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Copyright 2014 Pearson Education, Inc. 431

CHAPTER 6 APPLICATIONS OF DEFINITE INTEGRALS

6.1 VOLUMES USING CROSS-SECTIONS

1. ( )( )22(diagonal)

2 2( ) 2 ;

x xA x x

− −= = =

44 20 0

0, 4; ( ) 2 16b

aa b V A x dx x dx x = = = = = =

2. ( ) ( ) ( )

2 22 2 2

2 2 2 1(diameter) 2 44 4 4

( ) 1 2 ;x x x

A x x xπ ππ π − − − = = = = − + 1, 1;a b= − =

( ) ( )5 11 2 4 3 162 2 13 5 3 5 151 1( ) 1 2 2 1b xa

V A x dx x x dx x x ππ π π− −

= = − + = − + = − + =

3. ( )2 2

2 2 2 2 2( ) (edge) 1 1 2 1 4 1 ;A x x x x x = = − − − − = − = − 1, 1;a b= − =

( ) ( )3 11 2 1613 3 31 1( ) 4 1 4 8 1b xa

V A x dx x dx x− −

= = − = − = − =

4. ( ) ( ) ( )

2 22 2 2

2 1 1 2 1(diagonal) 22 2 2

( ) 2 2 1 ;x x x

A x x

− − − − − = = = = − 1, 1;a b= − =

( ) ( )3 11 2 813 3 31 1( ) 2 1 2 4 1b xa

V A x dx x dx x− −

= = − = − = − =

5. (a) STEP 1) ( ) ( ) ( )( )1 12 3 2 3( ) (side) (side) sin 2 sin 2 sin sin 3 sinA x x x xπ π= ⋅ ⋅ = ⋅ ⋅ = STEP 2) 0,a b π= =

STEP 3) 0 0

( ) 3 sin 3 cos 3(1 1) 2 3b

aV A x dx x dx x

ππ = = = − = + = (b) STEP 1) ( ) ( )2( ) (side) 2 sin 2 sin 4 sin xA x x x= = = STEP 2) 0,a b π= =

STEP 3) [ ]00( ) 4 sin 4cos 8b

aV A x dx x dx x

π π= = = − =

6. (a) STEP 1) ( )2(diameter) 2 2 24 4 4( ) (sec tan ) sec tan 2sec tanA x x x x x x xπ π π= = − = + − ( ) 22 2 sin4 cossec sec 1 2 xxx xπ = + − −

STEP 2) 3 3

,a bπ π= − =

STEP 3) ( ) ( )( ) ( ) ( )

2

1 12 2

/3/3 2 2sin 14 4 cos/3 cos /3

21 14 3 3 4 3

( ) 2sec 1 2 tan 2

2 3 2 2 3 2 4 3

b xxa x

V A x dx x dx x xππ π π

π π

π π π π π

− − = = − − = − + −

= − + − − − + + − = −

432 Chapter 6 Applications of Definite Integrals

Copyright 2014 Pearson Education, Inc.

(b) STEP 1) ( )22 2 2 sincos( ) (edge) (sec tan ) 2sec 1 2 xxA x x x x= = − = − − STEP 2)

3 3,a bπ π= − =

STEP 3) ( ) ( )2/3 2 2sin 23 3/3 cos( ) 2sec 1 2 2 3 4 3b xa xV A x dx x dxπ π ππ−= = − − = − = − 7. (a) STEP 1) ( ) (length) (height) (6 3 ) (10) 60 30A x x x= ⋅ = − ⋅ = − STEP 2) 0, 2a b= =

STEP 3) 22 2

0 0( ) (60 30 ) 60 15 (120 60) 0 60

b

aV A x dx x dx x x = = − = − = − − =

(b) STEP 1) ( )20 2(6 3 ) 22( ) (length) (height) (6 3 ) (6 3 )(4 3 ) 24 6 9xA x x x x x x− −= ⋅ = − ⋅ = − + = + − STEP 2) 0, 2a b= =

STEP 3) ( ) 22 2 2 30 0( ) 24 6 9 24 3 3 (48 12 24) 0 36b

aV A x dx x x dx x x x = = + + = + − = + − − =

8. (a) STEP 1) ( )12 2( ) (base) (height) (6) 6 3xA x x x x= ⋅ = − ⋅ = − STEP 2) 0, 4a b= =

STEP 3) ( ) 44 1/2 3/2 2320 0( ) 6 3 4 (32 24) 0 8b

aV A x dx x x dx x x = = − = − = − − =

(b) STEP 1) ( ) ( )3/2 21

2 4

22 3/2 2diameter1 1 1

2 2 2 2 2 4 8 4( )

xx x x xA x x x xπ ππ π

− − + = ⋅ = ⋅ = ⋅ = − +

STEP 2) 0, 4a b= =

STEP 3) ( ) ( )44 3/2 2 2 5/2 3 64 161 1 2 18 4 2 5 12 8 5 3 8 150 0( ) 8 (0)b

aV A x dx x x x dx x x xπ π π π = = − + = − + = − + − =

9. ( )22 2 454 4 4( ) (diameter) 5 0 ;A y y yπ π π= = − = 0, 2; ( )

d

cc d V A y dy= = =

( ) ( )52

2 4 55 54 4 5 40

02 0 8yy dyπ π π π = = = − =

10. ( )2 2

2 2 2 21 1 12 2 2

( ) (leg)(leg) 1 1 2 1 2 1 ;A y y y y y = = − − − − = − = − 1, 1;c d= − =

( ) ( )311 2 81

3 3 31 1( ) 2 1 2 4 1

d y

cV A y dy y dy y

− −

= = − = − = − =

Section 6.1 Volumes Using Cross-Sections 433


11. The slices perpendicular to the edge labeled 5 are triangles, and by similar triangles we have 343 4

.bh

h b= =

The equation of the line through (5, 0) and (0, 4) is 45

4,y x= − + thus the length of the base 45

4x= − + and

the height ( )3 344 5 54 3.x x= − + = − + Thus ( ) ( )31 1 42 2 5 5( ) (base) (height) 4 3A x x x= ⋅ = − + ⋅ − + 26 12

25 56x x= − + and ( ) 55 2 3 26 612 225 5 25 50 0( ) 6 6 (10 30 30) 0 10

b

aV A x dx x x dx x x x = = − + = − + = − + − =

12. The slices parallel to the base are squares. The cross section of the pyramid is a triangle, and by similar

triangles we have 3 35 5

.bh

b h= = Thus ( )2 52 2 23 9 95 25 250( ) (base) ( )d

cA y y y V A y dy y dy= = = = =

53325 0

15 0 15y = = − =

13. (a) It follows from Cavalieri’s Principle that the volume of a column is the same as the volume of a right prism with a square base of side length s and altitude h. Thus,

STEP 1) 2 2( ) (sidelength) ;A x s= = STEP 2) 0, ;a b h= =

STEP 3) 2 20

( )b h

aV A x dx s dx s h= = =

(b) From Cavalieri’s Principle we conclude that the volume of the column is the same as the volume of the

prism described above, regardless of the number of turns 2V s h =

14. 1) The solid and the cone have the same altitude of 12.

2) The cross sections of the solid are disks of

diameter ( )2 2 .x xx − = If we place the vertex of the cone at the origin of the coordinate system and make its axis or symmetry coincide with the -axisx then the cone’s cross sections will

be circular disks of diameter ( )4 4 2x x x− − = (see accompanying figure).

3) The solid and the cone have equal altitudes and identical parallel cross sections. From Cavalier’s Principle we conclude that the solid and the cone have the same volume.

15. [ ] ( ) ( )2 2 3 222 2 222 2 4 2 120 0 0 0( ) 1 ( ) 1 1x x x x xR x y V R x dx dx x dx xπ π π π = = − = = − = − + = − + ( )8 242 12 32 ππ= − + =

16. [ ] ( )2 22 2 223 3 2 39 3 32 2 4 4 40 0 0 0( ) ( ) 8 6y yR y x V R y dy dy y dy yπ π π π π π = = = = = = = ⋅ ⋅ = 17. ( )4 4 4( ) tan ; 4 ;R y y u y du dy du dyπ π π π= = = = 40 0, 1 ;y u y u π= = = =

[ ] ( ) ( ) [ ]21 1 /4 /42 /42 2 040 0 0 0( ) tan 4 tan 4 1 sec 4 tanV R y dy y dy u du u du u uπ π πππ π = = = = − + = − +

( )44 1 0 4π π= − + − = −



18. ( ) sin cos ; ( ) 0 0R x x x R x a= = = and 2

b π= are the limits of integration;

[ ]2/2 /2 /22 (sin 2 )2

40 0 0( ) (sin cos ) ;xV R x dx x x dx dx

π π ππ π π= = = 8 42 2 ;du dxu x du dx = = =

20 0,x u x uπ π = = = = ( )

221 18 8 2 4 8 2 160 0

sin sin 2 0 0uV u du uππ π π π ππ → = = − = − − =

19. [ ]2 220

( ) ( )R x x V R x dxπ= =

( ) 5 222 22 4 325 50 0 0xx dx x dx ππ π π = = = =

20. [ ]2 230

( ) ( )R x x V R x dxπ= =

( ) 7 222 23 6 1287 70 0 0xx dx x dx ππ π π = = = =

21. [ ]3 223

( ) 9 ( )R x x V R x dxπ−

= − =

( ) 3 33 2 33 39 9 xx dx xπ π− − = − = − 273

2 9(3) 2 π 18 36ππ = − = ⋅ ⋅ =

22. [ ]1 220

( ) ( )R x x x V R x dxπ= − =

( ) ( )21 12 2 3 40 0 2x x dx x x x dxπ π= − = − + ( )3 4 5 12 1 1 13 4 5 3 2 50

x x xπ π = − + = − +

30 30(10 15 6)π π= − + =

23. [ ]/2 20

( ) cos ( )R x x V R x dxπ

π= =

[ ]/2 /200 cos sin (1 0)x dx xπ ππ π π π= = = − =



24. [ ]/4 2/4

( ) sec ( )R x x V R x dxππ

π−

= =

[ ]/4 /42 /4/4sec tan [1 ( 1)]x dx xπ π

πππ π π−−= = = − − 2π=

25. 1 12 20 0

( ) [ ( )] ( )x xR x e V R x dx e dx− −= = = π π 11 2 2 2

2 20 0( 1)x xe dx e e− − −= = − = − − π ππ

( ) 22 2( 1)12 21 ee e −= − = ππ

26. ( )( )

2/2 /2 /2 /22 /2cos/6sin/6 /6 /6 /6

12

( ) cot [ ( )] cot cot [ln(sin )]

ln1 ln ln 2

xx

R x x V R x dx x dx x dx dx x= = = = = =

= − =

π π π π π

ππ π π ππ π π π π

π π

27. ( ) ( )24 4 42 41 1 1 11/44 4 4 4 21/4 1/4 1/42 2( ) [ ( )] [ln ] ln 4 ln ln 4xx xR x V R x dx dx dx x= = = = = = − = π π π ππ π

28. 3 3 31 2 1 2 2 2 2 2 3 4

12 21 1 1( ) [ ( )] ( ) [ ] ( 1) 84.19x x x xR x e V R x dx e dx e dx e e− − − −= = = = = = − ≈ π ππ π π

29. [ ]/4 20

( ) 2 sec tan ( )R x x x V R x dxπ

π= − =

( )2/40 2 sec tanx x dxπ

π= −

( )/4 2 20 2 2 2 sec tan sec tanx x x x dxπ

π= − + /4 /4

0 02 2 2 sec tandx x x dx

π ππ = −

/4 2 20

(tan ) secx x dxπ +

[ ] [ ] 3/4/4 /4 tan

0 0 3 02 2 2 sec xx x

ππ ππ = − +

( ) ( ) 312 30 2 2 2 1 (1 0)ππ = − − − + − ( )112 32 2ππ= + −



30. [ ]/2 20

( ) 2 2sin 2(1 sin ) ( )R x x x V R x dxπ

π= − = − =

( )/2 /22 20 04(1 sin ) 4 1 sin 2sinx dx x x dxπ π

π π= − = + − /2 1

204 1 (1 cos 2 ) 2sinx x dx

ππ = + − −

( )/2 3 cos 22 204 2sinx xπ

π= − − /23 sin 2

2 4 04 2cosxx x

ππ = − +

( )344 0 0 (0 0 2) (3 8)ππ π π = − + − − + = −

31. [ ]1 122 41 1

( ) 5 ( ) 5R y y V R y dy y dyπ π− −

= = = 15

1[1 ( 1)] 2yπ π π

− = = − − =

32. [ ]2 223/2 30 0

( ) ( )R y y V R y dy y dyπ π= = = 4 2

40

4yπ π = =

33. [ ]/2 20

( ) 2sin 2 ( )R y y V R y dyπ

π= =

[ ]/2 /200 2sin 2 cos 2y dy yπ ππ π= = −

[1 ( 1)] 2π π= − − =

34. [ ]0 24 2( ) cos ( )yR y V R y dyπ π

−= =

( ) 00 4 42 2cos 4 sin 4[0 ( 1)] 4y ydyπ ππ − − = = = − − =

35. [ ]3 322 110 01( ) ( ) 4 yyR y V R y dy dyπ π ++= = =

[ ]30

4 ln 1 4 ln4 ln1 4 ln 4yπ π π= + = − =

0 1 2

1

2

3

x

y

x

y

=+21



36. [ ]21 22

01( ) ( )y

yR y V R y dyπ

+= =

( ) 21 20 2 1 ;y y dyπ−

= + 2[ 1 2 ;u y du y dy= + = 0 1, 1 2]y u y u= = = =

22 2 1 12 21 1

( 1)u

V u du ππ π π− → = = − = − − − =

37. For the sketch given, [ ] [ ]( )2 22 2, ; ( ) 1, ( ) cos ; ( ) ( )baa b R x r x x V R x r x dxπ π π= − = = = = − [ ] ( )/2 /2 /2 20 2/2 0(1 cos ) 2 (1 cos ) 2 sin 2 1 2x dx x dx x x

π π π ππ

π π π π π π−

= − = − = − = − = −

38. For the sketch given, [ ] [ ]( )2 240, ; ( ) 1, ( ) tan ; ( ) ( )dcc d R y r y y V R y r y dyπ π= = = = = − ( ) ( ) [ ] ( ) 2/4 /4 /42 2 0 2 20 01 tan 2 sec 2 tan 1y dy y dy y y

π π π π ππ π π π π= − = − = − = − = −

39. [ ] [ ]( )1 2 20( ) and ( ) 1 ( ) ( )r x x R x V R x r x dxπ= = = − ( ) ( )3 11 2 213 3 30 01 1 0xx dx x ππ π π = − = − = − − =

40. [ ] [ ]( )1 2 20( ) 2 and ( ) 2 ( ) ( )r x x R x V R x r x dxπ= = = − ( )2 11 12 20 0(4 4 ) 4 4 1 2

xx dx xπ π π π = − = − = − =

41. 2( ) 1 and ( ) 3r x x R x x= + = +

[ ] [ ]( )2 2 21 ( ) ( )V R x r x dxπ− = − ( )22 2 21 ( 3) 1x x dxπ − = + − +

( ) ( )2 2 4 21 6 9 2 1x x x x dxπ − = + + − + + ( ) 5 3 2 22 4 2 65 3 21 16 8 8x x xx x x dx xπ π− − = − − + + = − − + +

( ) ( )32 8 624 1 15 3 2 5 3 216 8π = − − + + − + + − ( ) ( )33 5 30 33 1175 5 53 28 3 8 ππ π ⋅ −= − − + − + = =



42. 2( ) 2 and ( ) 4r x x R x x= − = −

[ ] [ ]( )2 2 21 ( ) ( )V R x r x dxπ− = − ( )22 2 21 4 (2 )x x dxπ − = − − − ( ) ( )2 2 4 21 16 8 4 4x x x x dxπ − = − + − − +

( ) 5 22 2 4 2 3 51 112 4 9 12 2 3 xx x x dx x x xπ π− − = + − + = + − + ( ) ( ) ( )32 33 10815 5 5 524 8 24 12 2 3 15 ππ π = + − + − − + + − = + =

43. ( ) sec and ( ) 2r x x R x= =

[ ] [ ]( )/4 2 2/4 ( ) ( )V R x r x dxππ π− = − ( ) [ ]/4 /42 /4/4 2 sec 2 tanx dx x x

π πππ

π π −−= − = −

( ) ( )2 21 1 ( 2)π ππ π π = − − − + = −

44. [ ] [ ]( )1 2 20( ) sec and ( ) tan ( ) ( )R x x r x x V R x r x dxπ= = = − ( ) [ ]1 1 12 2 00 0sec tan 1x x dx dx xπ π π π= − = = =

45. [ ] [ ]( )1 2 20( ) 1 and ( ) 1 ( ) ( )r y R y y V R y r y dyπ= = + = − ( )1 12 20 0(1 ) 1 1 2 1y dy y y dyπ π = + − = + + −

( ) ( )311 2 2 41

3 3 30 02 1yy y dy y ππ π π = + = + = + =

46. [ ] [ ]( )1 2 20( ) 1 and ( ) 1 ( ) ( )R y r y y V R y r y dyπ= = − = − ( )1 12 20 01 (1 ) 1 1 2y dy y y dyπ π = − − = − − +

( ) ( )311 2 2 21

3 3 30 02 1yy y dy y ππ π π = − = − = − =



47. [ ] [ ]( )4 2 20( ) 2 and ( ) ( ) ( )R y r y y V R y r y dyπ= = = − 2 44

20 0(4 ) 4 (16 8) 8yy dy yπ π π π = − = − = − =

48. 2( ) 3 and ( ) 3R y r y y= = −

[ ] [ ]( )3 2 20 ( ) ( )V R y r y dyπ = − ( ) 3

33 32 230 0 0

3 3 3yy dy y dyπ π π π = − − = = =

49. [ ] [ ]( )1 2 20( ) 2 and ( ) 1 ( ) ( )R y r y y V R y r y dyπ= = + = − ( ) ( )210 4 1 1 2y dy y y dyπ π

1

0 = − + = 4 − − −

( ) 211 3/24

3 20 03 2 3 yy y dy y yπ π = − − = − −

( ) ( )18 8 3 74 13 2 6 63 ππ π − −= − − = =

50. [ ] [ ]( )1 2 21/3 0( ) 2 and ( ) 1 ( ) ( )R y y r y V R y r y dyπ= − = = − ( ) ( )21 11/3 1/3 2/30 02 1 4 4 1y dy y y dyπ π = − − = − + −

( ) 5/311 31/3 2/3 4/3

50 03 4 3 3 yy y dy y yπ π = − + = − +

( )3 35 53 3 ππ= − + =

51. (a) ( ) and ( ) 2r x x R x= =

[ ] [ ]( )4 2 20 ( ) ( )V R x r x dxπ = − 2 44

20 0(4 ) 4 (16 8) 8xx dx xπ π π π = − = − = − =

(b) [ ] [ ]( ) 5 22 22 22 4 325 50 0 0( ) 0 and ( ) ( ) ( )yr y R y y V R y r y dy y dy ππ π π = = = − = = =

(c) [ ] [ ]( ) ( )24 42 20 0( ) 0 and ( ) 2 ( ) ( ) 2r x R x x V R x r x dx x dxπ π= = − = − = − ( ) ( )3/2 2 44 8 64 16 83 2 3 2 30 04 4 4 16

x xx x dx x ππ π π = − + = − + = − + =



(d) [ ] [ ]( ) ( )22 22 22 20 0( ) 4 and ( ) 4 ( ) ( ) 16 4r y y R y V R y r y dy y dyπ π = − = = − = − − ( ) ( ) ( )5

22 22 4 2 4 38 64 32 2243 5 3 5 150 0 0

16 16 8 8 yy y dy y y dy y ππ π π π = − + − = − = − = − =

52. (a) 2

( ) 0 and ( ) 1 yr y R y= = −

[ ] [ ]( )2 2 20 ( ) ( )V R y r y dyπ = − ( ) 222 22 40 01 1y ydy y dyπ π = − = − +

( )2 32

8 242 12 2 12 3

02y yy ππ π = − + = − + =

(b) 2

( ) 1 and ( ) 2 yr y R y= = −

[ ] [ ]( ) ( ) 222 2 22 2 2 40 0 0( ) ( ) 2 1 4 2 1y yV R y r y dy dy y dyπ π π = − = − − = − + −

( ) ( )2 322 2 8 82

4 12 12 3 30 03 2 3 6 4 2y yy dy y y ππ π π π = − + = − + = − + = + =

53. (a) [ ] [ ]( )1 2 22 1( ) 0 and ( ) 1 ( ) ( )r x R x x V R x r x dxπ−= = − = − ( ) ( )21 12 2 41 11 1 2x dx x x dxπ π− −= − = − +

( ) ( )3 5 12 15 10 3 162 13 5 3 5 15 151 2 1 2x xx ππ π π − +

− = − + = − + = =

(b) [ ] [ ]( ) ( )21 12 22 21 1( ) 1 and ( ) 2 ( ) ( ) 2 1r x R x x V R x r x dx x dxπ π− − = = − = − = − − ( ) ( ) ( )5 11 12 4 2 4 34 4 13 5 3 51 1 14 4 1 3 4 3 2 3xx x dx x x dx x xπ π π π− − − = − + − = − + = − + = − +

2 5615 15

(45 20 3)π π= − + =

(c) [ ] [ ]( ) ( )21 12 22 21 1( ) 1 and ( ) 2 ( ) ( ) 4 1r x x R x V R x r x dx x dxπ π− − = + = = − = − + ( ) ( ) ( )5 11 12 4 2 4 32 2 13 5 3 51 1 14 1 2 3 2 3 2 3xx x dx x x dx x xπ π π π− − − = − − − = − − = − − = − −

2 6415 15

(45 10 3)π π= − − =

54. (a) ( ) 0 and ( ) hb

r x R x x h= = − +

[ ] [ ]( )2 20 ( ) ( )bV R x r x dxπ = − ( )20

b hb

x h dxπ= − + ( )2 22 2 220b h hbb x x h dxπ= − + ( )3 2 222 2 3 33 0

bx x b h b

bbh x h b b ππ π = − + = − + =



(b) ( ) [ ] [ ]( ) ( )22 2 20 0( ) 0 and ( ) 1 ( ) ( ) 1h hy yh hr y R y b V R y r y dy b dyπ π= = − = − = − ( )2 2 3 22 222 2 2 3 30 3 01

hh y y y y h b hh hh h

b dy b y b h h ππ π π = − + = − + = − + =

55. 2 2 2 2( ) and ( )R y b a y r y b a y= + − = − −

[ ] [ ]( )2 2( ) ( )aaV R y r y dyπ− = − 2 2

2 2 2 2a

ab a y b a y dyπ

−

= + − − − −

2 2 2 24 4a a

a ab a y dy b a y dyπ π

− −= − = −

4bπ= ⋅ area of semicircle of radius 2 2 2

24 2aa b a bππ π= ⋅ =

56. (a) A cross section has radius 2r y= and area 2 2 .r yπ π= The volume is 55 2

0 02 25 .ydy yπ π π = =

(b) ( ) ( ) , so ( ).dVdh

V h A h dh A h= = Therefore 1( )( ) , so .dV dV dh dh dh dVdt dh dt dt dt A h dtA h= ⋅ = ⋅ = ⋅

For 4,h = the area is 3 3units 3 units1

8 sec 8 sec2 (4) 8 , so 3 .dh

dt π ππ π= = ⋅ = ⋅

57. (a) ( ) ( )3 3 3( )2 2 2 2 2 2 3 33 3 3( )h ah a y h a a

a aR y a y V a y dy a y a h a aπ π π

−− −− −

= − = − = − = − − − − +

( ) ( ) 23 3 (3 )2 3 2 2 3 2 2 213 3 3 33 3 h a ha ha h h h a ha a a h h a ha ππ π − = − − + − − = − + − = (b) Given 30.2 m /sec and 5 m,dV

dta= = find

4.dh

dt h= From part (a),

2 3(15 ) 23 3

( ) 5h h hV h hπ ππ−= = −

2 0.2 1 14 (10 4) (20 )(6) 1204

10 (10 ) m/sec.dV dV dV dh dh dhdh dt dh dt dt dt h

h h h h π π ππ π π −= = − = ⋅ = − = = =

58. Suppose the solid is produced by revolving 2y x= − about the -axis.y Cast a shadow of

the solid on a plane parallel to the -plane.xy

Use an approximation such as the Trapezoid Rule,

to estimate [ ] ˆ2

22

1

( ) .kn db

ak

R y dy yπ π=

≈ Δ

59. The cross section of a solid right circular cylinder with a cone removed is a disk with radius R from which a

disk of radius h has been removed. Thus its area is 2 2 2 21 ( ).A R h R hπ π π= − = − The cross section of the

hemisphere is a disk of radius 2 2 .R h− Therefore its area is ( )2

2 2 2 22 .A R h R hπ π

= − = −

We can see that 1 2.A A= The altitudes of both solids are R. Applying Cavalieri’s Principle we find

Volume of Hemisphere = (Volume of Cylinder) − (Volume of Cone) ( ) ( )2 2 31 23 3 .R R R R Rπ π π= − =



60. [ ] ( ) ( )26 6 622 2 2 412 144 1440 0 0( ) 36 ( ) 36 36x xR x x V R x dx x dx x x dxππ π= − = = − = − ( ) ( ) ( ) ( )5 5 363 3 36 6 36 196 60 36 36144 5 144 5 144 5 144 5 5012 12 6 12 cm .xxπ π π π π⋅ − = − = ⋅ − = − = =

The plumb bob will weigh about ( )365(8.5) 192 gm,W π= ≈ to the nearest gram.

61. [ ] ( ) 377 722 2

316 16 16( ) 256 ( ) 256 256 yR y y V R y dy y dy y

−− −

− − −

= − = = − = − π π π

( ) ( )3 3 3 3 3 37 16 7 163 3 3 3(256)( 7) (256)( 16) 256(16 7) 1053 cm 3308 cmπ π π = − + − − + = + − − = ≈

62. (a) [ ] ( )2 2 2 20 0 0( ) | sin |, so ( ) ( sin ) 2 sin sinR x c x V R x dx c x dx c c x x dxπ π π

π π π= − = = − = − +

( ) ( ) ( )2 2 21 cos 2 cos 2 sin 21 12 2 2 2 40 0 02 sin 2 sin 2 cosx x xc c x dx c c x dx c x c xππ π

π π π− = − + = + − − = + + −

( ) ( )2 22 22 0 (0 2 0) 4 .c c c c cπ ππ π π π = + − − − + − = + − Let ( )2 2( ) 4 .V c c cππ π= + − We find the extreme values of 2( ) : (2 4) 0dV

dcV c c c ππ π= − = = is a critical point, and ( ) ( )82 4 2V ππ π ππ= + −

( ) 242 2 4;π πππ= − = − Evaluate V at the endpoints: 2

2(0)V π= and ( ) 232 2(1) 4 (4 ) .V ππ π π π= − = − −

Now we see that the function’s absolute minimum value is 2

24,π − taken on at the critical point 2 .c π=

(See also the accompanying graph.) (b) From the discussion in part (a) we conclude that the function’s absolute maximum value is

2

2,π taken on

at the endpoint 0.c = (c) The graph of the solid’s volume as a function

of c for 0 1c≤ ≤ is given at the right. As c moves away from [0,1] the volume of the solid

increases without bound. If we approximate the solid as a set of solid disks, we can see that the radius of a typical disk increases without bounds as c moves away from [0,1].

63. Volume of the solid generated by rotating the region bounded by the -axisx and ( )y f x= from x a= to

x b= about the -axisx is 2[ ( )] 4 ,b

aV f x dxπ π= = and the volume of the solid generated by rotating the same

region about the line 21 is [ ( ) 1] 8 .b

ay V f x dxπ π= − = + = Thus [ ] [ ]2 2( ) 1 ( ) 8 4

b b

a af x dx f x dxπ π π π+ − = −

[ ] [ ]( )2 2( ) 2 ( ) 1 ( ) 4 (2 ( ) 1) 4 2 ( ) 4b b b ba a a af x f x f x dx f x dx f x dx dxπ π + + − = + = + = 41

2 2( ) ( ) 2 ( )

b b b aa a

f x dx b a f x dx − + + − = =

64. Volume of the solid generated by rotating the region bounded by the -axisx and ( )y f x= from tox a x b= =

about the -axisx is [ ]2( ) 6 ,ba

V f x dxπ π= = and the volume of the solid generated by rotating the same

region about the line 2y = − is [ ]2( ) 2 10 .ba

V f x dxπ π= + = Thus

Section 6.2 Volumes Using Cylindrical Shells 443


[ ] [ ] [ ] [ ]( )2 2 2 2( ) 2 ( ) 10 6 ( ) 4 ( ) 4 ( ) 4b b ba a af x dx f x dx f x f x f x dxπ π π π π π+ − = − + + − = (4 ( ) 4) 4 4 ( ) 4 4 ( ) ( ) 1 ( ) 1

b b b b b

a a a a af x dx f x dx dx f x dx b a f x dx b a + = + = + − = = − +

6.2 VOLUMES USING CYLINDRICAL SHELLS

1. For the sketch given, 0, 2;a b= =

( ) ( ) ( ) ( ) ( )2 3 2 4 22 2shell shell 164radius height 4 4 2 16 2 160 0 02 2 1 2 2 2b x x x xa

V dx x dx x dxπ π π π π = = + = + = + = + 2 3 6π π= ⋅ =


( )( ) ( ) ( ) ( )2 3 4 22 2shell shell 2radius height 4 4 160 0 02 2 2 2 2 2 2 4 1 6b x x xa

V dx x dx x dx xπ π π π π π = = − = − = − = − =

3. For the sketch given, 0, 2;c d= =

( ) ( ) ( ) 422 2shell shell 2 3

radius height 40 0 02 2 2 2 2

d y

cV dy y y dy y dyπ π π π π = = ⋅ = = =

4. For the sketch given, 0, 3;c d= =

( ) ( ) ( ) 433 3shell shell 2 3 9

radius height 4 20 0 02 2 3 3 2 2

d y

cV dy y y dy y dy ππ π π π = = ⋅ − − = = =


( ) ( ) 3shell shell 2radius height 02 2 1 ;baV dx x x dxπ π = = ⋅ + 2 1 2 ; 0 1, 3 4u x du x dx x u x u = + = = = = =

( ) ( )44 1/2 3/2 3/22 2 1423 3 3 31 1 4 1 (8 1)V u du u π π ππ π → = = = − = − =


( ) ( ) 33shell shell 9radius height 0 92 2 ;b xa x

V dx x dxπ π+

= =

3 2 2[ 9 3 3 9 ; 0 9, 3 36]u x du x dx du x dx x u x u= + = = = = = =

( )3636 1/2 1/29 92 3 6 2 12 36 9 36V u du uπ π π π− → = = = − =

7. 0, 2;a b= =

( ) ( ) ( )2shell shellradius height 202 2b xaV dx x x dxπ π = = − − 22 22 2 33

20 0 02 3 8x dx x dx xπ π π π = ⋅ = = =



8. 0, 1;a b= =

( ) ( ) ( )1shell shellradius height 202 2 2b xaV dx x x dxπ π= = − ( )2 11 1 2 3320 0 02 3x dx x dx xπ π π π = = = =

9. 0, 1;a b= =

( ) ( ) 1shell shell 2radius height 02 2 (2 )baV dx x x x dxπ π = = − − ( ) 3 4 11 2 3 2 3 40 02 2 2 x xx x x dx xπ π = − − = − −

( ) ( )12 4 3 10 51 13 4 12 12 62 1 2 π ππ π − −= − − = = = 10. 0, 1;a b= =

( )( ) ( )1shell shell 2 2radius height 02 2 2baV dx x x x dxπ π = = − − ( ) ( )1 12 30 02 2 2 4x x dx x x dxπ π= − = −

( )2 4 1 1 12 4 2 404 4x xπ π π = − = − =

11. 0, 1;a b= =

( ) ( ) 1shell shellradius height 02 2 (2 1)baV dx x x x dxπ π = = − − ( ) 11 3/2 2 5/2 3 22 2 15 3 20 02 2 2x x x dx x x xπ π = − + = − +

( ) ( )12 20 15 72 2 15 3 2 30 152 2 ππ π − += − + = =

12. 1, 4;a b= =

( ) ( ) ( )4shell shell 1/23radius height 212 2baV dx x x dxπ π −= = ( )44 1/2 3/2 3/2231 13 3 2 4 1x dx xπ π π = = = −

2 (8 1) 14π π= − =

13. (a) sin , 0 sin , 0

( ) ( ) ;0, 0, 0

xx

x x x xx f x x f x

xx x

π π ⋅ < ≤ < ≤= = == since sin 0 0= we have

sin , 0( ) ( ) sin , 0

sin , 0

x xx f x x f x x x

x x

ππ

< ≤= = ≤ ≤ =



(b) ( )( )shell shellradius height 02 2 ( )baV dx x f x dxππ π= = ⋅ and ( ) sin , 0x f x x x π⋅ = ≤ ≤ by part (a) [ ]002 sin 2 cos 2 ( cos cos 0) 4V x dx x

π ππ π π π π = = − = − + =

14. (a) 2 2tan

4, 0 tan , 0 /4

( ) ( ) ;0, 00, 0

xx

x x x xxg x xg x

xx x

π π ⋅ < ≤ < ≤ = = =⋅ =

since tan 0 0= we have

22

2

tan , 0 /4( ) ( ) tan , 0 /4

tan , 0

x xxg x xg x x x

x x

ππ

< ≤= = ≤ ≤=

(b) ( ) ( ) /4shell shellradius height 02 2 ( )baV dx x g x dxππ π= = ⋅ and 2( ) tan , 0 /4x g x x x π⋅ = ≤ ≤ by part (a) ( ) [ ] ( ) 2/4 /4 /42 2 40 4 20 02 tan 2 sec 1 2 tan 2 1V x dx x dx x x

π π π π π ππ π π π − = = − = − = − =

15. 0, 2;c d= =

( ) ( ) 2shell shellradius height 02 2 ( )dcV dy y y y dyπ π = = − − ( ) 5/2 3

22 23/2 25 30 0

2 2 y yy y dyπ π = + = +

( ) ( ) ( )35 8 2 282 2 15 3 5 3 5 32 2 2 16π π π = + = + = + ( )1615 3 2 5π= +

16. 0, 2;c d= =

( )( ) 2shell shell 2radius height 02 2 ( )dcV dy y y y dyπ π = = − − ( ) ( )4 3

22 3 2 2 14 3 4 30 0

2 2 16y yy y dyπ π π = + = + = +

( )5 406 316 ππ= =

17. 0, 2;c d= =

( ) ( ) ( )2shell shell 2radius height 02 2 2dcV dy y y y dyπ π= = − ( ) ( )3 4

22 22 3 16 163 4 3 40 0

2 2 2 2y yy y dyπ π π = − = − = −

( ) 32 81 13 4 12 332 π ππ= − = =



18. 0, 1;c d= =

( ) ( ) ( )1shell shell 2radius height 02 2 2dcV dy y y y y dyπ π= = − − ( ) ( )1 12 2 30 02 2y y y dy y y dyπ π= − = −

( )3 41

1 13 4 3 4 6

02 2y y ππ π = − = − =

19. 0, 1;c d= =

( )( ) [ ]1shell shellradius height 02 2 ( )dcV dy y y y dyπ π= = − − 11 2 34 4

3 30 02 2y dy yπ ππ = = =

20. 0, 2;c d= =

( ) ( ) ( )2shell shellradius height 202 2d ycV dy y y dyπ π= = − 2 22 3 8

2 3 30 02 y dy yπ ππ = = =

21. 0, 2;c d= =

( ) ( ) 2shell shell 2radius height 02 2 (2 )dcV dy y y y dyπ π = = + − ( ) 3 4

22 2 3 23 40 0

2 2 2 y yy y y dy yπ π = + − = + −

( )8 16 163 4 6 32 4 (48 32 48)π ππ= + − = + − = 22. 0, 1;c d= =

( )( ) 1shell shell 2radius height 02 2 (2 )dcV dy y y y dyπ π = = − − ( ) 3 4

11 2 3 23 40 0

2 2 2 y yy y y dy yπ π = − − = − −

( ) 51 13 4 6 62 1 (12 4 3)π ππ= − − = − − =

23. (a) ( )( ) 22 2shell shell 2 3radius height 0 0 02 2 (3 ) 6 2 16b

aV dx x x dx x dx xπ π π π π = = = = =

(b) ( )( ) ( ) 22 2shell shell 2 2 31radius height 30 0 02 2 (4 ) (3 ) 6 4 6 2b

aV dx x x dx x x dx x xπ π π π = = − = − = −

( )836 8 32π π= − =



(c) ( )( ) ( ) 22 2shell shell 2 3 21 1radius height 3 20 0 02 2 ( 1) (3 ) 6 6b

aV dx x x dx x x dx x xπ π π π = = + = + = +

( )836 2 28π π= + = (d) ( )( ) ( ) ( ) 66 6shell shell 2 2 31 1 1radius height 3 3 90 0 02 2 2 2 2 2

d

cV dy y y dy y y dy y yπ π π π = = − = − = −

2 (36 24) 24π π= − =

(e) ( )( ) ( ) ( )6 6shell shell 2131 1radius height 3 3 30 02 2 (7 ) 2 2 14dcV dy y y dy y y dyπ π π= = − − = − + 62 313 1

6 9 02 14 2 (84 78 24) 60y y yπ π π = − + = − + =

(f ) ( )( ) ( ) ( )6 6shell shell 21 4 1radius height 3 3 30 02 2 ( 2) 2 2 4dcV dy y y dy y y dyπ π π= = + − = + − 62 32 1

3 9 02 4 2 (24 24 24) 48y y yπ π π = + − = + − =

24. (a) ( )( ) ( ) ( ) 22 2shell shell 3 4 2 51radius height 50 0 02 2 8 2 8 2 4b

aV dx x x dx x x dx x xπ π π π = = − = − = −

( )32 965 52 16 ππ= − = (b) ( )( ) ( ) ( )2 2shell shell 3 3 4radius height 0 02 2 (3 ) 8 2 24 8 3baV dx x x dx x x x dxπ π π= = − − = − − +

( )22 4 53 32 26414 5 5 502 24 4 2 48 16 12x x x x ππ π = − − + = − − + = (c) ( )( ) ( ) ( )2 2shell shell 3 3 4radius height 0 02 2 ( 2) 8 2 16 8 2baV dx x x dx x x x dxπ π π= = + − = + − −

( )22 4 5 32 3361 12 5 5 502 16 4 2 32 16 8x x x x ππ π = + − − = + − − = (d) ( )( ) 88 8shell shell 1/3 4/3 7/36 6 768radius height 7 7 70 0 02 2 2 (128)

d

cV dy y y dy y dy yπ π ππ π π = = ⋅ = = = =

(e) ( )( ) ( ) 88 8shell shell 1/3 1/3 4/3 4/3 7/33radius height 70 0 02 2 (8 ) 2 8 2 6d

cV dy y y dy y y dy y yπ π π π = = − = − = −

( )384 5767 72 96 ππ= − =

(f ) ( )( ) ( ) 88 8shell shell 1/3 4/3 1/3 7/3 4/33 3radius height 7 40 0 02 2 ( 1) 2 2d

cV dy y y dx y y dy y yπ π π π = = + = + = +

( )384 936π7 72π 12= + =

25. (a) ( ) ( ) ( ) ( )2 2shell shell 2 2 3radius height 1 12 2 (2 ) 2 2 4 3baV dx x x x dx x x dxπ π π− −= = − + − = − + ( )23 4 271 14 4 212 4 2 (8 8 4) 2 4 1x x x ππ π π− = − + = − + − − + + =

(b) ( )( ) ( ) ( )2 2shell shell 2 3radius height 1 12 2 ( 1) 2 2 2 3baV dx x x x dx x x dxπ π π− −= = + + − = + − ( )22 43 3 271 12 4 2 4 212 2 2 (4 6 4) 2 2x x x ππ π π− = + − = + − − − + − =



(c) ( )( ) ( )( ) ( )1 4shell shellradius height 0 12 2 2 ( 2)dcV dy y y y dy y y y dyπ π π= = − − + − − ( ) 1 41 43/2 3/2 2 5/2 5/2 3 28 2 15 5 30 1 0 14 2 2 2y dy y y y dy y y y yππ π π = + − + = + − +

( ) ( )8 64 64 722 15 5 3 5 3 5(1) 2 16 2 1π ππ π= + − + − − + = (d) ( )( ) ( )( ) ( )1 4shell shellradius height 0 12 2 (4 ) 2 (4 ) ( 2)dcV dy y y y dy y y y dyπ π π= = − − − + − − −

( ) ( )1 43/2 2 3/20 14 4 2 6 4 8y y dy y y y y dyπ π= − + − − + + 1 43/2 5/2 3 5/2 2 3/28 82 1 2

3 5 3 5 30 14 2 3 8y y y y y y yπ π = − + − − + +

( ) ( ) ( )8 64 64 64 8 1082 1 23 5 3 5 3 3 5 3 54 2 48 32 2 3 8 .ππ π π= − + − − + + − − − + + =

26. (a) ( )( ) ( ) ( )1 1shell shell 2 4 5 4 3 2radius height 1 12 2 (1 ) 4 3 2 3 3 4 4baV dx x x x dx x x x x x dxπ π π− −= = − − − = − + − − + ( ) ( )16 5 4 3 23 3 3 561 1 1 1 1 16 5 4 6 5 4 6 5 4 512 2 4 2 1 2 4 2 1 2 4x x x x x x ππ π π− = − + − − + = − + − − + − + + + − − =

(b) ( )( ) ( )( )1 4 4 4shell shell 4 4radius height 3 30 12 2 2d y ycV dy y y y dy y dyπ π π − − = = − − + − − 1 45/4 40 13

4 4y dy y ydyππ= + − [ 4 4 ; 1 3, 4 0]u y y u du du y u y u= − = − = − = = = =

( )1 30 39/4 3/2 3/2 5/216 4 16 4 16 4 8 29 9 9 3 53 03 3 30 0(4 ) (1) 4y u u du u u du u uπ π π π π π = − − = + − = + − ( )16 4 18 16 88 8729 5 9 5 453 8 3 3π π π π π= + − = + =

27. (a) ( )( ) ( ) ( ) 4 511 1shell shell 2 3 3 4

radius height 4 50 0 02 2 12 24 24

d y y

cV dy y y y dy y y dyπ π π π = = ⋅ − = − = −

( ) 24 61 14 5 20 524 π ππ= − = = (b) ( )( ) ( ) ( )1 1shell shell 2 3 2 3radius height 0 02 2 (1 ) 12 24 (1 )dcV dy y y y dy y y y dyπ π π = = − − = − −

( ) ( ) ( )3 4 511 2 3 4 41 1 1 1

3 2 5 3 2 5 30 50 024 2 24 24 24y y yy y y dy ππ π π π = − + = − + = − + = =

(c) ( )( ) ( ) ( ) ( ) ( )1 1shell shell 2 3 2 38 8radius height 5 50 02 2 12 24dcV dy y y y dy y y y dyπ π π = = − − = − − ( ) ( )5

11 2 3 4 3 48 13 8 13 8 13 15 5 15 20 5 15 20 50 0

24 24 24yy y y dy y yπ π π = − + = − + = − + 24 2460 12

(32 39 12) 2π π π= − + = =

(d) ( )( ) ( ) ( ) ( ) ( )1 1shell shell 2 3 2 32 2radius height 5 50 02 2 12 24dcV dy y y y dy y y y dyπ π π = = + − = + − ( ) ( ) 5

11 13 4 2 3 2 3 4 3 43 32 2 2 25 5 5 5 15 20 50 0 0

24 24 24 yy y y y dy y y y dy y yπ π π = − + − = + − = + −

( )3 24 242 115 20 5 60 1224 (8 9 12) 2π ππ π= + − = + − = =



28. (a) ( )( ) 2 4 2 4 52 2 2shell shell 2 3radius height 2 4 2 4 40 0 02 2 2 2d y y y y ycV dy y dy y y dy y dyπ π π π = = − − = − = −

( ) ( ) ( ) ( )4 6 4 62 82 2 1 4 1 1 24 24 4 24 4 24 4 6 24 30

2 2 32 32 32y y ππ π π π π = − = − = − = − = =

(b) ( )( ) 2 4 2 42 2shell shell 2radius height 2 4 2 40 02 2 (2 ) 2 (2 )d y y y ycV dy y dy y y dyπ π π = = − − − = − −

( )4 5 3 5 4 622 22 3 16 32 16 64 8

2 4 3 10 4 24 3 10 4 24 50 02 2 2 2y y y y y yy y dy ππ π π = − − + = − − + = − − + =

(c) ( )( ) 2 4 2 42 2shell shell 2radius height 2 4 2 40 02 2 (5 ) 2 (5 )d y y y ycV dy y dy y y dyπ π π = = − − − = − −

( )5 3 5 4 622 5 52 4 35 40 160 16 64

4 4 3 20 4 24 3 20 4 240 02 5 2 2 8y y y y yy y y dyπ π π π = − − + = − − + = − − + =

(d) ( )( ) ( ) ( )2 4 2 42 2shell shell 25 5radius height 8 2 4 2 8 40 02 2 2d y y y ycV dy y dy y y dyπ π π = = + − − = + −

( )5 4 6 3 522 5 53 2 45 5 16 64 40 160

4 8 32 4 24 24 160 4 24 24 1600 02 2 2 4y y y y yy y y dyπ π π π = − + − = − + − = − + − =

29. (a) About -axis:x ( )( )shell shellradius height2dcV dyπ= ( ) ( )1 1 3/2 20 02 2y y y dy y y dyπ π= − = −

( )15/2 3 22 1 2 15 3 5 3 1502 2y y ππ π = − = − = About -axis:y ( )( )shell shellradius height2baV dxπ=

( ) ( )1 12 2 30 02 2x x x dx x x dxπ π= − = − ( )3 4 1 1 13 4 3 4 602 2

x x ππ π = − = − =

(b) About -axis:x ( )R x x= and [ ] [ ]( ) ( )12 22 2 40( ) ( ) ( )bar x x V R x r x dx x x dxπ π= = − = − ( )3 5 1 21 13 5 3 5 150

x x ππ π = − = − =

About -axis:y ( )R y y= and [ ] [ ]( ) ( )12 2 20( ) ( ) ( )dcr y y V R y r y dy y y dyπ π= = − = − ( )2 3

11 1

2 3 2 3 60

y y ππ π = − = − =

30. (a) [ ] [ ]( ) ( )242 2 220( ) ( ) 2b xaV R x r x dx x dxπ π = − = + − ( ) 3 44 2 234 40 02 4 4xx x dx x xπ π = − + + = − + + 16 16 16 16π π= (− + + ) =



(b) ( )( ) ( ) ( ) ( )24 4 4shell shellradius height 2 2 20 0 02 2 2 2 2 2 2b x x xaV dx x x dx x dx x dxπ π π π= = + − = − = − ( )3 42 64 326 6 302 2 16

xx ππ π = − = − =

(c) ( )( ) ( ) ( ) ( )24 4 4shell shellradius height 2 2 20 0 02 2 (4 ) 2 2 (4 ) 2 2 8 4b x x xaV dx x x dx x dx x dxπ π π π= = − + − = − − = − − ( )3 42 64 646 6 302 8 2 2 32 32

xx x ππ π = − + = − + =

(d) [ ] [ ]( ) ( ) ( ) ( )224 42 2 2 22 40 0( ) ( ) (8 ) 6 64 16 36 6b x xaV R x r x dx x dx x x x dxπ π π = − = − − − = − + − − + ( ) 3 44 2 234 40 010 28 5 28 [16 (5) (16) (7) (16)] (3) (16) 48xx x dx x xπ π π π π − + = − + = − + = =

31. (a) ( )( ) 2shell shellradius height 12 2 ( 1)dcV dy y y dyπ π= = − ( ) 3 2

22 23 21 1

2 2 y yy y dyπ π = − = −

( ) ( )8 4 1 13 2 3 22π = − − − ( )7 513 2 3 32 2 (14 12 3)π ππ= − + = − + =

(b) ( )( ) ( ) 3 22 2shell shell 2 2radius height 31 1 12 2 (2 ) 2 2 2b xa

V dx x x dx x x dx xπ π π π = = − = − = −

( ) ( ) ( ) ( ) ( )8 12 8 3 1 41 4 23 3 3 3 3 3 32 4 1 2 2 ππ π π− − = − − − = − = − = (c) ( )( ) ( ) ( )2 2shell shell 210 20 16radius height 3 3 31 12 2 (2 ) 2baV dx x x dx x x dxπ π π= = − − = − +

( ) ( ) ( )22 320 8 40 32 8 20 8 31 13 3 3 3 3 3 3 3 3 312 2 2 2x x xπ π π π = − + = − + − − + = =

(d) ( )( ) 322 2 ( 1)shell shell 2 2

radius height 3 31 1 12 2 ( 1)( 1) 2 ( 1) 2

d y

cV dy y y dy y ππ π π π − = = − − = − = =

32. (a) ( )( ) ( )2shell shell 2radius height 02 2 0dcV dy y y dyπ π= = − ( )4 422 3 24 40 02 2 2 8

yy dyπ π π π = = = =

(b) ( )( )shell shellradius height2baV dxπ= ( ) ( )4 4 3/20 02 2 2 2x x dx x x dxπ π= − = −

( )542 5/22 2 25 502 2 16x xπ π ⋅ = − = − ( )64 2 325 5 52 16 (80 64)π ππ= − = − =

(c) ( )( ) ( ) ( )4 4shell shell 1/2 3/2radius height 0 02 2 (4 ) 2 2 8 4 2baV dx x x dx x x x dxπ π π= = − − = − − + ( )43/2 2 5/28 64 64 2 2 22423 5 3 5 15 15 1502 8 2 32 16 (240 320 192) (112)x x x x π π ππ π = − − + = − − + = − + = =



(d) ( )( ) ( ) ( ) 422 2shell shell 2 2 3 32

radius height 3 40 0 02 2 (2 ) 2 2 2

d y

cV dy y y dy y y dy yπ π π π = = − = − = −

( )16 16 32 83 4 12 32 (4 3)π ππ= − = − =

33. (a) ( )( ) ( )1shell shell 3radius height 02 2dcV dy y y y dyπ π= = − ( ) ( )3 5

11 2 4 41 13 5 3 5 150 0

2 2 2y yy y dy ππ π π = − = − = − =

(b) ( )( ) ( )1shell shell 3radius height 02 2 (1 )dcV dy y y y dyπ π= = − − ( ) 2 3 4 5

11 2 3 42 3 4 50 0

2 2 y y y yy y y y dyπ π = − − + = − − +

( ) 2 71 1 1 12 3 4 5 60 302 (30 20 15 12)π ππ= − − + = − − + =

34. (a) ( )( ) ( )1shell shell 3radius height 02 2 1dcV dy y y y dyπ π = = − − ( ) 2 3 5

11 2 42 3 50 0

2 2 y y yy y y dyπ π = − + = − +

( ) 2 111 1 12 3 5 30 152 (15 10 6)π ππ= − + = − + =

(b) Use the washer method:

[ ] [ ]( ) ( ) ( )21 12 2 2 3 2 6 40 0( ) ( ) 1 1 2dcV R y r y dy y y dy y y y dyπ π π = − = − − = − − + ( )3 7 5

12 971 1 2

3 7 5 3 7 5 105 1050

1 (105 35 15 42)y y yy π ππ π = − − + = − − + = − − + =

(c) Use the washer method:

[ ] [ ]( ) ( ) ( ) ( )2 21 12 2 3 3 30 0( ) ( ) 1 0 1 2dcV R y r y dy y y dy y y y y dyπ π π = − = − − − = − − + −

( ) 3 7 4 511 22 6 3 4 2

3 7 2 50 01 2 2 2 y y y yy y y y y dy y yπ π = + + − + − = + + − + −

( ) 1211 1 1 23 7 2 5 210 2101 1 (70 30 105 2 42)π ππ= + + − + − = + + − ⋅ = (d) ( )( ) ( ) ( )1 1shell shell 3 3radius height 0 02 2 (1 ) 1 2 (1 ) 1dcV dy y y y dy y y y dyπ π π = = − − − = − − +

( ) ( ) 3 4 511 13 2 4 2 3 4 2

3 4 50 0 02 1 2 1 2 2 y y yy y y y y dy y y y y dy y yπ π π = − + − + − = − + + − = − + + −

( ) 2 231 1 13 4 5 60 302 1 1 (20 15 12)π ππ= − + + − = + − =



35. (a) ( )( ) ( )1shell shell 2radius height 02 2 8dcV dy y y y dyπ π= = − ( ) 4

22 3/2 3 5/24 25 40 0

2 2 2 2 yy y dy yπ π = − = −

( ) ( )5

4 34 2 2 2 4 2 4 45 4 5 4

2 2π π⋅ ⋅ ⋅

= − = −

( )8 8 245 5 52 4 1 (8 5)π ππ= ⋅ − = − =

(b) ( )( ) ( ) ( )2 3 4 44 4shell shell 3/2 5/22radius height 8 8 5 320 0 02 2 2 2b x x xa

V dx x x dx x dx xπ π π π = = − = − = −

( ) ( ) 7 9 45 4 6 8 2 2 3 2 3 482 2 4 2 25 32 5 32 160 160 5 52 2 (32 20)π π π ππ π ⋅ ⋅ ⋅ ⋅ ⋅⋅= − = − = − = = =

36. (a) ( )( ) ( )1shell shell 2radius height 02 2 2baV dx x x x x dxπ π = = − − ( ) ( )1 12 2 30 02 2x x x dx x x dxπ π= − = −

( )3 4 1 1 13 4 3 4 602 2x x ππ π = − = − =

(b) ( )( ) ( ) ( ) ( )( )1 1shell shell 2 2radius height 0 02 2 1 2 2 1baV dx x x x x dx x x x dxπ π π = = − − − = − − ( ) ( )2 4 11 2 3 3 22 1 2 12 3 4 2 3 4 12 60 02 2 2 2 (6 8 3)x xx x x dx x π ππ π π = − + = − + = − + = − + =

37. (a) [ ] [ ]( ) ( )12 2 1/21/16( ) ( ) 1baV R x r x dx x dxπ π −= − = − ( )11/2 1 14 161/162 (2 1) 2x xπ π = − = − − ⋅ −

( )7 916 161 ππ= − = (b) ( )( ) 41shell shell 1 1radius height 1602 2dc yV dy y dyπ π = = −

( ) 222 3 21

16 2 321 12 2y yy dy yπ π− − = − = − −

( ) ( ) ( )1 1 1 18 2 32 322 2π π1 18 4 = − − − − − = + 2 9(8 1)π π32 16= + =

38. (a) [ ] [ ]( ) 422 2 1 1161( ) ( )dc yV R y r y dy dyπ π = − = − ( ) ( )231 1 1 1 13 16 24 8 3 161

yyπ π− = − − = − − − − −

1148 48

( 2 6 16 3)π π= − − + + =



(b) ( )( ) ( ) ( ) 2 11 1shell shell 1/2 3/21 2radius height 3 21/4 1/4 1/42 2 1 2 2b xa x

V dx dx x x dx xπ π π π = = − = − = −

( ) ( ) ( ) 112 1 2 1 1 4 1 13 2 3 8 32 3 6 16 48 482 1 (4 16 48 8 3)π ππ π = − − ⋅ − = − − + = ⋅ − − + =

39. (a) Disk: 1 2V V V= −

[ ]11

21 1( )

b

aV R x dxπ= and [ ]2

2

22 2 ( )

b

aV R x dxπ= with 21 3( ) xR x += and 2 ( ) ,R x x=

1 1 2 22, 1; 0, 1a b a b= − = = = two integrals are required (b) Washer: 1 2V V V= +

[ ] [ ]( )11

2 21 1 1( ) ( )

b

aV R x r x dxπ= − with 21 3( ) xR x += and 1 1( ) 0; 2r x a= = − and 1 0;b =

[ ] [ ]( )22

2 22 2 2( ) ( )

b

aV R x r x dxπ= − with 22 3( ) xR x += and 2 2( ) ; 0r x x a= = and 2 1b = two integrals are required

(c) Shell: ( )( ) ( )shell shell shellradius height height2 2d dc cV dy y dyπ π= = where shell height ( )2 2 23 2 2 2 ;y y y= − − = − 0c = and 1.d = Only one integral is required. It is, therefore preferable to use the shell method.

However, whichever method you use, you will get .V π=

40. (a) Disk: 1 2 3V V V V= − −

[ ]2( ) ,ii

di ic

V R y dyπ= 1, 2, 3i = with 1( ) 1R y = and 1 1 21, 1; ( )c d R y y= − = = and 2 0c = and 2 1;d = 1/4

3( ) ( )R y y= − and 3 31, 0c d= − = three integrals are required (b) Washer: 1 2V V V= +

[ ] [ ]( )2 2( ) ( ) ,ii

di i ic

V R y r y dyπ= − 1, 2i = with 1 1 1( ) 1, ( ) , 0R y r y y c= = = and 1 1;d =

2 ( ) 1,R y = 1/4

2 ( ) ( ) ,r y y= − 2 1c = − and 2 0d = two integrals are required

(c) Shell: ( ) ( ) ( )shell shell shellradius height height2 2b ba aV dx x dxπ π= = , where shell height ( )2 4 2 4 ,x x x x= − − = + 0a = and 1b = only one integral is required. It is, therefore preferable to use the shell method. However, whichever method you use, you will get 5

6.V π=

41. (a) [ ] [ ]( ) ( ) ( )24 4 42 2 2 2 2 24 4 4( ) ( ) 25 (3) 25 9 16baV R x r x dx x dx x dx x dxπ π π π− − − = − = − − = − − = −

( ) ( )43 64 64 25613 3 3 3416 64 64x x ππ π π− = − = − − − + = (b) Volume of sphere 3 5004

3 3(5) ππ= = Volume of portion removed 500 256 244

3 3 3π π π= − =

42. ( )( ) ( )1shell shell 2radius height 12 2 sin 1 ;baV dx x x dxππ π+= = − 2[ 1 2 ;u x du x dx= − = [ ]001 0, 1 ] sin cos ( 1 1) 2x u x u u du u

π ππ π π π π π= = = + = → = − = − − − =



43. ( )( ) ( ) ( )shell shell 2 3 2radius height 3 20 0 02 2 2 2rb r rh h h h

r r raV dx x x h dx x h x dx x xπ π π π = = − + = − + = − +

( )2 2 213 2 32 r h r h r hπ π= − + =

44. ( )( )shell shell 2 2 2 2 2 2radius height 0 02 2 4d r rcV dy y r y r y dy y r y dyπ π π = = − − − − = − 22

2

02 2 2 1/2 3 2430 0

[ 2 ; 0 , 0] 2 2rr

ru r y du y dy y u r y r u u du u du uππ π = − = − = = = = → − = =

343

rπ=

45. ( ) 1 2 2( )

( ) [( ( )) ] 0 2 [ ( ) ( )] ( );f a a

f a aW a f y a dy x f a f x dx S a−= − = = − = π π

1 2 2 2 2( ) [ ( ( ))) ] ( ) ( ) ( );W t f f t a f t t a f t−′ ′ ′= − = −π π also 2 2( ) 2 ( ) 2 ( ) [ ( ) ( ) ] 2 ( )

t t t

a a aS t f t x dx xf x dx f t t f t a xf x dx= − = − − π π π π π 2 2 2 2( ) ( ) 2 ( ) ( ) 2 ( ) ( ) ( )S t t f t tf t a f t tf t t a f t′ ′ ′ ′= + − − = −π π π π π ( ) ( ).W t S t′ ′= Therefore, W(t) = S(t) for all t ∈ [a, b].

46. ( )/3 2 2 /3 40 30 [2 (sec ) ] [4 tan ] 3V y dy y y= − = − = −π π ππ π π

47. ( )2 211 1 0 1

0 0

shell shell2 2 ( ) 1

radius height

b x xea

V dx xe dx e e e− − −

= = = − = − − = −

π π π π π

48. Use washer cross sections. A washer has inner radius

r = 1, outer radius /2 ,xR e= and area 2 2( ) ( 1).xR r e− = −π π The volume is

ln 3ln 3

0 0( 1) [ ]

(3 ln 3 1) (2 ln 3)

x xV e dx e x= − = −

= − − = −

π ππ π

6.3 ARC LENGTH

1. ( ) ( )1/22 2313 2 2 2 2dydx x x x x= ⋅ + ⋅ = + ⋅ ( )3 32 2 2 40 01 2 1 2L x x dx x x dx = + + = + +

( ) ( ) 3 323 32 2 30 0 01 1 xx dx x dx x = + = + = + 273

3 12= + =

Section 6.3 Arc Length 455


2. 43 9

2 401 ;dy

dxx L x dx= = + 9 9 44 4 9

1 ;u x du dx du dx = + = =

0 1; 4 10]x u x u= = = =

( ) ( )1010 1/2 3/2 84 4 29 9 3 271 1 10 10 1L u du u → = = = −

3. ( )2 422 41 1 124 16dx dxdy dyy yy y= − = − + 4 4

3 34 41 1 1 12 21 116 16

1y y

L y dy y dy = + − + = + +

2 2

23 32 21 11 14 4y y

y dy y dy = + = +

( ) ( )3 13

27 1 1 1 1 1 13 4 3 12 3 4 12 3 4

19y y

− = − = − − − = − − +

( 1 4 3) ( 2) 5312 12 6

9 9− − + −= + = + =

4. ( ) ( )21/2 1/21 1 1 12 2 4 2dx dxdy dy yy y y−= − = − + ( ) ( )9 91 1 1 14 41 11 2 2y yL y dy y dy = + − + = + +

( )2

9 9 1/2 1/21 1 12 21 1y

y dy y y dy− = + = +

3/2 993/2 1/2 1/21 22 3 31 1

2 yy y y = + = +

( ) ( )33 321 13 3 3 33 1 11= + − + = − =

5. ( )3 623 61 1 124 16dx dxdy dyy yy y= − = − + 6 6

2 26 61 1 1 12 21 116 16

1y y

L y dy y dy = + − + = + +

3 3 4 22 22 23 34 4 4 81 1 1

y y y yy dy y dy− − − = + = + = −

( ) ( )16 1 1 1 1 1 14 (16)(2) 4 8 32 4 84= − − − = − − + 128 1 8 4 123

32 32− − += =



6. ( ) ( )2 2 2 4 41 12 42 2ydx dxdy dyy y y−= − = − + ( )3 4 4142 1 2L y y dy− = + − +

( )3 4 4142 2y y dy−= + + ( ) ( )23 32 2 2 21 12 22 2y y dy y y dy− −= + = +

( ) ( )33

1 27 81 1 1 12 3 2 3 3 3 2

2

y y− = − = − − −

( ) ( )26 8 131 1 1 12 3 3 2 2 2 46= − + = + =

7. ( ) 2/321/3 1/3 2/31 14 2 16dy dy xdx dxx x x −−= − = − + 2/38 2/3 1

2 1611 xL x dx

− = + − +

( )2/3 28 82/3 1/3 1/31 12 16 41 1xx dx x x dx− −= + + = +

( ) 88 1/3 1/3 4/3 2/33 314 4 81 1x x dx x x− = + = + ( )84/3 2/3 4 23 38 812 2 2 2 (2 1)x x = + = ⋅ + − +

3 998 8

(32 4 3)= + − =

8. 2 22 24 1 1

4(4 4) (1 )2 1 2 1dy

dx x xx x x x

+ += + + − = + + −

( )2 422 41 1 1 14 2(1 ) 16(1 )(1 ) (1 )dydxx xx x+ += + − = + − + 42 (1 )4 1

2 1601 (1 ) xL x dx

−+ = + + − + 42 (1 )4 1

2 160(1 ) xx dx

−+= + + + 2 22 (1 )2

40(1 ) xx dx

−+ = + +

22 (1 )2

40(1 ) ;xx dx

−+ = + + [ 1 ; 0 1, 2 3]u x du dx x u x u= + = = = = = ( )3 2 21

41L u u du−→ = +

( ) ( )3 31 108 1 4 3 106 531 1 1 13 4 12 3 4 12 12 61 9u u− − − + = − = − − − = = =



9. ( ) ( ) 222 21 1 1 14 4 2 16dx x dx x xdy x dy x x= − = = − = − + 2 2

2 2

2 21 1 1 11 12 16 2 16

1 x xx x

L dx dx = + − + = + + =

( ) ( ) 2222 21 1

4 4 81 1 1lnx x x

x xdx dx x + = + = + =

( ) ( ) 34 18 8 8ln 2 ln1 ln 2+ − − = + 0 1 2

0.5

0.5y x x

x

y

= −ln2

8

10. ( ) ( ) 22 2 21 1 1 14 4 2 16dy dydx x dx x xx x x= − = − = − + 2

3 2 1 121 16

1x

L x dx = + − + =

( )2 23 32 1 1 12 41 116 xxx dx x dx= + + = + =

( ) 2 33 1 14 2 41 1lnx

xx dx x + = + =

( ) ( )9 1 1 1 12 4 2 4 4ln3 ln1 4 ln3+ − + = +

0 1 2 3

12345

yx x

x

y

= −2

2 4ln

11. ( ) ( )2 2 4222 2 41 1 1 124 4 16dy dydx dxx x xx x x= − = − = − + 4

3 4 1 121 16

1x

L x dx = + − + =

( )4 2 23 34 21 1 121 116 4x xx dx x dx+ + = + = ( ) 32 33 2 1 13 41 4 1x xxx dx + = − = ( ) ( ) 531 1 112 3 3 69 + − + =

0 1 2 3

246810

y xx

x

y

= +3

314

12. ( ) ( )4 4 8224 4 81 1 1 124 4 16dy dydx dxx x xx x x= − = − = − + 8

1 8 1 121/2 16

1x

L x dx = + − + =

( )8 4 21 18 41 1 121/2 1/216 4x xx dx x dx+ + = + = ( ) 54 3 11 4 1 151/2 4 12 1/2xx xx dx + = − =

( ) ( ) 3731 1 1 25 12 160 3 480− − − =

0 0.5 1 1.5

0.5

1y x

x

x

y

= +5

35112

13. ( )24 4sec 1 sec 1dx dxdy dyy y= − = − ( )/4 /44 2/4 /41 sec 1 secL y dy y dy

π ππ π− −

= + − =

[ ] /4/4tan 1 ( 1) 2yπ

π−= = − − =



14. ( )24 43 1 3 1dy dydx dxx x= − = − ( )1 14 22 21 3 1 3L x dx x dx

− −

− − = + − =

3 1 33 3 7 33 3 3 32

3 1 ( 2) ( 1 8)x−

− = = − − − = − + =

15. (a) ( )2 22 4dy dydx dxx x= = ( )221 1 dydxL dx− = +

2 21

1 4x dx−

= + (c) 6.13L ≈

(b)

16. (a) ( )22 4sec secdy dydx dxx x= = 0 4

/31 secL x dx

π− = +

(c) 2.06L ≈

(b)

17. (a) ( )2 2cos cosdx dxdy dyy y= = 2

01 cosL y dy

π = +

(c) 3.82L ≈

(b)

18. (a) ( ) 2 222

11

y ydx dxdy dy yy −−

= − =

2

2 2

1/2 1/2 11/2 1/21 1

1 yy y

L dy dy− −− −

= + =

( ) 1/21/2 21/2 1 y dy−

−= −

(c) 1.05L ≈

(b)



19. (a) ( )2 22 2 2 ( 1)dx dxdy dyy y+ = = + 3 21

1 ( 1)L y dy−

= + + (c) 9.29L ≈

(b)

20. (a) ( )2 2 2cos cos sin sindy dydx dxx x x x x x= − + = 2 2

01 sinL x x dx

π = +

(c) 4.70L ≈

(b)

21. (a) ( )2 2tan tandy dydx dxx x= = 2 2

2

/6 /62 sin cos0 0 cos

1 tan x xx

L x dx dxπ π + = + =

/6 /6

cos0 0secdx

xx dx

π π= =

(c) 0.55L ≈

(b)

22. (a) ( )22 2sec 1 sec 1dx dxdy dyy y= − = − ( )/4 /42/3 /31 sec 1 | sec |L y dy y dy

π ππ π− −

= + − = /4

/3sec y dy

ππ−

= (c) 2.20L ≈

(b)

23. (a) ( )2dydx corresponds to 14x here, so take dydx as 12 .x Then y x C= + and since (1,1) lies on the curve, 0.C = So y x= from (1,1) to (4, 2).

(b) Only one. We know the derivative of the function and the value of the function at one value of x.

24. (a) ( )2dxdy corresponds to 41y here, so take dydx as 21 .y Then 1yx C= − + and, since (0, 1) lies on the curve, 1.C = So 1

1.

xy −=

(b) Only one. We know the derivative of the function and the value of the function at one value of x.



25. 2/4 /4 /4 2

0 0 0 0cos 2 cos 2 1 cos 2 1 cos 2 2cos

x dydx

y t dt x L x dx x dx x dxπ π π = = = + = + =

[ ] ( )/4 /40 40 2 cos 2 sin 2 sin 2 sin(0) 1x dx xπ π π= = = − =

26. ( ) ( ) ( ) ( ) ( )1/2 1/22/3 2/3

1/3 1/3

21 13/2 1/2 12/3 2/3 1/32 3 2

4 2 3 2 /41 , 1 1 1

x xdydx x x

y x x x x L dx− −−

= − ≤ ≤ = − − = − = + −

2/3

2/3 2/3 2/3 1/3

11 1 1 1 1 1/3 2/31 31 1 122 /4 2 /4 2 /4 2 /4 2 /4 2 /4

1 1 1xx x x x

dx dx dx dx x dx x−− = + = + − = = = =

( ) ( )2/32/3 23 3 3 3 312 2 4 2 2 2 4(1)= − = − = total length ( )348 6= =

27. 22 22

0 0 03 2 ,0 2 2 1 ( 2) 5 5 2 5.dy

dxy x x L dx dx x = − ≤ ≤ = − = + − = = =

2 2(2 0) (3 ( 1)) 2 5d = − + − − =

28. Consider the circle 2 2 2 ,x y r+ = we will find the length of the portion in the first quadrant, and multiply our result by 4.

2 2

2 2 2 22 2 2 2

22 2

0 0 0, 0 4 1 4 1 4

r r rdy x x x rdx r x r xr x r x

y r x x r L dx dx dx− −− −− −

= − ≤ ≤ = = + = + =

2 2 2 20 04 4

r r dxr

r x r xdx r

− −= =

29. 2 2 2 2 29 ( 3) 9 ( 3) 18 2 ( 3) ( 3) 3( 3)( 1)d d dxdy dy dy

x y y x y y x y y y y y = − = − = − + − = − −

( 3)( 1) ( 3)( 1)6 6

;y y y ydxdy x x

dx dy− − − − = = 2 2

2

2( 3)( 1) ( 3) ( 1)2 2 2 2 2 26 36

y y y yx x

ds dx dy dy dy dy dy− − − − = + = + = +

2 2 2 2 2

2

( 3) ( 1) ( 1) 2 1 4 ( 1)2 2 2 2 24 4 44 ( 3)

1y y y y y y yy y yy y

dy dy dy dy dy− − − − + + +−

= + = + = =

30. [ ]2 2 2 2 4 44 64 4 64 8 2 0 ;dy dyd d x xdx dx dx dx y yx y x y x y dy dx − = − = − = = = 2 22 2 2 2

2 2 2 2

2 162 2 2 2 2 2 2 2 24 16 16 4 64 161 y xx x x x xy y y y y

ds dx dy dx dx dx dx dx dx dx+ − + = + = + = + = + = =

2

2 22 2 220 64 4 (5 16)x

y ydx x dx−= = −

31. ( ) ( )2 202 1 , 0 2 1 1 ( )x dy dy dydt dx dxx dt x y f x x C= + ≥ = + = ± = = ± + where C is any real number.



32. (a) From the accompanying figure and definition of the differential (change along the tangent line) we see that 1( )k kdy f x x−′= Δ length of kth tangent fin is

( ) ( ) [ ]2 2 22 1( ) ( ) .k k k kx dy x f x x−′Δ + = Δ + Δ

(b) Length of curve ( ) [ ]2 211 1

lim (length of th tangent fin) lim ( )n n

k k kn nk k

k x f x x−→∞ →∞= =

′= = Δ + Δ

[ ] [ ]2 211

lim 1 ( ) 1 ( )n b

k k an k

f x x f x dx−→∞ =

′ ′= + Δ = +

33. 2 2 21 1 ;x y y x+ = = − { } ( ) ( )4

2 231 11 14 2 4

1

0, , , , 1 i i i ik

P L x x y y− −=

= ≈ − + −

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2 22 22 215 3 15 7 3 73 31 1 1 14 4 2 4 2 4 4 2 4 2 4 40 1 1 0= − + − + − + − + − + − + − + − 1.55225≈

34. Let 1 1 2 2( , ) and ( , ),x y x y with 2 1,x x> lie on ,y mx b= + where 2 12 1

,y yx x

m−−= then

dydx

m=

[ ] ( ) ( ) ( )2 2 121 2 11

22 2 2

2 1 2 11 1 1 1x y yx

x x xxL m dx m x m x x x x

−− = + = + = + − = + −

( ) ( )( )

( ) ( ) ( )( ) ( ) ( ) ( )2 22 2

2 1 2 12 1 2 12

2 12 1

2 22 1 2 1 2 1 2 1 .

x x y yx x y y

x xx xx x x x x x y y

− + −− + −−−

= − = − = − + −

35. 3/2 1/22 3 ;dydx

y x x= = ( )21/20 0( ) 1 3 1 9 ;x x

L x t dt t dt= + = + 1 91 9 3/2 3/21 2 2 2

9 27 27 271 1[ 1 9 9 ; 0 1, 1 9 ] (1 9 ) ;

xxu t du dt t u t x u x u du u x

++ = + = = = = = + → = = + − 2(10 10 1)3 22 2

27 27 27(1) (10)L −= − =

36. 3

2 22 2 21 1 1

3 4 4 4( 1) 4( 1)2 1 ( 1) ;dyx

x dx x xy x x x x x+ + +

= + + + = + + − = + −

4 4 2

2 2 4

2 24( 1) 1 [4( 1) 1]2 1

0 0 04( 1) 4( 1) 16( 1)( ) 1 ( 1) 1 1

x x xt t

t t tL x t dt dt dt+ − + −

+ + + = + + − = + = +

4 8 4 8 4 4 2 4

4 4 4 2

16( 1) 16( 1) 8( 1) 1 16( 1) 8( 1) 1 [4( 1) 1] 4( 1) 1

0 0 0 016( 1) 16( 1) 16( 1) 4( 1)

x x x xt t t t t t t

t t t tdt dt dt dt+ + + − + + + + + + + + + +

+ + + += = = =

22 1

0 4( 1)( 1) ;

x

tt dt

+ = + +

1 2 2141

[ 1 ; 0 1, 1]x

u t du dt t u t x u x u u du+ − = + = = = = = + → +

( ) ( )13 1 3 31 1 1 1 1 1 1 1 13 4 3 4( 1) 3 4 3 4( 1) 121 ( 1) ( 1) ;x

x xu u x x

+−+ +

= − = + − − − = + − − 8 591 13 8 12 24

(1)L = − − =



37-42. Example CAS commands:

Maple: with( plots );

with( Student[Calculus1] );

with( student );

f : - sqrt(1-x^2);a : -1;x= > = b : 1;= N : [2, 4, 8];= for n in N do xx : [seq( a i*(b-a)/n, i 0..n )];= + = pts : [seq([x, f (x)], x xx)];= = L : simplify(add( distance(pts[i 1], pts[i]), i 1..n ));= + = # (b) T : sprintf("#37(a) (Section 6.3)\nn %3d L %8.5f \n", n, L );= = = P[n] : plot( [f (x), pts], x a..b, title T ):= = = # (a) end do: display( [seq(P[n], n N)], insequence true, scaling constrained );= = = L : ArcLength( f(x), x a..b, output integral ):= = = L evalf ( L );= # (c)

Mathematica: (assigned function and values for a, b, and n may vary) Clear[x, f ]

2_{a, b} { 1, 1}; f[x ] Sqrt[1 x ]= − = − p1 Plot[f[x], {x, a, b}]= n 8;= pts Table[{xn, f[xn]}, {xn, a, b, (b a)/n}]/ / N= − Show[p1,Graphics[{Line[pts]}]}]

2 2Sum[ Sqrt[ (pts[[i 1,1]] pts[[i, 1]]) (pts[[i 1, 2]] pts[[i, 2]]) ], {i, 1, n}]+ − + + −

2NIntegrate[Sqrt[ 1 f '[x] ], {x, a, b}]+

6.4 AREAS OF SURFACES OF REVOLUTION

1. (a) ( )22 4sec secdy dydx dxx x= = /4 4

02 (tan ) 1 secS x x dx

ππ = +

(c) 3.84S ≈

(b)

Section 6.4 Areas of Surfaces of Revolution 463


2. (a) ( )2 22 4dy dydx dxx x= =

2 2 20

2 1 4S x x dxπ = + (c) 53.23S ≈

(b)

3. (a) ( )2 421 1 11 dx dxy dy dyy yxy x= = = − = 2 41

12 1

yS y dyπ − = +

(c) 5.02S ≈

(b)

4. (a) ( )2 2cos cosdx dxdy dyy y= = 2

02 (sin ) 1 cosS y y dy

ππ = +

(c) 14.42S ≈

(b)

5. (a) ( )21/2 1/2 1/23 3x y y x+ = = − ( ) ( )1/2 1/2122 3dydx x x− = − −

( ) ( )2 21/21 3dydx x− = − ( ) ( )2 24 1/2 1/212 3 1 1 3S x x dxπ − = − + −

(c) 63.37S ≈

(b)

6. (a) ( ) ( )2 21/2 1/21 1dx dxdy dyy y− −= + = + ( ) ( )22 1/212 2 1 1S y y y dxπ − = + + +

(c) 51.33S ≈

(b)



7. (a) ( )2 2tan tandx dxdy dyy y= = /3 2

0 02 tan 1 tan

yS t dt y dy

ππ = +

/3

0 02 tan sec

yt dt y dy

ππ =

(c) 2.08S ≈

(b)

0 0.2 0.4 0.6 0.8

0.5

1

x t dt

x

y

y= ∫ tan0

8. (a) ( )22 21 1dy dydx dxx x= − = − ( )5 2 21 12 1 1 1

xS t dt x dxπ = − + −

5 2

1 12 1

xt dt x dxπ = −

(c) 8.55S ≈

(b)

0 1 2 3

1

2

3y t dt

x

y

x= −∫ 21 1

9. 12 2

;dyxdx

y = = ( ) ( ) 2 42 4 45 512 4 2 2 20 0 02 1 2 1 4 5;b dy x x

dxaS y dx S dx x dxπ ππ π π = + = + = = =

Geometry formula: base circumference 2 (2),π= slant height 2 24 2 2 5= + =

Lateral surface area ( )12 (4 ) 2 5 4 5π π= = in agreement with the integral value

10. 2

2 2;x dxdy

y x y= = = ( )2 22 22 20 0 02 1 2 2 1 2 4 5 2 5d dx

dycS x dy y dy y dy yπ π π π = + = ⋅ + = =

2 5 4 8 5;π π= ⋅ = Geometry formula: base circumference 2 (4),π= slant height 2 24 2 2 5= + =

Lateral surface area ( )12 (8 ) 2 5 8 5π π= = in agreement with the integral value

11. 12

;dxdy

= ( ) ( ) 2 32 23 3( 1) 5 512 2 2 2 21 1 12 1 2 1 ( 1)b dy x x

dxaS y dx dx x dx xπ ππ π + = + = + = + = +

( ) ( )5 59 12 2 2 23 1 (4 2) 3 5;π π π = + − + = + = Geometry formula: 1 11 2 2 1,r = + = 3 12 2 2 2,r = + = slant height 2 2(2 1) (3 1) 5= − + − = Frustum surface area ( )1 2r rπ= + × slant height (1 2) 5 3 5π π= + = in

agreement with the integral value

12. 12 2

2 1 2;x dxdy

y x y= + = − = ( )2 2 21 12 1 2 (2 1) 1 4 2 5 (2 1)d dxdycS x dy y dy y dyπ π π= + = − + = − [ ]

22

12 5 2 5 (4 2) (1 1) 4 5;y yπ π π = − = − − − = Geometry formula: 1 21, 3,r r= =

slant height 2 2(2 1) (3 1) 5= − + − = Frustum surface area (1 3) 5 4 5π π= + = in agreement with the integral value



13. ( )2 4 3 42 2 23 9 9 90 1 ;dy dyx x x xdx dx S dxπ= = = + 4 334 1

9 9 4 91 ;x xu du x dx du dx = + = =

25/9 1/225 19 41

0 1, 2 2x u x u S u duπ= = = = → = ⋅

( ) ( )25/93/2 125 125 27 9822 3 3 27 3 27 811 1uπ π π π− = = − = =

14. ( )21/21 12 4dy dydx dx xx−= = 15/4 15/41 1

4 43/4 3/42 1 2

xS x dx x dxπ π = + = +

( ) ( ) ( )15/4 3/2 3/23/2 4 15 32 1 1 1

3 4 3 4 4 4 43/4

2 x ππ = + = + − +

( )34 4 2843 2 3 31 (8 1)π π π = − = − =

15. ( ) 222 22(2 2 ) (1 )11

2 22 2

dy x dy xxdx dx x xx x x x

− −−−− −

= = =

2

2

1.5 (1 )20.5 2

2 2 1 xx x

S x x dxπ −−

= − + 2 2

2

1.5 2 2 1 20.5 2

2 2 x x x xx x

x x dxπ − + − +−

= −

[ ]1.5 1.50.50.52 2 2dx xπ π π= = =

16. ( )21 14( 1)2 1dy dydx dx xx ++= = 5 51 1

4( 1) 41 12 1 1 2 ( 1)

xS x dx x dxπ π+ = + + = + +

( )53/25 5 52

4 3 411

2 dx 2x xπ π = + = +

( ) ( ) ( ) ( )3/2 3/2 3/2 3/24 5 5 4 25 93 4 4 3 4 45 1π π = + − + = −

( )3 33 34 5 3 98 493 6 6 32 2 (125 27)π π π π= − = − = =

17. ( ) 32 1 22 4 430 1 ;ydx dxdy dyy y S y dyπ= = = + 4 3 31

41 4 ;u y du y dy du y dy = + = =

] ( ) ( )2 1/21 13 410 1, 1 2 2y u y u S u duπ= = = = → = 22 1 2 3/22

6 6 3 91 1( 8 1)u du uπ π π = = = −



18. ( )3/2 1/213 0,x y y= − ≤ when 1 3.y≤ ≤ To get positive area, we take ( )3/2 1/213x y y= − −

( ) ( ) ( )21/2 1/2 11 12 4 2dx dxdy dyy y y y− − = − − = − + ( ) ( )3 3/2 1/2 11 13 41 2 1 2S y y y y dyπ − = − − + − +

( ) ( )3 3/2 1/2 11 13 412 2y y y y dyπ −= − − + +

( ) ( ) ( ) ( )21/2 1/2

1/2

3 3 33/2 1/2 1/2 1/21 1 1 13 2 3 31 1 1

2 1 1 ( 1)y y

yy y dy y y y dy y y dyπ π π

−+ = − − = − − + = − − +

( ) ( ) ( ) ( )3 333 2 27 91 2 1 1 1 1

3 3 9 3 9 3 9 3 9 31 11 3 1 3 1y yy y dy yπ π π π = − − − = − − − = − − − − − − = − − − + +

169 9

( 18 1 3)π π= − − − + =

19. ( )2 15/4 15/41 1 14 40 04 2 2 4 1 4 (4 ) 1dx dxdy dy y yy S y dy y dyπ π− − −−= = = ⋅ − + = − + ( ) ( )15/4 3/2 3/215/4 3/2 3/2 3/28 15 8 523 3 4 3 40 04 5 4 (5 ) 5 5 5y dy y π ππ π = − = − − = − − − = − −

( ) ( )5 5 40 5 5 5 35 58 83 8 3 8 35 5 ππ π −= − = =

20. ( )2 1 1 1 1/21 1 12 1 2 15/8 5/8 5/82 1 2 2 1 1 2 (2 1) 1 2 2dx dxdy dy y yy S y dy y dy y dyπ π π− −−= = = − + = − + = ( ) ( ) ( ) ( )1 3/23/2 3/2 5 5 8 2 2 5 54 2 4 2 4 2523 3 8 3 3 128 8 8 2 25/82 2 1 1 16 2 5 5y π π π ππ ⋅ −⋅ = = − = − = = −

21. ( ) ( ) ( )2ln 2 ln 2 2 212 2 2 40 02 1 2 1 ( 2 )y y y y y y y ye e e e e eS dy e e dy− − − −+ − += + = + − + π π ( ) ( ) ( )2 2ln 2 ln 2 ln 2 2 22 2 2 20 0 02 2 ( 2 )y y y y y y y ye e e e e edy dy e e dy− − − −+ + += = = + + ππ π

( ) ( )ln 22 2 2ln 2 2 ln 21 1 1 1 1 12 2 2 2 2 2 2 202 2ln 2 0y ye y e e e− − = + − = + − − + − π π ( ) ( ) ( )151 1 1 12 2 2 4 2 8 164 2ln 2 2 2ln 2 ln 2= ⋅ + − ⋅ = − + = +π π π

22. ( ) ( )3/2 22 2 2 4 2 413 02 2 1 2 2 1 2y x dy x x dx ds x x dx S x x x dxπ= + = + = + + = + +

( ) ( ) ( ) ( )4 2 222 2 22 2 3 4 24 2 4 20 0 0 02 1 2 1 2 2 2 4x xx x dx x x dx x x dxπ π π π π π = + = + = + = + = + =



23. 3 6 6

22 2 3 6 61 1 1 1 1

2 24 16 161 1

y y yds dx dy y dy y dy y dy = + = − + = − + + = + +

3 3

23 31 1

4 4;

y yy dy y dy = + = +

( )32 2 23 4 21 141 1 142 2 2πyS y ds y y dy y y dyπ π − = = + = +

( ) ( ) ( )52

1 32 31 2 2531 1 1 1 15 4 5 8 5 4 5 8 40 20

12 2 2 (8 31 5)y y π ππ π π− = − = − − − = + = ⋅ + =

24. ( )2 /22 2/2cos sin sin 2 (cos ) 1 sindy dydx dxy x x x S x x dxπππ −= = − = = +

25. ( ) ( ) 22 22 221/22 2 2 21

2( 2 )dy dyx x

dx dx a xa xy a x a x x

− −−−

= − = − − = =

( ) [ ]22 22 2 2 2 22 1 2 2 2a a a ax aa a aa xS a x dx a x x dx a dx a xπ π π π −− − −− = − + = − + = = 22 [ ( )] (2 )(2 ) 4a a a a a aπ π π= − − = =

26. ( ) 2 2 2 22 22 2 2 22 20 0 02 1 2h h hdy dy h r r h rr r r r r rh dx h dx h h hh h h hy x S x dx x dx x dxππ π + += = = = + = = ( )2 22 22 2 2 2 2 22 22 20

hr x r h

h hh r h r r h rπ π π = + = + = +

27. The area of the surface of one wok is ( )22 1 .d dxdycS x dyπ= + Now, 2 2 2 2 216 16x y x y+ = = − ( ) 22 22 2

2

1616;y ydx dx

dy dy yy

−−−

= = ( )22 27 72 2 2 2 216 16162 16 1 2 16y yS y dy y y dyπ π− −− −−

= − + = − + 7 216

2 16 32 9 288 904.78 cm .dyπ π π−−

= = ⋅ = ≈ The enamel needed to cover one surface of one wok is 3 30.5 mm 0.05 cm (904.78)(0.05) cm 45.24 cm .V S S= ⋅ = ⋅ = = For 5000 woks, we need

35000 5000 45.24 cm (5)(45.24) 226.2 226.2V L L⋅ = ⋅ = = liters of each color are needed.

28. ( ) 22 22 2 2 22

2 2 212

;dy x x dx xdx dy r xr x r x

y r x −−− −

= − = − = = 2

2 22 22 1

a h xa r x

S r x dxπ+

−= − +

( )2 2 22 2 2 ,a h a ha ar x x dx r dx rhπ π π+ +

= − + = = which is independent of a.

29. ( ) 22 22 2 2 22

2 2 212

;dy x x dx xdx dy R xR x R x

y R x −−− −

= − = − = = 2

2 22 22 1

a h xa R x

S R x dxπ+

−= = − +

( )2 2 22 2 2a h a ha aR x x dx R dx Rhπ π π+ +

= − + = =



30. (a) ( ) 22 22 22

2 2 2 2 2

454545 45 ;y ydx dx

dy dy yyx y x y −

−−+ = = − = =

( )22 245 45 452 2 2 2 222.5 22.5 22.5452 45 1 2 45 2 45y yS y dy y y dy dyπ π π− − −−= − + = − + = ⋅ (2 )(45)(67.5) 6075π π= = square feet

(b) 19,085 square feet

31. (a) An equation of the tangent line segment is (see figure) ( )( ) ( ) .k k ky f m f m x m′= + − When

1kx x −= we have

1 1( ) ( )( )k k k kr f m f m x m−′= + −

( )2 2( ) ( ) ( ) ( ) ;k kx xk k k kf m f m f m f mΔ Δ′ ′= + − = −when kx x= we have

( )2 ( ) ( )k k k kr f m f m x m′= + −

2( ) ( ) ;k

xk kf m f m

Δ′= +

(b) ( ) ( ) ( ) ( ) ( ) [ ]22 2 2 222 2 1 2 2( ) ( ) ( )k kx xk k k k k k k kL x r r x f m f m x f m xΔ Δ ′ ′ ′= Δ + − = Δ + − − = Δ + Δ ( ) [ ]2 2( ) ,k k k kL x f m x′ = Δ + Δ as claimed

(c) From geometry it is a fact that the lateral surface area of the frustum obtained by revolving the tangent line

segment about the -axisx is given by ( ) [ ] ( ) [ ]2 21 2 2 ( ) ( )k k k k k kS r r L f m x f m xπ π ′Δ = + = Δ + Δ using

parts (a) and (b) above. Thus, [ ]22 ( ) 1 ( ) .k k k kS f m f m xπ ′Δ = + Δ

(d) [ ] [ ]2 21 1

lim lim 2 ( ) 1 ( ) 2 ( ) 1 ( )n n b

k k k k an nk k

S S f m f m x f x f x dxπ π→∞ →∞= =

′ ′= Δ = + Δ = +

32. ( ) ( ) ( ) ( ) ( )1/22/3

2/3

1/3 2/3 2/3

213/2 1/22/3 2/3 1/33 12 12 3

1 1 1xdy dy x

dx dxx x xy x x x

−− −= − = − − = − = = −

( ) ( ) ( ) ( )2/33/2 3/2 3/21 1 12/3 2/3 2/3 2/3 1/310 0 02 2 1 1 1 4 1 4 1 ;xS x dx x x dx x x dxπ π π− − = − + − = − = − [ 2/3 1/3 1/3323 21 ;u x du x dx du x dx

− −= − = − − = ]0 1, 1 0x u x u= = = =

( ) ( )00 3/2 5/23 122 22 5 5 51 14 6 6 0S u du u ππ π π → = − = − = − − =

6.5 WORK AND FLUID FORCES

1. The force required to stretch the spring from its natural length of 2 m to a length of 5 m is ( ) .F x kx=

The work done by F is 33 3 2 9

2 20 0 0( ) .k kW F x dx k x dx x = = = = This work is equal to 1800 J

92

1800 400 N/mk k = =

Section 6.5 Work and Fluid Forces 469


2. (a) We find the force constant from Hooke’s Law: 8004

200 lb/in.Fx

F kx k k= = = =

(b) The work done to stretch the spring 2 inches beyond its natural length is 2 2

0 0200W kx dx x dx= =

2 2

2 0200 200(2 0) 400 in-lb 33.3 ft-lbx = = − = =

(c) We substitute 1600F = into the equation 200F x= to find 1600 200 8x x= = in.

3. We find the force constant from Hooke’s law: .F kx= A force of 2 N stretches the spring to 0.02 m Nm

2 (0.02) 100 .k k = ⋅ = The force of 4 N will stretch the rubber band y m, where Fk

F ky y= =

Nm

4 N

1000.04 m 4 cm.y y = = = The work done to stretch the rubber band 0.04 m is

0.04

0W kx dx=

22 0.040.04 (100)(0.04)2 20 0

100 100 0.08 Jxx dx = = = =

4. We find the force constant from Hooke’s law: 90 N1 m

90 .Fx

F kx k k k= = = = The work done to

stretch the spring 5 m beyond its natural length is ( )2 55 5 252 20 0 090 90 (90) 1125 JxW kx dx x dx = = = = =

5. (a) We find the spring’s constant from Hooke’s law: 21,714 21,714 lb8 5 3 in

7238Fx

F kx k k−= = = = =

(b) The work done to compress the assembly the first half inch is 0.5 0.5

0 07238W kx dx x dx= =

22 0.5 (0.5) (7238)(0.25)2 2 20

7238 (7238) 905 in-lb.x = = = ≈ The work done to compress the assembly the

second half inch is: 2 1.01.0 1.0 (7238)(0.75)27238

2 2 20.5 0.5 0.57238 7238 1 (0.5) 2714 in-lbxW kx dx x dx = = = = − = ≈

6. First, we find the force constant from Hooke’s law: ( )116150 lb

in16 150 2,400 .F

xF kx k= = = = ⋅ = If someone

compresses the scale 18

x = in, he/she must weigh ( )182,400 300 lb.F kx= = = The work done to compress the scale this far is

2 1/81/8 2400 2.52 2 64 160 0

2400 18.75 lb in. ft-lbxW kx dx ⋅ = = = = ⋅ =

7. The force required to haul up the rope is equal to the rope’s weight, which varies steadily and is proportional

to x, the length of the rope still hanging: ( ) 0.624 .F x x= The work done is: 50 50

0 0( ) 0.624W F x dx x dx= =

2 50

2 00.624 780 Jx = =

8. The weight of sand decreases steadily by 72 lb over the 18 ft, at 4 lb/ft. So the weight of sand when the

bag is x ft off the ground is ( ) 144 4 .F x x= − The work done is: 18

0( ) (144 4 )

b

aW F x dx x dx= = −

182

0144 2 1944 ft-lbx x = − =



9. The force required to lift the cable is equal to the weight of the cable paid out: ( ) (4.5)(180 )F x x= −

where x is the position of the car off the first floor. The work done is: 180 180

0 0( ) 4.5 (180 )W F x dx x dx= = −

( )2 2 2180 2 180 4.51802 2 204.5 180 4.5 180 72,900 ft-lbxx ⋅ = − = − = =

10. Since the force is acting toward the origin, it acts opposite to the positive -direction.x Thus 2( ) .kx

F x = −

The work done is ( )2 2 ( )1 1 1 1bb b k a bk x b a aba a ak xW dx k dx k k− = − = − = = − =

11. Let r = the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a constant rate, the amount of water in the bucket is proportional to (20 ),x− the distance the bucket is being raised. The leakage rate of the water is 0.8 lb/ft raised and the weight of the water in the bucket is

0.8(20 ).F x= − So: 2 2020

20 00.8 (20 ) 0.8 20 160 ft-lb.xW x dx x = − = − =

12. Let r = the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a constant rate, the amount of water in the bucket is proportional to (20 ),x− the distance the bucket is being raised. The leakage rate of the water is 2 lb/ft raised and the weight of the water in the bucket is 2(20 ).F x= −

So: 2 2020

20 02(20 ) 2 20 400 ft-lb.xW x dx x = − = − =

Note that since the force in Exercise 12 is 2.5 times the force in Exercise 11 at each elevation, the total work is also 2.5 times as great.

13. We will use the coordinate system given. (a) The typical slab between the planes at y and y y+ Δ

has a volume of (10)(12) 120V y yΔ = Δ = Δ ft3. The force F required to lift the slab is equal to its weight:

62.4 62.4 120 lb.F V y= Δ = ⋅ Δ The distance through which F must act is about y ft, so the work done lifting the slab is about WΔ = force × distance

62.4 120 ft-lby y= ⋅ ⋅ ⋅ Δ The work it takes to lift all the water is approximately

20 20

0 0

62.4 120 ft-lb.W W y y≈ Δ = ⋅ ⋅ Δ

This is a Riemann sum for the function 62.4 120y⋅ over the interval 0 20.y≤ ≤ The work of pumping the tank empty is the limit of these sums:

( )22020 400

2 20 062.4 120 (62.4)(120) (62.4)(120) (62.4)(120)(200) 1,497,600 ft-lbyW y dy = ⋅ = = = =

(b) The time t it takes to empty the full tank with ( )511 − hp motor is ft-lb ft-lbsec sec

1,497,600 ft lb250 250

5990.4Wt ⋅= = = sec

1.664 hr 1t= ≈ hr and 40 min (c) Following all the steps of part (a), we find that the work it takes to lower the water level 10 ft is

( )21010 100

2 20 062.4 120 (62.4)(120) (62.4)(120) 374,400 ft-lbyW y dy = ⋅ = = = and the time is ft-lbsec250

Wt =

1497.6 sec 0.416 hr 25= = ≈ min



(d) In a location where water weighs 62.26 3lbft

:

a) (62.26)(24,000) 1,494,240 ft-lbW = = .

b) 1,494,240250

5976.96 sec 1.660 hr 1t t= = ≈ ≈ hr and 40 min

In a location where water weighs 62.59 3lbft

a) (62.59)(24,000) 1,502,160 ft-lbW = = ⋅

b) 1,502,160250

6008.64 sec 1.669 hr 1 hrt t= = ≈ ≈ and 40.1 min

14. We will use the coordinate system given. (a) The typical slab between the planes at y and y y+ Δ has

a volume of 3(20)(12) 240 ft .V y yΔ = Δ = Δ The force F required to lift the slab is equal to its weight:

62.4 62.4 240F V y= Δ = ⋅ Δ lb. The distance through which F must act is about y ft, so the work done lifting the slab is about force distanceWΔ = ×

62.4 240 ft-lb.y y= ⋅ ⋅ ⋅ Δ The work it takes to lift all the water is approximately 20

10

W W≈ Δ 20

10

62.4 240 ft-lb.y y= ⋅ ⋅ Δ This is a Riemann sum for the function 62.4 240y⋅ over the interval

10 20.y≤ ≤ The work it takes to empty the cistern is the limit of these sums:

2 2020 y210 10

62.4 240 (62.4)(240) (62.4)(240)(200 50) (62.4)(240)(150)W y dy = ⋅ = = − = 2,246,400 ft-lb=

(b) ft-lbsec

2,246,400 ft-lb275275

8168.73 sec 2.27 hours 2 hrWt = = ≈ ≈ ≈ and 16.1 min

(c) Following all the steps of part (a), we find that the work it takes to empty the tank halfway is

( ) ( )21515 225 100 125

2 2 2 210 1062.4 240 (62.4)(240) (62.4)(240) (62.4)(240) 936,000 ft.yW y dy = ⋅ = = − = =

Then the time is ft-lbsec

936,000275275

3403.64 sec 56.7 minWt = = ≈ ≈

(d) In a location where water weighs 62.26 3lbft

:

a) (62.26)(240)(150) 2,241,360 ft-lb.W = =

b) 2,241,360275

8150.40 sec 2.264t = = = hours 2 hr≈ and 15.8 min

c) ( )1252(62.26)(240) 933,900 ft-lb;W = = 933,900275 3396 sec 0.94 hours 56.6 mint = = ≈ ≈ In a location where water weighs 62.59 3

lbft

:

a) (62.59)(240)(150) 2,253,240 ft-lb.W = =

b) 2,253,240275

8193.60 sec 2.276 hourst = = = ≈ 2 hr and 16.56 min

c) ( )1252(62.59)(240) 938,850 ft-lb;W = = 938,850275 3414 sec 0.95t = ≈ ≈ hours 56.9≈ min



15. The slab is a disk of area ( )22 2 ,yxπ π= thickness ,yΔ and height below the top of the tank (10 ).y− So the work to pump the oil in this slab, ,WΔ is 57 ( )22(10 ) π .yy− The work to pump all the oil to top of the tank is

( ) 3 41010 102 357 57π

4 4 3 40 010 11,875 ft lb 37,306 ft-lby yW y y dyπ π = − = − = ⋅ ≈

16. Each slab of oil is to be pumped to a height of 14 ft. So the work to pump a slab is ( )22(14 )( ) yy π− and since the tank is half full and the volume of the original cone is ( )2 2 32501 13 3 35 (10) ft ,V r h ππ π= = = half the volume 3250π

6ft ,= and with half the volume the cone is filled to a height y,

23250 1

6 3 4500 ft.y y yπ π= =

So ( )3

3 3 4 500500 142 357 57π4 4 3 40 0

14 60,042 ft-lb.y yW y y dyπ = − = − ≈

17. The typical slab between the planes at y and y y+ Δ has a volume of 2(radius) (thickness)V πΔ =

( )2 3202 100 ft .y yπ π= Δ = ⋅ Δ The force F required to lift the slab is equal to its weight: 51.2 51.2 100 lb 5120 lbF V y F yπ π= Δ = ⋅ Δ = Δ The distance through which F must act is about

(30 ) ft.y− The work it takes to lift all the kerosene is approximately 30 30

0 0

5120 (30 ) ft-lbW W y yπ≈ Δ = − Δ

which is a Riemann sum. The work to pump the tank dry is the limit of these sums:

( )23030 900

2 20 05120 (30 ) 5120 30 5120 (5120)(450 ) 7,238,229.48 ft-lbyW y dy yπ π π π = − = − = = ≈

18. (a) Follow all the steps of Example 5 but make the substitution of 64.5 3lbft

for 57 3lbft

. Then,

( ) ( )( )( )3 4 3 4 388 102 364.5 64.5 64.5 10 8 8 64.5 10 64.5 84 4 3 4 4 3 4 4 3 30 0(10 ) 8 2y yW y y dyπ π π π π⋅ ⋅ = − = − = − = − =

321.5 8 34,582.65 ft-lbπ= ⋅ ≈ (b) Exactly as done in Example 5 but change the distance through which F acts to distance (13 )y≈ − ft.

Then ( ) ( ) ( )( )3 4 3 4 388 132 357 57 57 13 8 8 57 13 57 8 74 4 3 4 4 3 4 4 3 3 40 0(13 ) 8 2y yW y y dyπ π π π π⋅ ⋅ ⋅⋅

= − = − = − = − = 2(19 )(8 )(7)(2) 53,482.5 ft-lbπ= ≈

19. The typical slab between the planes at y and y y+ Δ has a volume of about 2(radius) (thickness)V πΔ =

( )2 3ft .y yπ= Δ The force ( )F y required to lift this slab is equal to its weight: ( ) 73F y V= ⋅ Δ ( )273 73 lb.y y y yπ π= Δ = Δ The distance through which ( )F y must act to lift the slab to the top of the

reservoir is about (4 )y− ft, so the work done is approximately 73 (4 ) ft-lb.W y y yπΔ ≈ − Δ The work done

lifting all the slabs from 0y = ft to 4y = ft is approximately ( )0

73 4 ft-lb.n

k kk

W y y yπ=

≈ − Δ Taking the limit



of these Riemann sums as ,n → ∞ we get ( )4 4 20 073 (4 ) 73 4W y y dy y y dyπ π= − = − ( )42 3 64 233613 3 3073 2 73 32 ft-lb 2446.25 ft-lb.y y ππ π = − = − = ≈

20. The typical slab between the planes at y and y y+ Δ has volume of about VΔ = (length)(width)(thickness)

2 32 25 (10) ft .y y = − Δ

The force ( )F y required to lift this slab is equal to its weight:

2 2( ) 53 53 2 25 (10) 1060 25 lb.F y V y y y y = ⋅ Δ = − Δ = − Δ

The distance through which ( )F y must act to

lift the slab to the level of 15 m above the top of the reservoir is about (20 ) ft,y− so the work done is

approximately 21060 25 (20 ) ft-lb.W y y yΔ ≈ − − Δ The work done lifting all the slabs from 5 fty = − to

5 fty = is approximately ( )20

1060 25 20 ft-lb.n

k kk

W y y y=

≈ − − Δ Taking the limit of these Riemann sums as

,n → ∞ we get 5 52 25 51060 25 (20 ) 1060 (20 ) 25W y y dy y y dy

− −= − − = − −

5 52 25 5

1060 20 25 25 .y dy y y dy− −

= − − − To evaluate the first integral, we use we can interpret 5 25

25 y dy−

− as the area of the semicircle whose radius is 5, thus 5 52 25 5

20 25 20 25y dy y dy− −

− = − 21

220 (5) 250 .π π = = To evaluate the second integral let

225 2 ;u y du y dy= − = − 5 0,y u= − =

5 0,y u= = thus 5 02 1

25 025 0.y y dy u du

−− = − = Thus,

5 25

1060 20 25 y dy−

− − 5 52 25 5

1060 20 25 25 1060(250 0) 265000 832522 ft-lby dy y y dy π π− −

− − − = − = ≈

21. The typical slab between the planes at y and y y+ Δ has a volume of about 2(radius) (thickness)V πΔ =

( )22 325 m .y yπ= − Δ The force ( )F y required to lift this slab is equal to its weight: ( )

22 2( ) 9800 9800 25 9800 25F y V y y y yπ π = ⋅Δ = − Δ = − Δ

N. The distance through which ( )F y must

act to lift the slab to the level of 4 m above the top of the reservoir is about (4 ) m,y− so the work done is

approximately ( )29800 25 (4 ) N m.W y y yπΔ ≈ − − Δ ⋅ The work done lifting all the slabs from 5 my = − to 0 my = is approximately ( )

02

5

9800 25 (4 ) N m.W y y yπ−

≈ − − Δ ⋅ Taking the limit of these Riemann sums,

we get ( ) ( )0 02 2 35 59800 25 (4 ) 9800 100 25 4W y y dy y y y dyπ π− −= − − = − − + ( )4

02 325 25 25 6254 4

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