The question is: For any positive real number p, define f(x) = $\frac{p^2}{x^p}$ Find the minimum value of the volume of the solid created by rotating this function around the x-axis over the interval [1,∞)
When I tried to solve this problem by using integration of f(x) from 1 to ∞, I am getting $\frac{p}{1-p}$ as my answer for the volume.
I'm not sure if this is correct. Appreciate the help, thanks!
For the integral to converge, we need $p\in[1/2,+\infty$).The volume equals
\begin{align} V&=\pi\int_1^{+\infty} \frac{p^2}{x^{2p}}\,dx \\ &=p^2 \pi \left[ \frac{x^{1-2p}}{1-2p}\Biggr|_1^{+\infty} \right] \\ &= \frac{\pi p^2}{2p-1} \end{align} Differentiating with respect to $p$ and setting $\frac{dV}{dp}=0$,
\begin{align} \frac{dV}{dp}=\pi\frac{(2p-1)2p-2p^2}{(2p-1)^2}=0 &\implies 2p^2-2p=0 \\ &\implies p(p-1)=0 \\ &\implies p=1 \end{align}
The minimum volume is therefore $$V_{min}=\pi$$
If you want to show that this is indeed a minimum, you can compute $V''(1)$: $$V''(1)=2\pi>0$$ Our result is thus a minimum.
EDIT:
If $f(x)=\frac{p^2}{x^p}$, then
\begin{align} V&=\pi\int_1^{+\infty} \frac{p^4}{x^{2p}}\,dx \\ \\ &=\frac{\pi p^4}{2p-1} \end{align} Then setting $V'(p)=0$ in order to minimize $V(p)$, we find that \begin{align} V'(p)&=\pi \frac{(2p-1)4p^3 - 2p^4}{(2p-1)^2}= 0 \\ \\ &\implies 6p^4 - 4p^3=0 \\ &\implies p^3(3p-2)=0 \\ &\implies p=\frac{2}{3} \quad \text{ since } p\in\mathbb{R}^+ \end{align}
The minimum volume is then
$$V_{min}=\frac{16\pi}{27}$$
And, again, you can check that this is indeed a minimum by evaluating $V''(2/3)$. Indeed,
$$V''(2/3)=16\pi > 0 $$ This means that our volume is indeed a minimum.