Answer
$e^{\pi}-1$
Work Step by Step
The length of the curve is given as: $L=\int_{p}^{q}\sqrt{r^2+(\dfrac{dr}{d\theta})^2}d\theta$
Thus, $L=\int_{0}^{\pi} \sqrt{(\dfrac{e^\theta}{\sqrt 2})^2+(\dfrac{e^\theta}{\sqrt 2})^2} d \theta$
Then, we have $L=\int_{0}^{\pi} \sqrt {e^{2\theta}} d\theta$
Then, we get $L =[e^{\theta}]_{0}^{\pi}$
Thus, $L=e^{\pi}-1$
