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Line integral in polar coordinates

  1. Sep 22, 2009 #1
    1. The problem statement, all variables and given/known data

    calculate:
    [tex]\oint \frac{2-y}{x^2+(y-2)^2} dx + \frac{x}{x^2+(y-2)^2} dy [/tex]

    where [itex] y = \sin{t} + 2, x = \cos{t}, 0 \leq t \leq \pi[/itex]

    2. Relevant equations

    Green's Theorem.

    3. The attempt at a solution

    In what order should I do everything?
    I need to find the derivaties of P and Q (what comes before the dx and dy), and doing it like this is a bit complicated. So I thought I'll change it to polar, but then I need to take dP/dy but now P is a function of t and not x,y so I can't do that.
    In that case, how do I convert dy to polar differentials? I know that dA or dxdy equals rdrdt, but what about only dy?

    Or another idea I thought of was to make the denominator 1 (because it is - it's cos2t + sin2t, and then take the derivatives but they are different than what I get if I don't do that.

    I don't know which method would be correct, so if someone can help me that'd be great.
     
  2. jcsd
  3. Sep 22, 2009 #2

    arildno

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    Why use green's theorem at all?

    Green's theorem states an IDENTITY between a calculated line integral and a calculated surface integral.

    It is no need to switch to a surface/area integral here, since the indicated line integral is very simple to calculate
    (As a general rule, don't ever bother with Green's theorem, the divergence theorem etc, unless the switch warranted by these theorems simplifies your calculations)
     
  4. Sep 22, 2009 #3
    Ok so if I do the line integral,

    [tex]2-y = -\sin{t}[/tex]

    [tex]x = \cos{t}[/tex]

    [tex]x^2+(y-2)^2 = \cos^2{t} + \sin^2{t} = 1[/tex]

    [tex]\frac{dx}{dt} = -\sin{t} , \frac{dy}{dt} = \cos{t}[/tex]

    The integral becomes:

    [tex]
    \int_{0}^{2\pi} \frac{-\sin{t}(-\sin{t})}{1} dt + \int_{0}^{2\pi} \frac{\cos{t}\cos{t}}{1} dt = \int_{0}^{2\pi} \cos^2{t} + \sin^2{t} dt = \int_{0}^{2\pi} dt = 2\pi[/tex]

    but let's assume I do want to use green's theorem.
    So:

    [tex]P = \frac{2-y}{x^2+(y-2)^2}, Q = \frac{x}{x^2+(y-2)^2}[/tex]

    [tex]\frac{\partial P}{\partial y} = \frac{(y-2)^2-x^2}{(x^2+(y-2)^2)^2}[/tex]
    [tex]\frac{\partial Q}{\partial x} = \frac{(y-2)^2-x^2}{(x^2+(y-2)^2)^2}[/tex]

    Snd since Green's theorem says that it's the double integral of the difference of those 2 derivaties, I get a double integral of 0, not [itex]2\pi[/itex].

    So what went wrong here?
     
  5. Sep 22, 2009 #4

    arildno

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    First off, t only goes to pie, not two-pie!

    Secondly, you cannot use Green's theorem, since the integrand contains a singularity at (0,2)
     
  6. Sep 22, 2009 #5
    oh, then it was a mistake in the first post. it should go to [itex]2\pi[/itex]. sorry.
    and I also understand why green's theorem fails. thanks. :)
     
  7. Sep 22, 2009 #6

    arildno

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    To nit-pick:

    Green's theorem did not fail, it was your application of it that was inappropriate
    :smile:
     
  8. Sep 22, 2009 #7
    So Green should've come with a better theorem ;)
     
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