I am trying to find a way to generate random points uniformly distributed on the surface of an ellipsoid.
If it was a sphere there is a neat way of doing it: Generate three $N(0,1)$ variables $\{x_1,x_2,x_3\}$, calculate the distance from the origin
$$d=\sqrt{x_1^2+x_2^2+x_3^2}$$
and calculate the point
$$\mathbf{y}=(x_1,x_2,x_3)/d.$$
It can then be shown that the points $\mathbf{y}$ lie on the surface of the sphere and are uniformly distributed on the sphere surface, and the argument that proves it is just one word, "isotropy". No prefered direction.
Suppose now we have an ellipsoid
$$\frac{x_1^2}{a^2}+\frac{x_2^2}{b^2}+\frac{x_3^2}{c^2}=1$$
How about generating three $N(0,1)$ variables as above, calculate
$$d=\sqrt{\frac{x_1^2}{a^2}+\frac{x_2^2}{b^2}+\frac{x_3^2}{c^2}}$$
and then using $\mathbf{y}=(x_1,x_2,x_3)/d$ as above. That way we get points guaranteed on the surface of the ellipsoid but will they be uniformly distributed? How can we check that?
Any help greatly appreciated, thanks.
PS I am looking for a solution without accepting/rejecting points, which is kind of trivial.
EDIT:
Switching to polar coordinates, the surface element is $dS=F(\theta,\phi)\ d\theta\ d\phi$ where $F$ is expressed as $$\frac{1}{4} \sqrt{r^2 \left(16 \sin ^2(\theta ) \left(a^2 \sin ^2(\phi )+b^2 \cos ^2(\phi )+c^2\right)+16 \cos ^2(\theta ) \left(a^2 \cos ^2(\phi )+b^2 \sin ^2(\phi )\right)-r^2 \left(a^2-b^2\right)^2 \sin ^2(2 \theta ) \sin ^2(2 \phi )\right)}$$