Chapter 11: Parametric Equations and Polar Coordinates - Section 11.5 - Areas and Lengths in Polar Coordinates - Exercises 11.5 - Page 671: 22

$e^{\pi}-1$

The length of the curve can be calculated as: $L=\int_{p}^{q}\sqrt{r^2+(\dfrac{dr}{d\theta})^2}d\theta$ Here, we have $L=\int_{0}^{\pi} \sqrt{(\dfrac{e^\theta}{\sqrt 2})^2+(\dfrac{e^\theta}{\sqrt 2})^2} d \theta= \int_{0}^{\pi} \sqrt {e^{(2\theta)}} d\theta$ This implies that $L =[e^{\theta}]_{0}^{\pi}=e^{\pi}-1$