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.

$r=2+\cos \theta$

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 $\theta =\pi$. $$\begin{eqnarray*} A& =&2\int_{0}^{\pi }\dfrac{1}{2}r^{2}d\theta =\int_{0}^{\pi }(2+\cos \theta )^{2}\,d\theta =\int_{0}^{\pi }(4+4\cos \theta +\cos ^{2}\theta )\,d\theta \notag \\[4pt] &=& \int_{0}^{\pi }\left[ 4+4\cos \theta +\dfrac{1+\cos ( 2\theta ) }{2}\right] d\theta \;{\color{#0066A7}{\cos ^{2}\theta =\dfrac{1+\cos ( 2\theta ) }{2}}} \\[4pt]\\[4pt] & =&\left[ 4\theta +4\sin \theta +\frac{\theta }{2}+\frac{1}{4}\sin ( 2\theta ) \right] _{0}^{\pi }=\frac{9\pi }{2} \end{eqnarray*}$$