Find the area enclosed by the graph of $r=2\cos ( 3\theta )$, a rose with three petals.

Solution Figure 52 shows the rose. The area of the blue shaded region in quadrant I equals one-sixth of the area $A$ enclosed by the graph.* The shaded region in quadrant I is swept out beginning with the ray $\theta =0.$ It ends at the point $( 0,\theta ) ,$ $0\lt\theta \lt\dfrac{\pi }{2},$ where $\theta$ is the solution of the equation. $$\begin{eqnarray*} 2\cos ( 3\theta ) &=&0\qquad 0\lt\theta \lt\dfrac{\pi }{2} \\[4pt] \cos ( 3\theta ) &=&0 \\[4pt] 3\theta &=&\dfrac{\pi }{2}+2k\pi \\[4pt] \theta &=&\dfrac{\pi }{6}+\dfrac{2k\pi }{3} \end{eqnarray*}$$

* We need to exploit symmetry here since there are intervals on which $r\lt0,$ and the area formula requires that $r>0.$

$r=2\cos ( 3\theta )$

Since $0\lt\theta \lt\dfrac{\pi }{2},$ we have $\theta =\dfrac{\pi }{6}.$ The area of the shaded region in quadrant I swept out by the rays $\theta =0$ and $\theta =\dfrac{\pi }{6}$ is given by $\int_{0}^{\pi /6}\dfrac{1}{2} r^{2}d\theta ,$ and the area $A$ of the region we seek is $6$ times this area. $$\begin{eqnarray*} A &=& 6\int_{0}^{\pi /6}\dfrac{1}{2}r^{2}\,d\theta =3\int_{0}^{\pi /6} 4\cos ^{2}( 3\theta ) ~d\theta =12\int_{0}^{\pi /6}\cos ^{2}( 3\,\theta )~d\theta\\[4pt] &=& 12\int_{0}^{\pi /6}\frac{1+\cos ( 6\theta ) }{2}~d\theta\qquad {\color{#0066A7}{\hbox{\(\cos ^{2}( 3\theta ) =\dfrac{1+\cos ( 6\theta ) }{2}\)}}}\\\\[4pt] &=&6 \left[ \theta +\dfrac{1}{6}\sin ( 6\theta ) \right] _{0}^{\pi /6}=6\left( \frac{\pi }{6}\right) =\pi \end{eqnarray*}$$