## Calculus (3rd Edition)

$$\text { The length } =\int_{0}^{\pi/2} \sqrt{2e^{2\theta}+2e^\theta +1} d \theta.$$
Since $r=f(\theta)=e^\theta+1$, then $f'(\theta)=e^\theta$ The length is given by $$\text { The length }=\int_{0}^{\pi/2} \sqrt{f(\theta)^{2}+f^{\prime}(\theta)^{2}} d \theta\\ =\int_{0}^{\pi/2} \sqrt{2e^{2\theta}+2e^\theta +1} d \theta.$$