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