Polar Graphs Math Example 4

Solution

1. 1
   One petal occurs between consecutive zeros. $r = 0$ when $\cos(3\theta) = 0$, i.e., $3\theta = \frac{\pi}{2}$ or $3\theta = -\frac{\pi}{2}$, giving $\theta \in [-\frac{\pi}{6}, \frac{\pi}{6}]$ for the first petal.
2. 2
   Area $= \frac{1}{2}\int_{-\pi/6}^{\pi/6} r^2\,d\theta = \frac{1}{2}\int_{-\pi/6}^{\pi/6} 16\cos^2(3\theta)\,d\theta = 8\int_{-\pi/6}^{\pi/6} \frac{1+\cos(6\theta)}{2}\,d\theta = 4\left[\theta + \frac{\sin(6\theta)}{6}\right]_{-\pi/6}^{\pi/6} = 4\left(\frac{\pi}{3}\right) = \frac{4\pi}{3}$.