The total charge/mass in space should be the same whether we consider it to be distributed as a volume density or idealize it as a surface or line density. See § {Densities}.

The Dirac delta function relates line and surface charge densities (which are really idealizations) to volume densities. For example, if the surface charge density on a rectangular surface is $\sigma(x,y)$, with dimensions $C/L^2$, then the total charge on the slab is obtained by chopping up the surface into infinitesimal areas $dA = dx\, dy$ and summing up (integrating) the charge $\sigma(x,y) dA$ on each piece, $\int\int \sigma(x,y) \, dx \, dy$. Equivalently, one can recognize that this surface charge density is actually a volume charge density, idealized to be concentrated at, say, $z=0$. Thus, \begin{equation} \rho(x,y,z) = \sigma(x,y) \,\delta(z) \end{equation} and integrating this over a solid region yields \begin{equation} \int\!\int\!\int \rho(x,y,z) \,dz\,dx\,dy = \int\!\int\!\int \sigma(x,y) \,\delta(z) \,dz\,dx\,dy = \int\!\int \sigma(x,y) \,dx\,dy \end{equation} which yields the same answer as before. Recall that $\delta(z)$ has dimensions of inverse length, so that $\rho$ has the correct dimensions, namely $CL^{-3}$.