设f(x,y)在闭区间D={(x,y)|x2+y2≤y,x≥0}上连续,且f(x,y)=√（1-x2-y2）-8/π∫∫f(x,y)dxdy求f(x,y)书后答案是√（1-x^2-y^2)+8/9π-2/3

Question

设f(x,y)在闭区间D={(x,y)|x2+y2≤y,x≥0}上连续,且
f(x,y)=√（1-x2-y2）-8/π∫∫f(x,y)dxdy
求f(x,y)
书后答案是√（1-x^2-y^2)+8/9π-2/3

Answer 1

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)