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

#### MarkFL

Staff member
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.

#### MarkFL

Staff member
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 $p=1$.