 Solution

The double-angle identity eliminates the parameter: $y=1+\cos 2\theta =2{\cos }^2\theta =2x^2$. As $\theta$ increases from zero to $\pi$, $x$ decreases from one to minus one. The positive area below this parabola is thereforeEquivalently, the parameter integral is $-{\int }_0^{\pi }(1+\cos 2\theta )(-\sin \theta )d\theta$. The leading minus sign corrects for traversal from right to left; the signed integral $\int ydx$ along that traversal would be negative.