How does one prove the following identity?

$$\int \_Vf(\pmb{r})\delta (g(\pmb{r})) \, d\pmb{r}=\int\_S \frac{f(\pmb{r})}{\left|\operatorname{grad} g(\pmb{r})\right|} \, d\sigma$$

where $S$ is the surface inside $V$ where $g(\pmb{r})=0$ and it is assumed that $\operatorname{grad} g(\pmb{r})\neq 0$. 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

$$\int\_{-\infty}^\infty f(x)\delta (g(x)) \, dx=\sum \_a \int \_{a-\varepsilon}^{a+\varepsilon }f(x)\delta ((x-a)g'(a)) \, dx$$

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.