The x- and y-coordinates of any Cartesian point can be written as the following parametric equations:

$\left\{\begin{array}{cc}x=r \cos \left(\mathrm{\theta }\right)& \\ y = r \sin \left(\mathrm{\theta }\right)& \end{array}\right. a\le \mathrm{\θ} \le b$.

Using the chain rule, note that $\frac{\mathrm{d}x}{\mathrm{d}\theta } \= \frac{\mathrm{d}r}{\mathrm{d}\theta } \cos \left(\mathrm{\theta }\right) - r \sin \left(\mathrm{\theta }\right)$ and $\frac{\mathrm{d}y}{\mathrm{d}\theta } \= \frac{\mathrm{d}r}{\mathrm{d}\theta } \sin \left(\mathrm{\theta }\right) \+ r \cos \left(\mathrm{\theta }\right)$.

For these parametric equations, the arc length of a curve on the interval  $a \le \mathrm{\θ} \le b$  is given by:

$L \= {\int }_a^b \sqrt{{\left(\frac{\mathrm{d}x}{\mathrm{d}\mathrm{\θ}}\right)}^2 \+ {\left(\frac{\mathrm{d}y}{\mathrm{d}\mathrm{\θ}}\right)}^2} \ⅆ\mathrm{\θ}$

$L \={\int }_a^b \sqrt{{\left(\frac{\mathrm{d}r}{\mathrm{d}\mathrm{\θ}} \cos \left(\mathrm{\θ}\right) - r \sin \left(\mathrm{\θ}\right)\right)}^2 \+ {\left(\frac{\mathrm{d}r}{\mathrm{d}\theta } \sin \left(\mathrm{\theta }\right) \+ r \cos \left(\mathrm{\theta }\right)\right)}^2} \ⅆ\mathrm{\θ}$

$L \={\int }_a^b \sqrt{{\left(\frac{\mathrm{d}r}{\mathrm{d}\mathrm{\θ}}\right)}^2{\cos }^2\left(\mathrm{\θ}\right) - 2 r \left(\frac{\mathrm{d}r}{\mathrm{d}\theta }\right)\cos \left(\mathrm{\θ}\right) \sin \left(\mathrm{\θ}\right) \+ r^2 {\sin }^2\left(\mathrm{\θ}\right) \+ {\left(\frac{\mathrm{d}r}{\mathrm{d}\mathrm{\θ}}\right)}^2{\sin }^2\left(\mathrm{\θ}\right) \+ 2 r \left(\frac{\mathrm{d}r}{\mathrm{d}\theta }\right)\cos \left(\mathrm{\θ}\right) \sin \left(\mathrm{\θ}\right) \+ r^2 {\cos }^2\left(\mathrm{\θ}\right)} \ⅆ\mathrm{\θ}$

$L \={\int }_a^b \sqrt{{\left(\frac{\mathrm{d}r}{\mathrm{d}\theta }\right)}^2\left({\cos }^2\left(\mathrm{\θ}\right) \+ {\sin }^2\left(\mathrm{\theta }\right)\right)\+ r^2 \left({\sin }^2\left(\mathrm{\θ}\right) \+{\cos }^2\left(\mathrm{\theta }\right)\right)} \ⅆ\mathrm{\θ}$

Remembering that ${\cos }^2\left(\mathrm{\theta }\right) \+ {\sin }^2\left(\mathrm{\theta }\right) \= 1$, this expression can now be simplified to:

$L \={\int }_a^b \sqrt{{\left(\frac{\mathrm{d}r}{\mathrm{d}\theta }\right)}^2\+ r^2} \ⅆ\mathrm{\θ}$

Therefore the arc length for a polar curve on the interval $a \le \mathrm{\θ} \le b$ is $L \={\int }_a^b \sqrt{r^2 \+ {\left(\frac{\mathrm{d}r}{\mathrm{d}\theta }\right)}^2} \ⅆ\mathrm{\θ}$.