Describing Domain of Integration (Triple Integral) I'm really struggling to go about starting 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.

I've tried sketching the projections of this integral's domain onto the $xy$-plane, the $yz$-plane and the $xz$-plane. I established that the projection onto the $xy$-plane is the portion of the circle $x^2+y^2=4$ on $0\le x\le2,\ y\ge0$; that the projection onto the $yz$-plane is the portion of the circle $y^2+z^2=8$ above the line $z=y$, on $0\le y\le2$; and that the projection onto the $xz$-plane is the portion of the circle $x^2+z^2=8$ above the line $z=x$, on $0\le x \le2$.
I'm not completely confident that these findings are useful in any way, and if they are, I am struggling to see the link which will allow me to solve the problem.
Do you care to get me started?
Thank you very much.
 A: The bounds on $z$ correspond to equations:
$$z^2=x^2+y^2\\x^2+y^2+z^2=8$$
The first is a cone, the second is a sphere.  They intersect above the $xy$-plane in a circle of radius $2$.  The bounds on $x$ and $y$ show that we are considering the region in between the cone and the sphere with $x>0$ and $y>0$.  Can you see it?
Let's first just consider the $xy$-plane.  As you've said, the given bounds correspond to the quarter circle of $x^2+y^2=4$ that lies in the first quadrant.  So whatever region we consider, it will lie completely above this quarter circle.  Now $z$ lies in between the cone and the sphere as I mentioned above.  You can think of this region as an ice cream cone.  However, we're only considering part of the ice cream cone above our quarter circle, so its like a quarter of an ice cream cone.
Hopefully this is helpful, along with Babak's picture.
A: +1 for the @Jared, cause he pointed out what you need to do the problem. I am posting a graph maybe you find out better that points. The circle projected on $xy$ plane is the same as the intersection of two equations (See another answer). 

