September 13th, 2018, 12:04 PM  #11 
Senior Member Joined: Sep 2015 From: USA Posts: 2,264 Thanks: 1198 
I think $2a \to a$ mistakenly somewhere along the way as well.

September 13th, 2018, 02:04 PM  #12  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,641 Thanks: 119  Quote:
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: 2,264 Thanks: 1198 
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: 20,107 Thanks: 1907 
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  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,641 Thanks: 119  Quote:
$\displaystyle I=2\int_{a}^{a}\sqrt{ax}dx= \frac{4}{3}\sqrt{(ax)^{3}}\biggr\rvert_{a}^{a} = \frac{8\sqrt{2}}{3}a^{\frac{3}{2}}$  

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double, integral, solve 
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