Let I be the smaller portion and II be the greater portion of the given figure then, 
Area of I=∫−20[4−x2−(x+2]dx =[2x4−x2+24sin−1(2x)]−20−[2x2+2x]−20=[2sin−1(−1)]−[−24+4]=2×2π−2=π−2 Now, area of II= Area of circle − area of I. =4π−(π−2)=3π+2 Hence, required ratio = area of II area of I =3π+2π−2