sol to a9

# sol to a9 - MAT 137Y 2008-09 Winter Session Solutions to...

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MAT 137Y 2008-09 Winter Session, Solutions to Problem Set 9 1 (SHE 6.3) 12. The volume is V = Z 1 0 2 π x · 2 x dx + Z 2 1 2 π x · 2 2 - x dx = 4 π Z 1 0 x 2 dx - 4 π Z 0 1 ( 2 - u ) u du ( u = 2 - x ) = 4 π x 3 3 1 0 - 4 π 4 3 u 3 / 2 - 2 5 u 5 / 2 0 1 = 76 15 π . 34. The volume is given by V = Z r 2 - a 2 0 2 π x ( p r 2 - x 2 - a ) dx = 2 π Z r 2 - a 2 0 h x ( r 2 - x 2 ) 1 / 2 - ax i dx = 2 π - 1 3 ( r 2 - x 2 ) 3 / 2 - a 2 x 2 r 2 - a 2 0 = 1 3 π ( 2 r 3 + a 3 - 3 ar 2 ) . 46. We let u = r 2 - x 2 so du = - 2 x dx . The volume is given by V = 2 Z r q r 2 - h 2 4 2 π x p r 2 - x 2 dx = - 2 π Z 0 h 2 / 4 u 1 / 2 du = 2 π 2 3 u 3 / 2 h 2 / 4 0 = π h 3 6 . 2 (a) The region and a typical shell is illustrated below. (b) Using shells, the volume is V = Z 1 0 2 π ( y + 1 )( y - y 2 ) dy = 2 π Z 1 0 y 3 / 2 + y 1 / 2 - y 3 - y 2 dy = 2 π 2 5 y 5 / 2 + 2 3 y 3 / 2 - 1 4 y 4 - 1 3 y 3 1 0 = 29 π 30 .

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(c) If we were to use washers, then the volume would be expressed as V = Z 1 0 π ( x + 1 ) 2 - π ( x 2 + 1 ) 2 dx = π Z 1 0 x + 2 x + 1 - x 4 - 2 x 2 - 1 dx = π Z 1 0 x + 2 x - x 4 - 2 x 2 dx = π 1 2 x 2 + 4 3 x 3 / 2 - 1 5 x 5 - 2 3 x 3 1 0 = 29 30 π 3 (SHE 7.1) 40. Since f 0 ( x ) = 2 ( x - 1 ) 2 , so ( f - 1 ) 0 ( 3 ) = 1 f 0 ( f - 1 ( 3 )) = 1 f 0 ( 3 ) = 1 1
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