Find the area of the region outside r=9+9sin(theta), but inside r=27sin(theta)

I did the integral of .5((27sin(theta))^2-(9+9sin(theta))^2), from pi/6 to 5pi/6, but got the wrong answer. Any hints?

Find the area inside the curve r^2 = 72cos(2theta)

Do I just take the square root of the entire equation? If so, how do I integrate the square root of cosine?

Find the area inside one loop of the curve r^2=4sin(2theta)

Same question as the previous one