I keep getting 0 for the area? dy/dt = 36sin^2t dx/dt = 36(1-cost)^2 integral from 0 -> 2pi of sqrt(36sin^2t + 36(1-cost)^2) = integral 6sqrt(2) sqrt(1-cost) from 0-->2pi = 6sqrt(2) u^-1/2 = 0??
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Originally Posted by lc99 I keep getting 0 for the area? dy/dt = 36sin^2t dx/dt = 36(1-cost)^2 Where are the 36's coming from? Also the derivatives don't have any squares in them. integral from 0 -> 2pi of sqrt(36sin^2t + 36(1-cost)^2) = integral 6sqrt(2) sqrt(1-cost) from 0-->2pi = 6sqrt(2) u^-1/2 = 0?? For rotation about the x axis you need a $2\pi y$ factor.
I have taken the limits from 0 to pi/2 according to your Question. You can do this integral easily without calculator, but it will be a little long. I will give you the answer directly to shorten the time!
Last edited by joshuaa; Apr 27th 2018 at 04:06 PM.