Dirac delta function of non-linear multivariable arguments How does one compute a dirac delta function with a multivariable argument? For example, compute:
$$
\int^{\infty}_{-\infty}{\rm d}x\,{\rm d}y\,
\delta\left(x^{2} + y^{2} - 4\right)
\delta\left(\left[x - 1\right]^{2} + y^{2} -4\right){\rm f}\left(x,y\right).
$$
If we constrain the two delta functions we'll get two intersecting circles, and it seems reasonable to state that we evaluate $f(x,y)$ at the intersecting points, but I feel like there should be some extra identities.
Since for single variable 
$$\delta(f(x))=\sum_i \frac{\delta(x-x_i)}{|f'(x)|},$$ is there a multivariable generalization to be aware of?
 A: I worked with similar objects during my Master's project, and we had to derive a formula for that...couldn't find it anywhere. The formula can be found in the links below. 
See section 3.7
or 
see section A.4.7. 
I've called it a $"\bf Sweet ~Dirac-\delta~ formula"$ or "A lemma on twofold Dirac delta functions".  
A: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}
&\color{#0000ff}{\large%
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{\rm d}x\,{\rm d}y\,
\delta\pars{x^{2} + y^{2} - 4}
\delta\pars{\bracks{x - 1}^{2} + y^{2} -4}\fermi\pars{x,y}}
\\[3mm]&=
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{\rm d}x\,{\rm d}y\,
\delta\pars{x^{2} + y^{2} - 4}
\delta\pars{\bracks{x^{2} - 2x + 1} + y^{2} -4}\fermi\pars{x,y}
\\[3mm]&=
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{\rm d}x\,{\rm d}y\,
\delta\pars{x^{2} + y^{2} - 4}
\delta\pars{2x - 1}\fermi\pars{x,y}
\\[3mm]&=
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{\rm d}x\,{\rm d}y\,
\delta\pars{x^{2} + y^{2} - 4}\,
{\delta\pars{x - 1/2} \over 2}\,\fermi\pars{x,y}
\\[3mm]&=
\half\int_{-\infty}^{\infty}{\rm d}y\,
\delta\pars{\bracks{\half}^{2} + y^{2} - 4}\,\fermi\pars{\half,y}
\\[3mm]&=
\half\int_{-\infty}^{\infty}{\rm d}y\,\bracks{%
{\delta\pars{y + \root{15}/2} \over \root{15}}
+ {\delta\pars{y - \root{15}/2} \over \root{15}}}\fermi\pars{\half,y}
\\[3mm]&=\color{#0000ff}{\large%
{\root{15} \over 30}\bracks{\fermi\pars{\half,-\,{\root{15} \over 2}} + \fermi\pars{\half,{\root{15} \over 2}}}}
\end{align}
A: *

*There exists an $n$-dimensional generalization
$$\tag{1} \delta^n({\bf f}({\bf x})) ~=~\sum_{{\bf x}_{(0)}}^{{\bf f}({\bf x}_{(0)})=0}\frac{1}{|\det\frac{\partial {\bf f}({\bf x}_{(0)})}{\partial {\bf x}} |}\delta^n({\bf x}-{\bf x}_{(0)}) $$
of the substitution formula for the Dirac delta distribution
under pertinent assumptions, such as e.g., that the function ${\bf f}:\Omega \subseteq \mathbb{R}^n \to \mathbb{R}^n$ has isolated zeros. Here the sum on the rhs. of eq. (1) extends to all the zeros ${\bf x}_{(0)}$ of the function ${\bf f}$.


*Example: The function
$$\tag{2} {\bf f}(x,y)~=~(x^2+y^2-4,(x-1)^2+y^2-4)$$
with Jacobian determinant
$$\tag{3}\det\frac{\partial {\bf f}(x,y)}{\partial (x,y)} ~=~4y,$$
has two zeros
$$\tag{4} (x,y)~=~(\frac{1}{2},\pm \frac{\sqrt{15}}{2}),$$
leading to
$$\tag{5} \delta^2({\bf f}(x,y)) ~\stackrel{(1)}{=}~\frac{1}{2\sqrt{15}}\delta(x-\frac{1}{2})\sum_{\pm} \delta(y\mp\frac{\sqrt{15}}{2}). $$
A: There is indeed a multivariable generalization of the identity you mentioned.  See this Wikipedia article, where it says "As in the one-variable case, it is possible to define the composition of $δ$..."
