So I figured out part (a) already, but I really need help on part (b):
(a) When the graph of y = x2 for 0 ≤ x ≤ 2 is rotated about the horizontal line y = L, the volume obtained depends on L:
V(L) = (integral from 0 to 2) of pi*(x^2L)^2dx
= pi * (32/5  16/3L + 2L^2) <my answer, which I know is correct.
(b) What value of L minimizes the volume in part (a)?
I thought I had to take the derivative of the integral, but either it doesn't work that way or I did it wrong.... 🙁
1 Answer

Assuming your expression for V(L) is correct,
V'(L)=pi(4L16/3)
So stationary value for V is when L=4/3
V''(L)=4pi which is >0
So V is minimum at L=4/3
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