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?
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?
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?
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.
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.