Answer
$Area = 9\pi$
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, the figures starts at $r = 0$, and also finishes at $r = 0$. Find the first 2 angles with that "r" value:
$r = 6sin(\theta)$
$0 = 6sin(\theta)$
$0 = sin(\theta)$
$\theta = sin^{-1}(0)$
Therefore, the first 2 angles where that happen are: $\theta = 0, $ and $ \theta = \pi$;
2. Apply the formula for the area:
$Area = \frac{1}{2} \int_\alpha^\beta(f(\theta)^2)d\theta$
$Area = \frac{1}{2} \int_{0}^{\pi}(6sin(\theta))^2d\theta$
$Area = \frac{1}{2} \int_{0}^{\pi}(36sin^2(\theta))d\theta$
$Area = 36 * \frac{1}{2} \int_{0}^{\pi}sin^2(\theta)d\theta$
$Area = 18 \int_{0}^{\pi}sin^2(\theta)d\theta$
** $sin^2\theta = \frac{1-cos(2\theta)}{2}$
$Area = 18 \int_{0}^{\pi}\frac{1-cos(2\theta)}{2}d\theta$
$Area = 9 \int_{0}^{\pi}(1-cos(2\theta))d\theta$
$Area = 9 [ \theta - \frac{ sin(2\theta)}{2}]^{\pi}_0$
$Area = 9 ( \pi - \frac{ sin(2\pi)}{2}) - 9(0 - \frac{ sin(2*0)}{2})$
$Area = 9 ( \pi - \frac{0}{2}) - 9(0 - \frac{0}{2})$
$Area = 9\pi$
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