4[4+4cos(x)+(1+cos2x)/2]
= 18 + 16cos(x) + 2cos(2x)
1/2 integral {18 + 16cos(x) + 2cos(2x)dx} = 9x + 8sin(x) +1/2 sin(2x)
also the graph is generated by letting theta vary from 0 to 2pi
so you can integrate from 0 to pi and double it
i'll use x for theta because i don't know how to make the theta symbol
find the area inside the oval limacon r=4+2cos(x)
so i've squared r to get 16 +16cos(x) +4cos(x)^2
factor out a 4 to get 4(4+4cos(x)+cos(x)^2)
then change to 4(4+4cos(x)+(1+cos2x)/2
factor out a 1/2 and integrate, which comes to
2(3x + 2sin(x)+(sin2x)/2
so i think the limits of integration are from 0 to pi. but i'm not getting the correct answer using those. can anyone help me out?