Question

这道高数题怎么做？

Answer 1

因为∫∫(D) √(1-x^2-y^2)dxdy

=∫(0,π/2)dθ∫(0,sinθ)√(1-r^2)rdr
=(-1/2)*∫(0,π/2)dθ∫(0,sinθ)√(1-r^2)d(1-r^2)
=(-1/3)*∫(0,π/2)dθ*(1-r^2)^(3/2)|(0,sinθ)
=(-1/3)*∫(0,π/2)[(cosθ)^3-1]dθ
=(-1/3)*∫(0,π/2)[(cos3θ+3cosθ)/4-1]dθ
=(1/12)*∫(0,π/2)(4-cos3θ-3cosθ)dθ
=(1/12)*[4θ-(1/3)*sin3θ-3sinθ]|(0,π/2)
=(1/12)*(2π+1/3-3)
=π/6-2/9
所以∫∫(D)f(x,y)dxdy=∫∫(D)√(1-x^2-y^2)dxdy-∫∫(D)(8/π)*[∫∫(D)f(u,v)dudv]dxdy
=π/6-2/9-(8/π)*π(1/2)^2*(1/2)*∫∫(D)f(u,v)dudv
=π/6-2/9-∫∫(D)f(u,v)dudv
即∫∫(D)f(u,v)dudv=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)