$\iint_{\Sigma} \frac{d\sigma}{\sqrt{x^2+y^2+(z+R)^2}}$ with $\Sigma$ the upper half of the sphere $x^2+y^2+z^2=R^2$ My attempt:
$K = \{(r,\theta): 0 \le r  \le R, 0 \le \theta \le 2\pi \}$. I chose following parameterization: $$ \vec{\varphi}(r,\theta)=(r\cos\theta, r\sin\theta,\sqrt{R^2-r^2}).$$ And after further calculations, I got $$ \left\|\frac{\partial{\vec{\varphi}}}{\partial r} \times\frac{\partial{\vec{\varphi}}}{\partial \theta}\right\| = \frac{rR}{\sqrt{R^2-r^2}}.$$
And now, $$\iint_{\Sigma} \frac{d\sigma}{\sqrt{x^2+y^2+(z+R)^2}} = \int_0^{2\pi}d\theta\int_0^R \frac{r^2R}{\sqrt{(2R^2+2R\sqrt{R^2-r^2})(R^2-r^2)}}dr.$$ Applying the substitution $t = \sqrt{R^2-r^2}$, we get $$ 2\pi R\int_R^0\frac{R^2-t^2}{\sqrt{(2R^2+2Rt)t^2}}\frac{-t}{\sqrt{R^2-t^2}}dt=\sqrt{2R}\pi\int_0^R\sqrt{R-t}dt=\frac23\sqrt2\pi R^2.$$
The correct answer, however, should be $2\pi R(2-\sqrt2)$. Can somebody spot my mistake? 
 A: When you wrote $\int_0^R \frac{r^2}{\sqrt{(2 R^2+2 R\sqrt{R^2-r^2} ) (R^2-r^2)}}$ the numerator should have been $rR$ rather than $r^2$.  You can evaluate the integral using your substitution $t=\sqrt{R^2-r^2}$ as follows:
$$\int_0^{2 \pi}d\theta\int_0^R \frac{rR}{\sqrt{(2 R^2+2 R\sqrt{R^2-r^2} ) (R^2-r^2)}}dr$$
$$
= 2 \pi \int_R^0 \frac{R \sqrt{R^2-t^2}}{\sqrt{(2 R^2+2 Rt ) t^2}} \frac{-t}{\sqrt{R^2-t^2}} dt
$$
$$
= 2 \pi \int_R^0 \frac{R }{\sqrt{(2 R^2+2 Rt ) t^2}} (-t) dt
$$
$$
= 2 \pi \int_0^R \frac{R }{\sqrt{2 R^2+2 Rt }} dt
$$
$$
= 2\sqrt2 \pi \int_0^R \frac{R }{2\sqrt{R^2+ Rt }} dt
$$
$$
= 2\sqrt2 \pi  (\sqrt{R^2+ R^2 } - R)
$$
$$
= 2\sqrt2 \pi R (\sqrt{2} - 1)
$$
$$
 = 2 \pi R (2 - \sqrt2).
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With
$\ds{\Sigma \equiv \braces{\pars{x,y,z} \mid
x^{2} + y^{2} + z^{2} = R^{2}\,,\ z > 0\,,\ R > 0}}$ and
spherical coordinates $\ds{\pars{R,\theta,\phi}}$:
\begin{align}
&\bbox[10px,#ffd]{\iint_{\Sigma}{\dd\sigma \over
\root{x^{2} + y^{2} + \pars{z + R}^{2}}}} =
\int_{0}^{2\pi}\int_{0}^{\pi/2}
{R^{2}\sin\pars{\theta}\,\dd\theta\,\dd\phi \over
\root{2R^{2} + 2R^{2}\cos\pars{\theta}}}
\\[5mm] = &\
\root{2}\pi R\int_{0}^{\pi/2}{\sin\pars{\theta}\,\dd\theta \over
\root{1 + \cos\pars{\theta}}} =
\root{2}\pi R
\bracks{\vphantom{\Large A}-2\root{1 + \cos\pars{\theta}}}_{\ \theta\ =\ 0}^{\ \theta\ =\ \pi/2}
\\[5mm] = &\
\root{2}\pi R\bracks{-2 -\pars{-2\root{2}}} =
\bbx{2\pi R\pars{2 - \root{2}}}
\end{align}
