Integral in 3 dimensions I am trying to integrate
$$ \iiint  \delta(|\mathbf r| -R)\:\mathrm{d}^{3}\mathbf{r} $$
I know that $ \int f(r) \delta(r-R) d^3 \mathbf r =f(R) $, but when I try to apply this here I end up confusing myself, as I seem to have $f(r) = 1 $. So is my answer just 1? This seems wrong...
I then attempted in spherical polars giving:
$$ \iiint \delta (r-R)\delta (\theta-R)\delta (\phi-R) \:\mathrm{d}r \:\mathrm{d}\theta \:\mathrm{d}\phi $$
I am confused as to where I go from here, and how the magnitude sign in the initial integral affects the problem?
Thanks
 A: Using the volume element in spherical coordinates
$$d^3 {\mathbf r}=r^2\sin\theta dr d\theta d\phi$$
the integral factorizes to
$$\int_{\mathbb{R}^3}d^3 {\mathbf r}\delta(|\mathbf r|-R)=
\underbrace{\left(\int_0^\infty dr r^2\delta(r-R)\right)}_{\displaystyle{=R^2}}\underbrace{\left(\int_0 ^\pi d\theta \sin\theta\right)}_{\displaystyle{=\cos0-\cos\pi=2}} 
\underbrace{\left(\int_0^{2\pi} d\phi \right)}_{\displaystyle{=2\pi}}=4\pi R^2.$$
A: $\newcommand{\+}{^{\dagger}}%
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$$
\int_{0}^{\infty}\dd r\,r^{2}\delta\pars{R - r}
\overbrace{\int\dd\Omega_{\vec{r}}}^{\ds{4\pi}} = \color{#00f}{\large 4\pi R^{2}}
$$
