Answer
$x^{2} + y^{2} \leq 4$
$0 \leq z \leq 8$
Work Step by Step
Since the cylinder must be on or above a disk in the xy-plane and on or below the plane $z = 8$, the cylinder must lie on or between the planes $z = 8$ and $z = 0$ (the xy-plane has equation $z = 0$). From this information, we can write:
$0 \leq z \leq 8$.
Now, the cylinder lies on or above a disk in the xy-plane with radius 2 and center the origin. Such disk, since it is a shaded-in circle in the plane $z = 0$, can be represented by:
$x^{2} + y^{2} \leq 4$, $z = 0$
By replacing the equation $z = 0$ with the inequality $0 \leq z \leq 8$, we replicate this disk in every plane parallel to the xy-plane and between (or on) the planes $z = 0$ and $z = 8$, creating the desired solid cylinder.
Therefore the region can be described by:
$x^{2} + y^{2} \leq 4$
$0 \leq z \leq 8$
