Quick Green's theorem question

[linearfish]

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}

[MrSnoopy]

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)

\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


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

[linearfish]

Thanks, that does help.

[MrSnoopy]

No problem :wink: