Cylindrical coordinates on elliptic paraboloids. I want to compute the volume bounded by:


*

*the cylinder $x^2+4y^2=4$.

*the $z=0$ plane.

*the elliptic paraboloid $z = x^2 + 6y^2$.


I would like to use cylindrical coordinates. However I have never dealt with a problem in which $r$ bounds depend on $\theta$.
Here there's a horrible sketch of the solid:

The base is an ellipse. So I take the transformation to cylindrical coordinates:
\begin{align}
x&=r\cos\theta,\\
y&=r\sin\theta, \\
z&=z.
\end{align}
Which implies $0\leq z\leq x^2+6y^2 = r^2\cos^2\theta + 6r^2\sin^2\theta$ and $0\leq\theta\leq2\pi$.
However I'm having difficulties in determining the $r$ bounds. Also I would like to know the equation of the curve of intersection between the cylinder and the paraboloid, but don't know how to do it.
For the $r$ bounds I tried considering the equation for the base:
$$4y^2+x^2=4\Rightarrow 4r^2\sin^2\theta+r^2\cos^2\theta = 4 \Rightarrow r=\frac{2}{\sqrt{4\sin^2\theta+\cos^2\theta}}.$$
I'm not sure if doing that is correct, also the integral looks absolutely grim. I really appreciate your help.

Consider the change of variables $x=2u\Rightarrow dx = 2\,du$. Now we are working on a circular cylinder: $u^2+y^2=1$. Now let's take cylindric coordinates. Note that the Jacobian is $2r$. And the volume:
\begin{align}
V=\int_0^1\int_0^{2\pi}\int_0^{4r^2\cos^2\theta+6r^2\sin^2\theta} 2r\,dz\,d\theta\,dr = \int_0^1\int_0^{2\pi} 8r^3\cos^2\theta+12r^3\sin^2\theta\,d\theta\,dr\\
=\int_0^1\int_0^{2\pi} \overbrace{ 4r^3+4r^3\cos2\theta+6r^3-6r^3\cos2\theta}^{2\theta = s\;\Rightarrow\; 2\,d\theta=ds} \,d\theta\,dr\\
=\int_0^2\left[8\pi r^3+\int_0^{4\pi}2r^3\cos s\,ds+12\pi r^3 - \int_0^{4\pi}3r^3\cos s\,ds\right]dr\\
=\int_0^1 20\pi r^3\, dr = 5\pi.
\end{align}
