= Solution
For $n=(a,b,c)$ and $r=(x,y,z)$, the integrand is $\frac12(n\times r)\cdot dr$. Since
$$
\nabla\times\left[\frac12(n\times r)\right]=n,
$$
the <Stokes theorem> gives the signed planar area, and its absolute value gives the ordinary area.
The $xy$ projection of the given curve is the astroid $x=2\cos^3t$, $y=\sin^3t$, whose area is
$$
\frac12\oint(x\,dy-y\,dx)=\frac{3\pi}{4}.
$$
The plane $z=\sqrt3y$ stretches area from its $xy$ projection by $\sqrt{1+3}=2$. Hence
$$
\boxed{\operatorname{Area}=\frac{3\pi}{2}}.
$$
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