So I'm practicing on how to convert cartesian to cylindrical and I'm not sure how to go about the $z$-coordinate. This is what I'm trying to convert:
$U = \{(x,y,z):0\leq x^2+y^2\leq 2, 0\leq z\leq 6-x-y\}$
Usually, $z$ would be something like this, "$0\leq z\leq 6-x^2-y^2$", so I can make $z = 0$ and $z = 6-r^2$. I'm not sure how to do that for the other one.
Sure, well the relation $0 \leq x^2 + y^2 \leq 2$ still becomes $0 \leq r^2 \leq 2$ as you say. Unfortunately the other inequality is slightly more yuck because, again as is implied by what you say, $x$ and $y$ are related to $r$ and $\theta$ in a non-linear way. The best we can do is write $x = r \cos \theta$ and $y = r \sin \theta$ so that the second relation becomes $0 \leq z \leq 6 - r (\cos \theta + \sin \theta)$.
Geometrically what you've got there is a solid cylinder of radius 2 which has been sliced up by a plane (defined by $z = 6 - x - y$), with the plane itself not parallel to the $xy$-plane. Converting the linear condition (in $(x,y,z)$-variables) of being under this plane into variables which include the radial distance $r$ and angle $\theta$ will necessarily require the introduction of some trigonometric functions.