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. --Bell 15:04, 29 September 2008 (UTC)
Proof. Notice that
$ V(c)= \int_a^b \pi (f(x)-c)^2\ dx = Ac^2-Bc+D $
where $ D=\pi \int_a^b f(x)^2\ dx $ is a constant which is positive if $ f $ is not identically zero, $ B=2\pi\int_a^b f(x)\ dx $ and $ A=\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 $.
Hence, we may assume that $ f $ is not identically zero and that $ D $ is a positive constant. In this case, the graph of $ V(c) $ is an upward opening parabola. Such a parabola has an absolute minimum at the value $ c_0 $ such that $ V'(c_0)=0 $. Since $ V'(c)=2Ac-B $, this value of $ c_0 $ is
$ c_0 = \frac{B}{2A} = \frac{1}{b-a}\int_a^b f(x)\ dx, $
which is the average value of $ f $ on the interval. To complete the argument, we must show that this value of $ c $ falls in the interval between $ m $ and $ M $. But this is easy because
$ m \le f(x) \le M $
on $ [a,b] $ implies that
$ \int_a^b m\ dx \le int_a^b f(x)\ dx \le \int_a^b M\ dx, $
and so
$ m(b-a) \le int_a^b f(x)\ dx \le M(b-a), $
and dividing by $ (b-q) $ yields
$ m\le c_0\le M. $
The proof is complete.
