The curves (cardioid in red and limacon in blue) intersect when 2-2cosθ=2+cosθ, so cosθ=0, and θ=π/2 and 3π/2 (r=2).
Consider the point (r,θ). If we rotate the line joining the point to the origin anticlockwise through a small angle dθ we create a sector with area ½r2dθ. The area of this first quadrant piece=∫½r2dθ, θ∈[0,π/2]. In this case, r=2-2cosθ, so r2=4(1-2cosθ+cos2θ). Since cos(2θ)=2cos2θ-1, cos2θ=(cos(2θ)+1)/2. In the first quadrant (0≤θ≤π/2) we have the same enclosed area as in the fourth quadrant (3π/2≤θ≤2π).
The integrand ½r2 becomes 2(1-2cosθ+(cos(2θ)+1)/2)=2-4cosθ+cos(2θ)+1=3-4cosθ+cos(2θ).
Area=∫[3-4cosθ+cos(2θ)]dθ=[3θ-4sinθ+½sin(2θ)]0π/2=3π/2-4, so the total area of the cardioid in quadrants 1 and 4 is 3π-8 square units.
Now consider the area enclosed by the limacon in quadrant 2 (π/2≤θ≤π). Area=∫½r2dθ, where r= 2+cosθ, r2=4+4cosθ+(cos(2θ)+1)/2.
Area=∫½r2dθ=
∫(9/4+2cosθ+cos(2θ)/4)dθ=[9θ/4+2sinθ+sin(2θ)/8]π/2π=9π/4-9π/8-2=9π/8-2.
The area in quadrant 3 is the same, so the total area enclosed by the limacon is 9π/4-4.
The area enclosed by the two curves is 3π-8+9π/4-4=21π/4-12 square units (about 4.4934).