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Math Help - How can I integrate this messy function?

  1. #1
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    How can I integrate this messy function?

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

    I am talking about the this function
    can someone help me evaluate it?
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  2. #2
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    Quote Originally Posted by Riazy View Post
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    I am talking about the this function
    can someone help me evaluate it?
    The square root part simplifies to 2R.
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    I don't really understand , how, could you show me how its done?

    thanks
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    Quote Originally Posted by Riazy View Post
    Click image for larger version. 

Name:	mathtex.gif 
Views:	19 
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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
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  5. #5
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    Quote Originally Posted by Riazy View Post
    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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