设f(x,y)在闭区间D={(x,y)|x2+y2≤y,x≥0}上连续,且

设∫∫f(x,y)dxdy＝a,则f(x,y)＝√(1-x^2-y^2)－8a/π,所以
区域D：0≤θ≤π/2,0≤ρ≤sinθ
a＝∫∫f(x,y)dxdy＝∫∫[√(1-x^2-y^2)－8a/π]dxdy＝∫(0～π/2)dθ∫(0～sinθ) √(1-ρ^2)ρdρ－8a/π×π/8＝1/3×∫(0～π/2) [1－(cosθ)^3]dθ＝π/6－2/9－a
所以,a＝π/(12)－1/9
所以,f(x,y)＝√（1-x^2-y^2)－8/π×[π/(12)－1/9]＝√（1-x^2-y^2)－2/3＋8/(9π)