Describing/sketching region of integration of triple integral I'm having great difficulty with the following problem:

This question concerns the integral $\int_{0}^{2}\int_{0}^{\sqrt{4-y^2}}\int_{\sqrt{x^2+y^2}}^{\sqrt{8-x^2-y^2}}\!z\ \mathrm{d}z\ \mathrm{d}x\ \mathrm{d}y$. Sketch or describe in words the domain of integration. Rewrite the integral in both cylindrical and spherical coordinates. Which is easier to evaluate?

Below is what I believe I have established so far...
The projection of this integral's domain onto the $xy$-plane is the portion of the circle $x^2+y^2=4$ on $0\le x\le2,\ y\ge0$.
The bounds on $z$ correspond to

$z^2=x^2+y^2$ (cone) and $x^2+y^2+z^2=8$ (sphere).

These bounds intersect at

$x^2+y^2=4$.

Below $z=2$ (where the bounds on $z$ intersect), I believe that the cone and cylinder, $x^2+y^2=4$, are completely inside the sphere.
Would it hence be correct to say that the region of integration is the solid lying between the cone and the cylinder, on $x\ge0$, $y\ge0$ and $0\le z\le2$? I'm struggling to visualize this problem.
When I attempt to move on, and evaluate the integral in cylindrical/spherical coordinates, my solutions differ by a factor of 2.
That is, I evaluated this integral as,

$\int_{0}^{\frac{\pi}{2}}\int_{0}^{2}\int_{0}^{\sqrt{8-r^2}}\!z\ r\ \mathrm{d}z\ \mathrm{d}r\ \mathrm{d}\theta=2\pi$

And,

$\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\sqrt{2}}\!\rho\ \cos\phi\ \rho^2 \sin \phi\ \mathrm{d}\rho\ \mathrm{d}\theta\ \mathrm{d}\phi=4\pi$

Can you please help me to identify where I am going wrong?
Thank you very much.

You are very close.
The integral in cylindrical coordinates has the right value, but the lower limit for $z$ should be $r$, not $0$. 
Notice that in spherical coordinates the angle $\phi$ should be integrated from $0$ to $\pi/4$. This restrict us to the volume inside the cone.
Addendum: 
The volume is the region between the cone and the sphere in octant I ($x,y,z\ge 0$). 
See the figure below. 

In cylindrical coordinates we have 
$$\int_{0}^{\frac{\pi}{2}}\int_{0}^{2}\int_{r}^{\sqrt{8-r^2}}\!z\ r\ \mathrm{d}z\ \mathrm{d}r\ \mathrm{d}\theta.$$
In spherical coordinates we have 
$$\int_{0}^{\frac{\pi}{4}}\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\sqrt{2}}\!\rho\ \cos\phi\ \rho^2 \sin \phi\ \mathrm{d}\rho\ \mathrm{d}\theta\ \mathrm{d}\phi.$$
