Surface integral - spherical 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. 

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\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}
