$n$ dimension integrals involving one dimension Dirac delta functions I want to calculate an integral 
$$\iint_D f(x,y)\delta(g(x,y))\,dx\,dy$$
or simply 
$$\iint_D \delta(g(x,y))\,dx\,dy$$
$\iint_D \,dx\,dy$ is the area of $D$, and$\iint_D \delta(g(x,y))\,dx\,dy$ seems like we pick out the area of curve $g(x,y)=0$? or something else like it ?
 A: If I understand correctly what you meant, then yes, it is something like that. Have a look at
this section of Wikipedia's Dirac delta function page for a formal definition.
A: $\newcommand{\+}{^{\dagger}}%
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First integrate respect, for example $y$, for $x$ fixed:
$$
\int\delta\pars{{\rm g}\pars{x,y}}\,\dd y
=
\int\sum_{n}
{\delta\pars{y - \fermi_{n}\pars{x}}
 \over \verts{\left.\partiald{{\rm g}\pars{x,y}}{y}\right\vert_{y = y_{n}}}}\,\dd y
\quad\mbox{where}\quad{\rm g}\pars{x,\fermi_{n}\pars{x}} = 0
$$
The result is a function of $x$. Then, integrate over $x$ ( No problem !!! ). 
For example:
\begin{align}
&\int_{1}^{2}\int_{1}^{2}\delta\pars{x^{2} - y^{2}}\,\dd y\,\dd x
=
\int_{1}^{2}\int_{1}^{2}\bracks{{\delta\pars{y + \verts{x}} \over 2\verts{x}} + {\delta\pars{y - \verts{x}} \over 2\verts{x}}}\,\dd y\,\dd x
\\[3mm]&=
\int_{1}^{2}\int_{1}^{2}{\delta\pars{y - \verts{x}} \over 2\verts{x}}\,\dd y\,\dd x
=
{1 \over 2}\int_{1}^{2}
{\Theta\pars{\verts{x} - 1}\Theta\pars{2 - \verts{x}}  \over \verts{x}}\,\dd x
=
{1 \over 2}\int_{1}^{2}\,{\dd x \over x} = \half\,\ln\pars{2}
\end{align}
A: If the argument of the delta function is a vector valued function, e.g.
$$ I = \int_{\mathbb{R}^n} \delta\left[ \mathbf{G}(\mathbf{x}) \right] f(\mathbf{x}), $$
then you must calculate the determinant $|\det\nabla \mathbf{G}| = |\det\partial_i G_j|$, as
$$I = \frac{f(\mathbf{x_0})}{\left|(\det\nabla\mathbf{G})|_{\mathbf{x}=\mathbf{x_0}} \right|} $$
