general relativity - Do black holes have infinite areas and volumes?

How to calculate the area / volume of a black hole? Is there a corresponding mathematical function such as rotating $1/x$ around the $x$-axis or likewise to find the volume?

Answer

The event horizon is a lightlike surface, and so its area is coordinate-invariant. For a Schwarzschild black hole, $$ds^2 = -\left(1-\frac{2m}{r}\right)dt^2 + \left(1-\frac{2m}{r}\right)^{-1}dr^2 + r^2(d\theta^2+\sin^2\theta\,d\phi^2)$$ The horizon suface is at $r = 2m$ of Schwarzschild radial coordinate, and so at any particular Schwarzschild time ($dt = 0$) has the metric $$dS^2 = (2m)^2(d\theta^2 + \sin^2\theta\,d\phi^2),$$ which is just the metric on a standard 2-sphere of radius $2m$. You can find the square area element explicitly as the determinant of the metric, $dA^2 = (2m)^4\sin^2\theta\,d\theta^2d\phi^2$, and integrating: $$A = 16\pi m^2 = \frac{16\pi G^2}{c^4}m^2.$$

The volume of the black hole is not invariant, as Jerry Schirmer says. If you try to apply anything the above Schwarzschild coordinates above, then since the coefficients of $dt^2$ and $dr^2$ switch signs across the horizon, $t$ is spacelike and $r$ is timelike. Therefore, since the black hole is eternal, it could be said to have infinite volume (classically, but a real astrophysical black hole would have a finite but still extraordinarily high lifetime), as you'll be integrating $dt$ across its lifetime.

Technically, the above argument is a bit flawed, because the Schwarzschild coordinate chart is not defined across the event horizon, so one should be more careful how they're continued across the horizon (e.g., with Kruskal-Szekeres coordinates). But this can be made more rigorous.

In another coordinate chart, e.g., the Gullstrand-Painlevé coordinates adapted to a family of freely falling observers, $$ds^2 = -\left(1-\frac{2m}{r}\right)dt^2-2\sqrt{\frac{2m}{r}}\,dt\,dr + \underbrace{dr^2 + r^2(d\theta^2 + \sin^2\theta\,d\phi^2)}_{\text{Euclidean in spherical coord.}},$$ at any instant of time ($dt = 0$), space is precisely Euclidean; since the horizon is still $r = 2m$ in these coordinates, "the" volume is $$V_{\text{GP}} = \frac{4}{3}\pi(2m)^3.$$

If you pick yet another coordinate chart, you may get a yet different answer.