Show that $\iiint_R \frac{dV}{r^2} = \iint_{S=\partial R}\frac{\vec{r}}{r^2}dS$

Question

Show that $$\iiint_R \frac{dV}{r^2} = \iint_{S=\partial R}\frac{\vec{r}}{r^2}dS$$ for $r=\|\vec{r}\|$.

Answer 1

Basically this is the Divergence Theorem.

You just need to prove that $\frac{1}{r^2} = div\ \frac{r}{r^2}$ and you're done. Computing the derivatives of the divergence operator : $$ \frac{\partial}{\partial x}\ \frac{x}{x^2 + y^2 + z^2} = \frac{(x^2 + y^2 + z^2) - x\left(2x \right)}{(x^2 + y^2 + z^2)^2} = \frac{r^2 -2x^2}{r^4}\\ \frac{\partial}{\partial y}\ \frac{y}{x^2 + y^2 + z^2} = \frac{(x^2 + y^2 + z^2) - y\left(2y \right)}{(x^2 + y^2 + z^2)^2} = \frac{r^2 -2y^2}{r^4}\\ \frac{\partial}{\partial z}\ \frac{z}{x^2 + y^2 + z^2} = \frac{(x^2 + y^2 + z^2) - z\left(2z \right)}{(x^2 + y^2 + z^2)^2} = \frac{r^2 -2z^2}{r^4}\\ $$ And then summing them : $$ \nabla \cdot (r/r^2) = \frac{3r^2 - 2x^2 - 2y^2 - 2z^2}{r^4} =\frac{3r^2 - 2r^2}{r^4} = \frac{1}{r^2} $$

Answer 2

Use the Divergence Theorem. Edit: Wait, which surface integral is this supposed to be? I know different notation.

Answer 3

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{{1 \over r^{2}}} & = \pars{{\vec{r} \over r}}\cdot \pars{{1 \over r^{2}}\,{\vec{r} \over r}} = \pars{\nabla r}\cdot\nabla\pars{-\,{1 \over r}} = -\nabla\cdot\bracks{r\,\nabla\pars{1 \over r}} + r\,\nabla\cdot\nabla\pars{1 \over r} \\[3mm] & = -\nabla\cdot\bracks{r\pars{-\,{1 \over r^{2}}\,{\vec{r} \over r}}} + r\,\nabla^{2}\pars{1 \over r} = \nabla\cdot\pars{\vec{r} \over r^{2}} + r\bracks{-4\pi\,\delta\pars{\vec{r}}} = \color{#f00}{\nabla\cdot\pars{\vec{r} \over r^{2}}} \end{align}