Volume of hyperbola revolved about the y -axis

Question

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.

Answer 1

Your solution is correct.

Method 2: Using double integrals.

Namely, by rotating the graph around the $y$-axis, we can define $y$ as a two-variable function $y(x,z)=\sqrt{x^2+z^2-1}$, for $y\ge 0$. Next, define a region

$$D=\{(x,z)\ |\ 1\le x^2+z^2 \le 9\}$$

To get the volume of the upper body, we evaluate the integral

$$\iint\limits_D y(x,z)\ \text dx\ \text dz = \iint\limits_D \sqrt{x^2+z^2-1}\ \text dx\ \text dz$$

and to get the total volume, we just multiply this by two. The above integral can be easily found using polar coordinates, and we have:

$$V = 2\int_0^{2\pi}\int_1^3 r\sqrt{r^2-1}\ \text dr\ \text d\theta$$

Method 3: The washer method.

Consider a horizontal washer (ring) with a thickness of $\text dy$, at a height $y$ from the $x$-axis. Its inner radius is $r_1 = \sqrt{1+y^2}$ and its outer radius is $r_2 = 3$. The volume of the washer is $\text dV = (r_2^2-r_1^2)\pi$. To get the total volume, integrate the volumes of all such washers:

$$V=\int\limits_{-2\sqrt2}^{2\sqrt2} \pi(9-y^2-1)\ \text dy$$

Answer 2

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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