Find the intercepts for both planes.
Plane 1, x + y + 2z = 2:
[tex]y=z=0\implies x=2\implies (2,0,0)[/tex]
[tex]x=z=0\implies y=2\implies(0,2,0)[/tex]
[tex]x=y=0\implies 2z=2\implies z=1\implies(0,0,1)[/tex]
Plane 2, 4x + 4y + z = 8:
[tex]y=z=0\implies4x=8\implies x=2\implies(2,0,0)[/tex]
[tex]x=z=0\implies4y=8\impliesy=2\implies(0,2,0)[/tex]
[tex]x=y=0\implies z=8\implies(0,0,8)[/tex]
Both planes share the same x- and y-intercepts, but the second plane's z-intercept is higher, so Plane 2 acts as the roof of the bounded region.
Meanwhile, in the (x, y)-plane where z = 0, we see the bounded region projects down to the triangle in the first quadrant with legs x = 0, y = 0, and x + y = 2, or y = 2 - x.
So the volume of the region is
[tex]\displaystyle\int_0^2\int_0^{2-x}\int_{\frac{2-x-y}2}^{8-4x-4y}\mathrm dz\,\mathrm dy\,\mathrm dx=\displaystyle\int_0^2\int_0^{2-x}\left(8-4x-4y-\frac{2-x-y}2\right)\,\mathrm dy\,\mathrm dx[/tex]
[tex]=\displaystyle\int_0^2\int_0^{2-x}\left(7-\frac72(x+y)\right)\,\mathrm dy\,\mathrm dx=\int_0^2\left(7(2-x)-\frac72x(2-x)-\frac74(2-x)^2\right)\,\mathrm dx[/tex]
[tex]=\displaystyle\int_0^2\left(7-7x+\frac74 x^2\right)\,\mathrm dx=\boxed{\frac{14}3}[/tex]
