Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.4 Area and Arc Length in Polar - Exercises - Page 624: 14

area of the shaded region $ \simeq 0.1024$

1. First, we compute the area of the petal on the right side of the $y$-axis. When $\theta=0$, we have $r=1$. So, the curve starts at $\left( {1,0} \right)$. When $\theta = \frac{\pi }{6}$, we have $r=0$. So, the curve reaches the origin at $\theta = \frac{\pi }{6}$. From Figure 20, we see that the curve completed the upper-half of the petal for the interval $0 \le \theta \le \frac{\pi }{6}$. By symmetry, the area of the petal is twice the area of the upper-half. Using Eq. (2) of Theorem 1, the area of the petal is then area of the petal $ = 2\cdot\frac{1}{2}\cdot\mathop \smallint \limits_0^{\pi /6} {\cos ^2}3\theta {\rm{d}}\theta $ Since ${\cos ^2}3\theta = \frac{1}{2}\left( {1 + \cos 6\theta } \right)$, the integral becomes area of the petal $ = \frac{1}{2}\cdot\mathop \smallint \limits_0^{\pi /6} \left( {1 + \cos 6\theta } \right){\rm{d}}\theta $ area of the petal $ = \frac{1}{2}\left( {\theta + \frac{1}{6}\sin 6\theta } \right)|_0^{\pi /6}$ area of the petal $ = \frac{\pi }{{12}}$. 2. Next, we compute the area of the region inside the petal but outside the circle. Since this region is between two polar curves: $r = \frac{1}{2}$ and $r = \cos 3\theta $, we find the intersection points by solving the equation: $r = \frac{1}{2} = \cos 3\theta $. The solution is $\theta = \pm \frac{\pi }{9}$. We call this region $A$. So, the area of $A$ is area of $A$ $ = \frac{1}{2}\cdot\mathop \smallint \limits_{ - \pi /9}^{\pi /9} \left( {{{\left( {\cos 3\theta } \right)}^2} - {{\left( {\frac{1}{2}} \right)}^2}} \right){\rm{d}}\theta $ area of $A$ $ = \frac{1}{2}\cdot\mathop \smallint \limits_{ - \pi /9}^{\pi /9} \left( {{{\cos }^2}3\theta - \frac{1}{4}} \right){\rm{d}}\theta $ area of $A$ $ = \frac{1}{2}\cdot\mathop \smallint \limits_{ - \pi /9}^{\pi /9} \left( {\frac{1}{2}\left( {1 + \cos 6\theta } \right) - \frac{1}{4}} \right){\rm{d}}\theta $ area of $A$ $ = \frac{1}{2}\left( {\frac{1}{4}\theta + \frac{1}{{12}}\sin 6\theta } \right)|_{ - \pi /9}^{\pi /9}$ area of $A$ $ = \frac{1}{2}\left( {\frac{\pi }{{18}} + \frac{1}{6}\sin \frac{{2\pi }}{3}} \right)$ So, the area of the shaded region is the area of the petal minus the area of $A$: area of the shaded region $ = \frac{\pi }{{12}} - \frac{1}{2}\left( {\frac{\pi }{{18}} + \frac{1}{6}\sin \frac{{2\pi }}{3}} \right) \simeq 0.1024$