The problem is with your current density $\vec{j}$, let's start with it's basic definition as current over area, $\vec{j} = \frac{Q\vec{v}}{4\pi R^2}$, now knowing that the angular velocity can be used to get the linear velocity of the charge on the spherical surface using $\vec{v}=\vec{\omega}\times R\hat{r} = \omega R \hat{z} \times \hat{r} = \omega R (\cos\theta\hat{r}-\sin\theta\hat{\theta})\times\hat{r} = \omega R\sin\theta \hat{\phi}$
So that your current density is:
$$\vec{j} = \frac{Q\omega}{4\pi R} \sin\theta \delta(r-R)\hat{\phi}$$
Notice that this makes sense since the charges should only be moving in the $\phi$ direction around the z-axis, you should never have had any $r$ or $\theta$ components before. Also, at the poles of the sphere the current density goes to zero as expected and with the maximal current density at the equator.
With this current density your integrand will have a $\sin ^2\theta$ term which will clear up your previous issues.
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
1) Maybe this could provide you with some insight. This is from Griffiths Electrodynamics 3rd Edition. I'm not sure why the solutions you've found require the cosθ′ but this one doesn't include it.
2) When using spherical coordinates and trying to outline a full sphere, theta outlines a half-circle (existing in the xz plane), phi then rotates and projects this half-circle aroud the z axis. θ then must range from 0 to 2π, ϕ from 0 to 2π for the entire sphere.