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bob's question at Yahoo! Answers regarding mimimizing a solid of revolution

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MarkFL

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Feb 24, 2012
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Here is the question:

Solid of Revolution Calculus 2 question.?

For a positive real number p, define f(x)=p/x^(p). Find the minimum value of the volume of the solid created by rotating this function around the x-axis over the interval [1,infinity).
I have posted a link there to this thread so the OP can view my work.
 
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MarkFL

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Feb 24, 2012
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Hello bob,

First, let's write the given function as:

\(\displaystyle f(x)=px^{-p}\)

Now, using the disk method, we find that the volume of an arbitrary disk is:

\(\displaystyle dV=\pi r^2\,dx\)

where:

\(\displaystyle r=f(x)=px^{-p}\)

And so we have:

\(\displaystyle dV=\pi p^2 x^{-2p}\,dx\)

Summing us all of the disks, we may state:

\(\displaystyle V=\pi p^2\int_1^{\infty}x^{-2p}\,dx\)

This is an improper integral with the unbounded upper limit, so we may write:

\(\displaystyle V=\pi p^2\lim_{t\to\infty}\left(\int_1^{\infty}x^{-2p}\,dx \right)\)

Applying the FTOC, we have:

\(\displaystyle V=\pi p^2\lim_{t\to\infty}\left(\left[\frac{x^{1-2p}}{1-2p} \right]_1^{t} \right)\)

\(\displaystyle V=\frac{\pi p^2}{1-2p}\lim_{t\to\infty}\left(\frac{1}{t^{2p-1}}-1 \right)\)

For \(\displaystyle \frac{1}{2}<p\) we have:

\(\displaystyle V(p)=\frac{\pi p^2}{2p-1}\)

To determine the critical value(s), we need to differentiate with respect to $p$ and equate the result to zero. Using the quotient rule, we find:

\(\displaystyle V'(p)=\frac{(2p-1)(2\pi p)-\left(\pi p^2 \right)(2)}{(2p-1)^2}=\frac{2\pi p(p-1)}{(2p-1)^2}\)

For \(\displaystyle \frac{1}{2}<p\) we have:

\(\displaystyle p=1\)

If we observe that for \(\displaystyle \frac{1}{2}<p<1\) we have $V'(p)<0$ and for $1<p$ we have $V'(p)>0$, we may therefore conclude that this critical value is at a minimum.

Thus, we may conclude that the described solid of revolution is minimized when $p=1$.