Friday

April 18, 2014

April 18, 2014

Posted by **Mark** on Sunday, July 7, 2013 at 3:47pm.

- Calculus -
**Steve**, Sunday, July 7, 2013 at 3:59pmv = ∫[1,3] 2πrh dx

where r = x and h = y = −3(x−1)(x−3) so

v = ∫[1,3] 2πx(−3(x−1)(x−3)) dx

= -6π ∫[1,3] x^3 - 4x^2 + 3x dx

= 16π

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