.
30. Reaches height x at time t = x/n, and at that time weighs w − 8.3pt = w − 8.3px/n pounds,
Therefore, work =∫ m
0
(w − 8.3p
x
n
)dx =
(wm− 4.15pm2
n
)ft-lb.
31. Let λ(x) be the mass density of the chain at the point x units above the ground. Let g be the
gravitational constant. the work done to pull the chain to the top of the building is given by:
W =∫ H
0
(H − x) g λ(x) dx = Hg
∫ H
0
λ(x) dx− g
∫ H
0
xλ(x) dx
= HgM − g xM = (H − x) gM
= (weight of chain) × (distance from the center of mass to top of building)
32. By the hint
W =∫ b
a
F (x) dx =∫ b
a
madx =∫ b
a
mvdv
dxdx =
∫ vb
va
mv dv =[12mv2
]vb
va
=12mv2
b −12mv2
a
33. The height of the object at time t is y(t) = − 12gt
2 + h; time at impact is t =√
2h/g; velocity at
time t is v(t) = −gt; velocity at impact is v(√
2h/g)
= −g√
2h/g = −√2gh.
34. If the speed of the second object is 3 times the speed of the first, then the mass of the first object is
9 times the mass of the second.
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JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
SECTION 6.5 331
35. Assume that va = 0 and vb = 95 mph.
mass of ball:5/1632
=1
512; speed in feet per second:
(95)(5280)3600
W =12
1512
[(95)(5280)
3600
]2
∼= 18.96 ft-lbs.
36. Mass of the vehicle:200032
= 62.5;
speed in ft/sec: 30 mph = 44 ft/sec; 55 mph ∼= 80.67 ft/sec.
W = 12 (62.5)(80.67)2 − 1
2 (62.5)(44)2 ∼= 142, 864 ft-lbs.
37. Assume that va = 0 and vb = 17, 000 mph.
mass of satellite:100032
= 31.25; speed in feet per second:(17, 000)(5280)
3600
W =12
(31.25)[(17, 000)(5280)
3600
]2
∼= 9.714 × 109 ft-lbs.
38. (a) Acceleration: a =8815
feet/sec2, Force: F = ma =w
ga =
300032
· 8815
= 550 lbs
v(t) = at, so p =dW
dt= F (x(t)) v(t) = 550 · 88
15t
The engine must be able to sustain this until t = 15,
so need p = 550(88) ft-lb/s = 88 horse power.
(b) Now there is a4√
10016· 3000 component of gravity acting against the motion, so the total force
needed is 550 +4 · 3000√
10016∼= 670 lbs, so the required power is:
p = 670(88) ft-lb/sec = 107.2 horse power.
39. (a) The work required to pump the water out of the tank is given by
W =∫ 10
5
(62.5)π 52x dx = 1562.5π[12x2
]10
5
∼= 184, 078 ft-lb
A 12 -horsepower pump can do 275 ft-lb of work per second. Therefore it will take
184, 078275
∼= 669 seconds ∼= 11 min, 10 sec, to empty the tank.
(b) The work required to pump the water to a point 5 feet above the top of the tank is given by
W =∫ 10
5
(62.5)π 52(x + 5) dx =∫ 10
5
(62.5)π 52x dx +∫ 10
5
(62.5)π 53 dx ∼= 306796 ft-lb
It will take a 12 -horsepower pump approximately 1, 116 sec, or 18 min, 36 sec, in this case.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
332 SECTION 6.6
40. (a)W =
∫ 8
0
60x16π dx +∫ 4
0
60(x + 8)π(16 − x2) dx
= 960π[x2
2
]8
0
+ 60π∫ 4
0
(128 + 16x− 8x2 − x3) dx
= 30, 720π + 60π[128x + 8x2 − 8x3
3− x4
4
]4
0
= 30, 720π + 24, 320π
= 55, 040π ∼= 172, 913 ft-lbs.
(b) Pump does12(550) = 275 ft-lbs/sec. Therefore, it will take
172, 913275
∼= 629 seconds, or 10.5
minutes.
41. KE = 12mv2;
dKE
dt= mv
dv
dt= mav = Fv = P .
SECTION 6.6
1. F =∫ 6
0
(62.5) · x · 8 dx = 250[x2
]60
= 9,000 lbs
2. F =∫ 7
1
62.5 · x · 6 dx = (62.5)(3)[x2
]71
= 9,000 lbs
3. The width of the plate x meters below the surface is given by w(x) = 60 + 2(20 − x) = 100 − 2x
(see the figure). The force against the dam is
F =∫ 20
0
9800x(100 − 2x) dx
= 9800∫ 20
0
(100x− 2x2) dx
= 9800[50x2 − 2
3x3]200
∼= 1.437 × 108 newtons
4. F =∫ 20
15
9800 · x · 5 dx = (4900)(5)[x2
]2015
= 4, 287, 500 N
5. The width of the gate x meters below its top is given by w(x) = 4 + 23x (see the figure).
The force of the water against the gate is
F =∫ 3
0
9800(10 + x)(
4 +23x
)dx
= 9800∫ 3
0
[23x2 +
323x + 40
]dx
= 9800[29x
3 + 163 x2 + 40x
]30
∼= 1.7052 × 106 Newtons
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
SECTION 6.6 333
6. (a) F =∫ 75
0
62.5x · 1000 dx = (62.5)(1000)(
752
2
)= 175, 781, 250 lbs.
(b) F =∫ 50
0
62.5x · 1000 dx = (62.5)(1000)(
502
2
)= 78, 125, 000 lbs.
7. F =∫ 3
0
(60)x
[12
√1 − x2
9
]dx
= 240∫ 3
0
x(9 − x2
)1/2dx
= 240[− 1
3
(9 − x2
)3/2]3
0= 2160 lb
ellipse:x2
32+
y2
62= 1
8. F =∫ 16
0
σxw(x) dx =∫ 16
0
70x2√
16 − x dx = 140∫ 16
0
x√
16 − x dx
= −140∫ 0
16
(16 − u)√u du = −40
[323u3/2 − 2
5u5/2
]0
16
=114, 688
3lb
9. By similar triangles
4√
24√
2=
y
4√
2 − xso y = 4
√2 − x.
F =∫ 4
√2
0
(62.5)x[2(4√
2 − x)]
dx
= 125∫ 4
√2
0
(4√
2 x− x2)dx =
80003
√2 lb
10. By similar triangles,W (x)
5=
5 − x
5=⇒ W (x) = 5 − x.
F =∫ 5
0
62.5x(5 − x) dx = 62.5[5x2
2− x3
3
]5
0
=15, 625
12lb.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
334 SECTION 6.6
11. F =∫ 2
0
(62.5) · x · 2√
4 − x2 dx
= 125∫ 2
0
x√
4 − x2 dx
= − 1253
[(4 − x2
)3/2]2
0= 333.33 lb
12. F =∫ 4
0
62.5y 2√
4 − y dy
13. F =∫ 4
0
60x(2√
16 − x2)dx
= 120∫ 4
0
x(16 − x2
)1/2dx
= 120[− 1
3
(16 − x2
)3/2]4
0= 2560 lb
14. F =∫ 8
0
60x 2√
16 − (x− 4)2 dx = 120∫ 4
−4
(u + 4)√
16 − u2 du
= 120∫ 4
−4
u√
16 − u2 du + 120 · 4∫ 4
−4
√16 − u2 du
= 0 + 120 · 4 [Area of half circle]=480 · 12· 16π = 3840π lb.
15. (a) The width of the plate is 10 feet and the depth of the plate ranges from 8 feet to 14 feet. Thus
F =∫ 14
8
62.5x (10) dx = 41, 250 lb.
(b) The width of the plate is 6 feet and the depth of the plate ranges from 6 feet to 16 feet. Thus
F =∫ 16
6
62.5x (6) dx = 41, 250 lb.
16. W (x) = 2πr = 30π. F =∫ 50
0
60x30π dx = 900π502 = 2, 250, 000π lb.
17. (a) Force on the sides:
F =∫ 1
0
(9800)x 14 dx +∫ 2
0
(9800)(1 + x) 7(2 − x) dx
= 68, 600[x2
]10
+ 68, 600∫ 2
0
[2 + x− x2
]dx
= 68, 600 + 68, 600[2x + 1
2x2 − 1
3x3]20
∼= 297, 267 Newtons
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
REVIEW EXERCISES 335
(b) Force at the shallow end:
F =∫ 1
0
(9800) · x · 8 dx = 39, 200[x2
]10
= 39, 200 Newtons
Force at the deep end:
F =∫ 3
0
(9800) · x · 8 dx = 39, 200[x2
]30
= 352, 800 Newtons
18. F =∫ b
a
σxw(x) dx = σ
∫ b
a
xw(x) dx = σxA
where A is the area of the submerged surface and x is the depth of the centroid.
19. From Exercise 18, F1 = σx1A1 = σh1A1, F2 = σh2A2. Since A1 = A2, F2 =h2
h1F1
20. Following the argument given for a vertical plate, the approximate force on the i th strip is
σx∗iw(xi) sec θΔxi
Therefore, the force F on the plate is given by
F =∫ b
a
σxw(x) sec θ dx
21. F =∫ 14
0
(9800)(
1 +17x
)· 85
√2
7dx = 392, 000
√2
7
[x +
114
x2
]14
0
∼= 2.217 × 106 Newtons
22. (a) F =∫ 100
0
62.5x(1000) sec(π/6) dx (b) F =∫ 75
0
62.5x(1000) sec(π/6) dx
=125, 000√
3
[x2
2
]100
0
∼= 361, 000, 000 lbs =125, 000√
3
[x2
2
]75
0
∼= 203, 000, 000 lbs
REVIEW EXERCISES
1. A =∫ 2
−1
[2 − x2 + x] dx
=∫ 2
1
2√
2 − y dy +∫ 1
−2
[√2 − y + y
]dy
A =∫ 2
−1
[2 − x2 + x] dx =92
−1 1 2x
−2
−1
1
2
y
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JWDD027-06 JWDD027-Salas-v1 November 25, 2006 15:39
336 REVIEW EXERCISES
2. A =∫ 1
0
[y1/3 + y
]dy =
∫ 0
−1
(1 + x) dx +∫ 1
0
(1 − x3) dx
A =∫ 1
0
[y1/3 + y] dy = 5/4
−1 1x
1
y
3. A =∫ 4
−2
[4 + y − y2
2
]dy
=∫ 3
1
2√
2(x− 1) dx +∫ 9
3
[√2(x− 1) − x + 5
]dx
A =∫ 4
−2
[4 + y − y2
2
]dy = 18
1 3 5 7 9x
−2
−1
1
2
3
4
y
4. A =∫ 8
−1
[x + 4
3− x2/3
]dx
=∫ 1
0
2y3/2 dy +∫ 4
1
[y3/2 − 3y + 4
]dy
A =∫ 8
−1
[x + 4
3− x2/3
]dx = 27/10
1 3 5 7x
−2
−1
1
2
3
4
y
5. Consecutive intersections of sinx and cosx occur at x = π4 , x = 5π
4 .
A =∫ 5π/4
π/4
(sinx− cosx) dx =[− cosx− sinx
]5π/4
π/4= 2
√2
6. A =∫ π/4
0
tan2 xdx =∫ π/4
0
(sec2 x− 1) dx =[tanx− x
]π/40
= 1 − π/4
7. By symmetry, A = 2∫ 1
0
(1 − x)√x dx = 2
∫ 1
0
(x1/2 − x3/2
)dx = 2
[23x
3/2 − 25x
5/2]1
0= 8
15
8. A =∫ a
0
(a + x− 2√ax) dx =
[ax + 1
2x2 − 4
3a1/2x3/2
]a0
= 16a
2
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REVIEW EXERCISES 337
9. (a) V =∫ r
−r
A(x)dx =∫ r
−r
12π(√
r2 − x2)2
dx =∫ r
−r
π
2(r2 − x2)dx =
π
2
[r2x− 1
3πx3
]r−r
=23πr3
(b) V =∫ r
−r
A(x) dx =∫ r
−r
12(2)
√r2 − x2
√r2 − x2 dx =
∫ r
−r
(r2 − x2) dx =[r2x− 1
3x3
]r−r
=43r3
10. V =∫ a
√3/2
0
43x2dx =
√3
6a3
11. V =∫ 3
0
12π
(3 − x
3
)2
dx =12π
∫ 3
0
(1 − 2
3x +
19x2
)dx =
12π
[x− 1
3x2 +
127
x3
]3
0
=12π
12. V =∫ 3
0
2√3x√
9 − x2 dx = − 23√
3
[(9 − x2)3/2
]3
0= 2
13. V =∫ 2
0
π[(x/2)2 − (x2/4)2
]dx = π
∫ 2
0
(x2
4− x4
16
)dx = π
[x3
12− x5
80
]2
0
=4π15
14. V =∫ 1
0
π[4y − 4y2] dy =2π3
or V =∫ 2
0
2πx(x
2− x2
4
)dx =
2π3
15. V =∫ 1
0
π[(1)2 − (x3)2]dx = π
∫ 1
0
(1 − x6) dx =6π7
16. V =∫ 1
0
π(y1/3
)2
dy = π
∫ 2
0
y2/3 dy =3π5
17. V =∫ π/4
0
π sec2 x dx = π[tanx
]π/40
= π
18. V =∫ π/2
−π/2
π cos2 x dx =π
2
∫ π/2
−π/2
[1 + cos 2x] dx =π
2
[x + 1
2 sin 2x]π/2−π/2
=π2
2
19. V =∫ √
π
0
2πx sinx2 dx = π
∫ √π
0
2x sinx2 dx = π[− cosx2
]√π
0= 2π
20. V =∫ √
π/2
0
2πx cosx2 dx = π
∫ √π/2
0
2x cosx2 dx = π[sinx2
]√π/2
0= π
21. By symmetry, V = 2∫ 3
0
2πx(3x− x2) dx = 4π∫ 3
0
(3x2 − x3) dx = 4π[x3 − 1
4x4]3
0= 27π
22. By symmetry, V = 2∫ 3
0
2π(4 − x)(3x− x2) dx = 4π∫ 3
0
(12x− 7x2 + x3) dx
= 4π[6x2 − 7
3x3 + 1
4x4]3
0= 45π
23. V =∫ 3
0
π[(x + 1)2 − (x− 1)4] dx = π[
13 (x + 1)3 − 1
5 (x− 1)5]3
0= 72
5 π
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338 REVIEW EXERCISES
24. V =∫ 5
0
2πx[3x− (x2 − 2x)] dx = 2π∫ 5
0
(5x2 − x3) dx = 2π[
53x
3 − 14x
4]5
0= 625
6 π
25. With respect to x: V =∫ 4
0
π[(22 − (
√x)2
]dx = π
∫ 4
0
(4 − x) dx;
with respect to y: V =∫ 2
0
2πy(y2) dy = 2π∫ 2
0
y3 dy = 8π
26. With respect to x: V =∫ 4
0
π(2 −
√x)2
dx;
with respect to y: V =∫ 2
0
2π(2 − y)y2 dy = 2π∫ 2
0
(2y2 − y3) dy = 83π
27. With respect to x: V =∫ 4
0
2π(x + 1)(√
x− 12x
)dx = 2π
∫ 4
0
(x3/2 + x1/2 − 1
2x2 − 1
2x)dx = 104
15 π
with respect to y: V =∫ 2
0
π{[2y + 1]2 − [y2 + 1]2} dy
28. With respect to x: V =∫ 4
0
2πx(√x− 1
2x) dx
with respect to y: V =∫ 2
0
π[(2y)2 − (y2)2]dy = π
∫ 2
0
(4y2 − y4) dy = 6415π
29. With respect to x: V =∫ 4
0
2πx(12x) dx = π
∫ 4
0
x2 dx = 643 π;
with respect to y: V =∫ 2
0
π[42 − (2y)2] dy
30. With respect to x: V =∫ 4
0
π[(
12x + 2
)2 − (2)2]dx;
with respect to y: V =∫ 2
0
2π(y + 2)(4 − 2y) dy = 2π∫ 2
0
(8 − 2y2) dy = 643 π
31. Since the region is symmetric about the y-axis, x = 0;
A =∫ 2
−2
(4 − x2
)dx =
323
;
yA =∫ 2
−2
12(4 − x2
)2dx =
12
∫ 2
−2
(16 − 8x2 + x4
)dx =
25615
; y =85
32. By symmetry, the centroid of the region is (0, 0). The centroid of the first quadrant part is:
A =∫ 2
0
(4x− x3)dx =[2x2 − 1
4x4
]2
0= 4
xA =∫ 2
0
x(4x− x3)dx =6415
=⇒ x =1615
, yA =∫ 2
0
12
[(4x)2 − (x3)2
]dx =
25621
=⇒ y =6421
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REVIEW EXERCISES 339
33. A =∫ 2
−1
[(2x− x2
)−
(x2 − 4
)]dx =
∫ 2
−1
(2x− 2x2 + 4
)dx = 9;
xA =∫ 2
−1
x(2x− 2x2 + 4
)dx =
∫ 2
−1
(2x2 − 2x3 + 4x
)dx =
92;
yA =∫ 2
−1
12[(
2x− x2)−
(x2 − 4
)]dx =
12
∫ 2
−1
[12x2 − 4x3 − 16
]dx = −27
2;
x =12; y = −3
2
34. Since the region is symmetric about the y axis, x = 0
A =∫ π
2
−π2
cosxdx = 2
yA =∫ π
2
−π2
12
cos2 xdx =14
∫ π2
−π2
(cos 2x + 1)dx =π
4=⇒ y =
π
8
35. A =∫ 1
0
[(2 − x2) − x
]dx =
∫ 1
0
(2 − x2 − x) dx =76;
xA =∫ 1
0
x(2 − x2 − x)dx =∫ 1
0
(2x− x3 − x2
)dx =
512
; x =514
yA =∫ 1
0
12[(2 − x2)2 − x2
]dx =
12
∫ 1
0
[4 − 5x2 + x4
]dx =
1915
; y =3835
around x-axis: V = 2π(
3835
)(76
)=
38π15
around y-axis: V = 2π(
514
)(76
)=
5π6
36. Since the region is symmetric about the line y = x, x = y
A =∫ 1
0
(x13 − x3)dx =
12
xA =∫ 1
0
x(x13 − x3)dx =
835
x = y =1635
Vy = Vx =∫ 1
0
π(x2/3 − x6)dx =16π35
37. W =∫ 3
0
F (x)dx =∫ 3
0
x√
7 + x2 dx =[
13 (7 + x2)3/2
]3
0= 1
3 (64 − 73/2) ft-lbs
38. F (x) = −kx; 8000 = −k(−1
2
)=⇒ k = 16,000
W =∫ −3
0
(−16,000 x) dx = 16,000∫ 0
−3
x dx = 72,000 in-lbs = 6,000 ft-lbs
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340 REVIEW EXERCISES
39. Assume the natural length is x0. Then,∫ 10
9
k(x− x0) dx =32
∫ 9
8
k(x− x0) dx
The solution of this equation is x0 = 132 inches.
40. (a)
W =∫ 5
0
σ(10 − x)π(
25x
)2
dx =4πσ25
∫ 5
0
(10x2 − x3
)dx
=4πσ25
[103x3 − 1
4x4
]5
0=
125πσ3
∼= 8181.23 ft-lbs
(b) W =∫ 5
0
σ(9 − x)π(
25x
)2
dx = 35πσ ∼= 6872.23 ft-lbs
41. W =∫ 25
0
4(25 − x) dx =[100x− 2x2
]250
= 1250 ft-lbs
42. W =∫ 20
0
(5 + 60 − x) dx =∫ 20
0
(65 − x) dx =[65x− 1
2x2]20
0= 1100 ft-lbs.
43. W =∫ 10
0
60(20 − x)π(20x− x2) dx = 60π∫ 10
0
(400x− 40x2 + x3) dx
= 60π[200x2 − 40
3 x3 + 14x
4]10
0= 550, 000π ft-lbs.
44. (a) The force on the 1 × 1/2 side is: F =∫ 1/2
0
9800x dx = 1225 newtons.
The force on the 1/2 × 1/2 side is: F =∫ 1/2
0
9800x( 12 ) dx = 612.5 newtons.
(b) The force on the bottom is: F = 9800 × 12 × 1
2 × 1 = 2450 newtons.
45. (a) F =∫ 50
0
9800x(100 − 2x) dx = 9800∫ 50
0
(100x− 2x2) dx
= 9800[50x2 − 2
3x3]50
0= 1225
3 × 106 newtons.
(b) F =∫ 50
10
9800x(100 − 2x) dx =10976
3× 105newtons.