求∫L(x^2-y)dx-(x+sin^2y)dy,L是y=根号下1-x^2以A(-1,0)到B(1,0)

在圆弧L下补一条线N：y = 0,反向.
∮(L+N) (x² - y)dx - (x + sin²y)dy
= - ∫∫D [ ∂/∂x (- x - sin²y) - ∂/∂y (x² - y) ] dxdy
= - ∫∫D [ - 1 - (- 1) ] dxdy
= 0
∫N (x² - y)dx - (x + sin²y)dy
= ∫(1→- 1) x² dx
= - 2∫(0→1) x² dx
= - 2/3
因此∫L (x² - y)dx - (x + sin²y)dy = 0 - (- 2/3) = 2/3