TITLE: Volume of hyperbola revolved about the y -axis
QUESTION [5 upvotes]: I'm trying to calculate the volume of the solid formed by revolving the hyperbola ${x^2} - {y^2} = 1$ bounded by $x=1$ and $x=3$ about the y axis, however I do not know if I'm going about this the right way using cylindrical shells. 
Using volume of a solid of revolution with cylindrical-shell method where the radius is ${x}$ and the height is ${2\sqrt{x^2 - 1}}$, I got the integral:
$$
\begin{eqnarray}
V &=& 2 \pi \int_1^{3} [x (2\sqrt{x^2 - 1})] \, \textrm{d}x \\
&=& 4 \pi \left[ \frac{(x^2 - 1)^{3/2}}{3} \right]_1^{3} \\
&=&  \frac{32\sqrt{8} \pi}{3} \\
\end{eqnarray}
$$
I would like to know if this is the correct way to solve this problem using cylindrical shells and if there are any other ways to solve the this problem.

REPLY [4 votes]: The way you set up the integral seems to be correct (that's the exact same way I would set it up), but I think you calculated it slightly wrong. You forgot that you also have the lower part of the hyperbola. So, the volume should be doubled.
$$
V=2\cdot 2\pi\int_{1}^{3}x\sqrt{x^2-1}\,dx=
\frac{4}{2}\pi\int_{1}^{3}\sqrt{x^2-1}\frac{d}{dx}(x^2-1)\,dx=\\
2\pi\int_{0}^{8}\sqrt{u}\,du=2\pi\frac{2\sqrt{u^3}}{3}\bigg|_{0}^{8}=
\frac{4}{3}\pi\left(\sqrt{8^3}-\sqrt{0}\right)=\frac{64\sqrt{2}\pi}{3}
$$
Wolfram Alpha check