Suppose that $ f(x) $ is a continuous function on the interval $ [a,b] $. Let $ V(c) $ denote the volume of the solid obtained by revolving the area between the graph of $ y=f(x) $ and the line $ y=c $ about the line $ y=c $.
As $ c $ ranges from $ m=\{\text{min }f(x):a\le x\le b\} $ and $ M=\{\text{Max }f(x):a\le x\le b\} $, prove that the minmum value of $ V(c) $ is attained at $ c $ equal to the average value of $ f(x) $ on the interval $ [a,b] $.
I will give Josh Hunsberger's proof here.
Proof. Notice that
$ V(c)= \int_a^b \pi (f(x)-c)^2\ dx = Ac^2-Bc+D $
where $ A=\pi \int_a^b f(x)^2 $ is a constant which is positive if $ f $ is not identically zero, $ B=2\pi\int_a^b f(x)\ dx $ and $ D=\pi(b-a) $.
The case where $ f(x) $ is identically zero is easy. The minimum volume is zero and occurs at $ c=0 $, which is the average value of $ f $.
