x v=x(1 +m?)[vex1)(x x2)dx= zt+m?)(x2x1)?

SECTION 6.4 || The Method of Cylindrical Shells 739

SOLUTION Let x; and x2 denote the x-coordinates of the points ofintersection between the circle x? + y? = 1 and theliney = mx + b with x1 < x. Rotating the region enclosed by the two curves about the x-axis produces a solid whose crosssectionsare washers with outer radius R = V1 x? and inner radius r = mx + b. The volumeoftheresulting solid is then

xBecause x1 and x2 are roots of the equation (1 x?) (mx + b)* = Oand (1 x*) (mx + b)is a quadratic polynomial in xwith leading coefficient (1 + m7), it followsthat (1 x?) (mx + b)? = -(I+ m?)(x x1)(x x2). Therefore,

V = x(1 +m?) [ve x1)(x x2) dx = zt + m?)(x2 x1)?.xFrom the diagram, we see that h = x2 x). Moreover, by the Pythagorean theorem, d? = h? + (mh)? = (1 + m?)h?. Thus,

V= =( +m2)h? = ah [a + mi?| = ha?

6.4 The Method of Cylindrical Shells

Preliminary Questions1. Considerthe region R underthe graph of the constant function f(x) = h overthe interval [0, r]. Give the height and the radius

of the cylinder generated when R is rotated about:(a) the x-axis (b) the y-axisSOLUTION(a) Whentheregion is rotated about the x-axis, each shell will have radius / and heightr.(b) Whenthe regionis rotated about the y-axis, each shell will have radius rand height h.2. Let V be the volumeofa solid of revolution about the y-axis.

(a) Doesthe Shell Method for computing Vleadto an integral with respect to x or y?(b) Does the Disk or Washer Method for computing V lead to an integral with respect to x or y?SOLUTION(a) The Shell method requires slicing the solid parallel to the axis ofrotation. In this case, that will mean slicing the solid in thevertical direction, so integration will be with respectto x.(b) The Disk or Washer method requires slicing the solid perpendicularto the axis ofrotation. In this case, that meansslicing thesolid in the horizontal direction, so integration will be with respectto y. ExercisesIn Exercises 1-6, sketch the solid obtainedbyrotating the region underneath the graph ofthefunctionoverthe given interval aboutthe y-axis, andfind its volume.1. f@) =x, [0,1]SOLUTION sketch ofthe solid is shown below.Each shell has radius x and height x3, so the volumeofthe solid is

1 11 12x [ x-x3 dx = 2m | x4 dx = 2x =x =

0 0 5 0 alt a

y

2. f(x) = yx, [0,4SOLUTION A sketchofthe solid is shown below. Eachshell has radius x and height ./x, so the volumeofthe solid is

2 128x5/24 42x | xJ/x dx = 2x | x3/2 dy = 2n {= = n.0 0 5 0 5

740 CHAPTER 6 | APPLICATIONS OF THE INTEGRAL

3. f(a)=x"*, [1,3]

1SOLUTION sketch of the solid is shown below. Each shell has radius x and height x , so the volumeofthesolid is3 3

2x | x(x7!) dx = an | 1 dx = 2z (x)1 1

3

14x.

0.840.6 7 0.2 -3 -2 I E Z 3

4. f(x) =4-x?, [0,2]SOLUTION sketch ofthe solid is shown below. Eachshell has radius x and height 4 x?, so the volumeofthesolidis

2 2 12x | x(4=x2) dx = 2x | (4x x3) dx = 27 (2?0 0 2) 2= 81. 0

y

25. f(x) = Vx2+09, [0,3]SOLUTION sketch ofthe solid is shown below. Eachshell has radius x and height x2 + 9, so the volumeofthe solid is

32x | xVx2+94x.0Let u = x + 9. Then du = 2x dx and

3zu | xVx2+9dx= 7x0

1

9

18 2Vu du = x (°°)

8 = 187(2V2 1). 9

6. f(x) = TE [1,4]x==,so the volumeofthe solid isVl+x3

4 7 4 22x | (55) dx = 2m | dx.1 V1+x3 1 v1+x3

SOLUTION sketch ofthesolid is shown below. Each shell has radius x and height

SECTION 6.4 | The Method of Cylindrical Shells 741

Let u = 1 + x3. Then du = 3x? dx and4 42 2 65 2 65 4

2x | ds = 5| uU? au = Sa (2u/?) == (16542).1 YV14+x? 3 J2 3 3

In Exercises 7-12, use the Shell Method to compute the volume obtained byrotating the region enclosed bythe graphs asindicated,about the y-axis.7. y =3x 2, p=6=x, x=0SOLUTION The region enclosed by y = 3x 2, y = 6 x and x = 0 is shown below. Whenrotating this region about they-axis, each shell has radius x and height 6 x (3x 2) = 8 4x. The volumeofthe resulting solid is

2 2 4 2? 32zu | x(8 4x) dx = 2x | (8x 4x?) dx = 27 (4 3°) = 1.0 0 3 0 3

8. y= Vx, y=x?SOLUTION The region enclosed by y = ./x and y = x? is shown below. Whenrotating this region about the y-axis, each shellhas radius x and height ./x x2. The volumeofthe resulting solid is

1 1 2 1 i 32x | x(VX x2) dx = 2x | (x3/2 x3) dx = In =x5/2 x40 0 5 4 Jo 10

9. y = x?, y=8-x?, x=0, fox>0SOLUTION The region enclosed by y = x?, y = 8 x? and the y-axis is shown below. Whenrotating this region about they-axis, each shell has radius x and height 8 x? x? = 8 2x?. The volumeofthe resulting solid is

2 2 21an | x(8 2x?) dx = 2x [ (8x 2x3) dx = 27 4x? ¿e = 16x.0 0 0

y


10. y=8-x3, y=8-4x, forx>0SOLUTION The region enclosed by y = 8 x3 and y = 8 4x is shown below. Whenrotating this region about the y-axis,each shell has radius x and height (8 x3) (8 4x) = 4x x3. The volumeofthe resultingsolid is

2 2 4 1 5\|? 1282x | x(4x x3) dx = 2x [ (4x? x4) dx = 27 219 =x% ==,0 0 3 5 0 15y

y=8-x3

0 0.5 1 1.5 2

1.y=@?+12, y=2-@?+1)2, x=2SOLUTION The region enclosed by y = (x? +1)2,y =2- (x? + 1)~? and x = 2 is shown below. Whenrotating this regionaboutthe y-axis, each shell has radius x and height 2 (x? + 1)~? (x? + 1)~* = 2 2(x? + 1)~*. The volumeofthe resultingsolid is

a 2 2x 1 2 3- -2 2 - Es 2an f x(2 2(x* + 1) )ax = 2x f (#- =) dx = 2(x +75) - 5

y=2- (+1)?

12, y=1-|x-1l|, y=0SOLUTION Theregion enclosed by y = 1 |x 1| and the x-axis is shown below. Whenrotating this region about the y-axis,two different shells are generated. For each x [0, 1], the shell has radius x and height x; for each x [1, 2], the shell has radius xand height 2 x. The volumeofthe resulting solid is

1 2 1 22x | x(x) dx +20 f x(2 x) dx = 2m | (x?) dx +2x | (2x x?) dx

0 1 0 11 1 1 ,\P22x (5) + 27 (+ _ 3°) = 27.0 3 1

In Exercises 13 and 14, use a graphing utility to find the points of intersection of the curves numerically and then compute thevolumeofrotation ofthe enclosed region about the y-axis.

2

y= 5x 5 y = sin(x?), x > 0SOLUTION Theregion enclosed by y = 4x? and y = sinx? is shown below. When rotating this region about the y-axis,each shell has radius x and height sinx? 4x2. Using a computer algebra system, we find that the x-coordinate of the point ofintersection on the right is x = 1.376769504. Thus, the volume of the resulting solid of revolution is

1.376769504 127 I x (sins? _ 3) dx = 1.321975576.0

SECTION 6.4 [| The Method of Cylindrical Shells 743

14. [GU] yp =e/2, y=x, x=0. _r2 . . . .SOLUTION Theregion enclosed by y = e * 12, y = x and the y-axis is shown below. Whenrotating this region about the. . . 2 . . .y-axis, each shell has radius x and height e * 2 _x. Using a computer algebra system, we find that the x-coordinate of the point

of intersection on the right is x = 0.7530891650. Thus, volume of the resulting solid of revolution is

0.7530891650 2Qn / x(e /? x) dx = 0.65685055510

y= een

0.80.60.40.2

0.2 0.4 0.6 0.8

In Exercises 15-20, sketch the solid obtained by rotating the region underneath the graph off(x) over the interval about the givenaxis, and calculate its volume using the Shell Method.

15. f(x) =x, [0,1], aboutx =SOLUTION sketch ofthe solid is shown below. Eachshell has radius 2 x and height x3, so the volume ofthe solid is

x41 1_ 3\ x BM 2 2 = PT2x | (2 x) (x ) ax an | (2x° x") dx 21 3 z =

16. f(x) =x3, [0,1], about x = 2SOLUTION A sketch ofthe solid is shown below. Each shell has radius x ( 2) = x + 2 andheight x3, so the volumeofthesolid is

14 x?1 1an | @+x (x) dx = 2m | (2x? + x4) dx = 2x StS ur.0 0 0

17. f(x) =x7*, [23,1], about x = 4


SOLUTION sketchofthe solid is shown below. Each shell has radius 4 x and height x *, so the volume ofthesolidis>

1

onE Ga) (7) dx = 2x [. (4x~4 = x73) dx = 2n (57 = 3x) 1 28072 3 a 81

118. f(x) = .,, [0,2], aboutx =0Vx2+11SOLUTION sketch ofthe solid is shown below. Eachshell has radius x and height ====,so the volume ofthe solid isVx? +1

2 1 Es 2 wszu | x (a) dr= 2 x*+1 = 2n(V5 1).0 Vx? +1 ( ) 0 )

19. f(x) =a x witha>0, [0,a], aboutx = 1SOLUTION sketchofthe solid is shown below. Eachshell has radius x ( 1) = x + 1 and height a x, so the volumeofthesolid is

anf+ 1) (=x) ax=2n f° (a+ (Dx) dx

Il 0

1 127 ax + 32°) 0aon (at Er) a) _ ears,

20. f(x) =1 x7, [-1,1], x=c withe >1SOLUTION sketch ofthe solid is shown below. Eachshell has radius c x and height 1 x , so the volumeofthesolidis

1 1 1 12x @-x)(1-x2) dx = 2m | (x3 -cx?-x +c) dx = 2x (3x4 S49 53? ox)

1 1

2c 1 2c+1

SECTION 64 || The Method of Cylindrical Shells 745

In Exercises 21-26, sketch the enclosed region and use the Shell Methodto calculate the volume ofrotation about the x-axis.2l.x=y, y=0, x=1SOLUTION When the region shown below is rotated about the x-axis, each shell has radius y and height 1 y. The volume ofthe resulting solid is

11 | 2 lo ly I2x | y= y)dy = 20 f (y y*) dy = 2m =y° - sy =>.0 0 2 3 0 3y

+ + x0 0.2 0.4 0.6 0.8 122.x=4y+1, x= ty, y=0SOLUTION Whenthe region shown belowis rotated about the x-axis, each shell has radius y and height 2 3 y. The volume ofthe resulting solid is

4 4 41 1 1 3272 2-=y) dy =2 2y =y?) dy =2n [y?-=y?)| = .r [> ( x | (2»

23.x=y(4 y), y=0SOLUTION Whenthe region shown belowis rotated about the x-axis, each shell has radius y and height y(4 y). The volumeof the resulting solid is

4 42x [ y4 y) dy = 2x | (y? y°) dy = 2x [=y 2] =0 0 4° /l 3x=y(4-y)

24. x=y(4 y), x=(1-2)SOLUTION Setting y(4 y) = (y 2?yields

y?-4y+2=0 or y=2+ V2.When the region shown below is rotated about the x-axis, each shell has radius y and height 2y? + 8y 4. The volumeoftheresulting solid is

2 2492 _ 647 /224/2 24/2 1 82x | y( 2y + 8y 4) dy = 27 ( 2y3 + 8y? 4y) dy = 2n (--y4 + =y3 2y/2 2/2 2 3 2/2 3

y

x= y4- y)

746 CHAPTER 6 || APPLICATIONS OF THE INTEGRAL

25. y =4-x*?, x=0, y=0SOLUTION Whenthe region shownbelowis rotated about the x-axis, each shell has radius y and height ./4 y. The volume ofthe resulting solid is

42x | yV4 yay.0

Let u = 4 y. Then du = dy, y = 4 u, and4 0 42x [yFay = 2x [ 6-0) Van = 2x [ (45-267) au

0 4 0

8 2 4 256= 2x <u?/? - Zu?/? = 7.3 5 0 15

y43 y=4-x?

21

x0 05 1 15 226. y=x1/83-2 y=0, x=27SOLUTION When the region shown below is rotated about the x-axis, each shell has radius y and height 27 (y + 2)?. Thevolumeofthe resulting solid is

1 127 / y: (27 -(y+ 2°) dy = 2x | (19y = ly? ay? = y) dy

0 019 3 1= 21 (3 4y3 _ =y4 - 3°)

y

pa x0 5 10 15 20 25 30

27. Use boththe Shell and Disk Methodsto calculate the volumeobtainedbyrotating the region under the graph of f(x) = 8 x3for 0 < x < 2 about:(a) the x-axis (b) the y-axisSOLUTION(a) x-axis: Using the disk method, the cross sections are disks with radius R = 8 x3; hence the volumeofthe solid is

2 576x2 1x | (8 ipdx=n (4x 4x4 4 33")0 0 77

With the shell method, each shell has radius yand height (8 y)!/ 3, The volumeof the solid is82x | y(8 y)/9 dy

0

Let u = 8 y. Then dy = du, y = 8 u and8 8 8

2x | y (8-y)13 dy=2% | 62) -uMdu = 22 | (8u1/3 y*/3) du0 0 08

= 27 (9 37/3 37677 0 7


(b) y-axis: With the shell method, each shell has radius x and height 8 x3. The volumeofthe solid is2 21 96an | x(8 x3) dx = 20 (4x2 =x5\| = 27.0 5 0 5

Using the disk method,the cross sections are disks with radius R = (8 y) 1/3, The volumeis then given by8 8nf @=dy = eya] = 20 5 à à

28. Sketch the solid of rotation about the y-axis for the region under the graph of the constant function f(x) = c (where c > 0)forO <x <r.(a) Find the volume without using integration.(b) Use the Shell Method to compute the volume.SOLUTION

(a) Thesolid is simply a cylinder with height c and radius r. The volumeis given by srr?c.(b) Each shell has radius x and height c, so the volumeis

2 029. The graphin Figure 1(A) can be described by both y = f(x) and x = h(y), where his the inverse of f. Let V be the volumeobtained by rotating the region under the graph about the y-axis.

r 1 r2x | cx dx = 2x (ez =x5r?c.0

(a) Describe the figures generated by rotating segments AB and CB aboutthe y-axis.(b) Set up integrals that compute V by the Shell and Disk Methods.

134 y=fQ)x=h(y)

(A) (B)FIGURE1

SOLUTION(a) Whenrotated aboutthe y-axis, the segment AB generates a disk with radius R = h(y) and the segment CB generates a shellwith radius x and height f(x).(b) Based on Figure 1(A) and the information frompart (a), when using the Shell Method,

2V= 2x | xf(x) dx;

0whenusing the Disk Method,

1.3V= xf (h(y))? dy.0

30. ES Let W be the volumeofthe solid obtained by rotating the region under the graph in Figure 1(B) aboutthe y-axis. (a) Describe the figures generated by rotating segments A B and A C aboutthe y-axis.(b) Set up an integral that computes W by the Shell Method.(c) Explain the difficulty in computing W by the Washer Method.

748 CHAPTER 6 J APPLICATIONS OF THE INTEGRAL

SOLUTION (a) When rotated aboutthe y-axis, the segment A B generates a washer and the segment C A generates a shell with radius x andheight g(x).(b) Using Figure 1(B) and the information from part (a),

2W = 2x | xg(x) dx.

0

(c) The function g(x) is not one-to-one, which makesit difficult to determine the inner and outerradius of each washer.31. Let R be the region under the graph of y = 9 x? for 0 < x < 2. Use the Shell Method to compute the volumeofrotation ofR aboutthe x-axis as a sum oftwointegrals along the y-axis. Hint: The shells generated depend on whether y [0, 5] or y [5, 9].SOLUTION The region R is sketched below. Whenrotating this region about the x-axis, we produce a solid with two differentshell structures. For 0 < y < 5, the shell has radius y and height 2; for 5 < y < 9,the shell has radius y and height ./9 y. Thevolumeofthe solid is therefore

5 9v=20 | 2ydy +2x | yy9-ydy

0 5Forthe first integral, we calculate

5 52x | 2y dy = 2xy?| = 50x.

0 0For the second integral, we make the substitution u = 9 y, du = dyand find

9 02x | yV9 y dy = on | (9 u)Yudu

5 44

an | (9u1/2 _ y3/2) du0

4= 20 Cu _ pen)

ll

064 35271= 2n (4g- ) = .5 5

Thus, the total volumeis352 602Kain=.5 5

NMEAmao

| 05 10 15 20

32. Let R be the region under the graph of y = 4x7? for 1 < y < 4. Use the Shell Method to compute the volumeofrotation ofR aboutthe y-axis as a sum of two integrals along the x-axis.SOLUTION The region R is sketched below. Whenrotating this region about the y-axis, we produce a solid with two differentshell structures. For 0 < x < 1, the shell has radius x and height 3; for 1 < x < 4, the shell has radius x and height 4x1 1. Thevolumeofthe solid is therefore

1 4v=2n f 3x dx +27 | x(4x71 1) dx0 1

1 4= 20 f 3x dx +27 | (4 x) dx0 1

1 43= 27 -x? +27 (4 12IT X X> | 27=3n +22 (8-3) = 127.

1


In Exercises 33-38, use the Shell Method tofind the volume obtained byrotating region A in Figure 2 about the given axis.

yyer+2

FIGURE 2

33. y-axisSOLUTION Whenrotating region A about the y-axis, each shell has radius x and height 6 (x? + 2) = 4 x?. The volume ofthe resulting solid is

2 2 212x | x(4 x?) dx = 2x [ (4x x?) dx = 2x ES _ dl) = 8x.0 0 4 034. x = 3SOLUTION Whenrotating region A about x = 3, each shell has radius x ( 3) = x +3 and height 6 (x? + 2) = 4 x2.The volumeof the resultingsolid is

2 2 1 22x | (x +3)(4 x?) dx = 2x | (4x x+12 3x2) dx = 27 ES = qe + 12x - 3) = 40x.0 0 0

35. x=SOLUTION Whenrotating region A about x = 2, each shell has radius 2 x and height 6 (x? + 2) = 4 x?. The volume ofthe resulting solid is

2 2 22 1 402x | (2 x) (4-2?) dx =2n | (8 2x? 4x + x3) dx = 27 8x =x3 2x2 + -x* = u0 0 3 4 0 3

36. x-axisSOLUTION Whenrotating region A about the x-axis, each shell has radius y and height ./y 2. The volumeofthe resultingsolid is

62x | yVy 2dy2

Let u = y 2. Then du = dy, y = u +2 and4 4 704u ee6 4 2

2x | »Vy=3y =2x | (u + 2) Vudu = 2x Kyl? 4.5 5 5 3 o 1537. y = 2SOLUTION Whenrotating region A about y = 2,each shell has radius y ( 2) = y + 2 and height ./y 2. The volumeofthe resulting solid is

62x | (y + 2)J/y 2dy2

Letu = y 2, Then du = dy, y +2=u-+4and$ a 2 52, 8 3/2\l 1024727 (y +2) Vy 2dy = 27 (u + 4) Vu du = 2x


38. y=6SOLUTION Whenrotating region A about y = 6, each shell has radius 6 y and height /y 2. The volumeofthe resultingsolid is

62x [ 6-5)5-24)

2:

Let u = y 2. Then du = dy,6 y = 4 u and$ 4 8 2 * 2562x (6 y)Vy 24dy = 2x (4 u) Vudu = 2x (Que - un) = ==2 0 0

In Exercises 39-44, use the most convenient method (Disk or Shell Method) to find the volume obtained by rotating region B inFigure 2 about the given axis.39. y-axisSOLUTION Because a vertical slice of region B will produce a solid with a single cross section while a horizontal slice willproduce solid with two different cross sections, we will use a vertical slice. Now, becausea vertical slice is parallel to the axis ofrotation, we will use the Shell Method. Each shell has radius x and height x2 + 2. The volumeofthe resulting solid is

2 2 1 2an f x(x2 +2) dx = 2x | (x? + 2x) dx = 27 (q +x12)| =16x.0 0 0

40. x = 3SOLUTION Because a vertical slice of region B will produce a solid with a single cross section while a horizontal slice willproduce solid with two different cross sections, we will use a vertical slice. Now, becausea verticalslice is parallel to the axis ofrotation, we will use the Shell Method. Each shell has radius x ( 3) = x + 3 and height x? + 2. The volumeofthe resultingsolid is

2 2 1 22x | (x + 3)(x? +2) dx = 2x | (3 + 3x2 + 2x + 6) dx = 2x (qe + 2x9 + x2 + 6x = 567.

0 0 041. x =2SOLUTION Because a vertical slice of region B will produce a solid with a single cross section while a horizontal slice willproducea solid with two different cross sections, we will use a vertical slice. Now, because a vertical slice is parallel to the axis ofrotation, we will use the Shell Method. Each shell has radius 2 x and height x? + 2. The volume ofthe resulting solid is

an [2-0(x? +2) ax=2m [(220442) dx = 2m (42. x-axis

223 1 322Il = 2,3 4 ÿ 2SOLUTION Because a vertical slice of region B will produce a solid with a single cross section while a horizontal slice willproduce a solid with two different cross sections, we will use a vertical slice. Now, because a vertical slice is perpendicular to theaxis of rotation, we will use the Disk Method. Each disk has outer radius R x? + 2 and inner radius r = 0. The volume ofthesolid is then

2 2.x | (+2? ax =a | (x4 + 4x? + 4) dx0 0

1 4 2=x 10423 +4x5 3 02,323 3761

= TI cad = .5 3 1543. y = 2SOLUTION Because a vertical slice of region B will produce a solid with a single cross section while a horizontal slice willproduce a solid with two different cross sections, we will use a vertical slice. Now, because a vertical slice is perpendicular tothe axis of rotation, we will use the Disk Method. Eachdisk has outer radius R = x? + 2 ( 2) = x? + 4 and inner radiusr = 0 (-2) = 2. The volumeofthe solid is then

2 2nf (0? +4?-2) dx =x | (x* + 8x? + 12) dx

0 02

Il 1 8x x° + =x3 + 12x5 3 032 64 776a)aIl 5 3 15


44. y=8SOLUTION Becausea vertical slice of region B will produce a solid with a single cross section while a horizontal slice willproducea solid with two different cross sections, we will use a vertical slice. Now, because a vertical slice is perpendicularto the axisofrotation, we will use the Disk Method. Each disk has outer radius R = 8 0 = 8 and innerradius r = 8 (x? + 2) = 6 x?.The volumeofthe solid is then 2 2nf (8 (6 2?) dx =1x / (28 +12x? x%) dx

0 01 2

=a {28x + 4x3 =x5S Jo56 + 32 32 4087= I _ |= .5 5

In Exercises 45-50, use the most convenient method (Disk or Shell Method) tofind the given volumeofrotation.45. Region between x = y(5 y) andx = 0, rotated about the y-axisSOLUTION Examinethe figure below, which showsthe region bounded by x = y(5 y) and x = 0. If the indicated region issliced vertically, then the top of the slice lies along one branch of the parabola x = y(5 y) and the bottom lies along the otherbranch. On the other hand, if the regionis sliced horizontally, then the right endpoint of the slice always lies along the parabola andleft endpoint always lies along the y-axis. Clearly, it will be easier to slice the region horizontally.

Now, suppose the regionis rotated about the y-axis. Because a horizontal slice is perpendicular to the y-axis, we will calculatethe volumeofthe resulting solid using the disk method. Each cross sectionis a disk of radius R = y(5 y), so the volumeis

5 5 25 5 1 5 625%xf ySy)dy = nf (25y? 10y + y*) dy =a (y? 2y4 4 y] E5 o 3 2 s)lo 6

x=y6 - y)

46. Region between x = y(5 y) and x = 0, rotated about the x-axisSOLUTION Examinethefigure from the previous exercise, which showsthe region bounded by x = y(5 y) and x = 0. If theindicated regionis sliced vertically, then the top of the slice lies along one branch of the parabola x = y(5 y) and the bottom liesalong the other branch. On the other hand, if the region is sliced horizontally, then the right endpoint ofthe slice alwayslies alongthe parabola and left endpoint always lies along the y-axis. Clearly, it will be easier to slice the region horizontally.

Now, suppose the region is rotated about the x-axis. Because a horizontalslice is parallel to the x-axis, we will calculate thevolumeofthe resulting solid using the shell method. Each shell has a radius of y and a height of y(5 y), so the volumeis

5 5 55 1 6252x | y6 y)dy = 2x | (59? - y?) dy = 2 =y? - -y* =,0 0 3 4 0 6

47. Region in Figure 3, rotated about the x-axis

FIGURE 3

SOLUTION Examine Figure3.Ifthe indicated regionis sliced vertically, then the top oftheslice lies along the curve y = x x?2and the bottom lies along the curve y = 0 (the x-axis). On the other hand, if the region is sliced horizontally, the equationy = x x!? must besolved for x in order to determine the endpoint locations. Clearly, it will be easier to slice the regionvertically.

Now, suppose the region in Figure 3 is rotated about the x-axis. Because a vertical slice is perpendicular to the x-axis, we willcalculate the volumeofthe resulting solid using the disk method. Each crosssection is a disk of radius R = x x!?, so the volumeis

' 2 1 1 1 1 210 3 7 25 0 525

752 CHAPTER 6 || APPLICATIONS OF THE INTEGRAL

48. Region in Figure 3, rotated about the y-axisSOLUTION ExamineFigure 3. Ifthe indicated regionis sliced vertically, then the top ofthe slice lies along the curve y = x x 12and the bottom lies along the curve y = 0 (the x-axis). On the other hand, if the region is sliced horizontally, the equationy = x x!? mustbe solved for x in order to determine the endpoint locations. Clearly, it will be easier to slice the regionvertically.

Now supposethe regionis rotated about the y-axis. Because a verticalslice is parallel to the y-axis, we will calculate the volumeofthe resulting solid using the shell method. Each shell has radius x and height x x!?, so the volumeis

1 11 1 112x | x(x x!*) dx =2n (-x3 - x]] = E5 3% 14 Jo 249. Region in Figure4, rotated about x = 4

FIGURE 4

SOLUTION Examine Figure 4. If the indicated regionis sliced vertically, then the top of the slice lies along the curve y = x3 +2and the bottom lies along the curve y = 4 x?. On the other hand,the left end of a horizontal slice switches from y = 4 x? toy=x?°-+2aty = 3. Here, vertical slices will be more convenient.

Now,suppose the region in Figure 4 is rotated about x = 4. Because verticalslice is parallel to x = 4, we will calculate thevolumeofthe resulting solid using the shell method. Each shell has radius 4 x and height x? + 2 (4 x?) = x3 + x2 2, sothe volumeis

E si wd 15, 34,43, 2 2.11 563%2x | (4 x)(x° + x° 2) dx = 2x 5* + 4x + 3% +x* 8x m

50. Region in Figure 4, rotated about y = 2SOLUTION Examine Figure 4. If the indicated regionis sliced vertically, then the top ofthe slice lies along the curve y = x3 42and the bottom lies along the curve y = 4 x?. On the other hand,theleft end ofa horizontal slice switches from y = 4 x? toy =x +42at y = 3. Here,verticalslices will be more convenient.

Now supposethe region is rotated about y = 2. Becausea vertical slice is perpendicular to y = 2, we will calculate thevolumeofthe resulting solid using the disk method. Each cross section is a washer with outer radius R = x3 + 2 ( 2) = x3 44and inner radius r = 4 x? ( 2) = 6 x?, so the volumeis

2 21 1 1748xf (x + 4)? (6) dx=x ax! _ ¿e + 2x4 4 4x3 20x ; =>

In Exercises 51-54, use the Shell Method tofind the given volume ofrotation.51. A sphere of radius r

SOLUTION A sphere of radius r can be generated by rotating the region underthe semicircle y = Vr? x2 about the x-axis.Each shell has radius y and height

Thus, the volume of the sphere is

0

Let u = r? y?. Then du = 2ydy and

r r 2 74an | 2y fr? = yay = 20 f Ju du = 2x =u3/? = nr.0 0 3 0 352. The bead formed by removinga cylinder of radius r from the center of a sphere of radius R (compare with Exercise 59 inSection 6.3)


SOLUTION Eachshell has radius x and height 2V R2 x?. The volumeof the beadis thenR2x / 2x V R2 x2 dx.

y

Let u = R? x?. Then du = 2x dx andR

an | axVR2 dx = 20 |r 0

53. Thetorus obtained byrotating the circle (x a)? + y* = b? about the y-axis, where a > b (compare with Exercise 56 inSection 6.3). Hint: Evaluate the integral by interpreting part ofit as the area ofa circle.

R2 y2 R2 p2 4zum? zur

0

SOLUTION Whenrotating the region enclosed bythe circle (x a)? + y? = b? about the y-axis each shell has radius x andheight

VB - (x - a)? -/B2 _ @-a2) = 2,/b (x a)?.The volumeofthe resulting torus is then

a+b2x | 2x/b2 (x a)? dx.a-b

Let u = x a. Then du = dx,xv=u-+aanda+b b2x | 2xea?dx = 2x | 2(u + a)Vb2 u2 dua b b

b b= 4x | u?u du + san | Vb2 u? du.

b bNow,

b/ uvb? u du =0

bbecause the integrand is an odd function and the integration interval is symmetric with respect to zero. Moreover, the other integralis one-half the area of a circle of radius b; thus,

b 1/ Vb? 1u2 du = =xb?.b 2Finally, the volumeofthe torus is

1Ax(0) + 4ax sub? = 2n?ab?.

54. The paraboloid obtained by rotating the region between y = x? and y = c (c > 0) aboutthe y-axisSOLUTION When werotate the region in the first quadrant bounded by y = x? and y = c about the y-axis, each shell has aradius ofx and a height of c x. The volume ofthe paraboloid is then

ve ve 1 12x | x(c x?) dx = 2x | (cx x3) dx = 2n =cx? -x4 = rc .0 0 Further Insights and Challenges

55. ES The surface area of a sphere of radius r is 4srr. Use this to derive the formula for the volume ofa sphere of radiusR in a new way.(a) Show that the volumeof a thin spherical shell of inner radius r and thickness Aris approximately 41012 Ar.(b) Approximate V by decomposingthe sphere of radius R into N thin spherical shells of thickness Ar = R/N.(c) Show that the approximation is a Riemann sum that convergesto an integral. Evaluate the integral.SOLUTION(a) The volumeof a thin spherical shell of inner radius r and thickness Aris given by the product of the surface area ofthe shell,4xr and the thickness. Thus, we have 47072 Ar.


(b) The volume of the sphere is approximated byR N

Ry = 47 (7) y (xx)?k=1

where xp = KR.R\ N R 1 R 4() V =4n lim (77) Dow? = +f x2 dx = 4x (5*) = - aR°.o 3

No

56. Show that the solid (an ellipsoid) obtained by rotating the region R in Figure 5 about the y-axis has volume 4na?b.

b

x

2

FIGURE 5 The ellipse Ey + (2) =1.SOLUTION Let'sslice the portionofthe ellipse in the first and fourth quadrants horizontally androtate the slices about the y-axis.Theresulting ellipsoid hascross sections that are disks with radius

Thus, the volumeof the ellipsoid isbUsb 2.2 2 2p 4

=f e dy = x dy Ps db _ a?b+ = zua b.

57. The bell-shaped curve y = f(x) in Figure 6 satisfies dy/dx = xy. Use the Shell Methodand the substitution u = f(x) toshow that the solid obtained by rotating the region R about the y-axis has volume V = 2x(1 c), where c = f(a). Observe thatas c 0, the region R becomesinfinite but the volume V approaches 27.

FIGURE6 Thebell-shaped curve.

SOLUTION Let y = f(x) bethe exponential function depicted in Figure 6. Whenrotating the region R about the y-axis, eachshell in the resulting solid has radius x and height f(x). The volume ofthe solid is then

V=2n[ xf(x) dx.0Now,let « = f(x). Then du = f (x) dx = xf(x) dx; hence, xf(x)dx = du, and

c 1V= 2x | (-du) = 2x | du

1 [42n(l-c).

x v=x(1 +m?)[vex1)(x x2)dx= zt+m?)(x2x1)?

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