Intuition for the charge density of a homogeneously charged disc The charge distribution
$$\rho(\vec{r})=\frac{q}{\pi a^2r}\theta{(a-r)}\delta{(\vartheta-\pi/2)}$$
describes a homogeneously charged disc which makes sense in terms of distributions. But if we plot the charge distribution it would decrease with greater radius.
I know that the $r$ should only compensate a part of the Jacobian but the charge distribution makes intuitive no sense.
 A: It can help to think of the delta-function as a limit of a "pulse" function with a fixed area under the curve.  Suppose that instead of that delta-function in $\theta$, we replaced it with a fixed-integral pulse of width $\epsilon$ around $\pi/2$:
$$
P_\epsilon(\theta) = \begin{cases} 1/\epsilon & |\theta - \pi/2| < \epsilon/2 \\
0 & \text{otherwise} \end{cases}
$$
In the limit as $\epsilon \to 0$, this becomes $\delta(\theta - \pi/2)$;  this is one of the definitions of the $\delta$-function.
But let's now think about the support of $P_\epsilon(\theta)$ in 3D space (the part of space where it's non-zero.)  This would be a  "wedge of space" including the equator, narrower near the origin and getting wider as $r$ gets bigger.  In fact, the width of this wedge is proportional to $r$.  So if we wanted to renormalize this function such that its integral over a sphere of constant $r$ was independent of $r$, we would have to use $P_\epsilon(\theta)/r$ instead of $P_\epsilon(\theta)$.  In the limit as $\epsilon \to 0$, this would become $\delta(\theta - \pi/2)/r$ instead of $\delta(\theta - \pi/2)$.
