Hello bob,
First, let's write the given function as:
$$f(x)=px^{-p}$$
Now, using the disk method, we find that the volume of an arbitrary disk is:
$$dV=\pi r^2\,dx$$
where:
$$r=f(x)=px^{-p}$$
And so we have:
$$dV=\pi p^2 x^{-2p}\,dx$$
Summing us all of the disks, we may state:
$$V=\pi p^2\int_1^{\infty}x^{-2p}\,dx$$
This is an improper integral with the unbounded upper limit, so we may write:
$$V=\pi p^2\lim_{t\to\infty}\left(\int_1^{\infty}x^{-2p}\,dx \right)$$
Applying the FTOC, we have:
$$V=\pi p^2\lim_{t\to\infty}\left(\left[\frac{x^{1-2p}}{1-2p} \right]_1^{t} \right)$$
$$V=\frac{\pi p^2}{1-2p}\lim_{t\to\infty}\left(\frac{1}{t^{2p-1}}-1 \right)$$
For $$\frac{1}{2}<p$$ we have:
$$V(p)=\frac{\pi p^2}{2p-1}$$
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:
$$V'(p)=\frac{(2p-1)(2\pi p)-\left(\pi p^2 \right)(2)}{(2p-1)^2}=\frac{2\pi p(p-1)}{(2p-1)^2}$$
For $$\frac{1}{2}<p$$ we have:
$$p=1$$
If we observe that for $$\frac{1}{2}<p<1$$ 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 $p=1$.