昨天做过:用极坐标代换:
=∫(0,2π)dθ∫(0,1)(cosθf'x+sinθf'y)dr
由于f=f(rcosθ,rsinθ)
所以:f'r=cosθf'x+sinθf'y
故I=∫(0,2π)dθ∫(0,1)(cosθf'x+sinθf'y)dr
=∫(0,2π)dθ∫(0,1)f'rdr
=∫(0,2π)[f(cosθ,sinθ)-f(0,0)]dθ
=∫(0,2π)[f(cosθ,sinθ)]dθ
所以:|I|《∫(0,2π)|f(cosθ,sinθ)|dθ《∫(0,2π)dθ=2π
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