Let's say we have some vector field $\vec C$ such that $$\operatorname{div}\vec C=-\mu_0\vec j=-\mu_0 I\,\delta(x)\,\delta(y)\,\vec e_z.$$ where $\mu_0$ and $I$ are constants.
I am interested in the value of $\int_V d^3r\operatorname{div}\vec C$ where $V$ is a cylindrical volume centered at the $z$-axis. On the one hand, I can simply plug it in and obtain $$\int_V d^3r\operatorname{div}\vec C=-\mu_0 I\,\int\delta(x)\,dx\int\delta(y)\,dy\int dz=-\mu_0 I z$$ but I should get the same result using polar coordinates. For the infinitesimal Volume I obtain $d^3 r=\rho \, d\rho \, d\varphi \, dz$. And (this is probably where my error is) for the current density we get $\vec j=I\,\delta(\rho)\,\vec e_z$. Using this though yields the integral $$\int_V d^3r\operatorname{div}\vec C=-\mu_0 I\int_0^\rho\int_0^{2\pi}\int _{-z/2}^{z/2}\rho'\,\delta(\rho') \,d\rho'\,d\varphi \,dz$$ in the $z$-component. But if I'm not mistaken that would be $0$ since $\int_0^\rho\rho'\,\delta(\rho')\,d\rho'=0$. What do I have to do to get the same result in polar coordinates?
