Finding the limits in the enclosed volume between a paraboloid and parabolic cylinder. I've been struggling trying to work out the following exercise:

Here are some $xy$ and $yz$ cuts I've made when considering the limits of the following shape:

Here's a $xz$ picture of these two objects. I don't know if these two lines are actually creating an enclosed area - as, for instance, I don't know if the general line for $z=4\alpha ^2 + y^2$ even encloses the blue line, but I'm assuming it does since this would make the problem solvable (not a good justification, but..)

Here is the image of the two objects with constant $z$.
Here, I don't know if the general equation for the ellipse is actually even cut by the lines $y = \pm \sqrt{(4/3) - (\alpha/3)}$, so I don't know if the $y$ limit is the upper and lower ends of the ellipse or the equation of the two horizontal lines I specified in the previous sentence.
If these lines are for some constant $alpha$, how can I deduce the volume I'm sketching in? Depending on alpha, this can potentially change the limits, I'd think, as in the case of the second picture. With $alpha$ being sufficiently large, the ellipse is cut by the two lines.
 A: Equate equations $z=4x^2+y^2=4-3y^2$. We get, $x^2+y^2=1$.
Use cylindrical cordinate system.
$r:0\to 1$
$\theta:0\to 2\pi$
$z:4-3r^2\sin^2\theta \to r^2+3 \cos^2\theta$
$dxdydz=rdrd\theta dz$
A: Let me provide a solution without using cylindrical coordinates. 
Firstly you compute the intersection between the paraboloid and the parabolic cylinder which is given by 
$$x^2+y^2=1$$
Now you stop for a while and try to make sense of it visually. 

And now it should be clear that this is your desired integral. 
$$\int_{-1}^{1}\left(\int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}\left(\int_{4x^{2}+y^{2}}^{4-3y^{2}}dz\right)dy\right)dx$$
Just to make sure you understood, can you see why the above integral is equivalent to the one below? 
$$\int_{-1}^{1}\left(\int_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}}\left(\int_{4x^{2}+y^{2}}^{4-3y^{2}}dz\right)dx\right)dy$$
A: Intersection : $4x^2+y^2=4-3y^2 \implies4y^2+4x^2=4 \implies x^2+y^2=1$
using Cylindrical coordinates:
$$
\left\{ 
\begin{array}{c}
x=rcos\theta\\
y=rsin\theta\\
z=z
\end{array}
\right. 
$$
$\displaystyle\int_{0}^{2\pi}\int_{0}^{1}\int_{4r^2cos^2\theta+r^2sin^2\theta}^{4-3r^2sin^2\theta}rdzdrd\theta=8\pi\int_{0}^{1}r-r^3dr=2\pi$
