Section 8.2 Area of a Surface of Revolution

Section 8.2 Area of a Surface of Revolution

28. Find the exact area of the surface obtained by rotating the given curve about the x-axis.

y =p

x²+ 1, 0 ≤ x ≤ 3

Solution:

758 ¤ CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION 24. =√

²+ 1 ⇒ 

 = 

√²+ 1 ⇒  =

 1 +





2

 =

 1 + ²

²+ 1 ⇒

 =

 3 0

2

²+ 1

 1 + ²

²+ 1 = 2

 3 0

2²+ 1  = 2√ 2 

3 0



²+

√1 2

2



= 221 √ 2

1 2

²+¹₂+¹₄ln

 +

²+¹₂³

0= 2√ 2

3 2



9 +¹₂ +¹₄ln 3 +

9 +¹₂

−¹4ln√¹ 2



= 2√ 2

3 2

19 2 +¹₄ln

3 +

19 2

+¹₄ln√ 2

= 2√ 2

3 2

√19

√2 +¹₄ln 3√

2 +√ 19

= 3√

19 +√^ 2ln

3√ 2 +√

19

25. = ³and 0 ≤  ≤ 1 ⇒ ⁰= 3²and 0 ≤  ≤ 1.

 =1 0 2

1 + (3²)² = 23 0

√1 + ^{2 1}₆

 = 3²,

 = 6 



= ^₃3 0

√1 + ²

=21 [or use CAS] ^₃1 2√

1 + ²+¹₂ln

 +√

1 + ²³

0=^₃3 2

√10 +¹₂ln 3 +√

10

=^₆ 3√

10 + ln 3 +√

10

26. = ln( + 1), 0 ≤  ≤ 1.  =

 1 +





2

 =

 1 +

 1

 + 1

2

, so

 =

 1 0

2



1 + 1

( + 1)² =

2

1 2( − 1)

 1 + 1

²  [ =  + 1,  = ]

= 2

 2 1



√1 + ²

  − 2

 2 1

√1 + ²

  = 2

 2 1

1 + ² − 2

2 1

√1 + ²

 

21, 23

= [or use CAS] 2₁

2√

1 + ²+¹₂ln

 +√

1 + ²2 1− 2

√

1 + ²− ln

1 +√ 1 + ²



²

1

= 2√

5 +¹₂ln 2 +√

5

−¹2

√2 −¹2ln 1 +√

2

− 2√

5 − ln

1 +√ 5 2



−√ 2 + ln

1 +√ 2

= 2

1 2ln

2 +√ 5

+ ln

1 +√ 5 2



+^√₂² −³2ln 1 +√

2

27. = 2

 _∞

1



 1 +





2

 = 2

 _∞

1

1



 1 + 1

⁴ = 2

_∞

1

√⁴+ 1

³ . Rather than trying to evaluate this integral, note that√

⁴+ 1 √

⁴= ²for   0. Thus, if the area is finite,

 = 2

_∞

1

√⁴+ 1

³   2

 _∞

1

²

³ = 2

 _∞

1

1

. But we know that this integral diverges, so the area  is infinite.

28. =_∞

0 2

1 + ()² = 2_∞

0 ^−

1 + (−^−)² [ = ^−, ⁰=−^−].

Evaluate  =

^−

1 + (−^−)²by using the substitution  = −^−,  = ^−:

 = √

1 + ²=^{21 1}₂√

1 + ²+¹₂ln

 +√ 1 + ²

+  = ¹₂(−^−)√

1 + ^−2+¹₂ln

−^−+√

1 + ^−2 + .

Returning to the surface area integral, we have

° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c

33. Gabriels Horn The surface formed by rotating the curve y = 1/x, x ≥ 1, about the x-axis is known as Gabriels horn. Show that the surface area is infinite (although the enclosed volume is finite; see Exercise 7.8.75).

556 Chapter 8 Further Applications of Integration

15–18 The given curve is rotated about the y­axis. Find the area of the resulting surface.

15. y − ¹₃ x^3y2, 0 < x < 12 16. x^2y31y^2y3− 1, 0 < y < 1 17. x −sa²2y² , 0 < y < ay2 18. y −¹₄x²2¹₂ ln x, 1 < x < 2

19–22 Use Simpson’s Rule with n − 10 to approximate the area of the surface obtained by rotating the curve about the x­axis. Compare your answer with the value of the integral pro­

duced by a calculator.

19. y −¹₅x⁵, 0 < x < 5 20. y − x 1 x², 0 < x < 1 21. y − xe^x, 0 < x < 1 22. y − x ln x, 1 < x < 2

23–24 Use either a CAS or a table of integrals to find the exact area of the surface obtained by rotating the given curve about the x­axis.

23. y − 1yx, 1 < x < 2 24. y −sx²11, 0 < x < 3

25–26 Use a CAS to find the exact area of the surface obtained by rotating the curve about the y­axis. If your CAS has trouble evaluating the integral, express the surface area as an integral in the other variable.

25. y − x³, 0 < y < 1 26. y − lnsx 1 1d, 0 < x < 1

27. If the region 5 −hsx, yd

|

x > 1, 0 < y < 1yxj is rotated about the x­axis, the volume of the resulting solid is finite (see Exercise 7.8.63). Show that the surface area is infi­

nite. (The surface is shown in the figure and is known as Gabriel’s horn.)

0 1

1x y=

y

x

28. If the infinite curve y − e^2x, x > 0, is rotated about the x­axis, find the area of the resulting surface.

29. (a) If a . 0, find the area of the surface generated by rotat­

ing the loop of the curve 3ay²− xsa 2 xd² about the x­axis.

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

CAS

CAS

30. A group of engineers is building a parabolic satellite dish whose shape will be formed by rotating the curve y − ax² about the y­axis. If the dish is to have a 10­ft diameter and a maximum depth of 2 ft, find the value of a and the surface area of the dish.

31. (a) The ellipse x² a² 1 y²

b² − 1 a . b

is rotated about the x­axis to form a surface called an ellipsoid, or prolate spheroid. Find the surface area of this ellipsoid.

(b) If the ellipse in part (a) is rotated about its minor axis (the y­axis), the resulting ellipsoid is called an oblate spheroid. Find the surface area of this ellipsoid.

32. Find the surface area of the torus in Exercise 6.2.63.

33. If the curve y − fsxd, a < x < b, is rotated about the hori­

zontal line y − c, where fsxd < c, find a formula for the area of the resulting surface.

34. Use the result of Exercise 33 to set up an integral to find the area of the surface generated by rotating the curve y −sx, 0 < x < 4, about the line y − 4. Then use a CAS to evalu­

ate the integral.

35. Find the area of the surface obtained by rotating the circle x²1y²− r² about the line y − r.

36. (a) Show that the surface area of a zone of a sphere that lies between two parallel planes is S − 2Rh, where R is the radius of the sphere and h is the distance between the planes. (Notice that S depends only on the distance between the planes and not on their location, provided that both planes intersect the sphere.)

(b) Show that the surface area of a zone of a cylinder with radius R and height h is the same as the surface area of the zone of a sphere in part (a).

37. Show that if we rotate the curve y − e^xy21e^2xy2 about the x­axis, the area of the resulting surface is the same value as the enclosed volume for any interval a < x < b.

38. Let L be the length of the curve y − fsxd, a < x < b, where f is positive and has a continuous derivative.

Let Sf be the surface area generated by rotating the curve about the x­axis. If c is a positive constant, define tsxd − f sxd 1 c and let St be the corresponding surface area generated by the curve y −tsxd, a < x < b. Express S_t in terms of Sf and L.

39. Formula 4 is valid only when fsxd > 0. Show that when fsxd is not necessarily positive, the formula for surface area becomes

S −

y

a^b 2

|

^fsxd

|

s1 1 f f 9sxdg² dx

CAS

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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Solution:

758 ¤ CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION 24. =√

²+ 1 ⇒ 

 = 

√²+ 1 ⇒  =

 1 +





2

 =

 1 + ²

²+ 1 ⇒

 =

 3 0

2

²+ 1

 1 + ²

²+ 1 = 2

3 0

2²+ 1  = 2√ 2 

 3 0



²+

√1 2

2



= 221 √ 2

1 2

²+¹₂+¹₄ln

 +

²+¹₂³

0= 2√ 2

3 2



9 +¹₂+¹₄ln 3 +

9 +¹₂

−¹4ln√¹ 2



= 2√ 2

3 2

19 2 +¹₄ln

3 +

19 2



+¹₄ln√ 2

= 2√ 2

3 2

√19

√2 +¹₄ln 3√

2 +√ 19

= 3√

19 +√^ 2ln

3√ 2 +√

19

25. = ³and 0 ≤  ≤ 1 ⇒ ⁰= 3²and 0 ≤  ≤ 1.

 =1 0 2

1 + (3²)² = 23 0

√1 + ^{2 1}₆

 = 3²,

 = 6 



=^₃3 0

√1 + ²

=21 [or use CAS] ^₃₁

2√

1 + ²+¹₂ln

 +√

1 + ²3 0= ^₃₃

2

√10 +¹₂ln 3 +√

10

= ^₆ 3√

10 + ln 3 +√

10

26. = ln( + 1), 0 ≤  ≤ 1.  =

 1 +





2

 =

 1 +

 1

 + 1

2

, so

 =

 1 0

2



1 + 1

( + 1)²  =

 2 1

2( − 1)

 1 + 1

² [ =  + 1,  = ]

= 2

 2 1



√1 + ²

  − 2

 2 1

√1 + ²

  = 2

 2 1

1 + ² − 2

 2 1

√1 + ²

 

21, 23

= [or use CAS] 21 2√

1 + ²+¹₂ln

 +√

1 + ²²

1− 2

√

1 + ²− ln

1 +√ 1 + ²



²

1

= 2√

5 +¹₂ln 2 +√

5

−¹2

√2 −¹2ln 1 +√

2

− 2√

5 − ln

1 +√ 5 2

−√ 2 + ln

1 +√ 2

= 2

1 2ln

2 +√ 5

+ ln

1 +√ 5 2



+^√₂²−³2ln 1 +√

2

27. = 2

 _∞

1



 1 +





2

 = 2

_∞

1

1



 1 + 1

⁴ = 2

 _∞

1

√⁴+ 1

³ . Rather than trying to evaluate this integral, note that√

⁴+ 1 √

⁴= ²for   0. Thus, if the area is finite,

 = 2

 _∞

1

√⁴+ 1

³   2

_∞

1

²

³ = 2

_∞

1

1

. But we know that this integral diverges, so the area  is infinite.

28. =∞

0 2

1 + ()² = 2∞ 0 ^−

1 + (−^−)² [ = ^−, ⁰=−^−].

Evaluate  =

^−

1 + (−^−)²by using the substitution  = −^−,  = ^−:

 = √

1 + ²=^{21 1}₂√

1 + ²+¹₂ln

 +√ 1 + ²

+  =¹₂(−^−)√

1 + ^−2+¹₂ln

−^−+√

1 + ^−2+ .

Returning to the surface area integral, we have

° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c

37. (a) The ellipse

x² a² +y²

b² = 1 a > b

is rotated about the x-axis to form a surface called an ellipsoid, or prolate spheroid. Find the surface area of this ellipsoid.

(b) If the ellipse in part (a) is rotated about its minor axis (the y-axis), the resulting ellipsoid is called an oblate spheroid. Find the surface area of this ellipsoid.

Solution:

1

760 ¤ CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION 31. (a) ²

² +²

² = 1 ⇒  ()

² = −

² ⇒ 

 = −²

² ⇒

1 +





2

= 1 + ⁴²

⁴² =⁴²+ ⁴²

⁴² = ⁴²+ ⁴²

1 − ²²

⁴²(1 − ²²) =⁴²+ ⁴²− ²²²

⁴²− ²²²

=⁴+ ²²− ²²

⁴− ²² = ⁴−

²− ²

²

²(²− ²)

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

Thus,

 = 2

 0

2

 1 +





2

 = 4

  0





²− ²

⁴− (²− ²)²

√

²− ²  =4

²

  0

⁴− (²− ²)²

=4

²

 √

²−² 0

⁴− ² 

√²− ²

 =√

²− ² 30

= 4

²√

²− ²

 2

⁴− ²+⁴

2 sin⁻¹ 

²

^√

²−²

0

= 4

²√

²− ²

√

²− ² 2

⁴− ²(²− ²) + ⁴ 2 sin⁻¹

√²− ²





= 2





²+

² sin⁻¹

√²− ²

√ 

²− ²







(b) ²

² +²

² = 1 ⇒  ()

² = −

² ⇒ 

 = −²

² ⇒

1 +





2

= 1 + ⁴²

⁴² = ⁴²+ ⁴²

⁴² =⁴²(1 − ²²) + ⁴²

⁴²(1 − ²²) = ²⁴− ²²²+ ⁴²

²⁴− ²²²

= ⁴− ²²+ ²²

⁴− ²² = ⁴− (²− ²)²

²(²− ²)

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

-axis. Thus,

 = 2

  0

2 

 1 +





2

 = 4

  0





²− ²

⁴− (²− ²)²



²− ² 

= 4

²

  0

⁴− (²− ²) ² = 4

²

  0

⁴+ (²− ²) ² 

since   

= 4

²

 √

²−² 0

⁴+ ² 

√²− ²

 =√

²− ²

21= 4

²√

²− ²

 2

√⁴+ ²+⁴ 2 ln

 +√

⁴+ ²√

²−²

0

= 4

²√

²− ²

√

²− ²

2 () +⁴ 2 ln

√

²− ²+ 

−

 0 +⁴

2 ln(²)



= 4

²√

²− ²

²√

²− ² 2 +⁴

2 ln√

²− ²+ 

²



= 2²+ 2²

√²− ² ln

√²− ²+ 

 32.The upper half of the torus is generated by rotating the curve ( − )²+ ²= ²,   0, about the -axis.



 = −( − ) ⇒ 1 +





2

= 1 +( − )²

² = ²+ ( − )²

² = ²

²− ( − )². Thus,

° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c

42. Zone of a Sphere A Zone of a Sphere is the portion of the sphere that lies between two parallel planes.

Show that the surface area of a zone of a sphere that lies between two parallel planes is S = 2πRh, where R is the radius of the sphere and h is the distance between the planes. (Notice that S depends only on the distance between the planes and not on their location, provided that both planes intersect the sphere.)

Solution:

2

SECTION 8.2 AREA OF A SURFACE OF REVOLUTION ¤ 761

 = 2+

−2

 1 +





2

 = 4

 +

−

 

²− ( − )² = 4



−

 + 

√²− ²  

 = − 

= 4

 

−

√ 

²− ² + 4



−

√ 

²− ² = 4 · 0 + 8

  0

√ 

²− ²

since the first integrand is odd and the second is even



= 8

sin⁻¹()

0= 8_

2

= 4²

33.The analogue of (^∗)in the derivation of (4) is now  − (^∗), so

 = lim

→∞



=12[ − (^∗)]

1 + [⁰(^∗_)]²∆ =

2[ − ()]

1 + [⁰()]² .

34. = ^12 ⇒ ⁰=¹₂^−12 ⇒ 1 + (⁰)²= 1 + 14, so by Exercise 33,  =4 0 2

4 −√

 

1 + 1(4) .

Using a CAS, we get  = 2 ln√

17 + 4 +^₆

31√ 17 + 1

≈ 806095.

35.For the upper semicircle, () =√

²− ², ⁰() = −√

²− ². The surface area generated is

1=

 

−

2

 −

²− ²

1 + ²

²− ²  = 4

  0

 −

²− ² 

√²− ²

= 4

  0

 ²

√²− ² − 





For the lower semicircle, () = −√

²− ²and ⁰() = 

√²− ², so 2= 4

 0

 ²

√²− ² + 



.

Thus, the total area is  = 1+ 2= 8

  0

 ²

√²− ²



 = 8

²sin⁻¹ 





0= 8²  2

= 4²².

36. (a) Rotate  =√

²− ²with  ≤  ≤  +  about the -axis to generate a zone of a sphere.  =√

²− ² ⇒

⁰= ¹₂(²− ²)^−12(−2) ⇒  =

 1 +

√ −

²− ²

2

. The surface area is

 =

 +



2  = 2

 +



²− ²



1 + ²

²− ²

= 2

 +



²− ²+ ² = 2

+



= 2( +  − ) = 2

(b) Rotate  =  with 0 ≤  ≤  about the -axis to generate a zone of a cylinder.  =  ⇒ ⁰= 0 ⇒

 =√

1 + 0² = . The surface area is  =

0 2  = 2

0   = 2



0 = 2.

37. = ^2+ ^−2 ⇒ ⁰ = ¹₂^2−¹2^−2 ⇒

1 + (⁰)²= 1 +

1

2^2−¹2^−22

= 1 +¹₄^−¹2+¹₄^−=¹₄^+¹₂ +¹₄^−=

1

2^2+¹₂^−22

. If we rotate the curve about the -axis on the interval  ≤  ≤ , the resulting surface area is

° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c

3