Rigorous explanation of integration involving delta distribution In a physics class, I saw the following:
The charge density of a uniformly charged circle (charge $Q$) of radius $R$ can be described in cylindrical coordinates using the delta distribution as
$$ \rho(\vec{r}')= \rho(r', \varphi',z')= \delta(r'-R)\delta(z')\frac{Q}{2\pi R} $$
Integration of the density should give us $Q$:
$$ \int \rho(\vec{r}') d^3\vec{r}'=\int_0^\infty d r' \int_0^{2\pi} d\varphi' \int_{-\infty}^{\infty} dz' \rho(\vec{r}') r'=Q $$
Now intuitively some of this makes sense to me, however mathematically I don't really get it at all, I can't make sense of it with what I know about distributions.


*

*How is $ \delta(r'-R)\delta(z')$ correctly interpreted? I don't know of a multiplication of distributions, so I guess this is in a sense a shifted $\delta$ distribution over a 2-dimensional domain: $\langle \delta^{(2)}_{(R,0)}, \phi(r,z) \rangle:=\phi(R,0)$. However the domain of the problem is 3-dimensional, so how is this correctly interpreted? I have troubles especially with the angular component.

*The integral is missing a test-function and the constant one function is not a test function, so what does the integral actually mean? I have often seen identities of the form $\int \delta(x)dx = 1$, what does this mean if written down in a mathematically correct way?

*Integration of singular distributions does not make sense to me, so I guess the integration is meant formally, like $\int \delta\phi:=\langle \delta,\phi\rangle$, at least the integrals concerning $r'$ and $\varphi'$. I know that a variable transformation works for distributions with the inverse jacobian determinant multiplied with the test functions, however the mixing of an "actual" integral here with the abuse of notation integral is not clear to me at all how this is meant rigorously.
In all mathematical texts I have looked into I have never seen something like this, however I found that even here on math.stackexchange some people use that notation, see for e.g. Dirac delta in polar coordinates, so I hope someone could shed light on this "calulation" and how it's done right mathematically. 
 A: *

*The product $\delta(r'-R) \, \delta(z')$ should be interpreted as the tensor product $\delta(r'-R) \otimes 1(\varphi') \otimes \delta(z') \in \mathscr{D}'((0,\infty)\times(0,2\pi)\times(-\infty,\infty)),$ where $1(\varphi')$ is the distribution corresponding to a function having value $1$ everywhere.

*We have a distribution with compact support. Then the test function can have non-compact support, and in this case the test function can be taken to be the function having value $1$ everywhere.

*Replace also the true integral with a distribution action $\langle 1(\varphi'),  \psi(\varphi') \rangle := \int_0^{2\pi} \psi(\varphi') \, d\varphi',$ where $\psi$ is a test function.

Calculating $\langle \rho, 1 \rangle$ in cylindrical coordinates
If $\rho$ were just a function then we would have
$$\begin{align}
\langle \rho, 1 \rangle 
&:= \iiint \rho(\mathbf{x}) \, dV(\mathbf{x}) \\
&= \int_{-\infty}^{\infty} dx \int_{-\infty}^{\infty} dy \int_{-\infty}^{\infty} dz \, \rho_{\text{Cartesian}}(x,y,z) \ \text{in Cartesian coordinates} \\
&= \int_0^\infty dr \, r \int_0^{2\pi} d\varphi \int_{-\infty}^{\infty} dz \, \rho_{\text{cylindrical}}(r,\varphi,z) \ \text{in cylindrical coordinates}
.
\end{align}$$
For a distribution $T$ we define expressing it in some coordinates $(u,v,w)$ so that
$$
\langle T, \phi \rangle
= \langle J(u,v,w) \, T(u,v,w), \phi(u,v,w) \rangle,
$$
where $J(u,v,w)$ is the Jacobian.
Therefore,
$$\begin{align}
\langle \rho, 1 \rangle
&= \langle r' \, \rho(r',\varphi',z'), 1(r',\varphi',z') \rangle \\
&= \langle r' \, \frac{Q}{2\pi R} \delta(r'-R) \otimes 1(\varphi') \otimes \delta(z'), 1(r') \otimes 1(\varphi') \otimes 1(z') \rangle \\
&= \langle r' \, \frac{Q}{2\pi R} \delta(r'-R), 1(r') \rangle \cdot \langle 1(\varphi'), 1(\varphi') \rangle \cdot \langle \delta(z'), 1(z') \rangle \\
&= R \frac{Q}{2\pi R} \cdot 2\pi \cdot 1 \\
&= Q
.
\end{align}$$
