Answer
The area of one petal formed by this function is equal to $\frac{\pi}{8}$.
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:
$r = sin(2\theta)$
$0 = sin(2\theta)$
$2\theta = sin^{-1}(0)$
First angle: $2\theta = 0$
$\theta = 0$
Second angle: $2\theta = \pi$
$\theta = \frac{\pi}{2}$
Therefore, the first 2 angles where that happen are: $\theta = 0, $ and $ \theta = \pi/2$
2. Apply the formula for the area:
$Area = \frac{1}{2} \int_\alpha^\beta(f(\theta)^2)d\theta$
$Area = \frac{1}{2} \int_{0}^{\pi/2}(sin(2\theta))^2d\theta$
$Area = \frac{1}{2} \int_{0}^{\pi/2}(sin^2(2\theta))d\theta$
** $sin^2\theta = \frac{1-cos(2\theta)}{2}$
** $sin^2(2\theta) = \frac{1-cos(4\theta)}{2}$
$Area = \frac{1}{2} \int_{0}^{\pi/2}(\frac{1-cos(4\theta)}{2})d\theta$
$Area = \frac{1}{4} \int_{0}^{\pi/2}({1-cos(4\theta)})d\theta$
$Area = \frac{1}{4} [\theta - \frac{sin(4\theta)}{4}]_{0}^{\pi/2}$
$Area = \frac{1}{4} (\pi/2 - \frac{sin(4(\frac{\pi}{2}))}{4}) - \frac{1}{4}( 0 - \frac{sin(4(0))}{4}) $
$Area = \frac{1}{4} (\pi/2 - \frac{sin(2\pi)}{4}) - \frac{1}{4}( 0 - \frac{sin(0)}{4}) $
$Area = \frac{1}{4} (\pi/2 -0) - 0 $
$Area = \frac{\pi}{8}$
