# Using Polar Integrals to find Volume of surface

Here's the Question and the work that I've done so far to solve it:

Use polar coordinates to find the volume of the given solid. Enclosed by the hyperboloid $−x^2 − y^2 + z^2 = 61$ and the plane $z = 8$

Ah MathJaX is confusing by the way. Not sure how to close the text without having to add something like theta.

$\int_0^{\sqrt{3}} 8r-\sqrt{61+r^2}r dr$ (with respect to $r$)

I evaluated this to be $12 - 512/2 + \dfrac{61^3}{3}$ What do I put here

I then took the integral with respect to $\theta$ from $0$ to $2\pi$ because it is a circle covering the whole $xy$ plane. This gave the above evaluation multiplied by $2\pi$.

However this answer is wrong? Can anyone see something wrong with my reasoning?

Notice that $-x^2-y^2+z^2=61$ can be rewritten as $x^2+y^2=z^2-61$ or $$x^2+y^2=\sqrt{z^2-61}^2$$ Given a horizontal slice of the upper sheet of the hyperboloid (i.e. given $z$), it should be plain to see from the graph (and the equation that we have a circle of radius $\sqrt{z^2-61}$. The vertex of the upper sheet occurs at $(0,0,\sqrt{61})$ because $0^2+0^2-\sqrt{61}^2=61$, so $z$ will range from $\sqrt{61}$ to 8. So what we would like to evaluate would be this integral, where $r(\theta)=\sqrt{z^2-61}$, which is a constant with respect to $\theta$ $$\int_{\sqrt{61}}^8\int_0^{2\pi}\frac{1}{2}r(\theta)^2d\theta dz$$ $$=\int_{\sqrt{61}}^8\int_0^{2\pi}\frac{1}{2}\sqrt{z^2-61}^2d\theta dz$$ $$=\int_{\sqrt{61}}^8\int_0^{2\pi}\frac{1}{2}(z^2-61)d\theta dz$$ $$=\int_{\sqrt{61}}^8\frac{1}{2}(z^2-61)dz\int_0^{2\pi}1d\theta$$ $$=\pi\int_{\sqrt{61}}^8(z^2-61)dz$$ Alternatively, I think what you're trying to do is compute a volume by cylindrical shells, in which case the integral should look like this (where $h(r)$ is the height of a cylinder). $$\int_0^{\sqrt{3}}2\pi r h(r)dr$$ $$=\int_0^{\sqrt{3}}2\pi r (8-\sqrt{r^2+61})dr$$ $$=2\pi\int_0^{\sqrt{3}} (8r-r\sqrt{r^2+61})dr$$ $$=2\pi\bigg( 12 -\int_0^{\sqrt{3}} (r\sqrt{r^2+61})dr\bigg)$$ It seem that where you went wrong was in evaluating $$\int_0^{\sqrt{3}} r\sqrt{r^2+61}dr$$ $$=\int_0^{\sqrt{3}} r\sqrt{r^2+61}dr\cdot\frac{\frac{d(r^2+61)}{dr}}{\frac{d(r^2+61)}{dr}}$$ $$=\int_0^{\sqrt{3}} \frac{r\sqrt{r^2+61}}{2r}d(r^2+61)$$ $$=\frac{1}{2}\int_0^{\sqrt{3}} \sqrt{r^2+61}d(r^2+61)$$ $$=\frac{1}{2}\int_{61}^{64} \sqrt{u}du,\:u=r^2+61$$ $$=\frac{\sqrt{u}^3}{3}\bigg|_{61}^{64}$$ $$=\frac{8^3}{3} - \frac{\sqrt{61}^3}{3}$$ so the final answer would be $$\int_0^{2\pi}\int_0^{\sqrt{3}}(8r-r\sqrt{r^2+61})drd\theta=2\pi\bigg(12-\frac{8^3}{3} + \frac{\sqrt{61}^3}{3}\bigg)$$