= Solution
<Stokes theorem> states that $\oint_C A\cdot dr=\iint_S(\nabla\times A)\cdot n\,dS$, with consistent boundary orientation. Here $\nabla\times A=e_z$, so the circulation around the positively oriented planar curve equals its area. Substituting the given parametrization gives
$$
\frac12(x\dot y-y\dot x)
=\frac12[\sin2t\cos t-2\sin t\cos2t]=\sin^3t.
$$
The curve closes at the origin, passes along the right side to $(0,1)$ and returns along the left side, so it is anticlockwise. Consequently
$$
\boxed{\operatorname{Area}=\int_0^\pi\sin^3t\,dt=\frac43.}
$$
Equivalently its two sides are $x=\pm2y\sqrt{1-y^2}$, $0\le y\le1$, whose horizontal width integrates to the same result.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2017-oriented-area.png]
{title=The enclosed lobe and its positive boundary orientation}
{description=Original analytic plot. Arrows follow increasing parameter from the origin along the right side and back along the left; horizontal widths provide an independent area integral.}
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