Calculus: find the value of a variable that minimizes a volume given by a definite integral?

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^2-L)^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(4L-16/3)

    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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