Surface integral of a partially constant Dirac delta I am trying to integrate the product of a function and a partially constant delta function over a sphere of constant radius $r$.
The integral is of the form 
$$\int^{2\pi}_0 \int^{\pi}_0 f(\mathbf{r})\delta(x-x_0)\delta(y-y_0) \sin\theta\,\mathrm{d}\theta \,\mathrm{d}\phi.  $$
I know that this integral is very simple when the delta function is fully three-dimensional, such as $$\delta^3(\mathbf{r}-\mathbf{r}_0)=\delta(x-x_0)\delta(y-y_0)\delta(z-z_0),$$ but what I have here is something a bit different. Does anyone have anyone have any suggestions?
 A: Here we give a rough sketch, and leave it to the reader to iron out various technical issues.

*

*Define the spherical coordinates $(r,\theta,\phi)$ as
$$\begin{align} x~=~&r\xi, \qquad \xi~=~\sin\theta\cos\phi, \cr y~=~&r\eta, \qquad \eta~=~\sin\theta\sin\phi, \cr z~=~&r\cos\theta. \end{align} \tag{1} $$


*Define reflection
$$\sigma: (x,y,z)\mapsto (x,y,-z)\tag{2}$$
in the $(x,y)$-plane.


*Define function
$$g({\bf r})~:=~f({\bf r})+f\circ\sigma({\bf r}), \qquad {\bf r}~=~(x,y,z).\tag{3}$$


*Define function
$$h({\bf r})~:=~\frac{g({\bf r})}{rz}.\tag{4}$$


*Define unit disk
$$  D~:=~\{(\xi,\eta)\in\mathbb{R}^2 \mid \xi^2+\eta^2~<~1\}. \tag{5}$$


*OP's integral then reads
$$\begin{align}  I:~=~& \int_0^{2\pi}\!d\phi~\int_0^{\pi}\!d\theta~\sin\theta ~f({\bf r}) \delta(x-x_0)\delta(y-y_0)\cr
~=~&\int_0^{2\pi}\!d\phi~\int_0^{\frac{\pi}{2}}\!d\theta~\sin\theta\cos\theta ~h({\bf r})\delta\left(\xi-\frac{x_0}{r}\right)\delta\left(\eta-\frac{x_0}{r}\right)\cr
~=~&\iint_D \!d\xi~d\eta ~h({\bf r})\delta\left(\xi-\frac{x_0}{r}\right)\delta\left(\eta-\frac{x_0}{r}\right)\cr
~=~&\left\{\begin{array}{ccc} h\left(x_0, y_0, \sqrt{r^2-x_0^2-y_0^2}\right) &{\rm for}& x_0^2+y_0^2<r^2, \cr
\frac{1}{2}h(x_0, y_0, 0) &{\rm for}& x_0^2+y_0^2=r^2, \cr 
0 &{\rm for}& x_0^2+y_0^2>r^2.
\end{array}\right. \end{align}\tag{6}$$
