Let $f,g: \mathbb{R}^n \to \mathbb{R}$.
Let $\delta$ denote the Dirac delta function.
How can I write the integral over $\mathbb{R}^n$ (on the left hand side) as an integral over $g^{-1}(0)$
$$ \int\limits_{\mathbb{R}^n} f(r) \delta(g(r))\ dr \ \ \ = \int\limits_{g^{-1}(0)} ?\ dr $$
$\newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ You can use $$ \int{\rm f}\pars{\vec{r}}\delta\pars{{\rm g}\pars{\vec{r}}}\,\dd^{n}\vec{r} = \int{\rm f}\pars{\vec{r}} \bracks{% \int_{-\infty}^{\infty}\expo{\ic k{\rm g}\pars{\vec{r}}} \,{\dd k \over 2\pi}}\dd^{n}\vec{r} = \int_{-\infty}^{\infty}{\dd k \over 2\pi}\bracks{% \int{\rm f}\pars{\vec{r}}\expo{\ic k{\rm g}\pars{\vec{r}}}\,\dd^{n}\vec{r}} $$
ADDENDUM:
In some particular cases ( it depends on the particular form of ${\rm g}$ ), the Dirac delta $\delta\pars{{\rm g}\pars{\vec{r}}}$ can be reduced to a product of "more simple" Dirac delta's. For example, with $n = 3$ and spherical coordinates it's like $$ \delta\pars{\vec{r}} = {1 \over r^{2}}\delta\pars{r}\delta\pars{\cos\pars{\theta}}\delta\pars{\phi} $$
See Gelfand and Shilov "Generalized Functions, vol. 1" page 222. The Delta Distribution with a function as its argument is discussed there. On page 230 there is the following definition (with $k=1$):
$\delta(P,\phi) = \int_{P=0}\phi dx$ (where $P$ is a hypersurface).
So yours is the integral of f over the hypersurface $g=0$. If by $g^{-1}(0)$ is meant $g=0$ then
$\int_{g^{-1}(0)}f(r)dx$
is the right half of your equation.
