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Evaluate this double integral

  1. Nov 9, 2014 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    Evaluate the integral by changing into polar coordinates.
    [itex] \displaystyle \int_0^{4a} \int_{y^2/4a}^y \dfrac{x^2-y^2}{x^2+y^2} dx dy [/itex]


    3. The attempt at a solution
    Substituting x=rcos theta and y=rsin theta , the integrand changes to [itex]cos 2 \theta r dr d \theta [/itex] . I know that the region of integration is the area bounded by parabola y^2=4ax and y=x. However, I don't know the limits of r and theta.
     
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  3. Nov 9, 2014 #2

    Ray Vickson

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    Perhaps you should not use ##x = r \cos \theta, \: y = r \sin \theta##. Why would you not just do the inner x-integration first, then do the y-integration?
     
  4. Nov 9, 2014 #3

    LCKurtz

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    Have you drawn a picture of that region? Have you written the equation of that parabola in polar coordinates? You need to do both. Then think of a point inside the region. It will move from ##r=0## to ##r## on the parabola. Once you draw that you should be able to see what values ##\theta## must vary between.
     
  5. Nov 10, 2014 #4

    utkarshakash

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    Because I'm required to solve it by changing into polar coordinates.
     
  6. Nov 10, 2014 #5

    utkarshakash

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    The polar equation of given parabola will be ##r=-2a/(1+cos \theta)##. The limit of theta is pi/4 to pi/2. Integrating the expression wrt r i'm left with ##\int_{\pi/4}^{\pi/2} \dfrac{-2a cos 2 \theta}{1+cos \theta} d \theta ##
     
  7. Nov 10, 2014 #6

    LCKurtz

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    It looks like you have figured out the limits now, but I don't get the same equation for the parabola. Surely ##r## shouldn't be negative. And let's see the double integral after you have put it in polar coordinates and before you integrate it.
     
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