积分曲线x^2+(y+1)^2=1所以参数方程是x=cost, y=-1+sint. t∈[0,2π]ds=√[(x't)^2+(y't)^2]dt= dt∫√（x^2+y^2）ds=∫√(-2y)ds=∫√[2(1-sint)] dt=√2 ∫(0->2π) |sin(t/2)-cos(t/2)|dt=2√2 ∫(0->2π) |sin(t...你答案好像算错了。。不过方法我懂啦，谢啦