令x=ρcosθ,y=ρsinθ,则D:{(x,y)|x^2+y^2<=2y}即为{(ρ,θ)|ρ<=2sinθ}
x^2+y^2 = 2y 化为极坐标是 r = 2Rsint, 0 ≤ t ≤ π I = ∫∫√(x^2+y^2)dxdy = ∫<0, π>dt∫<0, 2Rsint> r rdr= (8/3)R^3∫<0, π>(sint)^3dt = -(8/3)R^3∫<0, π>[1-(cost)^2]dcost= -(8/3)R^3[cost-(1/3)(cost)^3]<0, π>= (32/9)R^3