<s>[INST] Integrating an Iterated Integral Given the iterated integral:
$$\int^{\sqrt{2}}_{-\sqrt{2}}\int^{\sqrt{2-x^2}}_{-\sqrt{2-x^2}}\int^{\sqrt{4-x^2-y^2}}_{\sqrt{x^2+y^2}}{(x^2+y^2+z^2)^{3/2}}dzdydx$$
Now, my question is, what are the two quadric surfaces that bound the region from above and below, and what are their equations? Also, how do I integrate this?
 [/INST] Ok, let's work out your quadrics. Given the order you must be integrating in (can you see why?), you have three constraints which characterise your volume: $$\sqrt{x^2+y^2}\le z\le\sqrt{4-x^2-y^2} \\-\sqrt{2-x^2}\le y \le \sqrt{2-x^2}\\-\sqrt{2}\le x \le \sqrt{2}$$
So the LHS of our first equation gives $z^{2}-x^{2}-y^{2}\ge 0$ whith $z \ge 0$, and the RHS gives $x^{2}+y^{2}+z^{2}\le 4$. Do you recognise these quadrics? You can get further constraints from the second equation. Now for the integral.   
Let's change to spherical polar co-ordinates $(r,\theta,\phi)$ before we get upset. Recall that: $$x=r\sin \theta \cos \phi, \quad y=r\sin \theta \sin \phi, \quad z=r\cos \theta$$ $$r^{2}=x^{2}+y^{2}+z^{2}, \quad dxdydz=r^{2}\sin\theta drd\theta d\phi $$
Now we just need to work out our region of integration. It's not hard to see that $x^{2}+y^{2}=r^{2}\sin^{2}\theta$, so we get the two constraints $r\le 2$ and $r\cos\theta\ge r\sin \theta$, i.e. $0 \le \theta \le \pi/4$. We also have $x^{2}+y^{2}\le 2$ (which is automatically true already) and $-\sqrt{2}\le r \cos\theta \sin \phi \le \sqrt{2}$ (which allows $\phi$ to take any value). Hence, the integral is $$\int_{0}^{\pi/4}\int_{0}^{2\pi}\int_{0}^{2}r^{5}\sin\theta drd\phi d\theta=\frac{2^{6}}{6}2\pi\frac{\sqrt{2}}{2}=\frac{32(2-\sqrt{2})}{3}\pi$$
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