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I'm struggling to visualize 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.

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 $0\le z\le2$? (This is what I am struggling to visualize.)

Thanks!

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