40 find the area of the surface obtained by rotating

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40. Find the area of the surface obtained by rotating y = cosh x over [− ln 2 , ln 2 ] around the x -axis. SOLUTION Let y = cosh x . Then y = sinh x , and 1 + ( y ) 2 = 1 + sinh 2 x = cosh 2 x = cosh x . Therefore, SA = 2 π ln 2 ln 2 cosh 2 x dx = π ln 2 ln 2 ( 1 + cosh 2 x ) dx = π x + 1 2 sinh 2 x ln 2 ln 2 = π ln 2 + 1 2 sinh ( 2 ln 2 ) + ln 2 1 2 sinh ( 2 ln 2 ) = 2 π ln 2 + π sinh ( 2 ln 2 ). We can simplify this answer by recognizing that sinh ( 2 ln 2 ) = e 2 ln 2 e 2 ln 2 2 = 4 1 4 2 = 15 8 . Thus, SA = 2 π ln 2 + 15 π 8 . 41. Prove that the surface area of a sphere of radius r is 4 π r 2 by rotating the top half of the unit circle x 2 + y 2 = r 2 about the x -axis. SOLUTION Let y = r 2 x 2 . Then y = − x r 2 x 2 , and 1 + ( y ) 2 = 1 + x 2 r 2 x 2 = r 2 r 2 x 2 . Finally, SA = 2 π r r r 2 x 2 r r 2 x 2 dx = 2 π r r r dx = 2 π r ( 2 r ) = 4 π r 2 . In Exercises 42–45, use a computer algebra system to find the exact or approximate surface area of the solid generated by rotating the curve about the x-axis. 42. y = 1 4 x 2 1 2 ln x for 1 x e , exact area SOLUTION Let y = 1 4 x 2 1 2 ln x . Then y = x 2 1 2 x , and 1 + ( y ) 2 = 1 + x 2 1 2 x 2 = 1 + x 2 4 1 2 + 1 4 x 2 = x 2 4 + 1 2 + 1 4 x 2 = x 2 + 1 2 x 2 . Thus, SA = 2 π e 1 x 2 4 ln x 2 x 2 + 1 2 x dx . Using a computer algebra system to evaluate the definite integral, we find SA = π ( e 4 9 ) 16 . 43. y = x 2 for 0 x 4, exact area
S E C T I O N 8.1 Arc Length and Surface Area 971 SOLUTION Let y = x 2 . Then y = 2 x , 1 + ( y ) 2 = 1 + 4 x 2 , and SA = 2 π 4 0 x 2 1 + 4 x 2 dx . Using a computer algebra system to evaluate the definite integral, we find SA = 129 65 4 π π 32 ln ( 8 + 65 ). 44. y = x 3 for 0 x 4, approximate area SOLUTION Let y = x 3 . Then y = 3 x 2 , 1 + ( y ) 2 = 1 + 9 x 4 , and SA = 2 π 4 0 x 3 1 + 9 x 4 dx . Using a computer algebra system to approximate the value of the definite integral, we find SA 12876 . 2 . 45. y = tan x for 0 x π 4 , approximate area SOLUTION Let y = tan x . Then y = sec 2 x , 1 + ( y ) 2 = 1 + sec 4 x , and SA = 2 π π / 4 0 tan x 1 + sec 4 x dx . Using a computer algebra system to approximate the value of the definite integral, we find SA 3 . 83908 . 46. Find the surface area of the torus obtained by rotating the circle x 2 + ( y b ) 2 = r 2 about the x -axis. SOLUTION y = b + r 2 x 2 gives the top half of the circle and y = b r 2 x 2 gives the bottom half. Note that in each case, 1 + ( y ) 2 = 1 + x 2 r 2 x 2 = r 2 r 2 x 2 . Rotating the two halves of the circle around the x -axis then yields SA = 2 π r r ( b + r 2 x 2 ) r r 2 x 2 dx + 2 π r r ( b r 2 x 2 ) r r 2 x 2 dx = 2 π r r 2 b r r 2 x 2 dx = 4 π br r r 1 r 2 x 2 dx = 4 π br · sin 1 x r r r = 4 π br π 2 π 2 = 4 π 2 br . 47. A merchant intends to produce specialty carpets in the shape of the region in Figure 15, bounded by the axes and graph of y = 1 x n (units in yards). Assume that material costs $50 / yd 2 and that it costs 50 L dollars to cut the carpet, where L is the length of the curved side of the carpet. The carpet can be sold for 150 A dollars, where A is the carpet’s area. Using numerical integration with a computer algebra system, find the whole number

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