The key to computing the length of a polar curve is to think of it as
a parametrized curve with parameter $\theta$. (When computing the
slope of a polar curve, we called the parameter $t$ and set
$\theta=t$. Calling the parameter $\theta$ is equivalent and saves a
step.) Then
\begin{eqnarray*}x&=& r \cos(\theta); \cr y &=& r
\sin(\theta). \end{eqnarray*}Taking derivatives we
get\begin{eqnarray*}\frac{dx}{d\theta} &=& r' \cos(\theta) - r
\sin(\theta); \cr \frac{dy}{d\theta} &=& r' \sin(\theta) + r
\cos(\theta), \end{eqnarray*}
where $r'$ is shorthand for $dr/d\theta$. Squaring gives
\begin{eqnarray*}\left (
\frac{dx}{d\theta}\right)^2 &=& (r')^2 \cos^2(\theta) + r^2
\sin^2(\theta) - 2 r r' \sin(\theta) \cos(\theta); \cr \left (
\frac{dy}{d\theta}\right)^2 &=& (r')^2 \sin^2(\theta) + r^2
\cos^2(\theta) + 2 r r' \sin(\theta)
\cos(\theta). \end{eqnarray*}
Adding then gives
$$\left
(\frac{dx}{d\theta}\right )^2 + \left ( \frac{dy}{d\theta} \right )^2
= r^2 + \left ( \frac{dr}{d\theta}\right)^2, \qquad \hbox{so}$$

The arc length of a polar curve $r=f(\theta)$ between $\theta=a$
and $\theta=b$ is given by the integral
$$L = \int_a^b \sqrt{r^2 + \left ( \frac{dr}{d\theta}\right )^2} \, d\theta.$$

In the following video, we derive this formula and use it to compute the arc length of a cardioid.