Evaluación de la integral ∫R0∫1−1r2/r2+L2+2Lα−−−−−−−−−−−−−√dαdr∫0R∫−11r2/r2+L2+2Lαdαdr\int_{0}^{R} \int_{-1}^{1} r^2/\sqrt{r^2+L^2+2L\alpha}\,d\alpha dr

Alquiler$A$denote la región correspondiente a la bola de radio$R$centrado en el origen y suponiendo que$L>R$, la integral de volumen sobre$A$se calcula en coordenadas esféricas de la siguiente manera:

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

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$

$$\begin{align}&\color{#66f}{\large\iiint_{A} {\dd x\,\dd y\,\dd z \over \root{x^2 + y^2 + \pars{L - z}^{2}}}} =\\[3mm]&\int_{0}^{2\pi}\dd\phi\int_{0}^{R}\dd r\,r^{2} \int_{0}^{\pi}\dd\theta\,\sin\pars{\theta} \sum_{\ell = 0}^{\infty} {r^{\ell} \over \verts{L}^{\ell + 1}}{\rm P}_{\ell}\pars{\cos\pars{\theta}} \\[3mm]&=2\pi\int_{0}^{R}\dd r\,r^{2} \sum_{\ell = 0}^{\infty} {r^{\ell} \over \verts{L}^{\ell + 1}}\ \overbrace{% \int_{0}^{\pi}{\rm P}_{\ell}\pars{\cos\pars{\theta}}\sin\pars{\theta}\,\dd\theta} ^{\ds{=\ 2\,\delta_{\ell 0}}} \\[3mm] = &\ {4\pi \over \verts{L}}\int_{0}^{R}r^{2}\,\dd r =\color{#66f}{\large{4\pi R^{3} \over 3\verts{L}}} \end{align}$$

$\ds{{\rm P}_{\ell}\pars{x}}$es un polinomio de Legendre .

Bueno, puedes considerar coordenadas esféricas$r,\theta,\phi$, por lo que obtienes

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

Considere la sustitución

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

Entonces obtenemos

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

entonces

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

Entonces podemos escribir la integral como

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

Usando los límites podemos escribir

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

Primero integrar$\phi$da

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

Siguiente integrar$\ell$da

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

Y por último integrar$r$da

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

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$Este es el potencial$\ds{\Phi\pars{L\,\hat{z}}}$debido a una esfera cargada uniforme en un punto fuera de ella. El punto está a distancia.$\ds{\verts{L}}$del centro de la esfera.

La carga total de la esfera es$\ds{q = \int_{A}1\,{\rm d}^{3}\vec{r} = {4 \over 3}\,\pi R^{3}}$.

Entonces,

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