Answer
The area of one petal formed by this equation is: $ \frac{\pi}{3}$
Work Step by Step
1. Find the limits of integration:
Using a graphing utility, graph this function, and determine the limits of integration.
As we can see by the graph, one petal starts at $r = 0$, and also finishes at $r = 0$. Find the first 2 angles with that "r" value:
$r = 2cos(3\theta)$
$0 = 2cos(3\theta)$
$0 = cos(3\theta)$
$3\theta = cos^{-1}(0)$
First angle: $3\theta = \pi/2$
$\theta = \frac{\pi}{6}$
Second angle: $3\theta = 3\pi/2$
$\theta = \frac{\pi}{2}$
Therefore, the first 2 angles where that happen are: $\theta = \pi/6, $ and $ \theta = \pi/2$
2. Apply the formula for the area:
$Area = \frac{1}{2} \int_\alpha^\beta(f(\theta)^2)d\theta$
$Area = \frac{1}{2} \int_{\pi/6}^{\pi/2}(2cos(3\theta))^2d\theta$
$Area = \frac{1}{2} \int_{\pi/6}^{\pi/2}(4cos^2(3\theta))d\theta$
$Area = \frac{4}{2} \int_{\pi/6}^{\pi/2}(cos^2(3\theta))d\theta$
** $cos^2\theta = \frac{1+cos(2\theta)}{2}$
** $cos^2(3\theta) = \frac{1+cos(6\theta)}{2}$
$Area = 2 \int_{\pi/6}^{\pi/2}(cos^2(3\theta))d\theta$
$Area = 2 \int_{\pi/6}^{\pi/2}(\frac{1+cos(6\theta)}{2})d\theta$
$Area = \int_{\pi/6}^{\pi/2}({1+cos(6\theta)})d\theta$
$Area = [\theta + \frac{sin(6\theta)}{6}]_{\pi/6}^{\pi/2}$
$Area = (\pi/2 + \frac{sin(6*(\pi/2))}{6}) - (\pi/6 + \frac{sin(6*(\pi/6))}{6})$
$Area = (\pi/2 + \frac{sin3\pi}{6}) - (\pi/6 + \frac{sin(\pi)}{6})$
$Area = (\pi/2 + \frac{1}{6}) - (\pi/6 + \frac{1}{6})$
$Area = (\pi/2 ) - (\pi/6 )$
$Area = (2\pi/6 ) = \frac{\pi}{3}$
