闭路积分假设是逆时针方向：A(0,0); B(1,0); C(0,1)AB: y = 0BC: x+y = 1, ds = √2 dxCA: x = 0 ∫AB + ∫BC + ∫CA= ∫[0,1] xdx + ∫[1,0](x+y)√2 dx + ∫[1,0]ydy= ∫[0,1] xdx + ∫[1,0]√2 dx + ∫[1,0]ydy= -√2
