Calculate the volume of the region trapped by $z^2=x^2+y^2, z=2(x^2+y^2)$ Task

Calculate the volume of the region trapped by $z^2=x^2+y^2, z=2(x^2+y^2)$ using a triple integral.

I'm kind of lost on this one, here's my (probably wrong) attempt:
Calculate the intersection, I get $z=0 \wedge z=\frac 12$. That means the region is: $E=\{(x,y,z): x^2+y^2=1/2, z=1/2\} \cup \{(x,y,z): x^2+y^2=0\}$. 
That means I should integrate the constant function $1$ over $E$ to get the volume of $E$, however, I don't think I can get a triple integral this way.
How do I proceed?
 A: Your definition of the region is incorrect (though your bounds for $z$ are correct). When $0\le z\le\frac{1}{2}$, the cone lies above the paraboloid; the circle of intersection at $z=\frac{1}{2}$ is $x^2+y^2 = \frac{1}{4}$. So one definition of the region is
\begin{equation*}
  -\frac{1}{2}\le x\le \frac{1}{2},\quad
  -\sqrt{\frac{1}{4}-x^2} \le y \le \sqrt{\frac{1}{4}-x^2},\quad
  2(x^2+y^2) \le z \le \sqrt{x^2+y^2}.
\end{equation*}
If you integrate $1$ over this region, you should get the answer. However, you may wish to use a different set of coordinates to make the integral simpler.

A: You have a function of two variables given implicitly to you. You can find the volume between the planes using a simple double integral.
A: If I may, I come up with a simpler solution. If using a triple integral is a must then, please, ignore my answer.
If $y=0$ then we have the following two equations
$$z^2=x^2\, \, \text{ and } \, \, z=2x^2.$$
I we look at the $xz$ plane we'll see the following intersection lines (drawn only for $x\ge0$):

Now, looking at the cross section of the body and the plane determined by $\color{red}z$ then we will see the ring whose area is 
$$A(z)=\pi\left(\frac z2-z^2\right).$$
The volume can be computed as follows
$$V=\int_0^{\frac12}A(z)\ dz=\frac{\pi}{48}.$$
