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Homework Help: Greens theorem on a circle

  1. Nov 26, 2011 #1
    1. The problem statement, all variables and given/known data
    Use greens theorem to solve the closed curve line integral:
    [itex]\oint[/itex](ydx-xdy)

    The curve is a circle with its center at origin with a radius of 1.

    2. Relevant equations
    x^2 + y^2 = 1

    3. The attempt at a solution
    Greens theorem states that:
    Given F=[P,Q]=[y, -x]=yi-xj

    [itex]\oint[/itex]F*dr=[itex]\oint[/itex]Pdx+Qdy=[itex]\int[/itex][itex]\int[/itex]([itex]\frac{dQ}{dx}[/itex]-[itex]\frac{dP}{dy}[/itex])dA

    From the circle equation i find:
    x=[itex]\sqrt{1-y^2}[/itex]
    y=[itex]\sqrt{1-x^2}[/itex]

    Which means that:
    [itex]\frac{dQ}{dx}[/itex]=0 and [itex]\frac{dP}{dy}[/itex]=0

    Obviously, I am doing something wrong... but which rules am I breaking??

    I did find the answer the "normal" way, without greens, which was -2[itex]\pi[/itex].

    I think a lot of my difficulties originates from lack of knowledge about different coordinate systems and the conversion between these.
    One could write:
    x=cos t and y=sin t, but do I have to? (btw, i used that for solving the normal way).
     
  2. jcsd
  3. Nov 26, 2011 #2
    Oh wait... I think I got the vector field F mixed up with the circle that I am integrating over.
    I shouldn't insert for x and y before I have applied greens theorem... I think.
     
  4. Nov 26, 2011 #3
    It's [itex]Q = -x, P= y[/itex], so [itex]\frac{dQ}{dx} = -1[/itex], etc...
     
  5. Nov 26, 2011 #4
    Solved, used polar coordinates on the integral after greens theorem.
     
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