# Math Help - Green's thm.

1. ## Green's thm.

1) $\oint_C x^3dx-y^3dy$, where C is the circle $x^2+y^2=1$ in an anticlockwise direction.

I got 0. Is that right? (Since $Q_x-P_y = 0$.)

2) Now this one I'm sort of stuck on: EDIT: Oops! Corrected sign.
$\oint_C (x-y)dx {\color{red}+} (x+2y)dy$ where C is closed curve $\frac{x^2}{4}+y^2=1$ in an anticlockwise direction.

So that equals:
$2\iint_D dA$ where D is the region bounded by C (the ellipse with radii 2 and 1).

Now the problem here is computation. I opted to go to polar coords (maybe that was a bad idea?), so I got:
$x=2\cos\theta, ~y=\sin\theta$

$\therefore r=\sqrt{4\cos^2\theta+sin^2\theta}$. So far so good?

So then I did:
$2\int_0^{2\pi}\int_0^{\sqrt{4\cos^2\theta+sin^2\th eta}} r dr d\theta$

$=\int_0^{2\pi} (4\cos^2\theta+sin^2\theta)d\theta$

Now.. do I really have to integrate that? Or is there a simpler way?

2. 1) Looks correct.

2) First of all, $Q_x - P_y=0$, so this one is zero as well.

As for your parametrization and the limits:
$x=2r\cos\theta,\ y=r\sin\theta$, $0 \leq r \leq 1$ and the Jacobian is $2r$

$\int_0^{2\pi}\int_0^1 (Q_x-P_y)\cdot 2r \cdot drd\theta$

4. So the answer is $4\pi$ for the second one then?