Answer
$L=\int_{a}^{b}\sqrt{r^2+(\dfrac{dr}{d\theta})^2}d\theta$
Work Step by Step
Here, we have $r=f(\theta)$
Then, $x=r \cos \theta \implies x=f(\theta) \cos \theta$ and $y=r \sin \theta \implies y=f(\theta) \sin \theta$
As we know that $L=\int_{a}^{b}\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2}dt$
or, $L=\int_{a}^{b} \sqrt{(f'(\theta) \cos \theta-f(\theta) \sin \theta)^2+(f'(\theta) \sin \theta+f(\theta) \cos \theta)^2}dt$
or, $L=\int_{a}^{b}\sqrt{(f'(\theta))^2 +f(\theta)^2}d\theta$
Thus, the length of the curve is given as:
$L=\int_{a}^{b}\sqrt{r^2+(\dfrac{dr}{d\theta})^2}d\theta$
