Calculus (3rd Edition)

$\pi$
Since $r=f(\theta)= \sin\theta$, then $f'(\theta)=\cos\theta$ The length is given by \begin{align*} \text { The length }&=\int_{0}^{\pi} \sqrt{f(\theta)^{2}+f^{\prime}(\theta)^{2}} d \theta\\ &=\int_{0}^{\pi}\sqrt{\sin^2\theta+\cos^2\theta} d \theta\\ &=\int_{0}^{\pi}1 d \theta\\ &=\pi. \end{align*}