∵x+y+z=1 ==>偏导数αz/αx=αz/αy=-1 ∴ds=√[1+(αz/αx)^2+(αz/αy)^2]dxdy=√3dxdy 故∫∫yds=∫∫y*√3dxdy =√3∫dx∫ydy =√3∫[(1-x)^2/2]dx =√3*(1/6) =√3/6.
