3c3 volums cylinshells stu

Some volume problems are very difficult to handle by the methods of Section 6.2. Forinstance, let’s consider the problem of finding the volume of the solid obtained by rotatingabout the -axis the region bounded by and . (See Figure 1.) If we sliceperpendicular to the y-axis, we get a washer. But to compute the inner radius and the outerradius of the washer, we would have to solve the cubic equation for x interms of y; that’s not easy.

Fortunately, there is a method, called the method of cylindrical shells, that is easier touse in such a case. Figure 2 shows a cylindrical shell with inner radius , outer radius ,and height . Its volume is calculated by subtracting the volume of the inner cylinderfrom the volume of the outer cylinder:

If we let (the thickness of the shell) and (the average radiusof the shell), then this formula for the volume of a cylindrical shell becomes

and it can be remembered as

Now let be the solid obtained by rotating about the -axis the region bounded by[where ], and , where . (See Figure 3.)

We divide the interval into n subintervals of equal width and let bethe midpoint of the ith subinterval. If the rectangle with base and height isrotated about the y-axis, then the result is a cylindrical shell with average radius , height

, and thickness (see Figure 4), so by Formula 1 its volume is

Therefore, an approximation to the volume of is given by the sum of the volumes ofthese shells:

V � �n

i�1 Vi � �

n

i�1 2�xi f �xi� �x

SV

Vi � �2�xi�� f �xi�� x

�xf �xi�xi

f �xi��xi�1, xi�xi�x�xi�1, xi ��a, b�

FIGURE 3

a bx

y

0

y=ƒ

x

y

a b0

y=ƒ

b � a � 0x � by � 0, x � a, f �x� � 0y � f �x�yS

V � [circumference][height][thickness]

V � 2�rh �r1

r � 12 �r2 � r1��r � r2 � r1

� 2� r2 � r1

2 h�r2 � r1�

� � �r2 � r1��r2 � r1�h

� �r22h � �r2

1h � � �r 22 � r 2

1�h

V � V2 � V1

V2

V1Vhr2r1

y � 2x 2 � x 3

y � 0y � 2x 2 � x 3y

FIGURE 1

FIGURE 2

rr¡

r™

Îr

h

y

x02

1

y=2≈-˛

xL=? x

R=?

1

Volumes by Cylindrical Shells

This approximation appears to become better as . But, from the definition of an inte-gral, we know that

Thus, the following appears plausible:

The volume of the solid in Figure 3, obtained by rotating about the y-axis theregion under the curve from a to b, is

The argument using cylindrical shells makes Formula 2 seem reasonable, but later wewill be able to prove it. (See Exercise 47.)

The best way to remember Formula 2 is to think of a typical shell, cut and flattened asin Figure 5, with radius x, circumference , height , and thickness or :

This type of reasoning will be helpful in other situations, such as when we rotate aboutlines other than the y-axis.

EXAMPLE 1 Find the volume of the solid obtained by rotating about the -axis the regionbounded by and .

SOLUTION From the sketch in Figure 6 we see that a typical shell has radius x, circumfer-ence , and height . So, by the shell method, the volume is

It can be verified that the shell method gives the same answer as slicing.

FIGURE 7

y

x

� 2� [ 12 x 4 �

15 x 5 ]0

2� 2� (8 �

325 ) � 16

5 �

V � y2

0 �2�x��2x 2 � x 3 � dx � 2� y

2

0 �2x 3 � x 4 � dx

f �x� � 2x 2 � x 32�x

y � 0y � 2x 2 � x 3y

FIGURE 5

2πxÎx

ƒ

y

xx

xƒ

heightcircumference

dx� f �x��2�x�yb

a

dx�xf �x�2�x

where 0 � a � b V � yb

a 2�x f �x� dx

y � f �x�2

lim n l �

�n

i�1 2�xi f �xi� �x � y

b

a 2�xf �x� dx

n l �

2 ■ VOLUMES BY CYL INDR ICAL SHELLS

FIGURE 4

x

y

a b0

y=ƒ

xi

–

a b0 x

y

xi-1

xi

y=ƒ

FIGURE 6

y

x

2≈-˛

x

x

2

■ ■ Figure 7 shows a computer-generated picture of the solid whose volume we computedin Example 1.

NOTE ■■ Comparing the solution of Example 1 with the remarks at the beginning of thissection, we see that the method of cylindrical shells is much easier than the washer methodfor this problem. We did not have to find the coordinates of the local maximum and we didnot have to solve the equation of the curve for in terms of . However, in other examplesthe methods of the preceding section may be easier.

EXAMPLE 2 Find the volume of the solid obtained by rotating about the -axis the regionbetween and .

SOLUTION The region and a typical shell are shown in Figure 8. We see that the shell hasradius x, circumference , and height . So the volume is

As the following example shows, the shell method works just as well if we rotate aboutthe x-axis. We simply have to draw a diagram to identify the radius and height of a shell.

EXAMPLE 3 Use cylindrical shells to find the volume of the solid obtained by rotatingabout the -axis the region under the curve from 0 to 1.

SOLUTION This problem was solved using disks in Example 2 in Section 6.2. To use shellswe relabel the curve (in the figure in that example) as in Figure 9. Forrotation about the x-axis we see that a typical shell has radius y, circumference , andheight . So the volume is

In this problem the disk method was simpler.

EXAMPLE 4 Find the volume of the solid obtained by rotating the region bounded byand about the line .

SOLUTION Figure 10 shows the region and a cylindrical shell formed by rotation about theline . It has radius , circumference , and height .

The volume of the given solid is

� 2�� x 4

4� x 3 � x 2�

0

1

��

2

V � y1

0 2� �2 � x��x � x 2 � dx � 2� y

1

0 �x 3 � 3x 2 � 2x� dx

FIGURE 10

0

y

x

y=x-≈

0

y

x

x

1 2 3 4

2-x

x=2

x � x 22� �2 � x�2 � xx � 2

x � 2y � 0y � x � x 2

� 2�� y 2

2�

y 4

4 �0

1

��

2

V � y1

0 �2�y��1 � y 2 � dy � 2� y

1

0 �y � y 3 � dy

1 � y 22�y

x � y 2y � sx

y � sxx

� 2�� x 3

3�

x 4

4 �0

1

��

6

V � y1

0 �2�x��x � x 2 � dx � 2� y

1

0 �x 2 � x 3 � dx

x � x 22�x

y � x 2y � xy

yx

VOLUMES BY CYL INDR ICAL SHELLS ■ 3

FIGURE 8

0 x

y

y=x

y=≈

x

shellheight=x-≈

FIGURE 9

1

y

y

shellradius=y

shell height=1-¥

0 x

x=1

1

x=¥


Exercises

1. Let be the solid obtained by rotating the region shown in the figure about the -axis. Explain why it is awkward to useslicing to find the volume of . Sketch a typical approxi-mating shell. What are its circumference and height? Use shellsto find .

2. Let be the solid obtained by rotating the region shown in thefigure about the -axis. Sketch a typical cylindrical shell andfind its circumference and height. Use shells to find the volumeof . Do you think this method is preferable to slicing? Explain.

3–7 Use the method of cylindrical shells to find the volume gen-erated by rotating the region bounded by the given curves about the-axis. Sketch the region and a typical shell.

3. , , ,

4. , ,

5.

6. ,

7. ,■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

8. Let be the volume of the solid obtained by rotating about the-axis the region bounded by and . Find both

by slicing and by cylindrical shells. In both cases draw a dia-gram to explain your method.

9–14 Use the method of cylindrical shells to find the volume ofthe solid obtained by rotating the region bounded by the givencurves about the -axis. Sketch the region and a typical shell.

9.

10.

11. , ,

12. ,

13. ,

14.■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

x � y � 3, x � 4 � �y � 1�2

2x � y � 6y � 4x 2

x � 0x � 4y 2 � y 3

x � 0y � 8y � x 3

x � sy, x � 0, y � 1

x � 1 � y 2, x � 0, y � 1, y � 2

x

Vy � x 2y � sxyV

y � x 2 � 4x � 7y � 4�x � 2�2

x � y � 3y � 3 � 2x � x 2

y � e�x2, y � 0, x � 0, x � 1

x � 1y � 0y � x 2

x � 2x � 1y � 0y � 1x

y

0 x

y

œ„π

y=sin{≈}

S

yS

0 x

y

1

y=x(x-1)@

V

SVy

S

15–20 Use the method of cylindrical shells to find the volumegenerated by rotating the region bounded by the given curves aboutthe specified axis. Sketch the region and a typical shell.

15. , ; about

16. , ; about the -axis

17. , ; about

18. , ; about

19. , ; about

20.■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

21–26 Set up, but do not evaluate, an integral for the volume of the solid obtained by rotating the region bounded by the givencurves about the specified axis.

21.

22. , ; about

23. , ; about

24. about

25. about

26. about ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

27. Use the Midpoint Rule with to estimate the volumeobtained by rotating about the -axis the region under the curve

, .

28. If the region shown in the figure is rotated about the -axis toform a solid, use the Midpoint Rule with to estimate thevolume of the solid.

29–32 Each integral represents the volume of a solid. Describethe solid.

29.

30.

31.

32.

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

y�4

0 2� �� x��cos x � sin x� dx

y1

0 2� �3 � y��1 � y2� dy

2� y2

0

y

1 � y 2 dy

y3

0 2�x 5 dx

0 x

y

1

1

2

3

4

5

2 3 4 5 6 7 8 9 10 11 12

n � 5y

0 � x � �4y � tan xy

n � 4

y � 5x 2 � y 2 � 7, x � 4;

y � 4x � ssin y, 0 � y � �, x � 0;

x � 2y � 1�1 � x 2 �, y � 0, x � 0, x � 2;

x � �1y � sin��x2�y � x 4

x � 7y � 4x � x 2y � x

y � ln x, y � 0, x � 2; about the y-axis

y � x 2, x � y 2; about y � �1

y � 3y � 0, x � 5y � sx � 1

x � �2y � 8x � 2x 2y � 4x � x 2

x � 4y � 0, x � 1, x � 2y � x 2

yy � 0, x � �2, x � �1y � x 2

x � 1y � 0, x � 1, x � 2y � x 2

Click here for answers.A Click here for solutions.S


; 33–34 Use a graph to estimate the -coordinates of the points ofintersection of the given curves. Then use this information to esti-mate the volume of the solid obtained by rotating about the -axisthe region enclosed by these curves.

33. ,

34. ,■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

35–36 Use a computer algebra system to find the exact volumeof the solid obtained by rotating the region bounded by the givencurves about the specified line.

35. , , ; about

36. , , ; about ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

37–42 The region bounded by the given curves is rotated aboutthe specified axis. Find the volume of the resulting solid by anymethod.



39. , ; about

40. , ; about

41. ; about the -axis

42. ; about the -axis■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

43–45 Use cylindrical shells to find the volume of the solid.

43. A sphere of radius

44. The solid torus (a donut-shaped solid with radii and )shown in the figure

rR

rR

r

xx 2 � �y � 1�2 � 1

yx 2 � �y � 1�2 � 1

x � 2x � 0x � 1 � y 4

x � �1y � x � �4�x�y � 5

yy � 0y � x 2 � 3x � 2

xy � 0y � x 2 � x � 2

x � �10 � x � �y � 0y � x 3 sin x

x � ��20 � x � �y � sin4 xy � sin2 x

CAS

y � 3x � x 3y � x 4

y � x � x 2 � x 4y � 0

y

x 45. A right circular cone with height and base radius ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

46. Suppose you make napkin rings by drilling holes with differentdiameters through two wooden balls (which also have differentdiameters). You discover that both napkin rings have the sameheight , as shown in the figure.(a) Guess which ring has more wood in it.(b) Check your guess: Use cylindrical shells to compute the

volume of a napkin ring created by drilling a hole withradius through the center of a sphere of radius andexpress the answer in terms of .

47. We arrived at Formula 2, , by usingcylindrical shells, but now we can use integration by parts toprove it using the slicing method of Section 6.2, at least for thecase where is one-to-one and therefore has an inverse func-tion . Use the figure to show that

Make the substitution and then use integration byparts on the resulting integral to prove that .

y

0 xa b

c

d

x=a

x=b

y=ƒx=g(y)

V � xba 2�x f �x� dx

y � f �x�

V � �b 2d � �a 2c � yd

c � �t�y��2 dy

t

f

V � xba 2�x f �x� dx

h

hRr

h

rh


Answers

1. Circumference , height ;

3.

5. 7. 16�� 1 � 1e�

x

y

0

1

x

x

y

x

2�

x(x-1)@

0

y

x1

x

y

x1

�15� x �x � 1�2� 2�x

9.

11. 13.15. 17. 19.21.23.

25. 27. 1.14229. Solid obtained by rotating the region ,about the -axis31. Solid obtained by rotating the region bounded by (i) , , and , or (ii) , , and about the line 33. 0, 1.32; 4.05 35. 37.39. 41. 43. 45. 1

3 �r 2h43 �r 34�38��3 � ln 4�

81�10132 � 3

y � 3y � 0x � 1x � y 2y � 0x � 0x � 1 � y 2

y0 � x � 30 � y � x 4

x�

0 2� �4 � y�ssin y dy

x10 2��x � 1��sin��x2� � x 4 � dx

x2

1 2�x ln x dx

24�67�617�6250�3768�7

x

y

01

1+¥

1

2

x

y

01

y

21�2Click here for solutions.S

1. If we were to use the “washer” method, we would first have

to locate the local maximum point (a, b) of y = x(x− 1)2using the methods of Chapter 4. Then we would have to

solve the equation y = x(x− 1)2 for x in terms of y to

obtain the functions x = g1(y) and x = g2(y) shown in the

first figure. This step would be difficult because it involves

the cubic formula. Finally we would find the volume using

V = πR b0

©[g1(y)]

2 − [g2(y)]2ªdy.

Using shells, we find that a typical approximating shell has radius x, so its circumference is 2πx. Its height is y, that

is, x(x− 1)2. So the total volume is

V =R 102πx

£x(x− 1)2¤ dx = 2π R 1

0

¡x4 − 2x3 + x2¢ dx = 2π·x5

5− 2x

4

4+x3

3

¸10

=π

15

3. V =

Z 2

1

2πx · 1xdx = 2π

Z 2

1

1 dx

= 2π [x]21 = 2π(2− 1) = 2π

5. V =R 102πxe−x

2

dx. Let u = x2.

Thus, du = 2xdx, so

V = πR 10e−u du = π

£−e−u¤10

= π(1− 1/e)


Solutions: Volumes by Cylindrical Shells

7. The curves intersect when 4(x − 2)2 = x2 − 4x+ 7 ⇔ 4x2 − 16x+ 16 = x2 − 4x + 7 ⇔

3x2 − 12x+ 9 = 0 ⇔ 3(x2 − 4x+ 3) = 0 ⇔ 3(x− 1)(x− 3) = 0, so x = 1 or 3.

V = 2πR 31

©x£¡x2 − 4x+ 7¢− 4(x− 2)2¤ª dx = 2π R 3

1

£x(x2 − 4x+ 7− 4x2 + 16x− 16)¤ dx

= 2πR 31

£x(−3x2 + 12x− 9)¤ dx = 2π(−3) R 3

1(x3 − 4x2 + 3x) dx = −6π£ 1

4x4 − 4

3x3 + 3

2x2¤31

= −6π£¡ 814− 36 + 27

2

¢− ¡ 14− 4

3+ 3

2

¢¤= −6π¡20− 36 + 12 + 4

3

¢= −6π¡− 8

3

¢= 16π

9. V =R 212πy

¡1 + y2

¢dy = 2π

R 21

¡y + y3

¢dy = 2π

£12y2 + 1

4y4¤21

= 2π£(2 + 4)− ¡ 1

2+ 1

4

¢¤= 2π

¡214

¢= 21π

2

11. V = 2π

Z 8

0

[y( 3√y − 0)] dy

= 2π

Z 8

0

y4/3 dy = 2πh37y7/3

i80

=6π

7(87/3) =

6π

7(27) =

768π

7


13. The curves intersect when 4x2 = 6− 2x ⇔ 2x2 + x− 3 = 0 ⇔ (2x+ 3)(x− 1) = 0 ⇔ x = − 32

or 1.

Solving the equations for x gives us y = 4x2 ⇒ x = ± 12

√y and 2x+ y = 6 ⇒ x = − 1

2y + 3.

V = 2π

Z 4

0

©y£¡

12

√y¢ − ¡−1

2

√y¢¤ª

dy + 2π

Z 9

4

©y£¡− 1

2y + 3

¢ − ¡−12

√y¢¤ª

dy

= 2π

Z 4

0

(y√y ) dy + 2π

Z 9

4

³−12y2 + 3y + 1

2y3/2

´dy = 2π

h25y5/2

i40+ 2π

h− 16y3 + 3

2y2 + 1

5y5/2

i94

= 2π¡25· 32¢+ 2π£¡− 243

2+ 243

2+ 243

5

¢− ¡−323+ 24 + 32

5

¢¤= 128

5π + 2π

¡43315

¢= 1250

15π = 250

3π

15. V =R 212π(x− 1)x2 dx = 2π£ 1

4x4 − 1

3x3¤21

= 2π£¡4− 8

3

¢− ¡ 14− 1

3

¢¤= 17

6π

17. V =R 212π(4− x)x2 dx = 2π£ 43x3 − 1

4x4¤21

= 2π£¡

323 − 4

¢− ¡ 43 − 14

¢¤= 67

6 π

19. V =R 202π(3− y)(5− x)dy

=R 202π(3− y)¡5− y2 − 1¢ dy

=R 202π¡12− 4y − 3y2 + y3¢ dy

= 2π£12y − 2y2 − y3 + 1

4y4¤20

= 2π(24− 8− 8 + 4) = 24π


21. V =R 212πx lnxdx 23. V =

R 102π[x− (−1)]¡sin π

2 x− x4¢dx

25. V =R π02π(4− y)√sin y dy

27. ∆x =π/4− 04

=π

16.

V =R π/40

2πx tan xdx ≈ 2π · π16

¡π32tan π

32+ 3π

32tan 3π

32+ 5π

32tan 5π

32+ 7π

32tan 7π

32

¢ ≈ 1.14229.

R 302πx5 dx = 2π

R 30x(x4) dx. The solid is obtained by rotating the region 0 ≤ y ≤ x4, 0 ≤ x ≤ 3 about the

y-axis using cylindrical shells.

31.R 102π(3− y)(1− y2) dy. The solid is obtained by rotating the region bounded by (i) x = 1− y2, x = 0, and

y = 0 or (ii) x = y2, x = 1, and y = 0 about the line y = 3 using cylindrical shells.

33. From the graph, the curves intersect at x = 0 and at x = a ≈ 1.32, with

x+ x2 − x4 > 0 on the interval (0, a). So the volume of the solid

obtained by rotating the region about the y-axis is

V = 2π

Z a

0

£x(x+ x2 − x4)¤ dx = 2π Z a

0

(x2 + x3 − x5) dx

= 2π£13x3 + 1

4x4 − 1

6x6¤a0≈ 4.05

35. V = 2π

Z π/2

0

£¡π2 − x

¢¡sin2 x− sin4 x¢¤ dx

CAS= 1

32π3


37. Use disks:

V =R 1−2 π

¡x2 + x− 2¢2 dx = π

R 1−2¡x4 + 2x3 − 3x2 − 4x+ 4¢ dx

= π£15x5 + 1

2x4 − x3 − 2x2 + 4x¤1−2 = π

£¡15+ 1

2− 1− 2 + 4¢− ¡− 32

5+ 8 + 8− 8− 8¢¤

= π¡335 +

32

¢= 81

10π

39. Use shells:

V =R 412π[x− (−1)][5− (x+ 4/x)] dx

= 2πR 41(x+ 1)(5− x− 4/x) dx

= 2πR 41

¡5x− x2 − 4 + 5− x− 4/x¢ dx

= 2πR 41

¡−x2 + 4x+ 1− 4/x¢ dx = 2π£− 13x

3 + 2x2 + x− 4 lnx¤41

= 2π£¡− 64

3+ 32 + 4− 4 ln 4¢− ¡− 1

3+ 2 + 1− 0¢¤

= 2π(12− 4 ln 4) = 8π(3− ln 4)

41. Use disks: V = π

Z 2

0

·q1− (y − 1)2

¸2dy = π

Z 2

0

¡2y − y2¢ dy = π

£y2 − 1

3y3¤20= π

µ4− 8

3

¶=4

3π

43. V = 2R r02πx

√r2 − x2 dx = −2π R r

0

¡r2 − x2¢1/2(−2x) dx = h−2π · 2

3

¡r2 − x2¢3/2ir

0

= − 43π¡0− r3¢ = 4

3πr3

45. V = 2πZ r

0

x

µ−hrx+ h

¶dx = 2πh

Z r

0

µ−x

2

r+ x

¶dx = 2πh

·−x

3

3r+x2

2

¸r0

= 2πhr2

6=

πr2h

3

Volume =R d0πb2 dy − R c

0πa2 dy − R d

cπ[g(y)]2 dy = πb2d− πa2c− R d

cπ[g(y)]2 dy. Let y = f(x), which

gives dy = f 0(x) dx and g(y) = x, so that V = πb2d − πa2c− πR bax2f 0(x) dx. Now integrate

by parts with u = x2, and dv = f 0(x) dx ⇒ du = 2x dx, v = f(x), andR bax2 f 0(x) dx =

£x2 f(x)

¤ba− R b

a2x f(x) dx = b2 f(b)− a2 f(a)− R b

a2x f(x) dx, but f(a) = c and f(b) = d

⇒ V = πb2d− πa2c− πhb2d− a2c− R b

a2xf(x) dx

i=R ba2πxf(x) dx.

47. Using the formula for volumes of rotation and the figure, we see that


3c3 volums cylinshells stu

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