计算曲面积分∫∫xdydz+zdxdy ,S是平面x+y+z=1在第一卦限部分的上侧
有1个网友回答
解：∵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∫<0,2π>dθ∫<0,2>(rsinθ+2)rdr
=√2∫<0,2π>dθ∫<0,2>(r²sinθ+2r)dr
=√2∫<0,2π>(8sinθ/3+4)dθ
=√2(4*2π-0)
=8√2π
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