Does Green's Theorem Apply to Clockwise Oriented Curves?

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  • Thread starter Thread starter linearfish
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    Green's theorem Theorem
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Discussion Overview

The discussion revolves around the application of Green's Theorem to curves oriented clockwise versus counterclockwise. Participants explore the implications of curve orientation on the theorem's formulation and its integrals.

Discussion Character

  • Technical explanation

Main Points Raised

  • One participant notes that Green's Theorem is typically stated for positively oriented (counterclockwise) curves and questions if a clockwise orientation simply results in a negative version of the theorem.
  • Another participant asserts that Green's formula accounts for the direction of the curve, indicating that the integral over a clockwise curve is the negative of that over a counterclockwise curve.
  • A later reply confirms that the explanation provided was helpful, suggesting that the points raised were understood.

Areas of Agreement / Disagreement

Participants appear to agree on the relationship between clockwise and counterclockwise orientations in the context of Green's Theorem, but the discussion does not resolve any deeper implications or nuances regarding the theorem's application.

Contextual Notes

The discussion does not address potential limitations or assumptions regarding the conditions under which Green's Theorem applies, nor does it explore any specific examples or counterexamples.

linearfish
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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?

\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}
 
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Hi :smile:
Greens formula tells whether you go clockwise or opposite direction on curve.
\oint_{K} \left(P(x,y)dx+Q(x,y)dy \right)=- \oint_{-K} \left(P(x,y)dx + Q(x,y)dy\right)
<br /> \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<br />

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

I hope I helped you :smile:

MrSnoopy
 
Thanks, that does help.
 
No problem :wink:
 

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