Triple integral between the paraboloids $z = x^2 + y^2$ and $z = 4 - x^2 - 3 y^2$. I have to compute $\int_D f$, where $D$ is the region in ${\mathbb{R}}^3$ between the paraboloids $z = x^2 + y^2$ and $z = 4 - x^2 - 3 y^2$, and $f : D \to \mathbb{R}$ is given by $f(x , y , z) = x^2 + 2 y^2$.
My attempt: If I use cylindrical coordinates $(\rho , \theta , z) \in (0 , \infty) \times (0 , 2 \pi) \times \mathbb{R}$ given by
$$
\left\{
\begin{array}
xx = \rho \cos \theta \\
y = \rho \sin \theta \\
z = z
\end{array}
\right.
$$
The bounds of $z$ with the paraboloids are
$$
{\rho}^2 = x^2 + y^2 \leq z \leq 4 - x^2 - 3 y^2 \leq 4 - {\rho}^2 (1 + 2 {\sin}^2 \theta).
$$
(why do we have $x^2 + y^2 \leq z \leq 4 - x^2 - 3 y^2$ and not for example $x^2 + y^2 \geq z \geq 4 - x^2 - 3 y^2$ without drawing?). From this we deduce that
$$
x^2 + y^2 \leq 4 - x^2 - 3 y^2 \quad \Longrightarrow \quad x^2 + 2 y^2 \leq 2 \quad \Longrightarrow \quad \rho \leq \sqrt{\frac2{1 + 2 {\sin}^2 \theta}}.
$$
Finally the integral is
$$
\int_{\theta = 0}^{2 \pi} \left(\int_{\rho = 0}^{\sqrt{\frac2{1 + 2 {\sin}^2 \theta}}} \left(\int_{z = {\rho}^2}^{4 - {\rho}^2 (1 + 2 {\sin}^2 \theta)} {\rho}^3 (1 + 2 {\sin}^2 \theta) \, dz\right) \, d \theta\right) \, d \theta = \frac{32 \pi}{9 \sqrt{3}}.
$$
Does it look good?
 A: In your working, you have $\rho \leq \sqrt{\frac2{1 + 2 {\sin}^2 \theta}}$. That is not correct as you can see. It should be,
$\rho \leq \sqrt{\frac2{1 + {\sin}^2 \theta}}$. Also the same correction is needed in integrand, $x^2+2y^2 = \rho^2 (1+\sin^2\theta)$ and not $\rho^2 (1 + 2 \sin^2\theta)$.
Having said that, here is what I would suggest as a simpler working. You found the intersection of surfaces as $x^2+2y^2 = 2$. So use the following substitution,
$x = \sqrt2 \ r \cos\theta, y = r \sin\theta$ then the ellipse $x^2+2y^2 = 2$ transforms to a circle of radius $1$ centered at origin. Jacobian of transformation is $\sqrt2 \ r$.
So equations of paraboloids become,
$z = x^2 + y^2 = r^2 + r^2 \cos^2\theta; z = 4 - x^2 - 3y^2 = 4 - 2 r^2 - r^2 \sin^2\theta$
Integrand $x^2 + 2y^2 = 2 r^2$.
Now we evaluate the integral,
$\displaystyle \iiint_D (x^2+2y^2) \ dx \ dy \ dz = 2 \sqrt2 \int_0^{2\pi} \int_0^1 \int_{r^2 + r^2 \cos^2\theta}^{4 - 2 r^2 - r^2 \sin^2\theta} r^3 \ dz \ dr \ d\theta $
$ \displaystyle  = \frac{4 \sqrt2 \ \pi}{3}$
