∵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∫(8s...你是不是看错题目了？