∵z=√(x²+y²),则αz/αx=x/√(x²+y²),αz/αy=y/√(x²+y²)
∴ds=√[1+(αz/αx)²+(αz/αy)²]dxdy
=√[1+(x/√(x²+y²))²+(y/√(x²+y²))²]dxdy
=√2dxdy
故 原式=∫∫(x²+y²)√2dxdy(S表示曲面∑在xy平面的投影：x²+y²=4)
=√2∫dθ∫r²*rdr(做极坐标变换)
=√2π/2(2^4-0^4)
=8√2π.