∑属于x/2+y/3+z/4=1的第一卦象,则曲面积分∫∫∑(x^2+y^2+z^2)^1/2dS

解：∵x²+y²+z²=1 ==>z=±√(1-x²-y²) 令s1：z=√(1-x²-y²)，s2：z=-√(1-x²-y²)。则s1和s2在xoy平面上的投影都是圆s：x²+y²=1 ∴球面∑=s1+s2 ∵αz/αx=±(-x/√(1-x²-y²))，αz/αy=±(-y/√(1-x²-y²)) ∴ds=√(1+(αz/αx)²+(αz/αx)²)dxdy=dxdy/√(1-x²-y²) 故∫∫<∑>(x+y+z+1)²ds=∫∫(x+y+z+1)²ds+∫∫(x+y+z+1)²ds =∫∫(x+y+√(1-x²-y²)+1)²dxdy/√(1-x²-y²)+∫∫(x+y-√(1-x²-y²)+1)²dxdy/√(1-x²-y²) =∫∫[(x+y+√(1-x²-y²)+1)²+(x+y-√(1-x²-y²)+1)²]dxdy/√(1-x²-y²) =4∫∫(xy+x+y+1)dxdy/√(1-x²-y²) =4∫<0,2π>dθ∫<0,1>[r²sinθcosθ+r(sinθ+cosθ)+1]rdr/√(1-r²) (作极坐标变换) =4∫<0,2π>[sin(2θ)/3+π(sinθ+cosθ)/4+1]dθ (中间运算省约) =4*(2π) =8π。