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<math>V(c)</math> is attained at <math>c</math> equal to the average value of <math>f(x)</math> | <math>V(c)</math> is attained at <math>c</math> equal to the average value of <math>f(x)</math> | ||
on the interval <math>[a,b]</math>. | on the interval <math>[a,b]</math>. | ||
| + | |||
| + | I will give Josh Hunsberger's proof here. | ||
| + | |||
| + | Proof. Notice that | ||
| + | |||
| + | <math>V(c)= \int_a^b \pi (f(x)-c)^2\ dx = Ac^2-Bc+D</math> | ||
| + | |||
| + | where <math>A=\pi \int_a^b f(x)^2</math> is a constant which is positive if <math>f</math> | ||
| + | is not identically zero, <math>B=2\pi\int_a^b f(x)\ dx</math> and <math>D=\pi(b-a)</math>. | ||
| + | |||
| + | The case where <math>f(x)</math> is identically zero is easy. The minimum volume is zero | ||
| + | and occurs at <math>c=0</math>, which is the average value of <math>f</math>. | ||
Revision as of 08:04, 29 September 2008
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 $.
