L为x^2+y^2=4,计算∮(x-yx^2)dx+(xy^2)dy的值?

题目

L为x^2+y^2=4,计算∮(x-yx^2)dx+(xy^2)dy的值?

答案

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