Answer
$\dfrac{2(1+\sqrt{2})}{3}$
Work Step by Step
$\int^1_0 \int^{\sqrt{2-x^2}}_{x} (x+2y) \space dy \space dx = \int^{\pi/2}_{\pi/4} \int^{\sqrt{2}}_0(r \cos\theta+2 r \sin\theta) \space r \space dr \space d\theta $
or, $=\int^{\pi/2}_{\pi/4}[\frac{r^3}{3} \cos\theta+\dfrac{2r^3}{3} \sin\theta]^{\sqrt{2}}_0 d\theta $
or, $=\int^{\pi/2}_{\pi/4}(\dfrac{2\sqrt{2} \cos\theta}{3}+\dfrac{4\sqrt{2}}{3} \sin\theta) d\theta $
or, $=[\dfrac{2\sqrt{2}}{3} \sin\theta-\dfrac{4\sqrt{2}}{3} \cos\theta]^{\pi/2}_{\pi/4}$
or, $=\dfrac {2\sqrt 2}{3}+\dfrac{2}{3}$
or, $=\dfrac{2(1+\sqrt{2})}{3}$
