Area Between Polar Curves

$\frac{1}{2} \int ((2+\sin\theta)^2 - 9\cos^2\theta) \, d\theta$

$\int ((2+\sin\theta) - 3\cos\theta) \, d\theta$

$\frac{1}{2} \int (9\cos^2\theta - (2+\sin\theta)^2) \, d\theta$

$\int (3\cos\theta - (2+\sin\theta)) \, d\theta$