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M2solns - Math 227.04 Solutions to Midterm Exam 2 1 6 Z p 1...

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Math 227.04 Solutions to Midterm Exam 2 1. 6 Z 2 1 p x ° 2 dx = lim a ! 2 + Z 6 a ( x ° 2) ° 1 = 2 dx = lim a ! 2 + 2 ( x ° 2) 1 = 2 j 6 a = lim a ! 2 + f 2(6 ° 2) 1 = 2 ° 2( a ° 2) 1 = 2 g = 4 ° 0 = 4 : 2. Let D be region in the °rst quadrant of the xy plane bounded by the curves y = x 2 and y = 0 and x = 2 : Suppose D is revolved around the line x = 2 . (a) Sketch the solid region. The sketches posted on a separate document. (b) Set up the volume integral using the disk/washer method. For each y between 0 and 4 , the cross section is a disk of radius 2 ° x where x = p y: Thus A ( y ) = ° (2 ° x ) 2 = ° ° 2 ° p y ± 2 so V = Z 4 0 ° (2 ° p y ) 2 dy: (c) Set up the volume integral using the shell method. The shells are centered on the vertical line x = 2 . A typical shell has radius r = 2 ° x and height h = x 2 and thickness dx: So V = Z 2 0 2 ° (2 ° x ) x 2 dx: (d) Compute the volume swept out using the your choice of the integrals in (b) or (c). Using the disk method, V = Z 4 0 ° (2 ° p y ) 2 dy = ° Z 4 0 ² 4 ° 4 y 1 = 2 + y ³ dy = ° ´ 4 y ° 8 3 y 3 = 2 + 1 2 y 2 µ 4 0 = ° ´ 16 ° 64 3 + 8 µ = 8 ° 3 :
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