which allows an exact solution.

We can take eq.(32) and substitute into it the product $g(\theta)U_{\theta}$ from eq.(31), and the product $g(\theta)V_{\theta}$ from eq.(30) to obtain an equation involving only $\rho_{\theta}$:

$$\frac{d^{2} \rho_{\theta}}{d \theta^{2}} + cot(\theta)\frac{d \rho_{\theta}}{d \theta} + 
\left(\frac{C_{c}C_{\theta}}{C_{r}} \right)\rho_{\theta} =0,$$

having solution:

$$\rho_{\theta} = c_{1} P_{m}(cos(\theta)) +c_{2} Q_{m}(cos(\theta)).$$

In the above equation $c_{1}$ and $c_{2}$ are normalisation constants, and $P_{m}$ and $Q_{m}$ are the Legendre polynomials of the first and second kinds, respectively. The index of these functions is given by the relation $2m=(4C_{T}+1)^{1/2} -1$, where $C_{T}=(C_{c} C_{\theta}/C_{r})$. As typical of separation of variables problems, we see the solution as an eigenvalue problem, with the separation constants determining the order of the solution function. The requirement of axial symmetry forces $m$ to be even. Any such desired angular distribution for $\rho_{\theta}$ can now be constructed as an infinite series of the above functions. For simplicity we analyse the case of $m=2$ $(C_{T}=6)$, $c_{1}=\bar{\rho}_{\theta}$, $c_{2}=0$:

$$\rho_{\theta} = \bar{\rho}_{\theta} (3 cos^{2}(\theta)-1 )$$

This will result in slightly less material along the plane of the disk, and slightly more along the poles, for the total density $\rho=\rho_{0}+\rho_{1}$, with respect to the background state $\rho_{0}$. At this point eqs.(30) and (31) can 