s=∫ab1+f′2(x)dx s = \int_a^b \sqrt{1 + f'^2(x)} dx s=∫ab1+f′2(x)dx
{x=x(t)y=y(t) \begin{cases}x = x(t) \\ y = y(t) \end{cases} {x=x(t)y=y(t)
s=∫abdx2+dy2=∫abx′2(t)+y′2(t)dt s = \int_a^b \sqrt{dx^2 + dy^2} = \int_a^b \sqrt{x'^2(t) + y'^2(t)} dt s=∫abdx2+dy2=∫abx′2(t)+y′2(t)dt
r=r(θ) r = r(\theta) r=r(θ)
转成参数方程
{x=r(θ)cosθy=r(θ)sinθ \begin{cases} x = r(\theta) \cos \theta \\ y = r(\theta) \sin \theta \end{cases} {x=r(θ)cosθy=r(θ)sinθ
再求弧长
s=∫αβx′2(θ),y′2(θ)dθ=∫αβr2(θ)+r′2(θ)dθ s = \int_\alpha^\beta \sqrt{x'^2(\theta), y'^2(\theta)} d\theta = \int_\alpha^\beta \sqrt{r^2(\theta) + r'^2(\theta)} d\theta s=∫αβx′2(θ),y′2(θ)dθ=∫αβr2(θ)+r′2(θ)dθ
