Yet another integration

paw

Oct 2014
15
1
Norway
Have 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
 

topsquark

Math Team
May 2013
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The Astral plane
Have 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
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
 
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paw

Oct 2014
15
1
Norway
Jul 2010
12,211
522
St. Augustine, FL., U.S.A.'s oldest city
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.
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? :D
 
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paw

Oct 2014
15
1
Norway
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.
 
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topsquark

Math Team
May 2013
2,532
1,051
The Astral plane
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.
As \(\displaystyle \arcsin(1) = \frac{\pi}{2}\) I'd say you've got it correct. :)

-Dan
 
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paw

Oct 2014
15
1
Norway
As \(\displaystyle \arcsin(1) = \frac{\pi}{2}\) I'd say you've got it correct. :)

-Dan
That was great! Just finished the rest of the integral, and got it right :)

Thank you topsquark.
 

topsquark

Math Team
May 2013
2,532
1,051
The Astral plane
That was great! Just finished the rest of the integral, and got it right :)

Thank you topsquark.
I live to serve. :spin:

-Dan
 
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