Solution

Use polar: $x=r\cos\theta,\; y=r\sin\theta$

Cylinder: $r^2=2r\cos\theta \Rightarrow r=2\cos\theta,\; -\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$

$V=\iint_D (x^2+y^2)\,dA=\int_{-\pi/2}^{\pi/2}\int_{0}^{2\cos\theta} r^2\cdot r\,dr\,d\theta$

$=\int_{-\pi/2}^{\pi/2}\left[\frac{r^4}{4}\right]_{0}^{2\cos\theta}d\theta=\int_{-\pi/2}^{\pi/2}4\cos^4\theta\,d\theta$

$=8\int_{0}^{\pi/2}\cos^4\theta\,d\theta=8\cdot\frac{3\pi}{16}=\frac{3\pi}{2}$