# Find the volume enclosed by the hyperboloid -x^2-y^2+z^2=1 and plane z=2. I correctly figure the bounds to be D={(r,theta)|0 leq theta leq 2pi, 0 leq r leq sqrt(3)}

Integrating over or under this object
There is clearly a kink in my understanding of double integrals with polar coordinates.
Our problem is to find the volume enclosed by the hyperboloid $-{x}^{2}-{y}^{2}+{z}^{2}=1$ and plane $z=2$. I correctly figure the bounds to be $D=\left\{\left(r,\theta \right)|0\le \theta \le 2\pi ,0\le r\le \sqrt{3}\right\}.$
But I have trouble finding if we are integrating between the section of the hyperboloid and the disk $r=1$ at $z=0$ or $z=2$.
While $r\sqrt{1-{r}^{2}}drd\theta$ would make some sense as an integrand, I essentially dont know if I am integrating (ie, finding volume) under or over the surface of the hyperboloid.
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blogpolisft
Step 1
Since you've said in the comments that you wish to find the volume, this can be done using a triple integral. The region in the xy plane can be written in polar coordinates as the set D you have said. As for the z limits, we'll have $\sqrt{1+{x}^{2}+{y}^{2}}\le z\le 2$, therefore
$V=\iiint dV={\int }_{0}^{\sqrt{3}}{\int }_{0}^{2\pi }{\int }_{\sqrt{1+{r}^{2}}}^{2}r\phantom{\rule{thinmathspace}{0ex}}dz\phantom{\rule{thinmathspace}{0ex}}d\theta \phantom{\rule{thinmathspace}{0ex}}dr.$
Step 2
This ensures that you are computing the volume between the hyperboloid and the plane $z=2$.
In general when you compute volumes you compute from the xy plane up to whatever height you wish. When we set $\sqrt{1+{x}^{2}+{y}^{2}}\le z\le 2$ we are essentially doing two computations: the volume of the cylinder of height 2 and subtracting the volume up to the hyperboloid, giving the volume desired.