How can I integrate this messy function?

**Riazy** · March 9th 2011, 12:48 AM

$How can I integrate this messy function?-mathtex.gif$

I am talking about the this function
can someone help me evaluate it?

**mr fantastic** · March 9th 2011, 01:36 AM

Originally Posted by Riazy

$Click image for larger version. Name: mathtex.gif Views: 19 Size: 2.8 KB ID: 21089$

I am talking about the this function
can someone help me evaluate it?

The square root part simplifies to 2R.

**Riazy** · March 9th 2011, 01:42 AM

I don't really understand , how, could you show me how its done?

thanks

**tonio** · March 9th 2011, 01:48 AM

Originally Posted by Riazy

$Click image for larger version. Name: mathtex.gif Views: 19 Size: 2.8 KB ID: 21089$

I am talking about the this function
can someone help me evaluate it?

Hmmm...either there's a mistake or this is trivial:

$\displaystyle{\left(2R\cos \phi\right)^2+\left(-2R\sin\phi\right)^2=\left(4R^2(\cos^2\phi+\sin^2\p hi\right))=4R^2}$ , and then

$\displaystyle{\sqrt{\left(2R\cos \phi\right)^2+\left(-2R\sin\phi\right)^2}=2R}$ , so we have

$\displaystyle{2\pi\int\limits^{\pi/2}_0 2R\cos \phi\sin\phi\sqrt{\left(2R\cos \phi\right)^2+\left(-2R\sin\phi\right)^2}\,d\phi=4R^2\cdot 2\pi\int\limits^{\pi/2}_0\cos\phi\sin\phi\,d\phi}$

$\displastyle{=8R^2\pi\cdot \frac{1}{2}\left[\sin^2\phi\left]^{\pi/2}_0=4R^2\pi}$

Tonio

**mr fantastic** · March 9th 2011, 01:49 AM

Originally Posted by Riazy

I don't really understand , how, could you show me how its done?

thanks

If you're attempting an integral like this you must surely have been taught the Pythagorean Identity ....

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