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?
 A: 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:
$$\text{vol}(E) = \int_0^{2\pi}\int_0^{1/\sqrt3}\int_{\sqrt2 r^2}^{\sqrt6 r^2}r\,dz\,dr\,d\theta + \int_0^{2\pi}\int_{1/\sqrt3}^{1/\sqrt2}\int_{\sqrt2 r^2}^{\sqrt{1- r^2}}r\,dz\,dr\,d\theta.$$
A: To expand on my comment, we first observe that the upper paraboloid $z=\sqrt6\,(x^2+y^2)$ meets the sphere $x^2+y^2+z^2=1$ in a cylinder with radius $\frac1{\sqrt3}$ :
$$\begin{cases}z = \sqrt6\,(x^2+y^2) \\ x^2 + y^2 + z^2 = 1 \end{cases} \implies \frac{z}{\sqrt6} + z^2 = 1 \implies z = \sqrt{\frac23} \\
\implies x^2 + y^2 + \frac23 = 1 \implies x^2 + y^2 = \left(\frac1{\sqrt3}\right)^2$$
Now, if $(x,y,z)$ belongs to the cylinder's interior $x^2+y^2\le\frac13$, then the paraboloids determine the "height" of the solid and we have $\sqrt2\,(x^2+y^2)\le z \le \sqrt6\,(x^2+y^2)$.
Otherwise, if $(x,y,z)$ is outside this cylinder, then $\sqrt2\,(x^2+y^2)\le z\le\sqrt{1-x^2-y^2}$.
Where the lower paraboloid and sphere meet will help set up the limits for $x$ and $y$. We have
$$\begin{cases} z = \sqrt2\,(x^2+y^2) \\ x^2+y^2+z^2 = 1 \end{cases} \implies \frac z{\sqrt2} + z^2 = 1 \implies z = \frac1{\sqrt2} \\
\implies x^2+y^2 + \frac12 = 1 \implies x^2+y^2 = \left(\frac1{\sqrt2}\right)^2$$
which is to say, the whole solid lies inside a cylinder of radius $\frac1{\sqrt2}$, so we can choose e.g. $-\frac1{\sqrt2}\le x\le\frac1{\sqrt2}$ and $|y| \le \sqrt{\frac12-x^2}$.
The volume is then given by the triple integral,
$$\iiint_E dV = \int_{-\frac1{\sqrt2}}^{\frac1{\sqrt2}} \int_{-\sqrt{\frac12-x^2}}^{\sqrt{\frac12-x^2}} \int_{\sqrt2\,(x^2+y^2)}^{\min\left\{\sqrt6\,(x^2+y^2),\sqrt{1-x^2}\right\}} \, dz \, dy \, dx$$
By symmetry, it's the same as
$$\iiint_E dV = 4 \int_0^{\frac1{\sqrt2}} \int_0^{\sqrt{\frac12-x^2}} \int_{\sqrt2\,(x^2+y^2)}^{\min\left\{\sqrt6\,(x^2+y^2),\sqrt{1-x^2}\right\}} \, dz \, dy \, dx$$
Converting to cylindrical coordinates from here should be straightforward. Expanding this any further in Cartesian is overkill.
