# Charge density as Dirac delta function

I little bit confused while I'm trying to convert a cylindrical charge distribution to spherical. The question is:

A uniformly charged thin disk of radius $a$ and surface charge density $\sigma$ placed at the $xy$-plane with its center at the origin. The disk is enclosed in a grounded conducting sphere of radius $b (b>a)$. Using the appropriate Green’s function, find the potential for all points inside the sphere.

First I need to convert that cylindrical distribution to spherical

$$\rho(x)=f(r)\delta(\cos\theta)\Theta(r-a)$$

$f(r)$ for r dependence and $\Theta$ is heavyside func. I'll integrate it over volume in spherical coordinates and equalize to $\sigma$$\pi$$a^2$ to decide $\rho(x).$

Now, am I have to add $1/r$ because of the metric from the beginning? And then that question came to my mind, what if I put a sphere with radius $a$ with $\rho$ inside a cylinder wiyh radius $b$, how can I convert this?

$$\int{\rm d}^3\mathbf{r}\;\rho(\mathbf{r}) = \int{\rm d}^2\mathbf{r}\;\sigma(\mathbf{r}) \tag{1}$$
I am going to call $x = \cos \theta$ so that the disk is located at $x = 0$, and then Eq. (1) becomes
$$\rho(r,\theta,\phi) = \frac{1}{r}\sigma(r,\phi)\delta(\cos \theta)$$