Consider the $d$-dimensional delta function $\delta^{(d)}(x)\equiv\prod_{i=1}^d\delta(x^i)$. Since $$ x^i=0,~\forall i\iff \sum_i(x^i)^2=0 $$ it should be possible to write $$ \delta^{(d)}(x)=f(x)\delta(r) $$ where $r=\sqrt{\sum_i(x^i)^2}$.
My question is what is $f(x)$? Is it $f(x)=\frac{1}{\Omega_{d}r^{d-1}}$? (where $\Omega_{d}$ is the $d$-dimensional solid angle)
