因為
xy+yz+zx
=(1/2)[(x+y+z)^2-(x^2+y^2+z^2)]
=-a^2/2
所以
∫(xy+yz+zx)ds=∫(-a^2/2)ds=(-a^2/2)∫ds
=(-a^2/2)*(2πa)=-πa^3
求函數(shù)xy+yz+zx對弧長的曲線積分,弧長為球面x^2+y^2+z^2=a^2與平面x+y+z
求函數(shù)xy+yz+zx對弧長的曲線積分,弧長為球面x^2+y^2+z^2=a^2與平面x+y+z
0的交線
0的交線
數(shù)學人氣：870 ℃時間：2020-06-21 04:56:13
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