设球面∑:x^2+y^2+z^2=1,则曲面积分∫∫(x+y+z+1)^2dS=
2个回答
∵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∫dθ∫[r²sinθcosθ+r(sinθ+cosθ)+1]rdr/√(1-r²) (作极坐标变换)
=4∫[sin(2θ)/3+π(sinθ+cosθ)/4+1]dθ (中间运算省约)
=4*(2π)
=8π.
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