Answer
$Area = \frac{1}{2} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}(3-2sin\theta)^2d\theta$
Work Step by Step
1. Find the limits of integration:
As we can see from the image, the shaded region starts at $\theta = \frac{\pi}{2}$, and ends at $\theta = \frac{3\pi}{2}$;
Notice: When $\theta = \pi/2:$
$r = 3 - 2sin(\pi/2) = 3 - 2*1 = 3-2 = 1$
Since this value is positive**, the start angle of the shaded region is equal to $\pi/2$
If "r" was negative, the point would appear on the opposite angle of $\pi/2$, which is $3\pi /2$.
When $\theta = 3\pi/2$
$r = 3 - 2sin(3\pi/2) = 3 - 2*-1 = 3 + 2 = 5$
Since this value is positive, the final angle of the shaded region is equal to $3\pi/2$
2. Apply the formula for the area:
$Area = \frac{1}{2} \int_\alpha^\beta(f(\theta)^2)d\theta$
$Area = \frac{1}{2} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}(3-2sin\theta)^2d\theta$
