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 September 13th, 2018, 12:04 PM #11 Senior Member     Joined: Sep 2015 From: USA Posts: 2,174 Thanks: 1143 I think $2a \to a$ mistakenly somewhere along the way as well.
September 13th, 2018, 02:04 PM   #12
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Quote:
 Originally Posted by romsek I think $2a \to a$ mistakenly somewhere along the way as well.

Quote:
 Originally Posted by zylo Take new coordinate system at center of circle. x'=x+a y'=y-a Drop primes and problem in new coordinate system becomes: $\displaystyle I=\int \int \frac{1}{\sqrt{a+x}}dxdy$ over disc $\displaystyle x^{2}+y^{2}=a^{2}$ $\displaystyle I=\int \frac{1}{\sqrt{a+x}}\int dydx$ $\displaystyle \int_{-\sqrt{a^{2}-x^{2}}}^{\sqrt{a^{2}-x^{2}}}dy=2\sqrt{a^{2}-x^{2}}$ $\displaystyle I=2\int_{-a}^{a}\frac{\sqrt{a^{2}-x^{2}}}{\sqrt{a+x}}dx=2\int_{-a}^{a}\sqrt{a-x}dx= -\frac{4}{3}\sqrt{(a-x)^{2}}{|^{a}}_{a}=\frac{8}{3}a$ EDIT: Acknowledgement. Equations typed with: https://www.codecogs.com/latex/eqneditor.php

Look at the beginning.
With new axes at the center of the circle:
x'=x+a
$\displaystyle \sqrt{2a+x} = \sqrt{2a+(x'-a))}= \sqrt{a+x'}$

Then I did calculation in the new coordinate system without primes because it would have been a pain to carry them through.

In above quote I corrected the minus sign at the end.

 September 13th, 2018, 02:48 PM #13 Senior Member     Joined: Sep 2015 From: USA Posts: 2,174 Thanks: 1143 afk in a pool of blood... don't call 911.... just let me rot here
 September 13th, 2018, 05:12 PM #14 Global Moderator   Joined: Dec 2006 Posts: 19,887 Thanks: 1836 Although romsek's method may seem to work with the aid of Mathematica, doing the integral in polar coordinates by hand is cumbersome and tricky. Special functions aren't needed. It's better to stick with Cartesian coordinates. Eventually, zylo got quite close to a correct answer, but is still not quite there.
September 14th, 2018, 11:48 AM   #15
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Quote:
 Originally Posted by zylo Take new coordinate system at center of circle. x'=x+a y'=y-a Drop primes and problem in new coordinate system becomes: $\displaystyle I=\int \int \frac{1}{\sqrt{a+x}}dxdy$ over disc $\displaystyle x^{2}+y^{2}=a^{2}$ $\displaystyle I=\int \frac{1}{\sqrt{a+x}}\int dydx$ $\displaystyle \int_{-\sqrt{a^{2}-x^{2}}}^{\sqrt{a^{2}-x^{2}}}dy=2\sqrt{a^{2}-x^{2}}$ $\displaystyle I=2\int_{-a}^{a}\frac{\sqrt{a^{2}-x^{2}}}{\sqrt{a+x}}dx=2\int_{-a}^{a}\sqrt{a-x}dx= -\frac{4}{3}\sqrt{(a-x)^{2}}{|^{a}}_{a}=\frac{8}{3}a$ EDIT: Acknowledgement. Equations typed with: https://www.codecogs.com/latex/eqneditor.php
Copied formula for integral in last line wrong. Corrected:

$\displaystyle I=2\int_{-a}^{a}\sqrt{a-x}dx= -\frac{4}{3}\sqrt{(a-x)^{3}}\biggr\rvert_{-a}^{a} = \frac{8\sqrt{2}}{3}a^{\frac{3}{2}}$

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