弧长

y=f(x)

$$s = \int_a^b \sqrt{1 + f'^2(x)} dx$$

参数方程

$$\begin{cases}x = x(t) \\ y = y(t) \end{cases}$$

$$s = \int_a^b \sqrt{dx^2 + dy^2} = \int_a^b \sqrt{x'^2(t) + y'^2(t)} dt$$

极坐标方程

$$r = r(\theta)$$

转成参数方程

$$\begin{cases} x = r(\theta) \cos \theta \\ y = r(\theta) \sin \theta \end{cases}$$

再求弧长

$$s = \int_\alpha^\beta \sqrt{x'^2(\theta), y'^2(\theta)} d\theta = \int_\alpha^\beta \sqrt{r^2(\theta) + r'^2(\theta)} d\theta$$