For what $p$ does the surface of revolution for $x^p$ have finite surface area? I am trying to investigate the surface of revolution of the $x^p$ functions, in the domain $[1,\infty)$
Using the formula for surface of revolution,
$$A=2\pi\int_1^\infty x^p \sqrt{1+(px^{p-1})^2}dx$$
I am trying to figure out for which values of p (if any) does the area converge. 
For the case of p=-1, I managed to show by comparison that it diverges:
$$A=2\pi\int_1^\infty x^{-1} \sqrt{1+x^{-4}}dx$$
Since I can't integrate that, I compared it to 
$2\pi\int_1^\infty x^{-1}dx$, which diverges.
So my question is, how do I do something similar for other cases of p 
So for p=-2, 
$$A=2\pi\int_1^\infty x^{-2} \sqrt{1+4x^{-6}}dx$$
I am too stupid to integrate this, so I tried to compare it with $2\pi\int_1^\infty x^{-2}dx$, but much to my dismay this converges. 
UPDATE
I have managed to show that the area diverges for the cases of p≥1, by making the approximation that $\sqrt{1+(px^{p-1})^2}\approx px^{p-1}$ and integrating from there.
However I am still stuck with the p≤-2 cases, so plz help ty.
 A: $$A=2\pi\int_1^\infty x^p \sqrt{1+p^2x^{2p-2}}dx=2\pi\int_1^\infty \sqrt{x^{2p}+p^2x^{4p-2}}dx$$
If $p>1$, then $x^{4p-2}\ge x^{2p}$ so $$\int_1^\infty \sqrt{x^{2p}+p^2x^{4p-2}}dx\geq \int_1^\infty \sqrt{x^{2p}+p^2x^{2p}}dx=\sqrt{1+p^2}\int_1^\infty x^p\,dx$$
which diverges.
If $0<p<1$, then $x^{2p}\geq x^{4p-2}$ so
\begin{align}
\int_1^\infty \sqrt{x^{2p}+p^2x^{4p-2}}dx&\geq \int_1^\infty \sqrt{x^{4p-2}+p^2x^{4p-2}}dx\\
&=\sqrt{1+p^2}\int_1^\infty \sqrt{x^{4p-2}}\,dx\\
&=\sqrt{1-p^2}\int_1^\infty x^{2p-1}\,dx
\end{align}
which diverges.

If $p<0$, say $p=-q$ where $q>0$, then $$\int_1^\infty \sqrt{x^{2p}+p^2x^{4p-2}}dx=\int_1^\infty \sqrt{x^{-2q}+(-q)^2x^{-4q-2}}dx=\int_1^\infty \sqrt{{1\over x^{2q}}+{q^2\over x^{4q+2}}}dx.$$
Now,
$${1\over x^{4q+2}}\le{1\over x^{2q}}\implies \sqrt{{q^2\over x^{4q+2}}+{1\over x^{2q}}}\le \sqrt{{q^2\over x^{2q}}+{1\over x^{2q}}}=\sqrt{{q^2+1\over x^{2q}}}={\sqrt{q^2+1}\over x^q}.$$ Thus,
$$
\int_1^\infty \sqrt{{1\over x^{2q}}+{q^2\over x^{4q+2}}}dx\leq \sqrt{q^2+1}\int_1^\infty {1\over x^q}\,dx,
$$
which converges for $q>1$ so the original integral converges for $p<-1$.
Finally, suppose $0<q\leq 1$. Then
$$
\sqrt{{1\over x^{2q}}+{q^2\over x^{4q+2}}}\geq \sqrt{{1\over x^2}+{1\over x^6}}=\sqrt{{x^4+1\over x^6}}\geq \sqrt{{x^4\over x^6}}={1\over x},
$$
so
$$
\int_1^\infty \sqrt{{1\over x^{2q}}+{q^2\over x^{4q+2}}}\geq \int_1^\infty {1\over x}\,dx
$$
which diverges. So the original integral diverges for $-1\leq p\leq 0$ (since $p=0$ is trivial).

In summary, the original integral converges if and only if $p<-1$.
