6. The graph of $r=2+\cos 2 \theta$ and its reflection over the line $y=x$ bound five regions in the plane. Find the area of the region containing the origin.
Solution:
$9 \pi / 2-8$
The original graph is closer to the origin than its reflection for $\theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right) \cup\left(\frac{5 \pi}{4}, \frac{7 \pi}{4}\right)$, and the region is symmetric about the origin. Therefore the area we wish to find is the polar integral
$$
\begin{aligned}
4 \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{1}{2}(2+\cos 2 \theta)^{2} d \theta & =2 \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}}\left(4+4 \cos 2 \theta+\cos ^{2} 2 \theta\right) d \theta \\
& =2 \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}}\left(4+4 \cos 2 \theta+\frac{1}{2}(1+\cos 4 \theta)\right) d \theta \\
& =\left[9 \theta+4 \sin 2 \theta+\frac{1}{4} \sin 4 \theta\right]_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \\
& =\left(\frac{27 \pi}{4}-4\right)-\left(\frac{9 \pi}{4}+4\right)=\frac{9 \pi}{2}-8 .
\end{aligned}
$$