计算∫∫﹙x+y+z﹚dS,其中∑为平面x+z=2被柱面X^2+Y^2=4所截得的有限部分上侧

∵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∫dθ∫(rsinθ+2)rdr
=√2∫dθ∫(r²sinθ+2r)dr
=√2∫(8sinθ/3+4)dθ
=√2(4*2π-0)
=8√2π