Setting up and computing the integral is straightforward. The two parabolic cylinders [tex]z=8-x^2[/tex] and [tex]z=x^2[/tex] meet when
[tex]x^2 = 8 - x^2 \implies 2x^2 = 8 \implies x = \pm2[/tex]
Then the volume is
[tex]\displaystyle \int_0^3 \int_{-2}^2 \int_{x^2}^{8-x^2} dz \, dx \, dy = 3 \int_{-2}^2 ((8-x^2) - x^2) \, dx \\\\ ~~~~~~~~ = 6 \int_0^2 (8 - 2x^2) \, dx = \boxed{64}[/tex]
