# Greens theorem

1. Jul 30, 2011

### Telemachus

Hi. I have a problem with this exercise. I wanted to verify the greens theorem for the vector field $$F(x,y)=(3x+2y,x-y)$$ over the path $$\lambda[0,2\pi]\rightarrow{\mathbb{R}^2},\lambda(t)=(\cos t, \sin t)$$

The Green theorem says: $$\displaystyle\int_{C^+}Pdx+Qdy=\displaystyle\int_{}\int_{D}\left (\frac{{\partial Q}}{{\partial x}}-\frac{{\partial P}}{{\partial y}}\right ) dxdy$$
and I have: $$P(x,y)=x-y,Q(x,y)=3x+2y$$

So then I've made the line integral:
$$\displaystyle\int_{0}^{2\pi}\left [-(3\cos t +2 \sin t)\sin t+ (cos t -\sin t)\cos t\right ]dt=-\pi$$
Creo que esto esta bien, la integral la resolví con la computadora para no tener problemas :p

And then the double integral:
$$\displaystyle\int_{-1}^{1}\int_{-\sqrt[ ]{1-x^2}}^{\sqrt[ ]{1-x^2}}(1-2y)dydx=\pi$$

The problem is clearly with the sign. The mistake I think I've committed was putting on the reverse the integral limits for the double integral. I think that x should go from 1 to -1, but the thing is I don't know why. So I'm not pretty sure on how to determine the integral limits on this cases. I've tried to think about the parametrization, but I don't know what to do so. Its clear to me that the parametrization plays an important role in the sign of the integral. But I don't know how to reason this, so I wanted some help and suggestions. I'm not pretty sure if green theorem only reefers to positive oriented paths, if it isn't its clear to me that the parametrization, which determines the orientation of the path plays a determinant role over the limits of integration, because inverting the limits I get the opposite sign for the integral, right?

Bye there.

Last edited: Jul 30, 2011
2. Jul 30, 2011

### dyem

You did something wrong: P=3x+2y, Q=x-y.
With that, if you do the calculus the answer is right.

3. Jul 31, 2011

### Telemachus

Thanks, I've found the mistake :D