## Thomas' Calculus 13th Edition

$L=\int_{a}^{b}\sqrt{r^2+(\dfrac{dr}{d\theta})^2}d\theta$
Let us consider that $r=f(\theta)$ Since $x=r \cos \theta$ then $x=f(\theta) \cos \theta \\ y=r \sin \theta \implies y=f(\theta) \sin \theta$ Use formula for the length curve: $L=\int_{a}^{b}\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2}dt$ $L=\int_{a}^{b} \sqrt{(f'(\theta) (\cos \theta)-f(\theta) (\sin \theta)^2+(f'(\theta) (\sin \theta)+f(\theta) (\cos \theta)^2}dt$ or, $L=\int_{a}^{b}\sqrt{[f'(\theta)]^2 +f(\theta)^2}d\theta$ Hence, the length of the curve can be written as: $L=\int_{a}^{b}\sqrt{r^2+(\dfrac{dr}{d\theta})^2}d\theta$