Find the area of the region that lies outside the cardioid $r=1+\cos \theta$ and inside the circle $r=3\cos \theta$.

Solution We begin by graphing each equation. See Figure 54(a). Then we find the points of intersection of the two graphs by solving the equation, $$\begin{eqnarray*} 3\cos \theta & =&1+\cos \theta \\[3pt] 2\cos \theta & =&1 \\[3pt] \cos \theta & =&\frac{1}{2} \end{eqnarray*}$$ $$\begin{equation*} \theta =-\frac{\pi }{3}\;\hbox{or}\;\theta =\frac{\pi }{3} \end{equation*}$$

The graphs intersect at the points $\left( \dfrac{3}{2},-\dfrac{\pi }{3} \right)$ and $\left( \dfrac{3}{2},\dfrac{\pi }{3}\right)$.

Figure 54

The area $A$ of the region that lies outside the cardioid and inside the circle is shown as the shaded portion in Figure 54(b). Notice that the area $A$ is the difference between the area of the region enclosed by the circle $r=3\cos \theta$ swept out by the rays $\theta =-\dfrac{\pi }{3}$ and $\theta =\dfrac{\pi }{3}$, and the area of the region enclosed by the cardioid $r=1+\cos \theta$ swept out by the same rays. So, $$\begin{array}{@rcl} A& =&\int_{-\pi /3}^{\pi /3}\frac{1}{2}(3\cos \theta )^{2}d\theta -\int_{-\pi /3}^{\pi /3}\frac{1}{2}(1+\cos \theta )^{2}d\theta =\frac{1}{2} \int_{-\pi /3}^{\pi /3}[ 9\cos ^{2}\theta -( 1+2\cos \theta +\cos^{2}\theta ) ] \,d\theta \notag \\[12pt] & =&\frac{1}{2}\int_{-\pi /3}^{\pi /3}(8\cos ^{2}\theta -1-2\cos \theta )~d\theta = \frac{1}{2}\int_{-\pi /3}^{\pi /3} \left[ 8\left( \frac{1+\cos ( 2\theta ) }{2}\right) -1-2\cos \theta \right] d\theta \notag \\[12pt] & =&\frac{1}{2}\int_{-\pi /3}^{\pi /3}[ 3+4\cos ( 2\theta ) -2\cos \theta ] ~d\theta =\frac{1}{2}\Big[ 3\theta +2\sin ( 2\theta ) -2\sin \theta \Big] _{-\pi /3}^{\pi /3}=\pi \end{array}$$