For the circle r = 2acos(theta) The upper half is generated as theta varies from 0 to pi/2 and the lower half as theta varies from -pi/2 to zero.

your outer limits should then be -pi/2 to pi /2

more simply go 0 to pi/2 and double it.

As for the integral let u = 4a^2 -r^2

after solving you'll end up with an integral involving sin^3(theta)

which can easily be solved as (1-cos^2(theta))sin(theta)