解：∵x+z=2 ==>z=2-x ∴αz/αx=-1，αz/αy=0 ==>ds=√[1+(αz/αx)²+(αz/αy)²]dxdy=√2dxdy 故原式=∫∫(x+y+2-x)√2dxdy =√2∫∫(y+2)dxdy =√2∫<0,2π>dθ∫<0,2>(rsinθ+2)rdr =√2∫<0,2π>dθ∫<0,2>(r²sinθ+2r)dr =√2∫<0,2π>(8sinθ/3+4)dθ =√2(4*2π-0) =8√2π