= Solution
Because $dx/dt>0$ along the arch, its area above the baseline is the <definite integral> $\int y\,dx$. Changing to the parameter gives
$$
\begin{aligned}
\mathcal A
&=a^2\int_0^{2\pi}(1-\cos t)^2dt\\
&=a^2\left[\int_0^{2\pi}1\,dt-2\int_0^{2\pi}\cos t\,dt
+\int_0^{2\pi}\cos^2t\,dt\right]\\
&=a^2(2\pi+\pi)=\boxed{3\pi a^2}.
\end{aligned}
$$
Adding or removing the two endpoint points does not change the area. This is the area formula in <area and length of a cycloid arch>.
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