1. ## Greens theorem

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

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 signi. 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. 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 ;D

Bye there.

2. ## Re: Greens theorem

Originally Posted by Ulysses
Creo que esto esta bien, la integral la resolví con la computadora para no tener problemas
This language sounds familiar to me.

And then the double integral: $\displaystyle\int_{-1}^{1}\int_{-\sqrt[ ]{1-x^2}}^{\sqrt[ ]{1-x^2}}(1-2y)dydx=\pi$
The left hand side is right so there must be a computation mistake. At any case you can find the double integral in the simple way:

$\iint_{D}(Q_x-P_y)\;dxdy=-\iint_{D}\;dxdy=-\textrm{Area\;}(D)=-\pi\cdot 1^2=-\pi$