Note: This is not a homework problem.
As the title somewhat eludes to, I have a density function for a sphere as a function of radius $\rho(r)$. I would like to then flatten the sphere slightly into an ellipsoid but keep the 'same' density function. In other words, I'd like to somehow adapt the density function to work with an ellipsoid.
I am somewhat familiar with basis transformations, matrices, and tensors.
Please let me know if I need to provide more information or if something is unclear. Thank you for your help.
By the "same" density function I assume you mean it should have a similar relationship to the ellipsoid as the original function had to the sphere; for starters, the new density function has the same value everywhere on the surface of the ellipsoid, just as the original had the same value everywhere on the surface of the sphere.
By "flattening" the sphere you obtain an ellipsoid with one axis of rotational symmetry. This ellipsoid is just a linear transformation of the sphere that preserves the distances of all points from that axis and multiplies all points' distances from an "equatorial" plane (perpendicular to that axis) by a factor less than $1$. Alternatively, you can increase all distances from the axis of symmetry by a constant factor in order to preserve volume. (There are other transformations from a sphere to an ellipsoid, but these are what I would consider the properties of a "flattening" transformation.)
If we use cylindrical coordinates $(u,\theta,z)$, using $u$ for the cylindrical radius coordinate since you've already used $r$ and $\rho$ for other purposes, and let the origin of these coordinates be the center of both the sphere and the ellipsoid and the $z$-axis be the ellipsoid's axis of symmetry, then the transformation from the sphere to the ellipsoid is $(u,\theta,z) \mapsto \left(\frac aR u, \theta, \frac bR z\right)$, where $R$ is the radius of the sphere and $a$ and $b$ are the equatorial and polar radii of the ellipsoid. (Note that if you just moved points closer to the equatorial plane, then $a=R$ and the transformation simplifies to $(u,\theta,z) \mapsto \left(u, \theta, \frac bR z\right)$.)
A reasonable way to make the density the "same" would be for the density at each point to be equal to the original density of the preimage of that point under the transformation. That is, when we "flatten" or "stretch" the sphere, a point where the density is $\rho_1$ "moves" to a point where the density is still $\rho_1$. A variation of this is that after this transformation, we multiply all densities by a constant factor in order to restore the total mass of the ellipsoid to the total mass of the sphere.
So we want the new density, $\rho'$, to obey the rule $$\rho'\left(\frac aR u, \theta, \frac bR z\right) = k\rho(u,\theta,z)$$
where $k$ is some constant (possibly $1$) and (for this equation) we treat $\rho$ as a function of the three cylindrical coordinates of points within the sphere. A more "user-friendly" version of these equations may be obtained by substitution of variables:
$$\rho'(u,\theta,z) = \rho\left(\frac Ra u, \theta, \frac Rb z\right).$$
Since the radial $r$ coordinate in spherical coordinates relates to the cylindrical coordinates via the equation $r^2 = u^2 + z^2$, this implies $$\rho'(u,\theta,z) = \rho\left(\sqrt{\frac{R^2}{a^2} u^2 + \frac{R^2}{b^2} z^2}\right).$$
You can also do this in Cartesian coordinates, where the transformation from sphere to ellipsoid is $(x,y,z) \mapsto \left(\frac aR x, \frac aR y, \frac bR z\right)$ and where $u^2 = x^2 + y^2$, so
$$\rho'(x,y,z) = \rho\left(\sqrt{\frac{R^2}{a^2} (x^2 + y^2) + \frac{R^2}{b^2} z^2}\right).$$
If you want to use spherical coordinates in general for both the ellipsoid and the sphere, then the relevant formula for the transformation from sphere to ellipsoid is $$(r,\theta,\phi) \mapsto \left(\frac rR\sqrt{a^2 \sin^2\phi + b^2 \cos^2\phi}, \theta, \cot^{-1}\left(\frac bR \cot\phi\right)\right).$$
This is not as simple to invert as the formula in cylindrical coordinates is, but if $r' = \frac rR\sqrt{a^2 \sin^2\phi + b^2 \cos^2\phi}$ and $\phi' = \cot^{-1}\left(\frac bR \cot\phi\right)$ then
\begin{align} \phi & = \cot^{-1}\left(\frac Rb \cot\phi'\right) \\ \sin^2\phi & = \frac{1}{1 + \left(\frac Rb \cot\phi'\right)^2} \\ \cos^2\phi & = \frac{1}{1 + \left(\frac Rb \cot\phi'\right)^{-2}} = 1 - \sin^2\phi \\ r & = \frac{R r'}{\sqrt{a^2 \sin^2\phi + b^2 \cos^2\phi}} \end{align}
Plugging into the last equation the known expressions for $\sin^2\phi$ and $\cos^2\phi$ in terms of $\phi'$, you can obtain a function $f$ such that $r = f(r', \phi')$, where $(r',\theta,\phi')$ are coordinates of a point within the ellipsoid and $r$ is the radius of the point within the sphere that was mapped to $(r',\theta,\phi')$, and so the density function you want is
$$\rho'(r',\theta,\phi') = \rho(f(r', \phi')).$$
If you mean to use geodetic coordinates on the ellipsoid, however, different formulas apply. I do not want even to start on that if you do not need it.