Notice you complete the outer loop between 0 and 2pi/3 (where r=0)
By integrating from 0 to pi you get the area inside the limacon plus the area of the innerloop
Find area of limacon.
Well I cant seem to get the latex to work for the integral that I set up so here it is rougly.
Upper limit Pi
Lower limit 0
1/2 int (4+2cos theta)^2 dtheta
by symmetry I multiply by 2 which gets rid of the 1/2 outside
int (16) + int (16 cos theta) + int (4 cos^2theta) dtheta
16 theta - 16 sin theta + 4 int 1 + cos(2theta)/2 dtheta
Am I going the right way with this integral? If not where am I going wrong?