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L为x^2+y^2=4,计算∮(x-yx^2)dx+(xy^2)dy的值?
人气:139 ℃ 时间:2020-04-01 10:13:47
解答
用格林公式:P=x-x²y,Q=xy²
∮(x-yx²)dx+(xy²)dy
=∫∫(y²+x²)dxdy
用极坐标
=∫∫ r²*rdrdθ
=∫[0--->2π] dθ ∫[0--->2] r³dr
=2π*(1/4)r⁴ |[0--->2]
=8π
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