Derive expression for surface area $(n-1)$-sphere in hyperbolic space The answer is $A(r) = n \frac{\pi^{n/2}}{\Gamma(n/2)}\sinh^{n-1}(r)$ as is given in (3.2) of this paper or in eqn III.4.2 of Riemannian Geometry: A Modern Introduction. However, neither source offers either a derivation of this or a reference as to where to look up a derivation.
I found this stackexchange post that gives a really beautiful derivation in the case of $n=2$ where they derive the surface area (circumference) of the 1-sphere as the limit of a polygon. Can we do something similar in the case of the $(n-1)$-sphere by considering some sort of tiling of the sphere by triangles?
 A: Here's a sketch of how it goes. The hyperbolic metric in the unit disk model is
$$ds^2 = \frac{4\|dx\|^2}{(1-r^2)^2} = \frac{4(dr^2 + r^2d\sigma^2)}{(1-r^2)^2},$$
where $d\sigma^2$ is the Euclidean metric on the unit sphere.
The hyperbolic distance $\delta$ from the origin to a point at Euclidean distance $R$ is
$$\delta=\int_0^R \frac{2\,dr}{1-r^2} = \log\left(\frac{1+R}{1-R}\right),$$
and so $e^{\delta} = \frac{1+R}{1-R}$ and $R=\tanh(\delta/2)$.
Here's where the $\sinh$ comes: Note next that
$$\frac{2R}{1-R^2} = \frac{2\tanh(\delta/2)}{\text{sech}^2(\delta/2)} = 2\sinh(\delta/2)\cosh(\delta/2) = \sinh(\delta).$$
Note, for example, that the circumference of the geodesic circle of radius $\delta$ will now be $2\pi\sinh{\delta}$ (in lieu of $2\pi \delta$ — and, to complete the circle of ideas, you can check that in the case of the unit sphere, this becomes $2\pi\sin \delta$).
The last step is to realize, given the form of the metric, that the surface area of the sphere of hyperbolic radius $\delta$ is given by multiplying the area of the Euclidean unit sphere by
$$\left(\frac{2R}{1-R^2}\right)^{n-1} = \sinh^{n-1}(\delta),$$
which is what I said should happen. :)
