area of a surface of revolution - cengage...actual surface area. by taking a limit, we can determine...

AREA OF A SURFACE OF REVOLUTION

A surface of revolution is formed when a curve is rotated about a line. Such a surface is

We want to define the area of a surface of revolution in such a way that it correspondsto our intuition. If the surface area is , we can imagine that painting the surface wouldrequire the same amount of paint as does a flat region with area .

Let’s start with some simple surfaces. The lateral surface area of a circular cylinder withradius and height is taken to be because we can imagine cutting the cylin-der and unrolling it (as in Figure 1) to obtain a rectangle with dimensions and .

Likewise, we can take a circular cone with base radius and slant height , cut it alongthe dashed line in Figure 2, and flatten it to form a sector of a circle with radius and cen-tral angle . We know that, in general, the area of a sector of a circle with radius

and angle is (see Exercise 67 in Section 6.2) and so in this case it is

Therefore, we define the lateral surface area of a cone to be .

What about more complicated surfaces of revolution? If we follow the strategy we usedwith arc length, we can approximate the original curve by a polygon. When this polygonis rotated about an axis, it creates a simpler surface whose surface area approximates theactual surface area. By taking a limit, we can determine the exact surface area.

The approximating surface, then, consists of a number of bands, each formed by rotat-ing a line segment about an axis. To find the surface area, each of these bands can be considered a portion of a circular cone, as shown in Figure 3. The area of the band (or frus-tum of a cone) with slant height and upper and lower radii and is found by sub-tracting the areas of two cones:

From similar triangles we have

which gives

or

Putting this in Equation 1, we get

or

where is the average radius of the band.r � 12 �r1 � r2 �

A � 2�rl2

A � � �r1l � r2l �

�r2 � r1�l1 � r1lr2l1 � r1l1 � r1l

l1

r1�

l1 � l

r2

A � �r2�l1 � l � � �r1l1 � � ��r2 � r1�l1 � r2l �1

r2r1l

l¨

2πr

F IGURE 2

l

r

cut

A � �rl

A � 12 l 2� � 1

2 l 2�2�r

l � � �rl

12 l 2��l

� � 2�r�ll

lrh2�r

A � 2�rhhr

AA

1

h

2πr

F IGURE 1

h

r

cut

r¡

r™

l¡

l

F IGURE 3

Thom

son

Broo

ks-C

ole

copy

right

200

7

the lateral boundary of a solid of revolution of the type discussed in Sections 7.2 and 7.3.

Propert

y of C

enga

ge Le

arning

Not for

repro

ducti

on

Now we apply this formula to our strategy. Consider the surface shown in Figure 4,which is obtained by rotating the curve , , about the -axis, where ispositive and has a continuous derivative. In order to define its surface area, we divide theinterval into n subintervals with endpoints and equal width , as wedid in determining arc length. If , then the point lies on the curve. Thepart of the surface between and is approximated by taking the line segment and rotating it about the -axis. The result is a band with slant height and aver-age radius so, by Formula 2, its surface area is

As in the proof of Theorem 7.4.2, we have

where is some number in . When is small, we have andalso , since is continuous. Therefore

and so an approximation to what we think of as the area of the complete surface of revo-lution is

This approximation appears to become better as and, recognizing (3) as a Riemannsum for the function , we have

Therefore, in the case where is positive and has a continuous derivative, we define thesurface area of the surface obtained by rotating the curve , , about the -axis as

With the Leibniz notation for derivatives, this formula becomes

If the curve is described as , , then the formula for surface areabecomes

and both Formulas 5 and 6 can be summarized symbolically, using the notation for arc

S � yd

c 2�y�1 � �dx

dy�2

dy6

c � y � dx � t�y�

S � yb

a 2�y�1 � �dy

dx�2

dx5

S � yb

a 2� f �x� s1 � � f ��x��2 dx4

xa � x � by � f �x�

f

lim n l �

n

i�1 2� f �xi*� s1 � � f ��xi*��2 �x � y

b

a 2� f �x� s1 � � f ��x��2 dx

t�x� � 2� f �x� s1 � � f ��x��2

n l �

n

i�1 2� f �xi*� s1 � � f ��xi*��2 �x3

2� yi�1 � yi

2 Pi�1Pi � 2� f �xi*� s1 � � f ��xi*��2 �x

fyi�1 � f �xi�1� � f �xi*�yi � f �xi� � f �xi*��x�xi�1, xi�xi*

Pi�1Pi � s1 � � f ��xi*��2 �x

2� yi�1 � yi

2 Pi�1Pi

r � 12 �yi�1 � yi�

l � Pi�1Pi xPi�1Pixixi�1

Pi�xi, yi�yi � f �xi ��xx0, x1, . . . , xn�a, b�

fxa � x � by � f �x�

2 ■ AREA OF A SURFACE OF REVOLUT ION

F IGURE 4

0 x

y y=ƒ

a b

(a) Surface of revolution

P¸Pi-1

Pi

Pn

yi

0 x

y

a b

(b) Approximating band

Thom

son

Broo

ks-C

ole

copy

right

200

7

Propert

y of C

enga

ge Le

arning

Not for

repro

ducti

on

length given in Section 7.4, as

For rotation about the -axis, the surface area formula becomes

where, as before, we can use either

or

These formulas can be remembered by thinking of or as the circumference of acircle traced out by the point on the curve as it is rotated about the -axis or -axis,respectively (see Figure 5).

EXAMPLE 1 The curve , , is an arc of the circle .Find the area of the surface obtained by rotating this arc about the -axis. (The surface isa portion of a sphere of radius 2. See Figure 6.)

SOLUTION We have

and so, by Formula 5, the surface area is

� 4� y1

�1 1 dx � 4� �2� � 8�

� 2� y1

�1 s4 � x 2

2

s4 � x 2 dx

� 2� y1

�1 s4 � x 2 �1 �

x 2

4 � x 2 dx

S � y1

�1 2�y �1 � �dy

dx�2

dx

dy

dx� 1

2 �4 � x 2 ��1�2��2x� ��x

s4 � x 2

xx 2 � y 2 � 4�1 � x � 1y � s4 � x 2

F IGURE 5 (a) Rotation about x-axis: S=j 2πy ds

(x, y)

y

circumference=2πy

x0

y

(b) Rotation about y-axis: S=j 2πx ds

(x, y)x

circumference=2πx

x0

y

yx�x, y�2�x2�y

ds � �1 � �dx

dy�2

dyds � �1 � �dy

dx�2

dx

S � y 2�x ds8

y

S � y 2�y ds7

AREA OF A SURFACE OF REVOLUT ION ■ 3

■ ■ Figure 6 shows the portion of the spherewhose surface area is computed in Example 1.

1 x

y

F IGURE 6

Thom

son

Broo

ks-C

ole

copy

right

200

7

Propert

y of C

enga

ge Le

arning

Not for

repro

ducti

on


EXAMPLE 2 The arc of the parabola from to is rotated about the -axis. Find the area of the resulting surface.

SOLUTION 1 Using

and

we have, from Formula 8,

Substituting , we have . Remembering to change the limits ofintegration, we have

SOLUTION 2 Using

and

we have

(where )

(as in Solution 1)

EXAMPLE 3 Find the area of the surface generated by rotating the curve ,, about the -axis.

SOLUTION Using Formula 5 with

anddy

dx� exy � ex

x0 � x � 1y � ex

��

6 (17s17 � 5s5)

u � 1 � 4y ��

4 y

17

5 su du

� 2� y4

1 sy �1 �

1

4y dy � � y

4

1 s4y � 1 dy

S � y 2�x ds � y4

1 2�x �1 � �dx

dy�2

dy

dx

dy�

1

2syx � sy

��

6 (17s17 � 5s5)

S ��

4 y

17

5 su du �

�

4 [ 2

3 u 3�2]517

du � 8x dxu � 1 � 4x 2

� 2� y2

1 x s1 � 4x 2 dx

� y2

1 2�x �1 � �dy

dx�2

dx

S � y 2�x ds

dy

dx� 2xy � x 2

y�2, 4��1, 1�y � x 2

■ ■ Figure 7 shows the surface of revolutionwhose area is computed in Example 2.

■ ■ As a check on our answer to Example 2,notice from Figure 7 that the surface area should be close to that of a circular cylinder withthe same height and radius halfway between the upper and lower radius of the surface:

. We computed that the surface area was

which seems reasonable. Alternatively, the sur-face area should be slightly larger than the areaof a frustum of a cone with the same top and bot-tom edges. From Equation 2, this is

.2� �1.5�(s10 ) � 29.80

�

6 (17 s17 � 5s5) � 30.85

2� �1.5��3� � 28.27

(2, 4)

y=≈

x0

y

1 2

F IGURE 7

■ ■ Another method: Use Formula 6 with.x � ln y

Thom

son

Broo

ks-C

ole

copy

right

200

7

Propert

y of C

enga

ge Le

arning

Not for

repro

ducti

on


we have

(where )

(where and )

(by Example 8 in Section 6.2)

Since , we have and

S � � [es1 � e 2 � ln(e � s1 � e 2 ) � s2 � ln(s2 � 1)]

sec2 � 1 � tan2 � 1 � e 2tan � e

� � [sec tan � ln�sec � tan � � s2 � ln(s2 � 1)] � 2� � 1

2 [sec � tan � � ln sec � � tan � ]��4

� tan�1eu � tan � � 2� y

��4 sec3� d�

u � e x � 2� ye

1 s1 � u 2 du

S � y1

0 2�y �1 � �dy

dx�2

dx � 2� y1

0 ex

s1 � e 2x dx

EXERCISES

1–4 Set up, but do not evaluate, an integral for the area of thesurface obtained by rotating the curve about the given axis.

1. , ; -axis

2. , ; -axis

3. , ; -axis

4. , ; -axis■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

5–12 Find the area of the surface obtained by rotating the curveabout the -axis.

5. ,

6. ,

7. ,

8. ,

9. ,

10. ,

11. ,

12. ,■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

13–16 The given curve is rotated about the -axis. Find the areaof the resulting surface.

13. ,

14. ,

15. ,

16. ,■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

�a � y � ax � a cosh�y�a�

0 � y � a�2x � sa 2 � y 2

0 � x � 1y � 1 � x 2

1 � y � 2y � s3 x

y

1 � y � 2x � 1 � 2y 2

1 � y � 2x � 13 �y 2 � 2�3�2

12 � x � 1y �

x 3

6�

1

2x

0 � x � 1y � cosh x

0 � x � ��6y � cos 2x

4 � x � 9y � sx

2 � x � 69x � y 2 � 18

0 � x � 2y � x 3

x

y1 � y � 2y � e x

y0 � x � ��4y � sec x

x0 � x � ��2y � sin2x

x1 � x � 3y � ln x

17–20 Use Simpson’s Rule with to approximate the areaof the surface obtained by rotating the curve about the -axis. Com-pare your answer with the value of the integral produced by your calculator.

17. ,

18. ,

19. ,

20. ,■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

21–22 Use either a CAS or a table of integrals to find the exactarea of the surface obtained by rotating the given curve about the -axis.

21. ,

22. ,■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

23–24 Use a CAS to find the exact area of the surface obtainedby rotating the curve about the -axis. If your CAS has troubleevaluating the integral, express the surface area as an integral in theother variable.

23. ,

24. ,■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

25. (a) If , find the area of the surface generated by rotatingthe loop of the curve about the -axis.

(b) Find the surface area if the loop is rotated about the -axis.

26. A group of engineers is building a parabolic satellite dishwhose shape will be formed by rotating the curve about the -axis. If the dish is to have a 10-ft diameter and amaximum depth of 2 ft, find the value of and the surface areaof the dish.

ay

y � ax 2

y

x3ay 2 � x�a � x�2a 0

0 � x � 1y � ln�x � 1�

0 � y � 1y � x 3

yCAS

0 � x � 3y � sx 2 � 1

1 � x � 2y � 1�x

x

CAS

0 � x � 1y � s1 � e x

0 � x � ��3y � sec x

1 � x � 2y � x � sx

1 � x � 3y � ln x

xn � 10

Click here for answers.A

■ ■ Or use Formula 21 in the Table of Integrals.

Click here for solutions.S

Thom

son

Broo

ks-C

ole

copy

right

200

7

Propert

y of C

enga

ge Le

arning

Not for

repro

ducti

on

27. The ellipse

is rotated about the -axis to form a surface called an ellipsoid.Find the surface area of this ellipsoid.

28. Find the surface area of the torus in Exercise 41 in Section 7.2.

29. If the curve , , is rotated about the horizon-tal line , where , find a formula for the area ofthe resulting surface.

30. Use the result of Exercise 29 to set up an integral to find thearea of the surface generated by rotating the curve ,

, about the line . Then use a CAS to evaluatethe integral.

31. Find the area of the surface obtained by rotating the circleabout the line .y � rx 2 � y 2 � r 2

y � 40 � x � 4y � sx

CAS

f �x� � cy � ca � x � by � f �x�

x

a bx 2

a 2 � y 2

b 2 � 1

32. Show that the surface area of a zone of a sphere that liesbetween two parallel planes is , where is the diam-eter of the sphere and is the distance between the planes.(Notice that depends only on the distance between the planesand not on their location, provided that both planes intersectthe sphere.)

33. Formula 4 is valid only when . Show that when is not necessarily positive, the formula for surface area becomes

34. Let be the length of the curve , , where is positive and has a continuous derivative. Let be the sur-

face area generated by rotating the curve about the -axis. If is a positive constant, define and let be thecorresponding surface area generated by the curve ,

. Express in terms of and .LSfSta � x � by � t�x�

Stt�x� � f �x� � ccx

Sffa � x � by � f �x�L

S � yb

a 2� f �x� s1 � � f ��x��2 dx

f �x�f �x� � 0

Sh

dS � �dh

6 ■ AREA OF A SURFACE OF REVOLUT IONTh

omso

n Br

ooks

-Col

e co

pyrig

ht 2

007

Propert

y of C

enga

ge Le

arning

Not for

repro

ducti

on


ANSWERS

1.

3.

5. 7.

9. 11. 21��2�[1 �14 �e 2 � e�2 �]

�(37s37 � 17s17)�6�(145s145 � 1)�27

y��4

0 2�xs1 � �sec x tan x�2 dx

y3

1 2� ln xs1 � �1�x�2 dx

13. 15.

17. 19.

21.

23.

25. (a) (b)

27.

29. 31. 4� 2r 2xba 2��c � f �x��s1 � � f ��x��2 dx

2�[b 2 � a 2b sin�1(sa 2 � b 2�a)�sa 2 � b 2 ]56�s3a 2�45�a 2�3

��6�[ln(s10 � 3) � 3s10]��4�[4 ln(s17 � 4) � 4 ln(s2 � 1) � s17 � 4s2]

13.5272969.023754

�a 2�(145s145 � 10s10)�27Click here for solutions.S

Thom

son

Broo

ks-C

ole

copy

right

200

7

Propert

y of C

enga

ge Le

arning

Not for

repro

ducti

on


SOLUTIONS

Thom

son

Broo

ks-C

ole

copy

right

200

7

1. y = lnx ⇒ ds = 1 + (dy/dx)2 dx = 1 + (1/x)2 dx ⇒ S =3

12π(lnx) 1 + (1/x)2 dx [by (7)]

2. y = sin2 x ⇒ ds = 1 + (dy/dx)2 dx = 1 + (2 sinx cosx)2 dx ⇒

S =π/2

02π sin2 x 1 + (2 sinx cosx)2 dx [by (7)]

3. y = secx ⇒ ds = 1 + (dy/dx)2 dx = 1 + (secx tanx)2 dx ⇒

S =π/4

02πx 1 + (secx tanx)2 dx [by (8)]

4. y = ex ⇒ ds = 1 + (dy/dx)2 dx =√1 + e2x dx ⇒ S =

ln 2

02πx

√1 + e2x dx [by (8)]

or 2

12π(ln y) 1 + (1/y)2 dy [by (6)]

5. y = x3 ⇒ y0 = 3x2. So

S =2

02πy 1 + (y0)2 dx = 2π 2

0x3√1 + 9x4 dx [u = 1 + 9x4, du = 36x3 dx]

= 2π36

145

1

√udu = π

1823u3/2

145

1= π

27145

√145− 1

6. The curve 9x = y2 + 18 is symmetric about the x-axis, so we only use its top half, given by

y = 3√x− 2. dy/dx =

3

2√x− 2 , so 1 + (dy/dx)

2 = 1 +9

4(x− 2) . Thus,

S =6

2

2π · 3√x− 2 1 +9

4(x− 2) dx = 6π6

2

x− 2 + 94dx = 6π

6

2

x+ 14

1/2dx

= 6π · 23

x+ 14

3/2 6

2= 4π 25

4

3/2 − 94

3/2= 4π 125

8− 27

8= 4π · 98

8= 49π

7. y =√x ⇒ 1 + (dy/dx)2 = 1 + [1/(2

√x )]

2= 1+ 1/(4x). So

S =9

4

2πy 1 +dy

dx

2

dx =9

4

2π√x 1 +

1

4xdx = 2π

9

4

x+ 14dx

= 2π 23x+ 1

4

3/2 9

4= 4π

318(4x+ 1)3/2

9

4= π

637√37− 17√17

8. y = cos 2x ⇒ ds = 1 + (dy/dx)2 dx = 1 + (−2 sin 2x)2 dx ⇒

S =π/6

02π cos 2x 1 + 4 sin2 2xdx = 2π

√3

0

√1 + u2 1

4du [u = 2 sin 2x, du = 4cos 2xdx]

21= π

212u√1 + u2 + 1

2 ln u+√1 + u2

√3

0= π

2

√32 · 2 + 1

2 ln√3 + 2 = π

√3

2+ π

4 ln 2 +√3

Propert

y of C

enga

ge Le

arning

Not for

repro

ducti

on


Thom

son

Broo

ks-C

ole

copy

right

200

79. y = coshx ⇒ 1 + (dy/dx)2 = 1 + sinh2 x = cosh2 x. So

S = 2π1

0coshx coshxdx = 2π

1

012 (1 + cosh 2x) dx = π x+ 1

2 sinh 2x1

0

= π 1 + 12sinh 2 or π 1 + 1

4e2 − e−2

10. y =x3

6+1

2x⇒ dy

dx=

x2

2− 1

2x2⇒

1 + (dy/dx)2 =x4

4+1

2+

1

4x4=

x2

2+

1

2x2

2

=x2

2+

1

2x2⇒

S =1

1/2

2πx3

6+1

2x

x2

2+

1

2x2dx = 2π

1

1/2

x5

12+

x

12+

x

4+

1

4x3dx

= 2π1

1/2

x5

12+

x

3+

x−3

4dx = 2π

x6

72+

x2

6− x−2

8

1

1/2

= 2π 172 +

16 − 1

8− 1

64 · 72 +124 − 1

2= 2π 263

512= 263

256π

11. x = 13y2 + 2

3/2 ⇒ dx/dy = 12y2 + 2

1/2(2y) = y y2 + 2 ⇒

1 + (dx/dy)2 = 1 + y2 y2 + 2 = y2 + 12. So

S = 2π2

1y y2 + 1 dy = 2π 1

4y4 + 1

2y2

2

1= 2π 4 + 2− 1

4− 1

2= 21π

2

12. x = 1 + 2y2 ⇒ 1 + (dx/dy)2 = 1 + (4y)2 = 1 + 16y2. So

S = 2π2

1y 1 + 16y2 dy = π

16

2

116y2 + 1

1/232y dy = π

162316y2 + 1

3/2 2

1

= π2465√65− 17√17

13. y = 3√x ⇒ x = y3 ⇒ 1 + (dx/dy)2 = 1 + 9y4. So

S = 2π2

1x 1 + (dx/dy)2 dy = 2π

2

1y3 1 + 9y4 dy = 2π

36

2

11 + 9y4 36y3 dy

= π18

231 + 9y4

3/2 2

1= π

27145

√145− 10√10

14. y = 1− x2 ⇒ 1 + (dy/dx)2 = 1 + 4x2 ⇒

S = 2π1

0x√1 + 4x2 dx = π

4

1

08x√4x2 + 1 dx = π

4234x2 + 1

3/2 1

0= π

65√5− 1

15. x = a2 − y2 ⇒ dx/dy = 12(a2 − y2)−1/2(−2y) = −y/ a2 − y2 ⇒

1 + (dx/dy)2 = 1 +y2

a2 − y2=

a2 − y2

a2 − y2+

y2

a2 − y2=

a2

a2 − y2⇒

S =a/2

0

2π a2 − y2a

a2 − y2dy = 2π

a/2

0

a dy = 2πa ya/2

0= 2πa

a

2− 0 = πa2. Note that this is

14the surface area of a sphere of radius a, and the length of the interval y = 0 to y = a/2 is 1

4the length of the

interval y = −a to y = a.

Propert

y of C

enga

ge Le

arning

Not for

repro

ducti

on


Thom

son

Broo

ks-C

ole

copy

right

200

716. x = a cosh(y/a) ⇒ 1 + (dx/dy)2 = 1 + sinh2(y/a) = cosh2(y/a). So

S = 2πa

−aa cosh

y

acosh

y

ady = 4πa

a

0

cosh2y

ady = 2πa

a

0

1 + cosh2y

ady

= 2πa y +a

2sinh

2y

a

a

0

= 2πa a+a

2sinh 2 = 2πa2 1 + 1

2 sinh 2 orπa2 e2 + 4− e−2

2

17. y = lnx ⇒ dy/dx = 1/x ⇒ 1 + (dy/dx)2 = 1 + 1/x2 ⇒ S =3

12π lnx 1 + 1/x2 dx.

Let f(x) = lnx 1 + 1/x2. Since n = 10, ∆x = 3− 110

= 15. Then

S ≈ S10 = 2π · 1/53 [f(1) + 4f(1.2) + 2f(1.4) + · · ·+ 2f(2.6) + 4f(2.8) + f(3)] ≈ 9.023754.

The value of the integral produced by a calculator is 9.024262 (to six decimal places).

18. y = x+√x ⇒ dy/dx = 1 + 1

2x−1/2 ⇒ 1 + (dy/dx)2 = 2 + x−1/2 + 1

4x−1 ⇒

S =2

1

2π x+√x 2 +

1√x+1

4xdx. Let f(x) = (x+

√x ) 2 +

1√x+1

4x.

Since n = 10,∆x = 2− 110 = 1

10 . Then

S ≈ S10 = 2π · 1/103 [f(1) + 4f(1.1) + 2f(1.2) + · · ·+ 2f(1.8) + 4f(1.9) + f(2)] ≈ 29.506566.


19. y = secx ⇒ dy/dx = secx tanx ⇒ 1 + (dy/dx)2 = 1 + sec2 x tan2 x ⇒

S =π/3

02π secx

√1 + sec2 x tan2 xdx. Let f(x) = secx

√1 + sec2 x tan2 x.

Since n = 10,∆x =π/3− 010

=π

30. Then

S ≈ S10 = 2π · π/303

f(0) + 4fπ

30+ 2f

2π

30+ · · ·+ 2f 8π

30+ 4f

9π

30+ f

π

3≈ 13.527296.


20. y = (1 + ex)1/2 ⇒ dy

dx=1

2(1 + ex)−1/2 · ex = ex

2(1 + ex)1/2⇒

1 +dy

dx

2

= 1 +e2x

4(1 + ex)=4 + 4ex + e2x

4(1 + ex)=(ex + 2)2

4(1 + ex)⇒

S =1

0

2π√1 + ex

ex + 2

2√1 + ex

dx = π1

0

(ex + 2) dx = π ex + 2x1

0= π[(e+ 2)− (1 + 0)] = π(e+ 1).

Let f(x) = 12(ex + 2). Since n = 10, ∆x = 1− 0

10 = 110 . Then

S ≈ S10 = 2π · 1/103 [f(0) + 4f(0.1) + 2f(0.2) + · · ·+ 2f(0.8) + 4f(0.9) + f(1)] ≈ 11.681330.


Propert

y of C

enga

ge Le

arning

Not for

repro

ducti

on


Thom

son

Broo

ks-C

ole

copy

right

200

721. y = 1/x ⇒ ds = 1 + (dy/dx)2 dx = 1 + (−1/x2)2 dx = 1 + 1/x4 dx ⇒

S =2

1

2π · 1x

1 +1

x4dx = 2π

2

1

√x4 + 1

x3dx = 2π

4

1

√u2 + 1

u212du [u = x2, du = 2xdx]

= π4

1

√1 + u2

u2du

24= π −

√1 + u2

u+ ln u+ 1 + u2

4

1

= π −√174+ ln 4 +

√17 +

√21− ln 1 +

√2 = π

√2−

√174+ ln 4+

√17

1+√2

22. y =√x2 + 1 ⇒ dy

dx=

x√x2 + 1

⇒ ds = 1 +dy

dx

2

dx = 1 +x2

x2 + 1dx ⇒

S =3

0

2π x2 + 1 1 +x2

x2 + 1dx = 2π

3

0

2x2 + 1 dx = 2√2π

3

0

x2 +1√2

2

dx

21= 2

√2π 1

2x x2 + 1

2+ 1

4ln x+ x2 + 1

2

3

0

= 2√2π 3

29 + 1

2 +14 ln 3 + 9 + 1

2− 1

4ln 1√

2= 2

√2π 3

2192+ 1

4ln 3 + 19

2+ 1

4 ln√2

= 2√2π 3

2

√19√2+ 1

4ln 3

√2 +

√19 = 3

√19π + π√

2ln 3

√2 +

√19

23. y = x3 and 0 ≤ y ≤ 1 ⇒ y0 = 3x2 and 0 ≤ x ≤ 1.

S =1

02πx 1 + (3x2)2 dx = 2π

3

0

√1 + u2 1

6 du [u = 3x2, du = 6xdx]

= π3

3

0

√1 + u2 du

21= [or use CAS] π

312u√1 + u2 + 1

2ln u+

√1 + u2

3

0

= π3

32

√10 + 1

2ln 3 +

√10 = π

63√10 + ln 3 +

√10

24. y = ln(x+ 1), 0 ≤ x ≤ 1. ds = 1 +dy

dx

2

dx = 1 +1

x+ 1

2

dx, so

S =1

0

2πx 1 +1

(x+ 1)2dx =

2

1

2π(u− 1) 1 +1

u2du [u = x+ 1, du = dx]

= 2π2

1

u

√1 + u2

udu− 2π

2

1

√1 + u2

udu = 2π

2

1

1 + u2 du− 2π2

1

√1 + u2

udu

21, 23= [or use CAS] 2π 1

2u 1 + u2 + 1

2ln u+ 1 + u2

2

1− 2π 1 + u2 − ln 1 +

√1 + u2

u

2

1

= 2π√5 + 1

2ln 2 +

√5 − 1

2

√2− 1

2ln 1 +

√2 − 2π √5− ln 1+

√5

2−√2 + ln 1 +

√2

= 2π 12 ln 2 +

√5 + ln 1+

√5

2+√22 − 3

2ln 1 +

√2

Propert

y of C

enga

ge Le

arning

Not for

repro

ducti

on


Thom

son

Broo

ks-C

ole

copy

right

200

7

25. Since a > 0, the curve 3ay2 = x(a− x)2 only has points with

x ≥ 0. (3ay2 ≥ 0 ⇒ x(a− x)2 ≥ 0 ⇒ x ≥ 0.) Thecurve is symmetric about the x-axis (since the equation is

unchanged when y is replaced by −y). y = 0 when x = 0 or a,

so the curve’s loop extends from x = 0 to x = a.

d

dx(3ay2) =

d

dx[x(a− x)2] ⇒ 6ay

dy

dx= x · 2(a− x)(−1) + (a− x)2 ⇒ dy

dx=(a− x)[−2x+ a− x]

6ay

⇒ dy

dx

2

=(a− x)2(a− 3x)2

36a2y2=(a− x)2(a− 3x)2

36a2· 3a

x(a− x)2the last fraction

is 1/y2 =(a− 3x)212ax

⇒

1 +dy

dx

2

= 1 +a2 − 6ax+ 9x2

12ax=12ax

12ax+

a2 − 6ax+ 9x212ax

=a2 + 6ax+ 9x2

12ax=(a+ 3x)2

12axfor x 6= 0.

(a) S =a

x=0

2πy ds = 2πa

0

√x(a− x)√3a

· a+ 3x√12ax

dx = 2πa

0

(a− x)(a+ 3x)

6adx

=π

3a

a

0

(a2 + 2ax− 3x2) dx = π

3aa2x+ ax2 − x3

a

0=

π

3a(a3 + a3 − a3) =

π

3a· a3 = πa2

3.

Note that we have rotated the top half of the loop about the x-axis. This generates the full surface.

(b) We must rotate the full loop about the y-axis, so we get double the area obtained by rotating the top half of

the loop:

S = 2 · 2πa

x=0

xds = 4πa

0

xa+ 3x√12ax

dx =4π

2√3a

a

0

x1/2(a+ 3x) dx

=2π√3a

a

0

(ax1/2 + 3x3/2) dx =2π√3a

2

3ax3/2 +

6

5x5/2

a

0

=2π√3

3√a

2

3a5/2 +

6

5a5/2

=2π√3

3

2

3+6

5a2 =

2π√3

3

28

15a2 =

56π√3a2

45

26. In general, if the parabola y = ax2, −c ≤ x ≤ c, is rotated about the y-axis, the surface area it generates is

2πc

0

x 1 + (2ax)2 dx= 2π2ac

0

u

2a1 + u2

1

2adu

u = 2ax,du = 2a dx

=π

4a2

2ac

0

1 + u21/22udu =

π

4a2231 + u2

3/2 2ac

0

=π

6a21 + 4a2c2

3/2 − 1

Here 2c = 10 ft and ac2 = 2 ft, so c = 5 and a = 225. Thus, the surface area is

S = π66254

1 + 4 · 4625

· 25 3/2 − 1 = 625π24

1 + 1625

3/2 − 1 = 625π24

41√41

125− 1

= 5π24

41√41− 125 ≈ 90.01 ft2

Propert

y of C

enga

ge Le

arning

Not for

repro

ducti

on


Thom

son

Broo

ks-C

ole

copy

right

200

7

27.x2

a2+

y2

b2= 1 ⇒ y (dy/dx)

b2= − x

a2⇒ dy

dx= − b2x

a2y⇒

1 +dy

dx

2

= 1 +b4x2

a4y2=

b4x2 + a4y2

a4y2=

b4x2 + a4b2 1− x2/a2

a4b2 (1− x2/a2)=

a4b2 + b4x2 − a2b2x2

a4b2 − a2b2x2

=a4 + b2x2 − a2x2

a4 − a2x2=

a4 − a2 − b2 x2

a2(a2 − x2)

The ellipsoid’s surface area is twice the area generated by rotating the first quadrant portion of the ellipse about the

x-axis. Thus,

S = 2a

0

2πy 1 +dy

dx

2

dx = 4πa

0

b

aa2 − x2

a4 − (a2 − b2)x2

a√a2 − x2

dx

=4πb

a2

a

0

a4 − (a2 − b2)x2 dx =4πb

a2

a√a2−b2

0

a4 − u2du√

a2 − b2[u = a2 − b2 x]

30=

4πb

a2√a2 − b2

u

2a4 − u2 +

a4

2sin−1

u

a2

a√a2−b2

0

=4πb

a2√a2 − b2

a√a2 − b2

2a4 − a2(a2 − b2) +

a4

2sin−1

√a2 − b2

a= 2π

b2 + a2b sin−1√a2 − b2

a√a2 − b2

28. The upper half of the torus is generated by rotating the curve (x−R)2 + y2 = r2, y > 0, about the y-axis.

ydy

dx= −(x−R) ⇒ 1 +

dy

dx

2

= 1 +(x−R)2

y2=

y2 + (x−R)2

y2=

r2

r2 − (x−R)2. Thus,

S = 2R+r

R−r2πx 1 +

dy

dx

2

dx = 4πR+r

R−r

rx

r2 − (x−R)2dx

= 4πrr

−r

u+R√r2 − u2

du [u = x−R]

= 4πrr

−r

udu√r2 − u2

+ 4πRrr

−r

du√r2 − u2

= 4πr · 0 + 8πRrr

0

du√r2 − u2

[since the first integrand is odd and the second is even]

= 8πRr sin−1(u/r)r

0= 8πRr π

2= 4π2Rr

29. The analogue of f(x∗i ) in the derivation of (4) is now c − f(x∗i ), so

S = limn→∞

n

i=1

2π[c− f(x∗i )] 1 + [f 0(x∗i )]2∆x =

b

a

2π[c− f(x)] 1 + [f 0(x)]2 dx.

30. y = x1/2 ⇒ y0 = 12x−1/2 ⇒ 1 + (y0)2 = 1 + 1/4x, so by Exercise 29,

S =4

02π(4−√x ) 1 + 1/(4x) dx. Using a CAS, we get S = 2π ln

√17 + 4 + π

631√17 + 1 ≈ 80.6095.

Propert

y of C

enga

ge Le

arning

Not for

repro

ducti

on


Thom

son

Broo

ks-C

ole

copy

right

200

7

31. For the upper semicircle, f(x) =√r2 − x2, f 0(x) = −x/√r2 − x2. The surface area generated is

S1 =r

−r2π r − r2 − x2 1 +

x2

r2 − x2dx = 4π

r

0

r − r2 − x2r√

r2 − x2dx

= 4πr

0

r2√r2 − x2

− r dx

For the lower semicircle, f(x) = −√r2 − x2 and f 0(x) = x√r2 − x2

, so S2 = 4πr

0

r2√r2 − x2

+ r dx.

Thus, the total area is S = S1 + S2 = 8πr

0

r2√r2 − x2

dx = 8π r2 sin−1x

r

r

0= 8πr2 π

2= 4π2r2.

32. Take the sphere x2 + y2 + z2 = 14d2 and let the intersecting

planes be y = c and y = c+ h, where − 12d ≤ c ≤ 1

2d− h.

The sphere intersects the xy-plane in the circle

x2 + y2 = 14d2. From this equation, we get x dx

dy+ y = 0,

so dxdy= −y

x. The desired surface area is

S = 2π xds = 2πc+h

cx 1 + (dx/dy)2 dy = 2π

c+h

cx 1 + y2/x2 dy = 2π

c+h

cx2 + y2 dy

= 2πc+h

c12d dy = πd

c+h

cdy = πdh

33. In the derivation of (4), we computed a typical contribution to the surface area to be 2π yi−1 + yi2

|Pi−1Pi|, the area

of a frustum of a cone. When f(x) is not necessarily positive, the approximations yi = f(xi) ≈ f(x∗i ) and

yi−1 = f(xi−1) ≈ f(x∗i ) must be replaced by yi = |f(xi)| ≈ |f(x∗i )| and yi−1 = |f(xi−1)| ≈ |f(x∗i )|. Thus,

2πyi−1 + yi

2|Pi−1Pi| ≈ 2π |f(x∗i )| 1 + [f 0(x∗i )]

2∆x. Continuing with the rest of the derivation as before, we

obtain S = b

a2π |f(x)| 1 + [f 0(x)]2 dx.

34. Since g(x) = f(x) + c, we have g0(x) = f 0(x). Thus,

Sg =b

a

2πg(x) 1 + [g0(x)]2 dx =b

a

2π[f(x) + c] 1 + [f 0(x)]2 dx

=b

a

2πf(x) 1 + [f 0(x)]2 dx+ 2πcb

a

1 + [f 0(x)]2 dx = Sf + 2πcL

Propert

y of C

enga

ge Le

arning

Not for

repro

ducti

on

area of a surface of revolution - cengage...actual surface area. by taking a limit, we can determine...

Documents