= Solution
The double-angle identity eliminates the parameter: $y=1+\cos2\theta=2\cos^2\theta=2x^2$. As $\theta$ increases from zero to $\pi$, $x$ decreases from one to minus one. The positive <area> below this <parabola> is therefore
$$
\boxed{\mathcal A=\int_{-1}^1 2x^2\,dx=\frac43.}
$$
Equivalently, the parameter integral is $-\int_0^\pi(1+\cos2\theta)(-\sin\theta)d\theta$. The leading minus sign corrects for traversal from right to left; the signed integral $\int y\,dx$ along that traversal would be negative.
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