Evaluation of the integral $\int_{0}^{R}\int_{-1}^{1} r^2/\sqrt{r^2+L^2+2L\alpha}\,d\alpha dr$

Question

I am trying to solve this integral

$$ \int_{0}^{R}\int_{-1}^{1}\frac{r^{2}\,{\rm d}\alpha\,{\rm d}r}{\, \sqrt{\vphantom{\Large A}\,r^{2} + L^{2} + 2L\alpha\,}\,} $$ where $L$ is some positive number.

The original question was to calculate the integral $$\iiint_A \frac{dxdydz}{\sqrt{x^2+y^2+(L-z)^2}}$$

Where $A$ is a sphere with radius $R$ and center at the origin, and $0 < R < L$, but after moving to spherical coordinates and then doing a variable switch i ended up with the double integral above. How would I do this?

Note: We can use Fubini's theorem to first integrate with respect to $r$ but i think that would be even harder.

Answer 1

Well you can consider spherical co-ordinates $r,\theta,\phi$, so you get

$$ \iiint \frac{ dx dy dz}{\sqrt{x^2+y^2+\big[z - L\big]^2}} = \iiint \frac{r^2 \sin(\theta) dr d\theta d\phi}{\sqrt{r^2 + L^2 - 2 r L \cos(\theta)}} $$

Consider the substitution

$$ \ell^2 = r^2 + L^2 - 2 r L \cos(\theta) $$

Then we get

$$ 2 \ell d\ell = 2 r L \sin(\theta) d\theta $$

so

$$ \frac{\sin(\theta) d\theta}{\ell} = \frac{d\ell}{rL} $$

So we can write the integral as

$$ \iiint \frac{r dr d\ell d\phi}{L} $$

Using the boundaries we can write

$$ \int_0^R dr \int_{|r-L|}^{|r+L|} d\ell \int_0^{2\pi} d\phi \frac{r}{L} $$

First integrate $\phi$ gives

$$ 2 \pi \int_0^R dr \int_{|r-L|}^{|r+L|} d\ell \frac{r}{L} $$

Next integrate $\ell$ gives

$$ 2 \pi \int_0^R dr \left[ \frac{r\ell}{L} \right]_{|r-L|}^{|r+L|} = 4 \pi \int_0^R dr \frac{r^2}{L} $$

And last integrate $r$ gives

$$ \frac{4 \pi R^3}{3 L} $$

Answer 2

Letting $A$ denote the region corresponding to the ball of radius $R$ centered at the origin and supposing that $L>R$, the volume integral over $A$ is computed in spherical coordinates as follows:

$$\begin{align} \iiint_A \frac{dxdydz}{\sqrt{x^2+y^2+(L-z)^2}}&=\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{R}\frac{r^2\sin{\theta}\,\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi}{\sqrt{r^2-2Lr\cos{\theta}+L^2}}\\ &=2\pi\int_{0}^{\pi}\int_{0}^{R}\frac{r^2\sin{\theta}\,\mathrm{d}r\mathrm{d}\theta}{\sqrt{r^2-2Lr\cos{\theta}+L^2}}\\ &=2\pi L^2\int_{0}^{\pi}\int_{0}^{R/L}\frac{x^2\sin{\theta}\,\mathrm{d}x\mathrm{d}\theta}{\sqrt{x^2-2x\cos{\theta}+1}}\\ &=2\pi L^2\int_{0}^{R/L}\mathrm{d}x\int_{0}^{\pi}\mathrm{d}\theta\frac{x^2\sin{\theta}}{\sqrt{x^2-2x\cos{\theta}+1}}\\ &=2\pi L^2\int_{0}^{R/L}\mathrm{d}x\int_{-1}^{1}\mathrm{d}u\frac{x^2}{\sqrt{x^2+2xu+1}}\\ &=2\pi L^2\int_{0}^{R/L}\mathrm{d}x\left(2x^2\right)\\ &=2\pi L^2\left(\frac23 \frac{R^3}{L^3}\right)\\ &=\frac{4\pi R^3}{3L}. \end{align}$$

Answer 3

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ This is the potential $\ds{\Phi\pars{L\,\hat{z}}}$ due to a uniform charged sphere at a point outside it. The point is at a distance $\ds{\verts{L}}$ from the sphere center.

The total charge of the sphere is $\ds{q = \int_{A}1\,{\rm d}^{3}\vec{r} = {4 \over 3}\,\pi R^{3}}$.

So, $$ \Phi\pars{L\,\hat{z}} = {q \over \verts{L}} = {4\pi R^{3} \over 3\verts{L}} $$

Answer 4

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align}&\color{#66f}{\large\iiint_{A} {\dd x\,\dd y\,\dd z \over \root{x^2 + y^2 + \pars{L - z}^{2}}}} =\\[3mm]&\int_{0}^{2\pi}\dd\phi\int_{0}^{R}\dd r\,r^{2} \int_{0}^{\pi}\dd\theta\,\sin\pars{\theta} \sum_{\ell = 0}^{\infty} {r^{\ell} \over \verts{L}^{\ell + 1}}{\rm P}_{\ell}\pars{\cos\pars{\theta}} \\[3mm]&=2\pi\int_{0}^{R}\dd r\,r^{2} \sum_{\ell = 0}^{\infty} {r^{\ell} \over \verts{L}^{\ell + 1}}\ \overbrace{% \int_{0}^{\pi}{\rm P}_{\ell}\pars{\cos\pars{\theta}}\sin\pars{\theta}\,\dd\theta} ^{\ds{=\ 2\,\delta_{\ell 0}}} \\[3mm] = &\ {4\pi \over \verts{L}}\int_{0}^{R}r^{2}\,\dd r =\color{#66f}{\large{4\pi R^{3} \over 3\verts{L}}} \end{align}

$\ds{{\rm P}_{\ell}\pars{x}}$ is a Legendre Polynomial.