Answer
$\sqrt 2 \pi$
Work Step by Step
Given: $r=2 \sin \theta+2\cos \theta$
$r'=2 \cos \theta-2 \sin \theta$
The length is given as follows:
$L= \int_{0}^{\pi/2} \sqrt{8(\cos^2 \theta+\sin^2 \theta)} d\theta$
Then, $L=[2\sqrt 2 \theta]_{0}^{\pi/2} $
After solving, we get $L=\sqrt 2 \pi$
