Find the volume of the following region $E= (x,y,z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 ≤1, \sqrt{2}(x^2 + y^2) ≤z≤ \sqrt{6}(x^2 + y^2) $ 
Find the volume of the following region $E= \{ (x,y,z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 ≤1, \sqrt{2}(x^2 + y^2) ≤z≤ \sqrt{6}(x^2 + y^2) \}  $

I figured that the region E is formed by the points that belong in the gap between the two paraboloids
$z=\sqrt{2}(x^2 + y^2)$ $\hspace{20pt}$ and $\hspace{20pt}$ $ z= \sqrt{6}(x^2 + y^2)$
that are as well inside of the sphere $x^2 + y^2 + z^2 = 1$.
$E$ can be obtained by making a whole rotation around the $z$-axis in the plane-$xy$ then using Guldino's theorem can be appropriate.
Considering now cylindrical coordinates we've that $E$ becomes the following:
$F= \{ (\rho, \theta, z): \rho^2 + z^2 ≤1, \sqrt{2}\rho^2 ≤z≤ \sqrt{6}\rho^2,  0≤ \theta≤ 2\pi \}$
Then we've that:
$vol \hspace{2pt} E = \int _{F} \rho \hspace{2pt} d\rho d\theta dz = 2\pi \int\int _{D}  \rho d\rho dz $
I'm struggling to find the region $D$. Any hint?

Sticking to cylindrical coordinates, you solve a quadratic to deduce that $z=\sqrt6r^2$ and $r^2+z^2=1$ intersect when $r=\dfrac1{\sqrt3}$. Similarly, $z=\sqrt2r^2$ and $r^2+z^2=1$ intersect when $r=\dfrac1{\sqrt2}$.

When $0\le r\le \dfrac1{\sqrt3}$, we have $\sqrt2 r^2\le z\le \sqrt6 r^2$. Then when $\dfrac1{\sqrt3}\le r\le \dfrac1{\sqrt2}$, we have $\sqrt 2r^2\le z\le \sqrt{1-r^2}$.
Thus, your volume is given by the sum of two iterated integrals:
$$\text{vol}(E) = \int_0^{2\pi}\int_0^{1/\sqrt3}\int_{\sqrt2 r^2}^{\sqrt6 r^2}r\,dz\,dr\,d\theta + \int_0^{2\pi}\int_{1/\sqrt3}^{1/\sqrt2}\int_{\sqrt2 r^2}^{\sqrt{1- r^2}}r\,dz\,dr\,d\theta.$$
