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

  • $\begingroup$ Not sure but I think you need to try integrating r three times. So first integral is 1. The second would be integral of 1. The third integral should be a function of r. Spherical coordinates probably need some trig functions to yield same results. $\endgroup$
    – CAGT
    Commented Jan 13, 2014 at 20:04
  • $\begingroup$ integrate r? But I dont have an r term? So integrate 1 three times, but with respect to what? $\endgroup$ Commented Jan 13, 2014 at 21:13
  • 1
    $\begingroup$ the delta dirac Is an infinite impulse at r=R and "loosely" or heuristically is \int_{-\infty}^\infty \delta(r) \, dr = 1. So you integrate in all the r domain, which probably is 0 to infinity. If you need polar angles, the angles must sweep the whole volume (sphere). So angles are unitless from 0 to pi or 2pi, where needed. $\endgroup$
    – CAGT
    Commented Jan 13, 2014 at 22:37
  • $\begingroup$ Ok but I dont understand why the answer isn't 1. We have the formula that defines the delta function, in 3D, and it says that the answer should be f(R). But since f(r)=1, f(R)=1? $\endgroup$ Commented Jan 14, 2014 at 13:31

2 Answers 2


Using the volume element in spherical coordinates

$$d^3 {\mathbf r}=r^2\sin\theta dr d\theta d\phi$$

the integral factorizes to

$$\int_{\mathbb{R}^3}d^3 {\mathbf r}\delta(|\mathbf r|-R)= \underbrace{\left(\int_0^\infty dr r^2\delta(r-R)\right)}_{\displaystyle{=R^2}}\underbrace{\left(\int_0 ^\pi d\theta \sin\theta\right)}_{\displaystyle{=\cos0-\cos\pi=2}} \underbrace{\left(\int_0^{2\pi} d\phi \right)}_{\displaystyle{=2\pi}}=4\pi R^2.$$

  • $\begingroup$ This exactly what I was looking for! Thanks a bunch! $\endgroup$ Commented Jan 14, 2014 at 16:23
  • $\begingroup$ @user1887919 no problem, you're welcome! $\endgroup$
    – flonk
    Commented Jan 14, 2014 at 17:02

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ \int_{0}^{\infty}\dd r\,r^{2}\delta\pars{R - r} \overbrace{\int\dd\Omega_{\vec{r}}}^{\ds{4\pi}} = \color{#00f}{\large 4\pi R^{2}} $$

  • $\begingroup$ ?? This was posted 6 days earlier in the other answer. Are you just trying to lower the signal/noise ratio? $\endgroup$
    – Did
    Commented Feb 24, 2014 at 8:00

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