Double integrals in polar coordinates -- Multivariable I've done some research on this topic but I am quite confused about finding the area under a specific volume in polar coordinates. 
Let's have an example, how would we find the volume of a hyperboloid $z=\sqrt{1+x^2+y^2}$ for $1 \le z \le 5$?
So I set up integral like so -- tell me if this is correct...
I know the region is from 0 to 2pi, so the outer integrand's limits are $0$ to $2\pi$. Would the inner limit be from $0$ to $\sqrt{24}$ since $z = \sqrt{1+x^2+y^2} = 5$ and $r=x^2+y^2$? Then would I just plug in the equation of the hyperboloid?
Would my integral be $$\int_0^{2\pi} {\int_0^{2\sqrt{6}} \! \sqrt{1+r^2}}\ \mathrm{d}r \ \mathrm{d}\theta$$ But then again, I heard something about top - bottom as the value being integrated, so I'm thoroughly confused.
Thanks for the help.
 A: Basically you are correct. In polar coordinates $(x, y) = (r\cos \theta, r\sin \theta)$. (Technically, since this is in 3 dimensions with $z$ as the third coordinate, it would be called "cylindrical coordinates", but in this case we are just treating $z$ as the function.) 
The hyperboloid function is just $z = \sqrt{ 1 + r^2}$, so $z = 1 \iff r = 0$ and $z = 5 \iff r = \sqrt{24}$ (since $r \ge 0$ by definition). So we can convert the limits from being on $z$ to being on $r$. There is no restriction on $\theta$, so it has its full range of $0 \le \theta \le 2\pi$. And lastly $dxdy = rdrd\theta$, so the integration becomes
$$\begin{align} V &= \iint z\ dxdy
\\&=\iint \sqrt{1+r^2} rdrd\theta
\\&= \int_0^{2\pi}\int_0^\sqrt{24}\sqrt{1+r^2}\ rdrd\theta
\\&= \int_0^{2\pi}d\theta\int_0^\sqrt{24}\sqrt{1+r^2}\ rdr
\\&= 2\pi\int_0^\sqrt{24}\sqrt{1+r^2}\ rdr
\end{align}$$
Where the last step (separating out the $\theta$ integration) works because the integrand is independent of $\theta$. A simple substitution should allow you to finish the integral.
