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

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

2) Now this one I'm sort of stuck on:EDIT: Oops! Corrected sign.

$\displaystyle \oint_C (x-y)dx {\color{red}+} (x+2y)dy$ where C is closed curve $\displaystyle \frac{x^2}{4}+y^2=1$ in an anticlockwise direction.

So that equals:

$\displaystyle 2\iint_D dA$ 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:

$\displaystyle x=2\cos\theta, ~y=\sin\theta$

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

So then I did:

$\displaystyle 2\int_0^{2\pi}\int_0^{\sqrt{4\cos^2\theta+sin^2\th eta}} r dr d\theta$

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

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