Surface integral - spherical

Question

I'm trying to calculate the following surface integral $$\int \int_{s_r} \frac{z-R}{(x^2+y^2+(z-R)^2)^{3/2}} dS $$, where $s_r=\{(x,y,z)\in \mathbb{R}^3 : x^2+y^2+z^2=r^2 \}. $

I've switched to spherical coordinates but don't really know how to do it.

Answer 1

Use parametric representation to solve the surface integral
\begin{array}{l} x = r\sin \phi \cos \theta \\ y = r\sin \phi \sin \theta \\ z = r\cos \phi \end{array} Where, $$0 \le \phi \le \pi ,\,\,\,0 \le \theta \le 2\pi $$ That is, $${\rm{r}}\left( {\phi ,\theta } \right) = r\sin \phi \cos \theta \widehat i + r\sin \phi \sin \theta \widehat j + r\cos \phi \widehat k$$ Compute the cross product of the Tangent Vectors

$${T_\phi } \times {T_\theta } = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr {{{\partial x} \over {\partial \phi }}} & {{{\partial y} \over {\partial \phi }}} & {{{\partial z} \over {\partial \phi }}} \cr {{{\partial x} \over {\partial \theta }}} & {{{\partial y} \over {\partial \theta }}} & {{{\partial z} \over {\partial \theta }}} \cr } } \right|$$ This will give, $$\left| {{T_\phi } \times {T_\theta }} \right| = {r^2}\sin \phi $$

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Now Calculate the integral,
$$\displaylines{ \int\!\!\!\int_{{s_r}} {{{z - R} \over {{{({x^2} + {y^2} + {{(z - R)}^2})}^{3/2}}}}dS} \cr = \int\!\!\!\int_D {\left( {{{r\cos \phi - R} \over {{{({{\left( {r\sin \phi \cos \theta } \right)}^2} + {{\left( {r\sin \phi \sin \theta } \right)}^2} + {{(r\cos \phi - R)}^2})}^{3/2}}}}} \right)\left| {{T_\phi } \times {T_\theta }} \right|dA} \cr = \int\!\!\!\int_D {\left( {{{r\cos \phi - R} \over {{{\left( {{r^2} + {R^2} - 2rR\cos \phi } \right)}^{3/2}}}}} \right)\left( {r^2\sin \phi } \right)dA} \cr = \int\limits_0^{2\pi } {\int\limits_0^\pi {\left( {{{r\cos \phi - R} \over {{{\left( {{r^2} + {R^2} - 2rR\cos \phi } \right)}^{3/2}}}}} \right)\left( {r^2\sin \phi } \right)d\phi d\theta } } \cr} $$

solve this integral.

Answer 2

The reason to use spherical coordinates is that the surface over which we integrate takes on a particularly simple form: instead of the surface $x^2+y^2+z^2=r^2$ in Cartesians, or $z^2+\rho^2=r^2$ in cylindricals, the sphere is simply the surface $r'=r$, where $r'$ is the variable spherical coordinate. This means that we can integrate directly using the two angular coordinates, rather than having to write one coordinate implicitly in terms of the others.

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So in spherical coordinates, $dS = r^2 \sin{\theta} \, d\theta \, d\phi$. The integral becomes $$ \int_{\theta=0}^{\pi} \int_{\phi=0}^{2\pi} \frac{r\cos{\theta}-R}{(r^2 -2 rR\cos{\theta + R^2})^{3/2}} r^2\sin{\theta} \, d\phi \, d\theta = r^2 \int_{\theta=0}^{\pi} \frac{r\cos{\theta}-R}{(r^2 -2 rR\cos{\theta + R^2})^{3/2}} \sin{\theta} \, d\theta. $$ Putting $u=\cos{\theta}$ gives $$ \int_{-1}^1 \frac{u-(R/r)}{(1-2u(R/r)+(R/r)^2)^{3/2}} \, du $$ It makes sense to write $R/r=a$ at this point, so $$ \int_{-1}^1 \frac{u-a}{(1-2ua+a^2)^{3/2}} \, du. $$ The easiest way to do this integral is by a sort of partial fractions idea: the integrand is $$ \frac{u-a}{(1-2ua+a^2)^{3/2}} = \frac{1}{2a}\left(\frac{a^2-1}{(1-2ua+a^2)^{3/2}} - \frac{1}{(1-2ua+a^2)^{1/2}} \right), $$ which has either $$ \frac{1-ua}{a^2(1-2ua+a^2)^{1/2}} \quad \text{or} \quad \frac{1-u/a}{a^2(1/a^2-2u/a+1)^{1/2}} $$ as continuous antiderivative, depending on whether $R$ is smaller or larger than $r$.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\left.\iint_{\Large s_{r}}{z - R \over \bracks{x^{2} + y^{2} + \pars{z - R}^{2}}^{3/2}}\,\dd S\, \right\vert_{\substack{x^{2} + y^{2} + z^{2}\ =\ r^{2} \\[1mm] z\ =\ r\cos\pars{\theta}}} \\[5mm] = &\ \partiald{}{R}\int_{0}^{2\pi}\int_{0}^{\pi}{1 \over \root{r^{2} -2rR\cos\pars{\theta} + R^{2}}}\,r^{2}\sin\pars{\theta}\,\dd\theta \,\dd\phi \\[5mm] & \stackrel{\xi\ =\ \cos\pars{\theta}}{=}\,\,\, \left.2\pi r^{2}\,\partiald{}{R}\int_{-1}^{1} {\dd\xi \over \root{r^{2} -2rR\xi + R^{2}}} = 2\pi r^{2}\,\partiald{}{R}{\root{r^{2} -2rR\xi + R^{2}} \over -rR} \,\right\vert_{\ \xi\ =\ -1}^{\ \xi\ =\ 1} \\[5mm] = &\ -2\pi r\,\partiald{}{R} {\root{r^{2} - 2rR + R^{2}} - \root{r^{2} + 2rR + R^{2}} \over R} = -2\pi r\,\partiald{}{R} {\verts{r -R} - \pars{r + R} \over R} \\[5mm] = &\ \bbox[#ffe,25px,border:1px dotted navy]{\left\{\begin{array}{lcl} \ds{\phantom{-\,}0} & \mbox{if} & \ds{R < r} \\[2mm] \ds{-\,{4\pi r^{2} \over R^{2}}} & \mbox{if} & \ds{R > r} \end{array}\right.} \end{align}