Let $C$ be the curve parametrized by the equation $r(t) = 2\cos^3(t) i + 2\sin^3(t) j$ for $t \in [0,2\pi]$. I want to find the line integral
$$ \oint_C y^2 \,dx + x \,dy . $$
I evaluated it directly and the answer appears to be $\frac{3\pi}{2}$. But I'm supposed to use Green's Theorem and I don't know how to set up the integral. If the region enclosed by $C$ is $D$, then the integral is
$$ \iint\limits_D 1 + 2y \, dy\, dx $$
But I have no idea how to find the region $D$. I tried to set $x = 2 \cos^3(t)$, which implies $ y \in [-k,k] $ for
$$ k = 2(1- (\frac{x}{2})^{2/3})^{3/2} $$
So the integral would become
$$ \int_{-2}^2 \int_{-2(1- (\frac{x}{2})^{2/3})^{3/2}}^{2(1- (\frac{x}{2})^{2/3})^{3/2}} 1 - 2y\, dy\, dx .$$
But this is too messy to evaluate by hand.
So, how do I evaluate the line integral using Green's Theorem?
Note that by symmetry $$I:=\int_{-2}^2 \int_{-2(1- (\frac{x}{2})^{2/3})^{3/2}}^{2(1- (\frac{x}{2})^{2/3})^{3/2}} (1 - 2y)\ dy dx=4\int_{0}^2 \int_{0}^{2(1- (x/2)^{2/3})^{3/2}} 1dy dx\\ =8\int_{0}^2 (1- (x/2)^{2/3})^{3/2} dx=48\int_{0}^{\pi/2} \cos^4(t) \sin^2(t) dt\\ =48\int_{0}^{\pi/2} \cos^4(t)dt-48\int_{0}^{\pi/2} \cos^6(t)dt$$ where we let $x=(2\sin(t))^3$. Now, for $n\geq 1$, by integration by parts, $$\int_{0}^{\pi/2} \cos^{2n}(t)dt=\frac{2n-1}{2n}\int_{0}^{\pi/2} \cos^{2n-2}(t)dt$$ which implies that $I=3\pi/2$.