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.
Best Answer
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.
$$ V=2\cdot 2\pi\int_{1}^{3}x\sqrt{x^2-1}\,dx= \frac{4}{2}\pi\int_{1}^{3}\sqrt{x^2-1}\frac{d}{dx}(x^2-1)\,dx=\\ 2\pi\int_{0}^{8}\sqrt{u}\,du=2\pi\frac{2\sqrt{u^3}}{3}\bigg|_{0}^{8}= \frac{4}{3}\pi\left(\sqrt{8^3}-\sqrt{0}\right)=\frac{64\sqrt{2}\pi}{3} $$
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