The problem states:
Find the volume of a region bounded by $x^2+y^2=4, \quad y=z, \quad y+z=4 $
So the idea is to triple integrate over the given region. Since sketching stuff in 3D is rather difficult, I tried to avoid it doing this:
Since we're dealing with an intersection of a cylinder and two planes, and those to planes only deal with $y$ and $z$ coordinates, I've decided to disregard the x coordinate for a second and see what happens in the $Oyz$ plane. It would look something like this.
So, treating those two planes as lines and the cylinder simply as two lines with the equations $y=2$ and $y=-2$ I've managed to work out that the bounds for z are $$-2 \leq z \leq 6$$ Now the next part is the one I'm not sure of how correct it is. It's pretty much Fubini's theorem, I'm just not sure how correct.
Since I have the bounds for $z$, now I need bounds for $x$ and $y$, which I've found by "looking from the top". Looking from the top we pretty much get a circle with radius 2, so I did a polar substitution $(x, y) = (r\cos\theta, r\sin\theta)$ with $0 \leq r \leq 2, 0 \leq \theta \leq 2\pi$
$$V(T) = \iiint_Tdxdydz = \int_{-2}^{6}dz\int_{0}^{2\pi}d\theta\int_{0}^{2}rdr =\\ \ \frac{1}{2}\int_{-2}^{6}dz\int_{0}^{2\pi}d\theta\cdot4 = 2\int_{-2}^{6}2\pi dz = 4\pi(6-(-2)) = 32 \pi$$
The thing I'm not sure of is whether I'm allowed to just "look from the top" as I said after finding the bounds for $z$ and treat everything from there as if it was in 2D.
Your integral is not correct because the $z$ limits depend upon $r$ and $\theta$:
If you're clever, you'll see how the solid can be split in half horizontally and that the top half can be placed on the bottom half to make a cylinder, which has a trivial volume calculation.
If you insist on using a triple integral:
$$\int\limits_{r=0}^2 \int\limits_{\theta = 0}^{2 \pi} \int\limits_{z = 0}^{2 - r \cos (\theta)}\ r\ dr\ d \theta\ dz = 16 \pi.$$