= Solution
On the loop, $y=tx$, so $dy=t\,dx+x\,dt$ and $x\,dy-y\,dx=x^2\,dt$. The traversal is anticlockwise. The <area> formula from <Green's theorem> becomes
$$
A_{\rm loop}=\frac12\int_0^\infty x(t)^2\,dt
=\frac{9a^2}{2}\int_0^\infty\frac{t^2}{(1+t^3)^2}\,dt.
$$
Set $u=1+t^3$ to obtain
$$
\boxed{A_{\rm loop}=\frac{3a^2}{2}\int_1^\infty u^{-2}\,du
=\frac{3a^2}{2}}.
$$
Although the parameter interval is unbounded, the return to the origin closes the loop and the integral converges.
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