= Solution
{c}
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 $\frac12\int(r_2^2-r_1^2)\,d\theta$.
For $-\pi/2\le\theta\le\pi/2$, the <cardioid> lies outside the <unit circle>. The right-hand region therefore has <area>
$$
A_R=\frac12\int_{-\pi/2}^{\pi/2}\left((1+\cos\theta)^2-1\right)\,d\theta
=\int_{-\pi/2}^{\pi/2}\cos\theta\,d\theta+\frac12\int_{-\pi/2}^{\pi/2}\cos^2\theta\,d\theta
=2+\frac\pi4.
$$
On $\pi/2\le\theta\le\pi$, the <unit circle> lies outside the <cardioid>. The upper left region has <area>
$$
A_U=\frac12\int_{\pi/2}^{\pi}\left(1-(1+\cos\theta)^2\right)\,d\theta
=-\int_{\pi/2}^{\pi}\cos\theta\,d\theta-\frac12\int_{\pi/2}^{\pi}\cos^2\theta\,d\theta
=1-\frac\pi8.
$$
By <reflection symmetry> the lower left region has the same <area>. Thus
$$
\boxed{A_R=2+\frac\pi4,\qquad A_U=A_L=1-\frac\pi8.}
$$
The 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 $A_R-A_U-A_L=\pi/2$.
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