Unformatted text preview: x –axis. So we obtain the following: Volume = π Z 1 ( f 1 ( x )) 2( f 2 ( x )) 2 dx = π Z 1 sec 2 xtan 2 xdx = π Z 1 1 dx = π where we used the fact that sec 2 xtan 2 x = 1. 3. Find the volume of the solid generated by revolving about the yaxis the region bounded by the curves y = 3 x +22 √ x 3 +11 x 2 +4 , the xaxis, and the lines x = 0 and x = 1. Solution: Let f ( x ) = 3 x +22 √ x 3 +11 x 2 +4 and observe that the volume of the solid obtained by revolving around the y –axis is most easily obtained by the shell method. Hence, Volume = Z 1 2 πxf ( x ) dx = 2 π Z 1 3 x 2 + 22 x √ x 3 + 11 x 2 + 4 dx = 2 π h 2 √ x 3 + 11 x 2 + 4 i 1 = 4 π (42) = 8 π. 1...
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 Spring '08
 Vershynin
 Calculus, Cos

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