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Sigrid O'Kon

Feb 20, 2025

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)
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Assuming your expression for V(L) is correct,V'(L)=pi(4L-16/3) So stationary value for V is when L=4/3V''(L)=4pi which is >0So V is minimum at L=4/3... Show More
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