<s>[INST] $\iint_{\Sigma} \frac{d\sigma}{\sqrt{x^2+y^2+(z+R)^2}}$ 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? 
 [/INST] 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:
$$\int_0^{2 \pi}d\theta\int_0^R \frac{rR}{\sqrt{(2 R^2+2 R\sqrt{R^2-r^2} ) (R^2-r^2)}}dr$$
$$
= 2 \pi \int_R^0 \frac{R \sqrt{R^2-t^2}}{\sqrt{(2 R^2+2 Rt ) t^2}} \frac{-t}{\sqrt{R^2-t^2}} dt
$$
$$
= 2 \pi \int_R^0 \frac{R }{\sqrt{(2 R^2+2 Rt ) t^2}} (-t) dt
$$
$$
= 2 \pi \int_0^R \frac{R }{\sqrt{2 R^2+2 Rt }} dt
$$
$$
= 2\sqrt2 \pi \int_0^R \frac{R }{2\sqrt{R^2+ Rt }} dt
$$
$$
= 2\sqrt2 \pi  (\sqrt{R^2+ R^2 } - R)
$$
$$
= 2\sqrt2 \pi R (\sqrt{2} - 1)
$$
$$
 = 2 \pi R (2 - \sqrt2).
$$
</s>