 Solution

The cardioid meets the unit circle where $1+\cos \theta =1$, namely at $\theta =\pi /2,3\pi /2$. The polar area formula gives the area between radii $r_1\le r_2$ as $\frac{1}{2}\int (r_2^2-r_1^2)d\theta$.

For $-\pi /2\le \theta \le \pi /2$, the cardioid lies outside the unit circle. The right-hand region therefore has areaOn $\pi /2\le \theta \le \pi$, the unit circle lies outside the cardioid. The upper left region has areaBy reflection symmetry the lower left region has the same area. ThusThe two left regions touch only at the cusp; their interiors are disconnected. As a check, the difference between the cardioid's total area $3\pi /2$ and the unit circle's area $\pi$ is $A_R-A_U-A_L=\pi /2$.