Volume of region in the first octant bounded by coordinate planes and a parabolic cylinder? Find the volume of the solid region in the first octant bounded by the coordinate planes, the plane $y + z = 2$ and the parabolic cylinder $x = 4 - y^2$.

I have a final answer, I would just like to make sure I am correct.

Because it is the first octant bounded by the coordinate planes: $x \geq 0, y \geq 0, z \geq 0$. This means that we can define the boundaries of the region as $0 \leq z \leq 2 - y, 0 \leq y \leq \sqrt{4 - x}, 0 \leq x \leq 4$ (from substituting $y = 0$ into $x = 4 - y^2$).

So E = { $(x,y,z) | 0 \leq x \leq 4, 0 \leq y \leq \sqrt{4 - x}, 0 \leq z \leq 2-y$} and the projected xy plane would be the function $y = \sqrt{4 - x}$.

V = $\int_0^4 \int_0^{\sqrt{4 - x}} \int_o^{2-y}dzdydx$  (I wanted to make sure this part was accurate in terms of order and logic).

V = $\int_0^4 \int_0^{\sqrt{4 - x}} (2-y)dydx$

V = $\int_0^4 [2 \sqrt{4-x} - \frac{4 - x}{2}]dx$

V = $2\int_0^4 \sqrt{4-x}dx - \frac{1}{2}\int_0^4(4 - x)dx$

V = $6\frac{2}{3}$ $units^3$
The solution appears to be correct. Personally, I used different construction of the integral, which is 

$\int_0^2dy\int_0^{2-y}dz\int_0^{4-y^2}dx = \frac{20}{3}$.

Hope this helps.
Find the volume of the region in the first octant 
bounded by the coordinate planes, the plane $ +
 = 2$, and the cylinder $
^2 + 4^
2 = 16$.