Expressing radiation flux density through a surface with Dirac delta What is the general way in which radiation density for a given surface is expressed using Dirac deltas?
Consider this surface expressed in cylindrical coordinates (for any $\phi$ and $r_0$ an oblateness parameter):
$\sqrt{r^2+z^2}-\sqrt{r_0^2-z^2}==0$
It isn't a cylinder, so, were it not for the fact that it is symmetric about the origin (inviting degenerate solutions), and were it not for the fact that it is an oblate spheroid (inviting coordinate transforms involving oblate spheroid coordinates) it would be a generic example as to how to express radiation flux density as a Dirac delta in cylindrical coordinates.
Taking a stab in the dark at this, is this correct?
$Q\frac{\delta(\sqrt{r^2+z^2}-\sqrt{r_0^2-z^2})}{4\pi(r^2+z^2)}$?
Note that all I did was to place the surface equation to 0 in the Dirac delta, multiply it by the quantity $Q$ and normalize it to the area of a sphere of the same radius as the norm of the vector to the point at which the radiation flux density is measured.
Is it that simple?
Before I get into a potentially misleading description of how I (probably mistakenly) came up with that example, here is the general challenge:
Pick  any 3D surface expressed in any coordinate system except one that matches the surface, and express that surface's radiation flux density as a $\delta$.  Examples of what I mean by "a coordinate system that matches the surface" would be spherical coordinates matching a sphere, Cartesian coordinates matching a box, cylindrical coordinates matching a cylinder, oblate spheroid coordinate system matching an oblate spheroid, etc.
Nor am I interested in using coordinate transforms to go from a shape that matches a coordinate system to a coordinate system that doesn't match the shape.
If necessary to avoid falling into the trap of thinking of this as a coordinate transform question, pick a surface that has no corresponding (in the above sense) orthogonal coordinate system.  (If, for example, the surface $x+y+\sqrt{x^2+y^2+z^2}==0$ has no matching orthogonal coordinate system, it might do.)
So, now to get to the meat of my question about "radiation flux density through a surface" and its relationship to Dirac delta:
p 32 Barton et al gives the strong definition of the 3D (spherical coordinate) radial Dirac delta as:
$$\delta^3(\vec{r}) = \frac{\delta(r)}{4\pi r^2}\tag{1}$$
and, correspondingly in 2D (polar coordinate):
$$\delta^2(\vec{r}) = \frac{\delta(r)}{2\pi r}.\tag{2}$$
Since $4\pi r^2$ and $2\pi r$ measure the area and length of sphere and circle respectively, and the surface and line integrals of these two Dirac deltas are 1 (by definition) it seems natural to, in appropriate circumstances, use the radial Dirac delta in modeling density distributions, normalized to 1, over radial surfaces and radial lines respectively.
However, I'm not looking for trivial examples.  A trivial example would be a spherical surface, $Q\frac{\delta(r)}{4\pi r^2}$ for a given total radiated quantity Q, in spherical coordinates.
At first I thought of using an oblate spheroid centered at the origin, in either Cartesian or spherical coordinates, with radiation coming from the origin but it's symmetric about the origin (inviting degenerate solutions for this case) and there is such a thing as "oblate spheroid coordinates" (inviting degenerate solutions for this case).
 A: The trick to switching coordinates with Dirac deltas is to equate two definite integrals (whose limits I won't show). You were at least subconsciously familiar with this in inferring (1) for spherically symmetric Schwartz functions $f$, viz.$$f(O)=\int f(x)\delta^{(3)}(x)\underbrace{dxdydz}_{4\pi r^2dr}=\int\delta(r)f(x)dr\implies\delta^{(3)}(x)=\frac{\delta(r)}{4\pi r^2},$$where $O$ denote the origin.
Now let's consider cylindrical coordinates. Since $dxdy=rdrd\theta$ in the $z=0$ plane that passes through $O$, $\theta$-symmetric $f$ satisfy$$\delta^{(2)}(x)=\frac{\delta(r)}{2\pi r}.$$That looks awfully $2$-dimensional. If you want to make the fact the space is $3$-dimensional explicit, multiply the above by $\delta(z)$ to get$$\delta^{(3)}(x)=\frac{\delta(r)\delta(z)}{2\pi r}.$$This follows in particular from $dxdydz=2\pi r drdz$.
A: For a general orthogonal coordinate system $\xi_i$, the delta function takes the form,
$$ \delta\left(\mathbf r-\mathbf r_0\right)=\frac{\delta\left(\xi_1-\xi_{10}\right)}{h_1}\,\frac{\delta\left(\xi_2-\xi_{20}\right)}{h_2}\,\frac{\delta\left(\xi_3-\xi_{30}\right)}{h_3} \tag{1}$$
where $h_i$ is the scale factor,
$$ h_i^2=\sum_j\left(\frac{\partial r_j}{\partial \xi_i}\right)^2.\tag{2}$$
Thus, you should be able to compute the delta function for any orthogonal coordinate system by using (1) and (2).
For oblate spheroids, the coordinate transformations $\{x,\,y,\,z\}\to\{\mu,\,\nu,\,\phi\}$ follow,
\begin{align}
    x &= a\cosh\mu\cos\nu\cos\phi \\
    y &= a\cosh\mu\cos\nu\sin\phi \\
    z &= a\sinh\mu\sin\nu
\end{align}
The scale factors and delta function will follow from this.
