Question

高数一道关于二重积分的题目，求解

可以画图吗，谢谢了

Answer 1

简单计算一下即可，答首肢芹历案如者首世图所示

Answer 2

设a=∬<D>f(u,v)dudv,则f(x,y)=√(1-x^2-y^2)-8a/π，
设u=rcosθ,v=rsinθ,则dudv=rdrdθ,于是D:r≤sinθ,0≤θ≤π/2，
a=∫<0,π/2>dθ∫<0,sinθ>[√(1-r^2)-8a/亩没π]rdr

=∫<0,π/2>dθ[(-1/3)(1-r^2)^(3/2)-4ar^2/π]|<0,sinθ>
=∫<0,π/2>{(-1/3)[(cosθ)^3-1)-2a(1-cos2θ)/π]dθ
=[(1/3)[θ-sinθ+(1/迅纳纳3)(sinθ)^3]-a(2θ-sin2θ）/π}|<0,π/2>
=(1/3)(π/2-2/3）-a
解得a=(1/36)(3π-4),
所以f(x,y)=√(1-x^2-y^2)-(6π-8）/（茄扰9π）。

Answer 3

Answer 4

令A=8/π ∫∫f(u，v)dudv
则f(x，y)=∨(1–x²–y²)–A
A=8/π ∫∫f(u，v)dudv
=8/π ∫∫ [∨(1–u²–v²)–A]dudv
=8/π ∫∫∨(1–戚蔽u²–v²)dudv–8A/π ∫纯坦∫dudv
令u=rcosθ，v=rsinθ，r∈[0，sinθ]，θ∈[0，π/2]，dudv=rdθdr
=8/π ∫高裤州(0，π/2) dθ ∫(0，sinθ) r∨(1–r²)dr–8A/π·1/2·π·(1/2)²
=–8/(3π) ∫(0，π/2) (cos³θ–1) dθ–A
=–8/(3π) ∫(0，π/2) cos³θdθ+8/(3π) ∫(0，π/2) dθ – A
=–8/(3π) ∫(0，π/2) (1–sin²θ)dsinθ+4/3– A
=–8/(3π) (sinθ–1/3 sin³θ)|(0，π/2) +4/3–A
=–8/(3π) (1–1/3)+4/3–A
=–16/(9π)+4/3–A
2A=(12π–16)/(9π)
A=(6π–8)/(9π)
所以f(x，y)=∨(1–x²–y²)–(6π–8)/(9π)