对坐标的曲面积分（未学高斯公式）∫∫∑ ydzdx+（x+z)dxdy,其中∑为圆柱面x^2+y^2=a^2(0

题目

对坐标的曲面积分（未学高斯公式）∫∫∑ ydzdx+（x+z)dxdy,其中∑为圆柱面x^2+y^2=a^2(0<=z<=1)外侧.
RT,求详解

答案

原式=∫∫√(a²-x²)dzdx-∫∫[-√(a²-x²)]dzdx+∫∫(x+1)dxdy-∫∫(x+0)dxdy
(S1：-a≤x≤a:,0≤z≤1.S2：x²+y²≤a²)
=2∫∫√(a²-x²)dzdx+∫∫dxdy
=2∫√(a²-x²)dx∫dz+∫dθ∫rdr (第二个积分作极坐标变换)
=2∫√(a²-x²)dx+πa²
=2∫a²cos²tdt+πa² (作变换x=asint)
=a²∫[1+cos(2t)]dt+πa² (应用倍角公式)
=a²[t+sin(2t)/2]│+πa²
=a²(π/2+π/2)+πa²
=2πa².