How does one prove the following identity?
$\int _Vf(\pmb{r})\delta (g(\pmb{r}))d\pmb{r}=\int _S\frac{f(\pmb{r})}{|\text{grad} g(\pmb{r})|}d\sigma$
where $S$ is the surface inside $V$ where $g(\pmb{r})=0$ and it is assumed that $\text{grad} g(\pmb{r})\neq 0$. Thanks.
Edit: I have proved a one-dimensional version of this formula:
\delta (g(x))=\sum _a \frac{\delta (x-a)}{\left|g'(a)\right|}
where $a$ goes through the zeroes of $g(x)$ and it is assumed that at those points g'(a)\neq 0. the integral can be divided into into a sum of integrals over small intervals containing the zeros of $g(x)$. In these intervals $g(x)$ can be approximated by g(a)+(x-a)g'(a)=(x-a)g'(a) since $g(a)=0$. Thus
\int _{-\infty }^{\infty }f(x)\delta (g(x))dx=\sum _a \int _{a-\epsilon }^{a+\epsilon }f(x)\delta \left((x-a)g'(a)\right)dx
Using the property $\delta (kx)=\frac{\delta (x)}{|k|}$, it follows that
\int _{-\infty }^{\infty }f(x)\delta (g(x))dx=\sum _a \frac{f(a)}{\left|g'(a)\right|}
This is the same result we would have obtained if we had written \sum _a \frac{\delta (x-a)}{\left|g'(a)\right|} instead of $\delta (g(x))$ as a factor of the integrand.