How do we calculate a volume in curved space?

Question

The volume of a sphere in flat space is:

$$V = \frac{4}{3}\pi r^3.$$

In curved space, $r$ itself is dependent on position, so, in spherical coordinates,

$$r=r(r, \phi, \theta) .$$

Assuming a spherical symmetric spacetime, for example the Schwarzschild metric:

$$ds^2=-Bc^2dt^2+Adr^2+r^2d\Omega.$$

$r$ is not dependent on $\phi$ and $\theta$, therefore, we can write

$$r=r(r).$$

Looks simple, but confuses me totally, as I do not see how to calculate $r$ from $r$ itself.

Ok, to calculate the volume of a sphere around a point one needs to calculate the integral of $V$, and as $r$ is dependent on the position, I would suggest to write down

Something like $V = \int\!f(A, r)\,dr$

But I do not know how this looks in detail... Could you help me please?

Answer 1

The volume of a radius-$R$ sphere is $\int_{S^2}d\Omega\int_0^Rr^2\sqrt{|A|}dr$. This simplifies in the case of spherical symmetry of $\sqrt{|A|}$, to $4\pi\int_0^Rr^2\sqrt{|A|}dr$. For example, if the sphere is centred on the mass in a Schwarzschild metric, $A=\frac{r}{r-r_s}$. For the empirically interesting case $R>r_s$, the resulting integral is quite fiendish, but if $R\gg r_s$ (so the region $r\in[0,\,r_s]$ is only a relatively small contribution) a first-order correction is easily obtained:$$r^2\left(\sqrt{|A|}-1\right)=r^2((1-r_s/r)^{-1/2}-1)\approx\tfrac12r_sr,$$which under the integration operator $4\pi\int_0^Rdr$ adds first-order excess volume$$2\pi r_s\int_0^Rrdr=\pi r_sR^2.$$In particular, the first-order relative excess volume is $\tfrac{3r_s}{4R}$. (The at-$r$ first-order relative excess infinitesimal volume, which is really a relative excess surface area, is $\tfrac{r_s}{2r}$, as discussed e.g. here.)