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.

The 'outer' radius, $r_2$, is the curve farther from the origin, while the 'inner' radius, $r_1$, is the curve closer to the origin. The area is calculated by integrating the difference of their squares.

The limits of integration, $a$ and $b$, are the angles at which the two polar curves intersect. Solve $r_1(\theta) = r_2(\theta)$ to find these intersection points.

It represents the area of an infinitesimally small sector of a circle with radius $r$ and angle $d\theta$. Integrating this gives the total area.

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

$A = \frac{1}{2} \int_a^b (r_2^2 - r_1^2) d\theta$, 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.