Calculus one and several variables 10E Salas solutions manual ch06

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



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



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



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



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



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



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.



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 → ∞.



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



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



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



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)



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



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.)



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



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



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).



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.



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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



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

)



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



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).



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)



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

)



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



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



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.



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.



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.



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.



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



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.



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



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



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




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




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


0

π(2 −

√x)2

dx;


0

2π(2 − y)y2 dy = 2π∫ 2

0

(2y2 − y3) dy = 83π


0

2π(x + 1)(√

x− 12x

)dx = 2π

∫ 4

0

(x3/2 + x1/2 − 1

2x2 − 1

2x)dx = 104

15 π


0

π{[2y + 1]2 − [y2 + 1]2} dy


0

2πx(√x− 1

2x) dx


0

π[(2y)2 − (y2)2]dy = π

∫ 2

0

(4y2 − y4) dy = 6415π


0

2πx(12x) dx = π

∫ 4

0

x2 dx = 643 π;


0

π[42 − (2y)2] dy


0

π[(

12x + 2

)2 − (2)2]dx;


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




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




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.

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Calculus one and several variables 10E Salas solutions manual ch06