OK, so the easiest way to get at this is to consider the unit sphere--this can be embedded in 3-space with equation:

$$x^{2} + y^{2} + z^{2} = r^{2}$$

Or alternately, with $r = {\rm cons}$. Using the normal Jacobian business, we can find that the measure of this space is $\int dV = \int r^{2}\sin\theta d\theta d\phi$

Now, consider the hyperbolic plane. this is given by the equation

$$x^{2} - y^{2} - z^{2} = r^{2}$$

Now, let's explicitly do our jacobian business after choosing $r= {\rm cons}$. And to save ourselves some work, we're going to do our best to make it nice and simple. First, we make the substitution $y = \rho \sin \theta$ and $z = \rho \cos \theta$, which means that $y^{2} + z^{2} = \rho^{2}$, and our constraint is

$$x^{2} - \rho^{2} = r^{2}$$

Now, to make our lives yet simpler, we make the substitution $x = r\cosh \phi$ and $\rho = r \sinh \phi$. Using the identity $\cosh^{2} \phi- sinh^{2}\phi = 1$, we see that our constraint is automatically satisfied, and simple algebra can convince you that any point $x,y,z$ for $x > 0$ can be solved for some $r, \phi, \theta$. All in all we have:

$$x = r \cosh \phi$$
$$y = r \sinh\phi \sin \theta$$
$$z = r \sinh\phi \cos\theta$$

Now, remembering that our constraint is $r={\rm cons}$, we don't have to take $r$ derivatives when taking differentials and substituting back into our Euclidean distance formula:

$$ds^{2} = dx^{2} + dy^{2} + dz^{2}$$

After all of this, we'll find a new metric tensor where

$$ds^{2} = g_{ab}dx^{a}dx^{b}$$

Finally, our volume element will be:

$$dV = \sqrt{{\rm det}\,g}d\theta d\phi$$

I'll leave it as an excersise for you to do all of the differentials and matrix algebra required to show that it comes out in the form you've given above.

This post imported from StackExchange Mathematics at 2014-06-02 20:20 (UCT), posted by SE-user Jerry Schirmer