$A = \frac{1}{2} \int_a^b r^2 d\theta$

$A = \frac{1}{2} \int_a^b (r_2^2 - r_1^2) d\theta$, where $r_2$ is the outer radius and $r_1$ is the inner radius.

$A = \frac{1}{2} \int_{\alpha}^{\beta} [f(\theta)^2 - g(\theta)^2] d\theta$, where $f(\theta)$ is the outer curve and $g(\theta)$ is the inner curve.

A curve defined using polar coordinates $(r, \theta)$, where $r$ is the distance from the origin and $\theta$ is the angle from the positive x-axis.

A two-dimensional coordinate system where each point is determined by a distance $r$ from a reference point (the pole) and an angle $\theta$ from a reference direction (the polar axis).

$r$ represents the radial distance from the origin to the point in the polar coordinate system.

$\theta$ represents the angle measured counterclockwise from the positive x-axis to the line segment connecting the origin to the point.

Set the two polar equations equal to each other, $r_1(\theta) = r_2(\theta)$, and solve for $\theta$. These $\theta$ values are the angles at the intersection points.

1. Identify $r_1 = 3$ and $r_2 = 3 - 2\sin(2\theta)$. 2. Determine bounds: $\frac{\pi}{2}$ to $\pi$. 3. Set up the integral: $A = \frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} [(3 - 2\sin(2\theta))^2 - 3^2] d\theta$. 4. Evaluate the integral.

Imagine yourself at the origin. The function you see first is the inner function.