Integral transform with Dirac delta 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 $g^{-1}(0)$
$$
\int\limits_{\mathbb{R}^n} f(r) \delta(g(r))\ dr 
\ \ \ = \int\limits_{g^{-1}(0)} ?\ dr
$$
 A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
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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
$$
\delta\pars{\vec{r}}
=
{1 \over r^{2}}\delta\pars{r}\delta\pars{\cos\pars{\theta}}\delta\pars{\phi}
$$
A: 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.
