x^2+y^2 = 1 与 x^2+y^2 = 2x 的交点 (1/2,√3/2)
极坐标：0 ≤θ≤ π/3,1 ≤p≤ 2 cosθ
∫∫ xy dxdy = ∫[0,π/3] cosθsinθ dθ ∫[1,2cosθ] p³ dp
= ∫[0,π/3] 【4(cosθ)^5 sinθ - (1/4)cosθsinθ】dθ
= (2/3) ( 1-1/64) - (1/8)* (3/4)
= 9/16非常正确，非常感谢！