How does one compute the following integral
$$\int _Vf(\pmb{r})\delta (g(\pmb{r}))d\pmb{r}$$
if $g: \mathbb{R}^n \to \mathbb{R}^m$. In this question and the Wikipedia page on the Dirac delta function an answer is given for $m=1$ as
$$\int _Vf(\pmb{r})\delta (g(\pmb{r}))d\pmb{r}=\int _{g^{-1}(0)}\frac{f(\pmb{r})}{|\nabla g(\pmb{r})|}d\sigma\ .$$
So my question concerns the case wheere $m>1$. Is there also some good reference for citing the corresponding result?
Pretend for now that $\delta$ is a bounded measurable function. Consider the measure $\mu$ with $d\mu/dr=f$ (here dr denotes the corresponding Lebesgue measure). By the change of variables formula, $$\int \delta\circ gd\mu=\int\delta d(g_*\mu)$$ where $(g_*\mu)(U)=\mu(g^{-1}U)$. By definition of the delta function, the right hand side is equal to ${\frac {d(g_*\mu)}{dr}}(0)$, i.e. the Radon-Nikodym derivative evaluated at the origin.
Using the Smooth Coarea formula, it's easy to check that
https://en.wikipedia.org/wiki/Smooth_coarea_formula
$${\frac {d(g_*\mu)}{dr}}(y)=\int_{g^{-1}y} {\frac {d\mu/dr}{\sqrt{det(dg(dg)^T)}}}dN$$ where $dN$ is the induced Riemannian measure on the preimage.
You also probably want g to be a submersion (i.e. differential is always surjective).