Volume of submanifold as integral of delta-function Let $M$ be an $n-m$ dimensional sub-manifold of $\mathbf R^n$ defined by the following set of equations:
\begin{equation}
f_1(\vec x)=0, \\
\vdots \\
f_m(\vec x)=0,
\end{equation}
(where $\vec x$ are coordinates in $\mathbf R^n$).
Is it true that the volume of $M$ is
\begin{equation}
\int d^nx\sqrt{\det (JJ^T)}\prod_{i=1}^m\delta(f_i(\vec x))
\end{equation}
where $\delta()$ is the Dirac-Delta and where $J$ is the rectangular matrix $J_{i\mu}\equiv \frac{\partial f_i(\vec x)}{\partial x^\mu}$ with $i=1,...,m$ and $\mu=1,...,n$?
 A: One should distinguish between the volume of the submanifold (a number that might be infinite) and  the volume form, an exterior differential form $\omega$ of degree $n{-}m$ on the (presumed regular) 0 level set $M = f^{-1}(0)\subset \mathbb{R}^n$ of the mapping $f:\mathbb{R}^n\to\mathbb{R}^m$.
The formula for $\omega$ is easy to write down:  If $f = (f^1,\ldots,f^m)$ and we set $J^{ij} = J^{ji} =  \nabla f^i\cdot\nabla f^j$, then
$$
\omega(v_1,\ldots,v_{n-m}) 
= \frac{\Omega(\nabla f^1,\ldots,\nabla f^m,v_1,\ldots,v_{n-m})}{\det(J)^{1/2}},
$$
where $\Omega = \mathrm{d}x^1\wedge\cdots\wedge\mathrm{d}x^n$ is the volume form on $\mathbb{R}^n$.  Regularity is equivalent to the condition that $\det(J)$ be nonvanishing on $M=f^{-1}(0)$. This assumes, of course, that $M$ is given the orientation for which $\omega$ is a positive $n{-}m$ form.
For example, when $n=2$ and $m=1$, one finds that
$$
\omega = \frac{f_x\,\mathrm{d}y-f_y\,\mathrm{d}x}{\sqrt{{f_x}^2+{f_y}^2}}\,.
$$
There is, of course, no explicit formula for $\int_M\omega$ in terms of $f$, even when $f$ is a polynomial.
