设函数f(u)连续，区或D={(x，y)|x²+y²≤2y}，则∫∫_Df(xy)dxdy等于()

设函数f(u)连续，区或D={(x，y)|x²+y²≤2y}，则∫∫_Df(xy)dxdy等于()

A.∫-11dx∫-√1-x2√1-x2f(xy)dy

B.∫₀^πdθ∫₀^2sinθf(r²sinθcosθ)dr

C.2∫₀²dy∫₀^√2y-y2f(xy)dx

D.∫₀^πdθ∫₀^2sinθf(r²sinθcosθ)rdr

正确答案是D