# Greens theorem

• Jul 30th 2011, 02:07 PM
Ulysses
Greens theorem
Hi. I have a problem with this exercise. I wanted to verify the greens theorem for the vector field $\displaystyle F(x,y)=(3x+2y,x-y)$ over the path $\displaystyle \lambda[0,2\pi]\rightarrow{\mathbb{R}^2},\lambda(t)=(\cos t, \sin t)$

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

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

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

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.
• Jul 31st 2011, 12:29 AM
FernandoRevilla
Re: Greens theorem
Quote:

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. :)

Quote:

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

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