Surface area of revolution PDF

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AREA OF A SURFACE OF REVOLUTION

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A surface of revolution is formed when a curve is rotated about a line. Such a surface is the lateral boundary of a solid of revolution of the type discussed in Sections 7.2 and 7.3. We want to define the area of a surface of revolution in such a way that it corresponds to our intuition. If the surface area is A , we can imagine that painting the surface would require the same amount of paint as does a flat region with areaA . Let’s start with some simple surfaces. The lateral surface area of a circular cylinder with radius r and height h is taken to be A 苷 2␲ rh because we can imagine cutting the cylinder and unrolling it (as in Figure 1) to obtain a rectangle with dimensions2␲ r andh . Likewise, we can take a circular cone with base radius r and slant height l , cut it along the dashed line in Figure 2, and flatten it to form a sector of a circle with radiusl and central angle ␪ 苷 2 ␲ r兾l . We know that, in general, the area of a sector of a circle with radius l and angle ␪ is 12 l 2␪ (see Exercise 67 in Section 6.2) and so in this case it is

h r

h

1

冉 冊

A 苷 21 l 2␪ 苷 2 l 2

2πr F I GURE 1

2␲r l

苷 ␲ rl

Therefore, we define the lateral surface area of a cone to beA 苷 ␲ rl . 2πr

cut l

¨

r

l

F I GURE 2

What about more complicated surfaces of revolution? If we follow the strategy we used with arc length, we can approximate the original curve by a polygon. When this polygon is rotated about an axis, it creates a simpler surface whose surface area approximates the actual 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 rotating 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 frustum of a cone) with slant height l and upper and lower radii r1 and r2 is found by subtracting the areas of two cones:

l¡

r¡

1 l

A 苷 ␲ r2共l1 ⫹ l 兲 ⫺ ␲ r1l1 苷 ␲ 关共r2 ⫺ r1兲l1 ⫹ r2 l 兴

From similar triangles we have l ⫹l l1 苷 1 r2 r1

r™

which gives r2 l1 苷 r1l1 ⫹ r1l

F I GURE 3

or

共r2 ⫺ r1兲l1 苷 r1l

Putting this in Equation 1, we get A 苷 ␲ 共r1l ⫹ r2 l 兲 Thomson Brooks-Cole copyright 2007

or 2

A 苷 2␲ rl

1 where r 苷 2 共r1 ⫹ r2 兲 is the average radius of the band.

1

2 ■ ARE A OF A S U RFAC E OF RE VOL U T I ON

y

y=ƒ

0

a

b

x

ⱍ

(a) Surface of revolution y

P¸ 0

Pi-1

Pi

a

Now we apply this formula to our strategy. Consider the surface shown in Figure 4, which is obtained by rotating the curve y 苷 f 共x兲, a 艋 x 艋 b, about the x -axis, where f is positive and has a continuous derivative. In order to define its surface area, we divide the interval 关a, b兴 into n subintervals with endpoints x0, x1, . . . , xn and equal width ⌬x , as we did in determining arc length. If y i 苷 f 共x i 兲 , then the point Pi 共x i, y i 兲 lies on the curve. The part of the surface between x i⫺1 and x i is approximated by taking the line segmentPi⫺1Pi and rotating it about the x -axis. The result is a band with slant heightl 苷 Pi⫺1Pi and aver1 age radius r 苷 2 共y i⫺1 ⫹ y i 兲 so, by Formula 2, its surface area is

yi

b

x

ⱍ

ⱍ

As in the proof of Theorem 7.4.2, we have

ⱍP (b) Approximating band

y i⫺1 ⫹ y i Pi⫺1Pi 2

2␲

Pn

ⱍ

ⱍ

Pi 苷 s1 ⫹ 关 f ⬘共xi*兲兴 2 ⌬x

i⫺1

where x*i is some number in 关x i⫺1, x i 兴. When ⌬x is small, we have y i 苷 f 共x i 兲 ⬇ f 共 xi*兲 and also y i⫺1 苷 f 共x i⫺1 兲 ⬇ f 共 xi*兲, since f is continuous. Therefore

F I GURE 4

2␲

y i⫺1 ⫹ y i Pi⫺1Pi ⬇ 2␲ f 共xi*兲 s1 ⫹ 关 f ⬘共xi*兲兴 2 ⌬x 2

ⱍ

ⱍ

and so an approximation to what we think of as the area of the complete surface of revolution is n

兺 2␲ f 共x *兲 s1 ⫹ 关 f ⬘共x *兲兴

3

i

i

2

⌬x

i苷1

This approximation appears to become better as n l ⬁ and, recognizing (3) as a Riemann sum for the function t共x兲 苷 2 ␲ f 共x兲 s1 ⫹ 关 f ⬘共x兲兴 2 , we have n

lim

兺 2␲ f 共x*兲 s1 ⫹ 关 f ⬘共x *兲兴

n l ⬁ i苷1

i

i

2

⌬x 苷 y 2␲ f 共x兲 s1 ⫹ 关 f ⬘共x兲兴 2 dx b

a

Therefore, in the case where f is positive and has a continuous derivative, we define the surface area of the surface obtained by rotating the curve y 苷 f 共x兲 , a 艋 x 艋 b , about the x-axis as

4

S 苷 y 2 ␲ f 共x兲 s1 ⫹ 关 f ⬘共x兲兴 2 dx b

a

With the Leibniz notation for derivatives, this formula becomes

5

S 苷 y 2␲ y b

a

冑 冉冊 1⫹

dy dx

2

dx

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If the curve is described as x 苷 t共y兲 , c 艋 y 艋 d , then the formula for surface area becomes

6

S 苷 y 2␲ y d

c

冑 冉冊 1⫹

dx dy

2

dy

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

ARE A OF A S U RFAC E OF RE VOL U T I ON ■ 3

length given in Section 7.4, as S 苷 y 2 ␲y ds

7

For rotation about the y-axis, the surface area formula becomes S 苷 y 2 ␲x ds

8

where, as before, we can use either ds 苷

冑 冉 冊 1⫹

dy dx

2

ds 苷

or

dx

冑 冉 冊 1⫹

dx dy

2

dy

These formulas can be remembered by thinking of 2␲ y or 2␲ x as the circumference of a circle traced out by the point 共x, y兲 on the curve as it is rotated about the x -axis or y -axis, respectively (see Figure 5). y

y (x,y)

y x 0

circumference=2π y

circumference=2πx 0

F I GURE 5

(x,y)

x

(a) Rotation about x-axis: S=j2πyds

x

(b) Rotation about y-axis: S=j2πxds

EXAMPLE 1 The curve y 苷 s4 ⫺ x 2, ⫺1 艋 x 艋 1 , is an arc of the circle x 2 ⫹ y 2 苷 4 .

Find the area of the surface obtained by rotating this arc about the x -axis. (The surface is a portion of a sphere of radius 2. See Figure 6.) SOLUTION We have

y

⫺x dy 苷 21 共4 ⫺ x 2 兲⫺1兾2共⫺2x兲 苷 s4 ⫺ x 2 dx and so, by Formula 5, the surface area is 1

x

S 苷 y 2 ␲y 1

⫺1

苷 2␲

y

1

y

1

y

1

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⫺1

苷 2␲

⫺1

F I GURE 6 Figure 6 shows the portion of the sphere whose surface area is computed in Example 1.

■ ■

苷 4␲

⫺1

冑 冉 冊 冑 1⫹

s4 ⫺ x 2

s4 ⫺ x 2

dy dx

2

1⫹

dx x2 dx 4 ⫺ x2

2 dx s4 ⫺ x 2

1 dx 苷 4 ␲共2兲 苷 8␲

4 ■ ARE A OF A S U RFAC E OF RE VOL U T I ON

Figure 7 shows the surface of revolution whose area is computed in Example 2.

■ ■

y

EXAMPLE 2 The arc of the parabola y 苷 x 2 from 共1, 1兲 to 共2, 4兲 is rotated about the y -axis. Find the area of the resulting surface. SOLUTION 1 Using

(2,4)

y 苷 x2

dy 苷 2x dx

and

y=≈

we have, from Formula 8,

0

1

2

S 苷 y 2 ␲ x ds

x

F I GURE 7

苷 y 2␲ x 2

1

冑 冉 冊 dy dx

1⫹

2

dx

苷 2 ␲ y x s1 ⫹ 4x 2 dx 2

1

Substituting u 苷 1 ⫹ 4x 2, we have du 苷 8x dx. Remembering to change the limits of integration, we have S苷

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 with the same height and radius halfway between the upper and lower radius of the surface: 2␲ 共1.5兲共3兲 ⬇ 28.27 . We computed that the surface area was

■ ■

␲ ( 17 s17 ⫺ 5 s5 ) ⬇ 30.85 6 which seems reasonable. Alternatively, the surface area should be slightly larger than the area of a frustum of a cone with the same top and bottom edges. From Equation 2, this is 2␲ 共1.5兲 ( s10 ) ⬇ 29.80.

苷

␲ 4

y

17

5

su du 苷

␲ 2 3兾2 17 [ 3 u ]5 4

␲ (17s17 ⫺ 5s5 ) 6

SOLUTION 2 Using

x 苷 sy

dx 1 苷 2 sy dy

and

we have S 苷 y 2 ␲ x ds 苷 y 2␲ x 4

1

冑

y

苷

␲ 4

y

苷

␲ ( 17s17 ⫺ 5s5) 6

4

17

5

sy

1⫹

1⫹

dx dy

2

dy

1 4 dy 苷 ␲ y s4y ⫹ 1 dy 1 4y

苷 2␲

1

冑 冉 冊

su du

(where u 苷 1 ⫹ 4y )

(as in Solution 1)

EXAMPLE 3 Find the area of the surface generated by rotating the curve y 苷 e x ,

0 艋 x 艋 1 , about the x -axis. Another method: Use Formula 6 with x 苷 ln y.

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■ ■

SOLUTION Using Formula 5 with

y 苷 ex

and

dy 苷 ex dx

ARE A OF A S U RFAC E OF RE VOL U T I ON ■ 5

we have S 苷 y 2 ␲y 1

0

冑 冉 冊 dy

1⫹

2

dx 苷 2␲ y e x s1 ⫹ e 2x dx 1

0

dx

苷 2 ␲ y s1 ⫹ u 2 du

(where u 苷 e x )

苷 2␲ y

(where u 苷 tan ␪ and ␣ 苷 tan⫺1e )

e

1

␣

␲ 兾4

■ ■

sec 3␪ d␪

[

ⱍ

1 苷 2 ␲ ⴢ 2 sec ␪ tan ␪ ⫹ ln sec ␪ ⫹ tan ␪

Or use Formula 21 in the Table of Integrals.

]ⱍ

␣ ␲兾4

(by Example 8 in Section 6.2)

[

]

苷 ␲ sec ␣ tan ␣ ⫹ ln共sec ␣ ⫹ tan ␣ 兲 ⫺ s2 ⫺ ln(s2 ⫹ 1) Since tan ␣ 苷 e , we have sec 2␣ 苷 1 ⫹ tan 2 ␣ 苷 1 ⫹ e 2 and

S 苷 ␲[ es1 ⫹ e 2 ⫹ ln (e ⫹ s1 ⫹ e 2 ) ⫺ s2 ⫺ ln(s2 ⫹ 1 )]

EXERCISES

Click here for answers.

A

S

17–20

Use Simpson’s Rule with n 苷 10 to approximate the area of the surface obtained by rotating the curve about thex -axis. Compare your answer with the value of the integral produced by your calculator.

Click here for solutions.

1– 4

Set up, but do not evaluate, an integral for the area of the surface obtained by rotating the curve about the given axis. 1. y 苷 ln x, 1 艋 x 艋 3;

2 2. y 苷 sin x, 0 艋 x 艋 ␲兾2;

x -axis

3. y 苷 sec x, 0 艋 x 艋 ␲兾4;

y-axis

4. y 苷 e x , 1 艋 y 艋 2 ; ■

■

■

■

■

17. y 苷 ln x,

x-axis

19. y 苷 sec x,

■ ■

■

■

■

■

■

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

6. 9 x 苷 y ⫹ 18, 7. y 苷 sx,

■

9. y 苷 cosh x,

0艋x艋1

x3 1 ⫹ , 6 2x

1 2

1

2 12. x 苷 1 ⫹ 2y , ■

■

■

CAS

艋x艋1

2 3兾2 11. x 苷 3 共 y ⫹ 2兲 ,

■

■

■

■

■

■

■

3 13. y 苷 s x , 1 艋 y 艋 2

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■

■

■

0 艋 y 艋 a兾2 ⫺a 艋 y 艋 a ■

■

■

■

■

■

■

■

1艋x艋2

■

■

0艋x艋3 ■

■

■

■

■

■

■

■

■

Use a CAS to find the exact area of the surface obtained by rotating the curve about the y -axis. If your CAS has trouble evaluating the integral, express the surface area as an integral in the other variable. 0艋y艋1

■

■

■

■

■

■

■

■

■

■

■

■

25. (a) If a ⬎ 0, find the area of the surface generated by rotating

the loop of the curve 3ay 2 苷 x共a ⫺ x兲2 about the x -axis. (b) Find the surface area if the loop is rotated about the y-axis. 26. A group of engineers is building a parabolic satellite dish

0艋x艋1

16. x 苷 a cosh共 y兾a兲, ■

■

■

The given curve is rotated about the y -axis. Find the area of the resulting surface.

15. x 苷 sa 2 ⫺ y 2,

■

24. y 苷 ln共 x ⫹ 1兲 , 0 艋 x 艋 1

13–16

2 14. y 苷 1 ⫺ x ,

■

23–24

■

■

■

23. y 苷 x 3,

1艋y艋2

1艋y艋2

■

0艋x艋1

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 x -axis. 22. y 苷 sx 2 ⫹ 1,

4艋x艋9 0 艋 x 艋 ␲兾6

■

21–22

21. y 苷 1兾x,

2艋x艋6

8. y 苷 cos 2x ,

10. y 苷

CAS

0艋x艋2 2

■

■

5–12

3 5. y 苷 x ,

■

1艋x艋2

0 艋 x 艋 ␲兾3

20. y 苷 s1 ⫹ e x,

y -axis ■

1艋x艋3

18. y 苷 x ⫹ sx,

■

■

■

■

■

■

■

whose shape will be formed by rotating the curvey 苷 ax 2 about the y -axis. If the dish is to have a 10-ft diameter and a maximum depth of 2 ft, find the value ofa and the surface area of the dish.

6 ■ ARE A OF A S U RFAC E OF RE VOL U T I ON

27. The ellipse

32. Show that the surface area of a zone of a sphere that lies

y2 x2 ⫹ 2 苷1 b a2

a⬎b

is rotated about the x -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 y 苷 f 共x兲, a 艋 x 艋 b , is rotated about the horizon-

tal line y 苷 c , where f 共x兲 艋 c , find a formula for the area of the resulting surface. CAS

30. Use the result of Exercise 29 to set up an integral to find the

area of the surface generated by rotating the curvey 苷 sx , 0 艋 x 艋 4, about the line y 苷 4 . Then use a CAS to evaluate the integral. 31. Find the area of the surface obtained by rotating the circle

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x 2 ⫹ y 2 苷 r 2 about the line y 苷 r.

between two parallel planes isS 苷 ␲ dh , whered is the diameter of the sphere andh 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.) 33. Formula 4 is valid only whenf 共x兲 艌 0 . Show that whenf 共x兲

is not necessarily positive, the formula for surface area becomes S 苷 y 2␲ f 共x兲 s1 ⫹ 关 f ⬘共x兲兴 2 dx b

a

ⱍ

ⱍ

34. Let L be the length of the curvey 苷 f 共x兲 ,a 艋 x 艋 b , where

f is positive and has a continuous derivative. LetSf be the surface area generated by rotating the curve about thex -axis. Ifc is a positive constant, definet共x兲 苷 f 共x兲 ⫹ c and letSt be the corresponding surface area generated by the curvey 苷 t共x兲 , a 艋 x 艋 b . Express St in terms of Sf and L.

ARE A OF A S U RFAC E OF RE VOL U T I ON ■ 7

ANSWERS

Click here for solutions.

S

13. ␲(145 s145 ⫺ 10 s10 )兾27 17. 9.023754

1.

y

3

1

3.

y

2␲ ln x s1 ⫹ 共1兾x兲 dx

␲ Ⲑ4

0

2 ␲x s1 ⫹ 共sec x tan x兲2 dx

9. ␲[1 ⫹ 4 共e 2 ⫺ e⫺2 兲]

Thomson Brooks-Cole copyright 2007

1

19. 13.527296

21. 共 ␲兾4兲 [4 ln(s17 ⫹ 4) ⫺ 4 ln(s2 ⫹ 1 ) ⫺ s17 ⫹ 4 s2]

2

5. ␲(145s145 ⫺ 1)兾27

2 15. ␲a

7. ␲ (37 s37 ⫺ 17 s17 )兾6 11. 21␲兾2

23. 共 ␲兾6兲 [ln( s10 ⫹ 3) ⫹ 3 s10 ] 2 25. (a) ␲ a 兾3

(b) 56␲ s3 a 2兾45

27. 2␲[b 2 ⫹ a 2b sin⫺1 (sa 2 ⫺ b 2兾a)兾sa 2 ⫺ b 2 ] 29.

xab 2␲ 关c ⫺ f 共x兲兴s1 ⫹ 关 f ⬘共x兲兴 2 dx

31. 4␲ 2r 2

8 ■ ARE A OF A S U RFAC E OF RE VOL U T I ON

SOLUT IONS

1. y = ln x ⇒ ds =

  3  1 + (dy/dx)2 dx = 1 + (1/x)2 dx ⇒ S = 1 2π(ln x) 1 + (1/x)2 dx [by (7)]

3. y = sec x ⇒ ds = S=

 π/4 0

5. y = x3

2πx

 1 + (sec x tan x)2 dx [by (8)]

⇒ y0 = 3x2 . So S= =

7. y =

  1 + (dy/dx)2 dx = 1 + (sec x tan x)2 dx ⇒

2

2πy

0

√   1 + (y0 )2 dx = 2π 02 x3 1 + 9x4 dx

 145 √ u du = 1

2π 36

π 18



2 3/2 u 3

145

=

1

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

 √  145 145 − 1

π 27

√ √ 2 x ⇒ 1 + (dy/dx)2 = 1 + [1/(2 x )] = 1 + 1/(4x). So

S=



9

2πy

4

= 2π



  2 x+ 3

1+



dy dx

 9 1 3/2 4 4

2

dx =

=

4π 1 (4x 3 8



9

2π 4



  9 √ 1 dx = 2π x 1+ x+ 4x 4

+ 1)

3/2

9

=

4

π 6

1 3

3/2  2 y +2

⇒ dx/dy =

1 2 y 2

+2

1/2

(2y) = y

   2 1 + (dx/dy)2 = 1 + y2 y2 + 2 = y2 + 1 . So S = 2π

13. y =

√ 3 x ⇒ x = y3

dx

 √ √  37 37 − 17 17

9. y = cosh x ⇒ 1 + (dy/dx)2 = 1 + sinh 2 x = cosh2 x. So    S = 2π 01 cosh x cosh x dx = 2π 01 12 (1 + cosh 2x) dx = π x +      or π 1 + 41 e2 − e−2 = π 1 + 12 sinh 2 11. x =

1 4

1 2

1 sinh 2x 0

 y2 + 2 ⇒

 2  2 2   y y + 1 dy = 2π 14 y4 + 12 y2 1 = 2π 4 + 2 − 1

1 4

 − 21 =

21π 2

⇒ 1 + (dx/dy)2 = 1 + 9y4 . So

    S = 2π 12 x 1 + (dx/dy)2 dy = 2π 12 y3 1 + 9y4 dy =    2  √ √  π 145 π 2 1 + 9y 4 3/2 145 − 10 10 = 18 = 27 3

2π 36