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If I draw a closed curve $C$ on this flat screen, then that will enclose a disk like region $D$. I can compute the area of $D$ by either one of the following two methods

$Area_1 = \int_D dx \wedge dy$

and

$Area_2 = \int_C x dy$

The two areas are the same, $Area_1 = Area_2$.

My question is the following. If I have a $(d+1)$-dimensional bulk region $M$ with a Riemannian curved metric tensor $g_{\mu\nu}$ and a $d$-dimensional boundary $\partial M$, then I know that the volume of $M$ is given by

$Vol_1 = \int_M d^{d+1} x \sqrt{g}$

Is there a corresponding boundary integral that gives the same answer,

$Vol_2 = \int_{\partial M} d^d x \sqrt{h} X[h]$

Here $h_{\mu\nu}$ denotes the boundary metric, but what is the quantity $X[h]$?

Here is another example. The volume of a ball. I write

$x = r \sin \theta \sin \varphi$

$y = r \sin \theta \cos \varphi$

$z = r \cos \theta$

Then I get

$dx^1 \wedge dx^2 = d\varphi \wedge d\theta r^2 \sin \theta \cos\theta$

and the volume can be computed as the surface integral at fixed radius $r$ (the boundary of the ball, which is a two-sphere)

$\int dx^1 \wedge dx^2 x^3 = \int d\varphi d\theta r^3 \sin \theta \cos^2 \theta = \frac{2\pi r^3}{3}$

Einstein
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    You can perturb the metric on the interior and change the interior volume without affecting the boundary metric. Therefore such a formula cannot exist. – Benoît Kloeckner Aug 23 '19 at 08:05
  • I am interested in a specific bulk metric. That one of a sphere. Then I am wondering what is the volume of a region on that sphere. More specifically, if one can find a coordinate independent expression in terms of a boundary integral of that region. –  Aug 23 '19 at 08:52
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    You can apply Stokes theorem: find a $d$-form $\eta$ on $M$ whose exterior derivative is the volume form on $M$ for the metric $g,$ and integrate it on the boundary. Note that for two different metrics on $M$ which coincide on a neighbourhood of the boundary, the corresponding forms $\eta_1$ and $\eta_2$ differ by a closed $d$ form, whose integral along $\partial M$ is non-vanishing in general, see the comment of Benoit. –  Aug 23 '19 at 10:10

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