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Quick Green's theorem question

  1. Aug 8, 2008 #1
    As far as I know, Green's Theorem is normally stated for positively oriented curves (counterclockwise). If a curve is oriented clockwise, is it just the negative version?

    [tex]\oint Pdx + Qdy = - \int\int \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \int\int \frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x}[/tex]
     
  2. jcsd
  3. Aug 10, 2008 #2
    Hi :smile:
    Greens formula tells whether you go clockwise or opposite direction on curve.
    [tex]\oint_{K} \left(P(x,y)dx+Q(x,y)dy \right)=- \oint_{-K} \left(P(x,y)dx + Q(x,y)dy\right)[/tex]
    [tex]
    \iint_{S}\left[\frac{\partial Q(x,y)}{\partial x} - \frac{\partial P(x,y)}{\partial y} \left]\;dxdy = (-)\iint_{S}\left[\frac{-\partial Q(x,y)}{\partial x} + \frac{\partial P(x,y)}{\partial y} \left]\;dxdy
    [/tex]

    The conclusion [tex]\oint_{K} f(x)\;dx= - \oint_{-K} f(x)\;dx[/tex] and yes it is the same.

    I hope I helped you :smile:

    MrSnoopy
     
  4. Aug 11, 2008 #3
    Thanks, that does help.
     
  5. Aug 11, 2008 #4
    No problem :wink:
     
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