You can use a graphing utility to see that the curves intersect when Θ = π/2 and when Θ = π, so those become the bounds of integration. You are subtracting the area of the cardioid (r = 1 + cosΘ) in QII from the area of the circle in QII. You will use the formula for polar area A = ∫ab 1/2r2dΘ. So ...
A = ∫π/2π 1/2 [(sinΘ)2 - (1 + cosΘ)2]dΘ ~ .4292