Question

高数二重积分计算，写一下过程？

Answer 1

解：∑: z=√(1-x^2-y^2) ；z'x=-x/√(1-x^2-y^2), z'y=-y/√(1-x^2-y^2)
√(1+z'x^2+z'y^2)=1/√(1-x^2-y^2)
原式=∫∫(∑)（x+y+z)ds=∫∫(∑){[x+y+√(1-x^2-y^2)]/√(1-x^2-y^2)}ds
=∫∫(∑)[(x+y)/√(1-x^2-y^2)+1]ds
=∫(0,1)dρ∫(0,2π)[(ρsinθ+ρcosθ)/√(1-ρ^2)+1]dθ
=∫(0,1)ρdρ∫(0,2π)[(ρsinθ+ρcosθ)/√(1-ρ^2)+1]dθ
=∫(0,1)ρdρ[p(-cosθ+sinθ)/√(1-ρ^2)+θ](0,2π)
=2π∫(0,1)ρdρ=2π*(ρ^2/2)(0,1)
=π