Here is an example from a textbook:
Suppose there is a unit charge or unit mass at the point $(x,y,z)=(-1,\sqrt{3},-2)$; then in rectangular coordinates, the density is
$$\rho=\delta(x+1)\,\delta(y-\sqrt{3})\,\delta(z+2)\tag{1}$$
In cylindrical coordinates the point is $(r,\theta,z)=\left(2,\frac{2\pi}{3},-2\right)$, so in cylindrical coordinates the density is
$$\rho=\frac{\delta(r-2)\,\delta\left(\theta -\frac{2\pi}{3}\right)\delta(z+2)}{r}\tag{2}$$ In spherical coordinates, the point is $(r,\theta,\phi)=\left(2\sqrt{2},\frac{3\pi}{4},\frac{2\pi}{3}\right)$, so in spherical coordinates the density is
$$\rho=\frac{\delta(r-2\sqrt{2})\,\delta\left(\theta -\frac{3\pi}{4}\right)\delta\left(\phi-\frac{2\pi}{3}\right)}{{r^2}\sin\theta}\tag{3}$$
When I substitute the point $(x,y,z)=(-1,\sqrt{3},-2)$ into equation $(1)$ I get a density of unity as expected.
When I substitute the point $(r,\theta,z)=\left(2,\frac{2\pi}{3},-2\right)$ into equation $(2)$ I get a density of $$\color{red}{\frac{1}{2}\ne 1}$$
When I substitute the point $(r,\theta,\phi)=\left(2\sqrt{2},\frac{3\pi}{4},\frac{2\pi}{3}\right)$ into equation $(3)$ I get a density of $$\color{red}{\frac{1}{4\sqrt{2}}\ne 1}$$
I thought that choice of coordinate system is irrelevant since all coordinate systems you use will always lead you to the same result.
So why are the parts marked $\color{red}{\mathrm{red}}$ not equal to unity?
EDIT:
In response to the comments below, since we
"cannot evaluate the Dirac Delta as if it has a value at a point; it
is not a function"
From $(1)$; Is it plausible to write $$\rho=\delta(0)\,\delta(0)\,\delta(0)$$ at the point $(x,y,z)=(-1,\sqrt{3},-2)$?
Notice that I did not evaluate the Dirac Delta, but I did substitute the coordinates in.
In other words; Is substitution allowed but evaluation not allowed?
Best Answer
By definition $Q=\iiint \rho \, dV$
In Cartesian: $$Q=\iiint \rho(x,y,z) \, dx \, dy \, dz=1$$
In Cylindrical: $$Q=\iiint \rho(r,\theta,z) \, r\, dr \, d\theta \, dz=1$$
In Spherical: $$Q=\iiint \rho(r,\theta,\phi) \, r^2\sin \theta \, d\theta \, dr \, d\phi=1$$
Hence for point charge:
\begin{align*} \rho(x,y,z) &= \delta (x-x_{0}) \, \delta (y-y_{0}) \, \delta (z-z_{0}) \\ \rho(r,\theta,z) &= \frac{1}{r} \delta (r-r_{0}) \, \delta (\theta-\theta_{0}) \, \delta (z-z_{0}) \\ \rho(r,\theta,\phi) &= \frac{1}{r^2\sin \theta} \delta (r-r_{0}) \, \delta (\theta-\theta_{0}) \, \delta (\phi-\phi_{0}) \\ \end{align*}