Green's Theorem; computing a double integral This is the last part of an exercise in Apostol Vol. II.  (p.385, 1 (e), to be precise.)  No doubt there's a trick I'm missing, because evaluating the double integral over the region involved seems unduly complicated.
We are supposed to use Green's Theorem to evaluate the line integral $\oint_CPdx + Qdy$ (where $C$ is traversed counterclockwise), with $P = y^2$, $Q = x$, and $C$ described by the parametric equations $x = 2\cos^3t$, $y = 2\sin^3t$ where $t$ ranges from $0$ to $2\pi$.
We have $\frac{\partial Q}{\partial x} = 1$, $\frac{\partial P}{\partial y} = 2y$.  So what we want to evaluate is  
\begin{align}
\iint\limits_R (1 - 2y)dydx
\end{align}
where $R$ is the interior of the the region bounded by $C$.  Trying to evaluate by iterated integration gives us
\begin{align}
\int_{-2}^2\left[\int_{-2\cos^3\left[\arcsin\left(\sqrt[3]{\frac{x}{2}}\right)\right]}^{2\cos^3\left[\arcsin\left(\sqrt[3]{\frac{x}{2}}\right)\right]}(1-2y)dy\right]dx
& = \int_{-2}^24\cos^3\left[\arcsin\left(\sqrt[3]{\frac{x}{2}}\right)\right]dx
\end{align}
and, I mean, give me a break.  There's got to be a better way, right?
 A: The curve $C$ is known as astroid. This is what it looks like

The identity $\cos^2t+\sin^2t=1$ gives us after a bit of tinkering that in $xy$-coordinates it has the equation
$$
x^{2/3}+y^{2/3}=2^{2/3}.
$$
This means that the region is bounded by $-2\le x\le2$, $-(2^{2/3}-x^{2/3})^{3/2}\le y\le(2^{2/3}-x^{2/3})^{3/2}$.
Thus the inner integral is
$$
\int_{y=-(2^{2/3}-x^{2/3})^{3/2}}^{(2^{2/3}-x^{2/3})^{3/2}}(1-2y)\,dy
=\mathop{\Bigg/}\nolimits_{\hspace{-2mm}y=-(2^{2/3}-x^{2/3})^{3/2}}^{\hspace{1mm}(2^{2/3}-x^{2/3})^{3/2}}(y-y^2)=2(2^{2/3}-x^{2/3})^{3/2},
$$
as the substitutions into $y^2$ cancel.
The function and this region are both symmetric w.r.t. the $y$-axis, so the answer is
$$
\iint_R(1-2y)\,dy\,dx=4\int_{x=0}^2(2^{2/3}-x^{2/3})^{3/2}\,dx.
$$
Here the substitution $x=2\cos^3t$ stands out. Then you get
$$
(2^{2/3}-x^{2/3})^{3/2}=(2^{2/3}(1-\cos^2t))^{3/2}=2\sin^3t
$$
and 
$$
dx=-6\cos^2t\sin t.
$$
This gives a standard trig integral that I'm sure you can manage. I got $3\pi/2$.
A: Hint:
$$\begin{align}
\iint_{R}\left(1-2y\right)\,\mathrm{d}A
&=\iint_{R}\mathrm{d}A-2\iint_{R}y\,\mathrm{d}A\\
&=\operatorname{area}{\left(R\right)}-2\,S_{y}{\left(R\right)}\\
&=\operatorname{area}{\left(R\right)}-2\,\bar{y}\cdot\operatorname{area}{\left(R\right)},\\
\end{align}$$
where $S_{y}{\left(R\right)}$ is the first moment of area of the region $R$ in the $y$-direction, and where $\bar{y}$ is the $y$-coordinate of the centroid of $R$. By symmetry, the coordinates of the centroid of $R$ are simply $(\bar{x},\bar{y})=(0,0)$, which should be immediately obvious from a plot of the region. Thus the value of the sought integral reduces to just the area of the region.
