A 3d integral over a ball

Question

Given $a \in \mathbb{R}^3$ and $r>0$, is it possible to compute $$\int_{B_r(a)} \frac{a\cdot x}{|x|^3} dx$$

where $B_r(a)$ is the ball in $\mathbb{R}^3$ with radius $r$ and centered at $a$.

Answer 1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

\begin{align} \color{#f00}{\iiint_{\verts{\vec{R} - \vec{a}} < r}\ {\vec{a}\cdot\vec{R} \over R^{3}}\,\dd^{3}\vec{R}} & = \vec{a}\cdot\iiint_{R < r}\ {\vec{R} + \vec{a}\over \verts{\vec{R} + \vec{a}}^{3}}\,\dd^{3}\vec{R} = -\vec{a}\cdot\iiint_{R < r}\ {\vec{R} - \vec{a} \over \verts{\vec{R} - \vec{a}}^{3}}\,\dd^{3}\vec{R} \\[3mm] & = \vec{a}\cdot\nabla_{\vec{a}}\iiint_{R < r}\ {\dd^{3}\vec{R} \over \verts{\vec{R} - \vec{a}}}\tag{1} \end{align}

In general; we can consider, for simplicity, the case of $\vec{a}$ along the $z$-axis $\pars{~\vec{a} = \verts{\vec{a}}\hat{z} = a\hat{z}~}$. We'll recover the general result at the very end. Then, \begin{align} \iiint_{R < r}\ {\dd^{3}\vec{R} \over \verts{\vec{R} - \vec{a}}} & = 2\pi\int_{0}^{r}\int_{0}^{\pi} {\sin\pars{\theta}\,\dd\theta \over \root{R^{2} - 2Ra\cos\pars{\theta} + a^{2}}} \,R^{2}\,\dd R \\[3mm] & = 2\pi\int_{0}^{r}\int_{-1}^{1} {\dd\xi \over \root{R^{2} - 2Ra\xi + a^{2}}}\,R^{2}\,\dd R \\[3mm] & = 2\pi\int_{0}^{r}\left.{\root{R^{2} - 2Ra\xi + a^{2}} \over -Ra} \right\vert_{\ \xi\ =\ -1}^{\ \xi\ =\ 1}\,R^{2}\,\dd R \\[3mm] & = {2\pi \over a}\int_{0}^{r}\bracks{% {\root{R^{2} + 2Ra + a^{2}} - {\root{R^{2} - 2Ra + a^{2}}}}}\,R\,\dd R \\[3mm] & = {2\pi \over a}\int_{0}^{r}\pars{R + a - \verts{R - a}}\,R\,\dd R = \left\lbrace\begin{array}{lcl} \ds{{4\pi r^{3} \over 3}\,{1 \over a}} & \mbox{if} & \ds{r < a} \\[2mm] \ds{2\pi\pars{r^{2} - {1 \over 3}\,a^{2}}} & \mbox{if} & \ds{r \geq a} \end{array}\right. \end{align} and $$ \nabla_{\vec{a}}\iiint_{R < r}\ {\dd^{3}\vec{R} \over \verts{\vec{R} - \vec{a}}} = \left\lbrace\begin{array}{lcl} \ds{-\,{4\pi r^{3} \over 3}\,{\vec{a} \over a^{3}}} & \mbox{if} & \ds{r < a} \\[2mm] \ds{-\,{4\pi \over 3}\,\vec{a}} & \mbox{if} & \ds{r \geq a} \end{array}\right. $$

---

With the result $\pars{1}$: $$\color{#f00}{\iiint_{\verts{\vec{R} - \vec{a}} < r}\ {\vec{a}\cdot\vec{R} \over R^{3}}\,\dd^{3}\vec{R}} = \color{#f00}{\left\lbrace\begin{array}{lcl} \ds{-\,{4\pi r^{3} \over 3}\,{1 \over a}} & \mbox{if} & \ds{r < a} \\[2mm] \ds{-\,{4\pi \over 3}\,a^{2}} & \mbox{if} & \ds{r \geq a} \end{array}\right.} $$