Posted by jump on Monday, December 3, 2012 at 10:41pm.
check to be sure that the graph lies all on one side of the axis for the interval. It does.
So, volume is
v = ∫[0,1] pi (R^2-r^2) dx
where R = 3, r = 3-y
= pi∫[0,1] 9 - (3-(x^3-x^5))^2 dx
= pi∫[0,1] -x^10 + 2x^8 - x^6 - 6x^5 + 6x^3 dx
= pi(-1/11 x^11 + 2/9 x^9 - 1/7 x^7 - x^6 + 3/2 x^4) [0,1]
= 677/1386
*pi
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