If we take projections on the XY plane, the area is bounded by the circle $x^2 + y^2 = 2$, the line y = x and the line x = 0.

We change the coordinate to cylindrical polar by putting x = $rcos \theta, y = rsin \theta$, z = z .

Eqn. of the cylinder becomes $x^2 + y^2 = 2 \Rightarrow r = \sqrt{2}$ then the line becomes $y= x \Rightarrow \theta = \pi/4$

$x=0 $ becomes $\theta = \frac{\pi}{2}$

If we consider a radial strip in the projection, r varies from $0 \hspace{0.1cm}\text{to} \sqrt{2}$, $\theta$ varies from $\theta = \frac{\pi}{4} \hspace{0.1cm}\text{to}\hspace{0.1cm}\theta = \frac{\pi}{2}$ then z varies from z = 0 to z = x + y

$V = \int_{\theta = \pi / 4}^{\pi/2}\int_{r = 0}^{\sqrt{2}}\int_{z=0}^{r(cos \theta+sin \theta)} r \hspace{0.1cm}dr\hspace{0.1cm}d\theta\hspace{0.1cm}dz\\ \hspace{0.2cm}= \int_{\theta = \pi / 4}^{\pi/2}\int_{r = 0}^{\sqrt{2}}r [z]_0^{r(cos \theta+sin \theta)}dr\hspace{0.1cm}d\theta\\ \hspace{0.2cm}= \int_{\theta = \pi / 4}^{\pi/2}\int_{r = 0}^{\sqrt{2}}r^2(cos \theta + sin \theta)dr\hspace{0.1cm}d\theta\\ \hspace{0.2cm}= \int_{\theta= \pi/4}^{\pi/2}(cos \theta + sin \theta)\Big[\frac{r^3}{3}\Big]_0^{\sqrt{2}}d \theta\\ \hspace{0.1cm} = \frac{2 \sqrt{2}}{3}\int_{\theta=\pi/4}^{\pi/2}(cos \theta + sin \theta)d \theta = \frac{2 \sqrt{2}}{3}\Big[sin \theta - cos \theta\Big]_{\pi/4}^{\pi/2}\\ \hspace{0.2cm} = \frac{2 \sqrt{2}}{3}\Big[(1-0) - \big(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\big)\Big] = \frac{2 \sqrt{2}}{3}$