# Quick Green's theorem question

1. Aug 8, 2008

### 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?

$$\oint Pdx + Qdy = - \int\int \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \int\int \frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x}$$

2. Aug 10, 2008

### MrSnoopy

Hi
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

MrSnoopy

3. Aug 11, 2008

### linearfish

Thanks, that does help.

4. Aug 11, 2008

No problem