Question

求解高数题，急

Answer 1

∑：z=1-x^2-y^2 （x>0，y>0，z>0）
则x=√(1-y^2-z)，y=√(1-x^2-z)
I=∫∫(∑) xdydz+ydzdx+2zdxdy

=∫∫(Dyoz) √(1-y^2-z)dydz+∫∫(Dzox) √(1-x^2-z)dzdx+∫∫(Dxoy) (1-x^2-y^2)dxdy
=∫(0,1)dy∫(0,1-y^2)√(1-y^2-z)dz+∫(0,1)dx∫(0,1-x^2)√(1-x^2-z)dz+∫(0,π/2)dθ∫(0,1)(1-r^2)rdr
=∫(0,1)dy*(-2/3)*(1-y^2-z)^(3/2)|(0,1-y^2)+∫(0,1)dx*(-2/3)*(1-x^2-z)|(0,1-x^2)+(π/2)*[(1/2)*r^2-(1/4)*r^4]|(0,1)
=(4/3)*∫(0,1) (1-t^2)^(3/2)dt+π/8
=3π/16+π/8
=5π/16