Finding the Area Enclosed by a Limaçon

Find the area of the region enclosed by the limaçon \(r=2+\cos \theta \).

Solution Figure 53 shows the graph of \(r=2+\cos \theta\), a limaçon without an inner loop.

We see that the region above the polar axis equals the region below it, so the area \(A\) of the region enclosed by the limaçon equals twice the area of the region enclosed by \(r=2+\cos \theta \) and swept out by the rays \(\theta =0\) and \(\theta =\pi \). \[ \begin{eqnarray*} A& =&2\int_{0}^{\pi }\dfrac{1}{2}r^{2}d\theta =\int_{0}^{\pi }(2+\cos \theta )^{2}\,d\theta =\int_{0}^{\pi }(4+4\cos \theta +\cos ^{2}\theta )\,d\theta \notag \\[4pt] &=& \int_{0}^{\pi }\left[ 4+4\cos \theta +\dfrac{1+\cos ( 2\theta ) }{2}\right] d\theta \;{\color{#0066A7}{\cos ^{2}\theta =\dfrac{1+\cos ( 2\theta ) }{2}}} \\[4pt]\\[4pt] & =&\left[ 4\theta +4\sin \theta +\frac{\theta }{2}+\frac{1}{4}\sin ( 2\theta ) \right] _{0}^{\pi }=\frac{9\pi }{2} \end{eqnarray*} \]