Green's Theorem and polar coordinates

[sailsinthesun]

Homework Statement

Using Green's Theorem, (Integral over C) -y^2 dx + x^2 dy=____________
with C: x=cos t y=sin t (t from 0-->2pi)

Homework Equations

(Integral over C) Pdx + Qdy=(Double integral over D) ((partial of Q w.r.t. x)-(partial of P w.r.t. y))dxdy

The Attempt at a Solution

I'm stuck from here. I remember the professor said to use polar coordinates which makes sense to get the limits on D, but how do I convert the integral (-y^2 dx + x^2 dy) to polar?

In my method I go from the original integral over C to (double integral over D) 2x+2y dxdy. I convert this to polar to get limits on D and I get (integral from 0 to 2pi)(integral from 0 to 1) 2rcos(theta)+2rsin(theta)*r*dr*d(theta). Once you calculate all of this you get 0 which I don't believe is correct. Any help?

 

[LCKurtz]

Why don't you think 0 is correct? It's quite a nice number, one of my favorites, actually.

 

[Mark44]

People sometimes want answers in round numbers. You can't get any rounder than zero.