Question

求曲面积分,∫∫zds,Σ:z=根号X^2+y^2在柱体x^2+y^2

Answer 1

∵z=√(x^2+y^2)

==>αz/αx=x/√(x^2+y^2),αz/αy=y/√(x^2+y^2)

∴ds=√[1+(αz/αx)^2+(αz/αy)^2]dxdy=√2dxdy

故 ∫∫zds=√2∫∫√(x^2+y^2)dxdy

=√2∫dθ∫r^2dr

=(8√2/3)∫(sinθ)^3dθ

=(8√2/3)∫[(cosθ)^2-1]d(cosθ)

=(8√2/3)(4/3)

=32√2/9.