As far as the integration is concerned R and x are constants. So let \(\displaystyle a^2 = R^2 - x^2\). Then your integral becomesHave tried to solve this one with substitution, but didn't quite get there (according to Wolfram Alpha):
$$ \int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} \left( \sqrt{R^2-x^2-y^2} \right) \, dy $$
Care to help?
-paw
Just watched this video about this integral, is this the way to do it?As far as the integration is concerned R and x are constants. So let \(\displaystyle a^2 = R^2 - x^2\). Then your integral becomes
\(\displaystyle \int_{-a}^a \sqrt{a^2 - y^2}~dy\)
How do you attack this kind of integral in general?
-Dan
Rather than post a link to a video, why not take what you gleaned from the video and post your attempt so that those helping do not have to follow a link, load a video, watch the video, and then guess what you got from it and tried?Just watched this video about this integral, is this the way to do it?
Integration of square root of (a^2- x^2)dx - YouTube
Even more confused.
As \(\displaystyle \arcsin(1) = \frac{\pi}{2}\) I'd say you've got it correct.The integral is the exact same integral topsquark presented to me, and should therefore give the same solution. The derivation was quite extensive and more tricky than I expected it to be.
The result he ends up with is: $ \frac{a^2}{2} \arcsin(\frac{y}{a})+\frac{y}{2} \sqrt{a^2-y^2} $
In my case:
$ \biggr[\frac{a^2}{2} \arcsin(\frac{y}{a})+\frac{y}{2} \sqrt{a^2-y^2} \biggr]_{-a}^{a} = a^2\arcsin(\frac{a}{a})+a \sqrt{a^2-a^2} = a^2\arcsin(1) $
This does not look like the answer I end up with when using Wolfram Alpha.
That was great! Just finished the rest of the integral, and got it rightAs \(\displaystyle \arcsin(1) = \frac{\pi}{2}\) I'd say you've got it correct.
-Dan
I live to serve. :spin:That was great! Just finished the rest of the integral, and got it right
Thank you topsquark.
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