You can avoid integration* - the region is simple, one right-angled triangle atop a square of area 2x2=4. On sweeping the square about $x=1$, the average radius of the sweep is $(1+3)/2$ =2, giving you a volume of $2*\pi*2*4$ = $16\pi$.
*Oops, a fix: The average radius for the triangle is not 2 - The danger of using geometric arguments carelessly.
The correct radius to use is the radius to the abscissa of the centroid of the triangle $10/3$ from $x=1$, which gives $r=7/3$. The volume of the sweep is $(1/2)*2*2*7/3*2\pi$, which is $28\pi / 3$.
The total volume is $16\pi + 28\pi /3$, which gives v=$76\pi / 3$.