# Volume of hyperbola revolved about the y -axis

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.

• do you have bounds on the $x$ values? – Andres Mejia Mar 9 '19 at 18:24
• Yes, sorry, the bounds on the x values are 1 and 3 – Ludwig Mar 9 '19 at 18:27

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:

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

• But when using cylindrical shells to integrate don't you need to take the integral of $2 \pi x f(x)$ where $x$ is the radius of the incremental cylinder and $f(x)$ is the height of that incremental cylinder. – Ludwig Mar 9 '19 at 18:47
• @Ludwig I apologize, I have misread your question. I have updated my answer. – Haris Gušić Mar 9 '19 at 19:12

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

• The OP's solution is equivalent, since $\sqrt8 =2\sqrt2$. – Haris Gušić Mar 9 '19 at 19:03
• I think the OP lost the $1/2$ somewhere that they should have gotten from doing the u-substitution. – Michael Rybkin Mar 9 '19 at 19:04