Consider a 3D spherical Gaussian distribution function that depends on radius only,
$f(r) = \frac{1}{N} e^{-(\frac{r-R_\mu}{\sigma})^2}$
where $R_\mu$ is the radial offset of the distribution and $\sigma$ is the usual variance parameter. $N$ is the normalization constant. To help visualize this function, see the plot below which shows $f(r)$ plotted in one quadrant of the x-y plane, using $R_\mu = 10$ and $\sigma^2 =0.5$.
What is the projection of the distribution function $f(r)$ on to the x-y plane, $f(x,y)$?
Note that it is obviously not as simple as what I have plotted above which is simply $f(x,y,0)$, the slice through the x-y plane. I'm asking for the projection of this distribution function in to the x-y plane, $f(x,y)$.
A simpler example
To try and solve this problem I have tried to evaluate the simpler case of a uniform distribution on the surface of a sphere with radius $R_\mu$, i.e. the limit of $f(r)$ above as $\sigma$ goes to zero. I believe this sample distribution would be defined as
$ f_s(r) = \begin{cases} \frac{1}{N}, & \text{if }r = R_\mu \\ 0, & \text{if }r \neq R_\mu \end{cases} $
The integral over the whole surface must equal 1. $ \int_\Omega f_s(r) d\mathbf{a} = \int_0^{\pi} \int_0^{2\pi} f_s(r) r^2 Sin(\phi) d\theta d\phi = 1 $
To find the projection of this much simpler function on to the xy plane, I express it in terms of an infinitesimal area element. In spherical coordinates the area element is $ d\mathbf{a} = r^2 Sin(\phi)$. But $r^2 = R_\mu$, and $Sin(\phi)$ can be re-written in terms of the x-y coordinates since $Sin (\phi ) = \frac{z}{R_\mu}$. e.g. from the equation for a sphere
$x^2 + y^2 + z^2 = R_\mu ^2$
$\therefore \frac{z^2}{R_\mu^2} = 1- \frac{x^2+y^2}{R_\mu^2}$
$\therefore Sin(\phi) = \sqrt{1- \frac{x^2+y^2}{R_\mu^2}}$
Hence, I find the projection of this distribution in to the x-y plane is:
$ f_s(x,y) = \begin{cases} \frac{1}{N} R_\mu^2 \sqrt{1- \frac{x^2+y^2}{R_\mu^2}}, & \text{if }x^2 + y^2 < R_\mu^2 \\ 0, & \text{if }x^2 + y^2 > R_\mu^2 \end{cases} $
Is this correct? If so, how do I do the more complex case above for a spherical Gaussian in 3D projected on to the xy plane? Do I need to express $f(r)$ in Cartesian coords and integrate the elemental volume over $z$? If I use this logic can I write it in 3D as
$f(r) = \frac{1}{N} e^{-(\frac{r-R_\mu}{\sigma})^2}$
$ dV = r^2 Sin(\phi) dr d\theta d\phi$
$ f(x,y) = \int (x^2 + y^2 +z^2) \frac{z}{\sqrt{x^2 + y^2 +z^2}} e^{-(\frac{r-R_\mu}{\sigma})^2} dz $
I don't think I've got this last step correct because it doesn't look right when I evaluate it numerically and plot it. I suspect it needs to visually converge on the previous result as $\sigma$ -> 0.
Any help you can give is appreciated. This is for a research project where I plan to use these distribution functions in another expression. I expect my question is a very specific case of someone's more general question.
EDIT: Extra graph inspired by Joriki's answer below
See Joriki's answer below. Its easy to see why it makes sense by generating 1000 random unit vectors and plotting their (x-y) components on a scatter plot. You get something like this. Its clear that my original result has the wrong behaviour.