The important thing you need to take into account is
$$
\int{\rm d}^3\mathbf{r}\;\rho(\mathbf{r}) = \int{\rm d}^2\mathbf{r}\;\sigma(\mathbf{r}) \tag{1}
$$
I am going to call $x = \cos \theta$ so that the disk is located at $x = 0$, and then Eq. (1) becomes
\begin{eqnarray}
\int{\rm d}^2\mathbf{r}\;\sigma(\mathbf{r}) &=& \int{\rm d}r{\rm d}\phi\; r\sigma(r,\phi) = \int {\rm d}r{\rm d}\phi\; r\sigma(r,\phi) \underbrace{\int {\rm d}x\delta(x)}_{=1} \\
&=& \int{\rm d}r{\rm d}\phi{\rm d}x\; r^2\frac{1}{r}\sigma(r,\phi)\delta(x) \\
&\stackrel{(1)}{=}& \int{\rm d}r{\rm d}\phi{\rm d}x\; r^2\rho(r, x, \phi) = \int{\rm d}^3\mathbf{r}\;\rho(\mathbf{r})
\end{eqnarray}
You can then conclude that
$$
\rho(r,\theta,\phi) = \frac{1}{r}\sigma(r,\phi)\delta(\cos \theta)
$$
Okay, finally I think I got the idea.
Here is the equation of a sphere of radius $R$:
$$R^2 = (x - a)^2 + (y - b)^2 + (z - c)^2$$
With $(a, b, c)$ the center of the sphere.
Let's say we take our 2 spheres and apply a displacement of $\frac{d}{2}$ on the z axis. We got for the positive and the negative sphere
$$a^2 = x^2 + y^2 + \left(z - \frac{d}{2}\right)^2\\a^2=x^2 + y^2 + \left(z + \frac{d}{2}\right)^2$$
Now, we want to compute the distance to the origin of both spheres' surfaces.
Let's take a rayon in any direction and use an intersection method to compute the distance to both spheres' surfaces.
$$\vec{r} = \left(\begin{array}{c}a + tx\\b+ty\\c+tz\end{array}\right)$$
Since we work at the origin, $(a, b, c) = \vec{0}$ and since we work with spherical coordinates, we got
$$\vec{r} = \left(\begin{array}{c}
t\cdot \sin(\theta)\cdot \cos(\phi)\\
t\cdot \sin(\theta)\cdot \sin(\phi)\\
t\cdot \cos(\theta)
\end{array}\right)$$
Let's replace these values into our prior equations.
$$a^2 = t^2 \cdot \cos^2(\phi) \cdot \sin^2(\theta) + t^2\cdot \sin^2(\phi)\cdot \sin^2(\theta) + t^2\cdot \cos^2(\theta) - d\cdot t\cdot \cos(\theta) + \frac{d^2}{4}$$
That gives
$$t^2 - d\cdot t \cdot \cos(\theta) + \frac{d^2}{4} - a^2 = 0$$
We deduce that $\Delta = d^2\cdot \cos^2(\theta) - 4(\frac{d^2}{4} - a^2)$
Hence, the distance (positive) is
$$t=\frac{d\cdot \cos(\theta) + \sqrt{d^2\cdot \cos^2(\theta) - 4(\frac{d^2}{4} - a^2)}}{2}$$
If we solve in the same manner for the second sphere, we got:
$$t=\frac{-d\cdot \cos(\theta) + \sqrt{d^2\cdot \cos^2(\theta) - 4(\frac{d^2}{4} - a^2)}}{2}$$
And so, the distance between the shell of the first sphere and the second sphere is $d\cdot \cos(\theta)$
So, we can deduce that (if $d$ is small enough):
$$\rho dV = \rho \cdot d\cdot \cos(\theta)dS$$
Taking $\rho = \sigma_0 / d$
We obtained as wanted $\rho dV = \sigma_0 \cos(\theta)dS$
Best Answer
You could solve the above problem by superposition.
* Potential due to inner conducting sphere only is $V_0$ for $r < a$ and $V_0*a/r$ for $r > a$
* Potential due to outer dielectric is $\sigma_0b^3\cos(\theta)/3\epsilon_0r^2$ for $r > b$ and $\sigma_0r\cos(\theta)/3\epsilon_0$ for $r<b$ (This can be calculated by treating the charge distribution as superposition of positively and negatively charged spheres. Ref. The Feynman Lectures on Physics Chapter 6)
Potential at $r=a$ is $V_0 + \sigma_0a\cos(\theta)/3\epsilon_0$
Now, as potential of conducting sphere must be constant, charges redistribute to give a potential of $-\sigma_0a\cos(\theta)/3\epsilon_0$ at $r=a$. We have seen that $\sigma = \sigma_0\cos(\theta)$ gives potential distribution of that kind.
Thus charge distribution on conducting sphere may be taken as $4\pi\epsilon_0aV_0 - \sigma_0\cos(\theta)$
Thus Potentials come out to be