Dirac delta function in polar coordinates It would be wonderful if someone could direct me to a proof or explanation on why we can express the delta function of $\delta(x-\xi)=\delta(x_{1}-\xi_{1})\delta(x_{2}-\xi_{2})$ as $\delta(x-\xi)=r^{-1}\delta(r-r')\delta(\theta-\phi)$ for $$x=(r\hspace{0.05cm}\cos\hspace{0.05cm}\theta,r\hspace{0.05cm}\sin\hspace{0.05cm}\theta)$$ and $$\xi=(r'\hspace{0.05cm}\cos\hspace{0.05cm}\phi,r'\hspace{0.05cm}\sin\hspace{0.05cm}\phi)$$
 A: In short: For any $\varphi \in C_c^\infty(\mathbb R^2)$ you want
$$
\varphi^{\text{rect}}(x_0, y_0)
= \int_{x=-\infty}^{\infty} \int_{y=-\infty}^{\infty} \delta_{(x_0,y_0)}^{\text{rect}}(x-x_0, y-y_0) \, \varphi^{\text{rect}}(x, y) \, dx \, dy \\
= \int_{r=0}^{\infty} \int_{\theta=0}^{2\pi} \delta_{(r_0,\theta_0)}^{\text{polar}}(r,\theta) \, \varphi^{\text{polar}}(r,\theta)\,r\,dr\,d\theta,
$$
where subperscript $\textit{rect}$ and $\textit{polar}$ denotes rectangular and polar representations.
Since
$$
\varphi^{\text{rect}}(x_0, y_0) = \varphi^{\text{polar}}(r_0, \theta_0) = \int_{r=0}^{\infty} \int_{\theta=0}^{2\pi} \delta(r-r_0)\,\delta(\theta-\theta_0)\,\varphi^{\text{polar}}(r,\theta)\,dr\,d\theta
$$
it follows that you should have 
$$
r\,\delta_{(r_0,\theta_0)}^{\text{polar}}(r,\theta) = \delta(r-r_0)\,\delta(\theta-\theta_0),
$$
i.e.
$$
\delta_{(r_0,\theta_0)}^{\text{polar}}(r,\theta) 
= r^{-1} \delta(r-r_0)\,\delta(\theta-\theta_0)
= r_0^{-1} \delta(r-r_0)\,\delta(\theta-\theta_0)
.
$$
