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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?

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  • $\begingroup$ 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 $r^2$. I think you made another error when you did the $t$ substitution, but I'm not sure yet. $\endgroup$ – irchans Mar 20 at 15:23
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    $\begingroup$ @zachary If you choose to use polar coordinates here, note that the projection of an elemental surface onto the $x-y$ plane results in the transformation $$d\sigma=\frac{dx\,dy}{\hat z\cdot\left( \frac{\hat xx+\hat yy+\hat zz}{\sqrt{x^2+y^2+z^2}}\right)}=\frac{R\,dx\,dy}{\sqrt{R^2-r^2}}=\frac{Rr\,dr\,d\theta}{\sqrt{R^2-r^2}}$$ $\endgroup$ – Mark Viola Mar 20 at 16:35
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ 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}

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  • $\begingroup$ (+1) This is the easier approach. $\endgroup$ – Mark Viola Mar 20 at 16:19
  • $\begingroup$ @MarkViola That's true. Thanks, Mark. $\endgroup$ – Felix Marin Mar 20 at 16:28
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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: $$\int_0^{2 \pi}d\theta\int_0^R \frac{rR}{\sqrt{(2 R^2+2 R\sqrt{R^2-r^2} ) (R^2-r^2)}}dr$$ $$ = 2 \pi \int_R^0 \frac{R \sqrt{R^2-t^2}}{\sqrt{(2 R^2+2 Rt ) t^2}} \frac{-t}{\sqrt{R^2-t^2}} dt $$ $$ = 2 \pi \int_R^0 \frac{R }{\sqrt{(2 R^2+2 Rt ) t^2}} (-t) dt $$ $$ = 2 \pi \int_0^R \frac{R }{\sqrt{2 R^2+2 Rt }} dt $$ $$ = 2\sqrt2 \pi \int_0^R \frac{R }{2\sqrt{R^2+ Rt }} dt $$ $$ = 2\sqrt2 \pi (\sqrt{R^2+ R^2 } - R) $$ $$ = 2\sqrt2 \pi R (\sqrt{2} - 1) $$ $$ = 2 \pi R (2 - \sqrt2). $$

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  • $\begingroup$ You might consider adding the reason that your first statement is correct. If you choose to use polar coordinates here, note that the projection of an elemental surface onto the $x-y$ plane results in the transformation $$d\sigma=\frac{dx\,dy}{\hat z\cdot\left( \frac{\hat xx+\hat yy+\hat zz}{\sqrt{x^2+y^2+z^2}}\right)}=\frac{R\,dx\,dy}{\sqrt{R^2-r^2}}=\frac{Rr\,dr\,d\theta}{\sqrt{R^2-r^2}}$$ $\endgroup$ – Mark Viola Mar 20 at 16:39
  • $\begingroup$ @MarkViola I assumed that Felix just had a typo. I liked how he computed the determinant of the Jacobian $ \left\|\frac{\partial{\vec{\varphi}}}{\partial r} \times\frac{\partial{\vec{\varphi}}}{\partial \theta}\right\| = \frac{rR}{\sqrt{R^2-r^2}}.$ Glad that you posted the spherical coordinate solution which is cleaner. $\endgroup$ – irchans Mar 20 at 16:51
  • $\begingroup$ Actually, the OP is "Zachary" and "Felix Marin" posted the solution that used spherical coordinates. $\endgroup$ – Mark Viola Mar 20 at 17:02
  • $\begingroup$ Oops! I apologize. $\endgroup$ – irchans Mar 20 at 17:07

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