= Solution
{c}
Use <polar coordinates> with $0\leq r\leq\sin^2\theta$ and $0\leq\theta\leq\pi$. The <Jacobian determinant> gives $dA=r\,dr\,d\theta$, so the <area> is
$$
A=\iint_R1\,dA
=\int_0^\pi\int_0^{\sin^2\theta}r\,dr\,d\theta
=\frac12\int_0^\pi\sin^4\theta\,d\theta.
$$
The <trigonometric identity>
$$
\sin^4\theta=\frac{3-4\cos(2\theta)+\cos(4\theta)}8
$$
reduces the <integral> to its constant term: both <cosine> terms have zero <integral> over this interval. Therefore
$$
\boxed{\operatorname{Area}(R)=\frac{3\pi}{16}}.
$$
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