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Green's Theorem over a circle

  1. Feb 10, 2013 #1
    1. The problem statement, all variables and given/known data
    Solve: ∫(-ydx+xdy)/(x2+y2) counterclockwise around x2+y2=4


    2. Relevant equations
    Greens Theorem:
    ∫Pdx + Qdy = ∫∫(dQ/dx - dP/dy)dxdy

    3. The attempt at a solution
    Using Greens Theorem variables, I get that:
    P = -y/(x2+y2) and
    Q=x/(x2+y2)

    and thus dQ/dx = (y2-x2)/(y2+x2)2

    and dP/dx = (y2-x2)/(y2+x2)2

    So, ∫∫(dQ/dx - dP/dy)dxdy = ∫∫( (y2-x2)/(y2+x2)2 - (y2-x2)/(y2+x2)2)dxdy

    ... which means I'm integrating 0 (which can't be right as that would equal 0 over a definite integral). Not sure where I've gone wrong! Can anyone spot an error?
     
  2. jcsd
  3. Feb 10, 2013 #2

    LCKurtz

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    What's wrong with zero? You have something against it??
     
  4. Feb 10, 2013 #3
    Not at all! Just doesn't seem like the correct solution in this instance
     
  5. Feb 10, 2013 #4

    LCKurtz

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    Well, looking more closely at your problem, I see that you have a singularity at (0,0). Perhaps you should try evaluating the integral directly.
     
  6. Feb 10, 2013 #5
    I don't understand what you mean by evaluating it directly. My method was to use greens theorem and then (providing it didn't give me 0!) convert to polar coordinates and integrate over r and theta, thus the singularity wouldn't affect anything. Is this wrong?
     
  7. Feb 10, 2013 #6

    LCKurtz

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    Since you are integrating over a region that encloses a singularity, the hypotheses of Green's theorem are not fulfilled. Your function isn't even defined at (0,0). So you can't use Green's theorem. So just work the line integral out. I would suggest using the polar angle ##\theta## as your parameter.
     
  8. Feb 10, 2013 #7
    Right, ok. Thank you very much for your help!

    Sorry to be such a pain but I'm still a little baffled about the zero I'm getting. Just as a side note, suppose I was integrating that function over a region with no singularities (say circle centre (3,0), rad 1) would I still not be able to use Green's Theorem? If I did, I'd still be integrating zero, which still seems wrong in context.
     
  9. Feb 10, 2013 #8

    LCKurtz

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    Yes, you could use Green's theorem in that case. And you would get zero. And there is nothing wrong with getting zero in a line integral. It happens all the time, for example when integrating around a closed loop in a conservative force field (under appropriate hypotheses). I don't know what "context" is bothering you.
     
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