Integrating Dirac delta function over two variables I am trying to evaluate one equation like this
$$
\int_{-\infty}^\infty \int_{-\infty}^\infty xy\delta \left[ (x-1)^2 + y^2-2 \right] \, dx\, dy
$$
but found my result is not the same compared with the numerical result using Lorentzian approximation.
Therefore, there must be something wrong with my mathematical evaluation in here. Does this equation can be changed into a line integral? I would like to simplify it to make my numerical integral more efficient.
The equation in the real problem for me is much more complicated. However, the function inside the Dirac delta function can also be solved as $y(x)$ in my problem.
I would appreciate your help or just giving me some hints. I tried for a long time during the new year. Thank you.
 A: The integrand, $xy \, \delta[ (x-1)^2 + y^2-2 ],$ is odd in $y$, and the domain of integration is symmetric in $y$, so the integral vanishes (i.e. equals $0$).
A: By definition, we see that
\begin{align}
\int_{\mathbb{R}^2} dxdy\ xy\ \delta((x-1)^2+y^2-2) = \frac{1}{2}\int_{S}d\sigma\  \frac{xy}{\sqrt{(x-1)^2+y^2}} = \frac{1}{2\sqrt{2}} \int_S xy\ d\sigma
\end{align}
where $S = \{(x, y)\mid (x-1)^2+y^2=2\}$. 
Using the parametrization $r(t)=(\sqrt{2}\cos(t)+1, \sqrt{2}\sin(t))$ , we see that
\begin{align}
I = \frac{1}{\sqrt{2}}\int^{2\pi}_0 (\sqrt{2}\cos(t)+1)\sin(t)\ dt = 0.
\end{align}
A: If we go from rectangular coordinates to (shifted) polar coordinates 
$$(x,y)~=~(1,0)+r(\cos\theta,\sin\theta),$$
then OP's integral becomes
$$ \begin{align}I~:=~&\iint_{\mathbb{R}^2} \!dx~dy~xy~\delta((x-1)^2+y^2-2)\cr
~=~&\int_{[0,2\pi]} \!d\theta \int_{\mathbb{R}_+} \!dr~r~(r\cos\theta+1)r\sin\theta ~\delta(r^2-2)\cr
~=~&\int_{[0,2\pi]} \!d\theta  \int_{\mathbb{R}_+} \!d(r^2)\frac{1}{2}(r\cos\theta+1)r\sin\theta \delta(r^2-2)\cr
~=~&\int_{[0,2\pi]} \!d\theta   \frac{1}{2}(\sqrt{2}\cos\theta+1)\sqrt{2}\sin\theta\cr
~=~& 0,
\end{align}$$
in agreement with Jacky Chong's answer. This result could actually have be seen from the very beginning since OP's integrand is odd under the transformation $(x,y)\mapsto(x,-y)$, cf. a comment by Pasha.
