Answer
$\sqrt 2 \pi$
Work Step by Step
As we are given that $r=2 \sin \theta+2\cos \theta$
This gives: $r'=2 \cos \theta-2 \sin \theta$
The length of the curve is: $L= \int_{0}^{(\pi/2)} \sqrt{8(\cos^2 \theta+\sin^2 \theta)} d\theta$
This implies that
$L=(2\sqrt 2) [\theta]_{0}^{(\pi/2)}=\sqrt 2 \pi$
