How to find Lebesgue measure of a set? How to find $\lambda_3(A)$, where
$$A = \{ (x,y,z) \in \mathbb{R}^3 | z^2 < \frac{x^2}{4} + \frac{y^2}{9} < 2z \}$$
I know that measure of set A is given by $\int_A 1 d\lambda_3$.
I suppose that here I should use  Fubini Theorem, however I can't figure out how to get the limits of integration.
When we integrate sum of intervals can we write it out as a sum of integrals of intervals?
What I mean is when we have $$\int_A 1 d\lambda_3$$ I want to rewrite it as
$$\int_{0}^{2} \int_a^b \int_c^d 1 dxdydz$$
where $c,d$ depend on $y$ and $z$ while $a,b$ depend on $z$.
However since in the definition of set $A$ there are squares, set of possible values of $x$ is not an interval, but a sum of few intervals, since we can write $\frac{36z^2 - 4y^2}{9} < x^2 < \frac{72z - 4y^2}{9}$.
So my question is is it true that in such situation we can write:
$$\int_{0}^{2} \int_{a_1}^{b_1} \int_{c_1}^{d_1} 1 dxdydz + ... + \int_{0}^{2} \int_{a_k}^{b_k} \int_{c_k}^{d_k} 1 dxdydz$$
where $a_i, b_i, c_i, d_i$ would be boundary points of intervals of possible value of variable.
 A: Go to these eliptic cylindrical coordinates: $F:(0, \infty) \times (0, 2\pi) \times \mathbb{R}$,
$$
F(r, \theta, u) := (2r\cos(\theta), 3r\sin(\theta), u)^\top.
$$
which have volume element
$$
\mathrm{det}DF(r, \theta, u) = \mathrm{det}
\begin{pmatrix}
2\cos(\theta) & -2r\sin(\theta) & 0 \\
3\sin(\theta) & 3r\cos(\theta) & 0 \\
0 & 0& 1
\end{pmatrix} = 6r.
$$
Further note that $z^2 < 2z$ for all $(x, y, z) \in A$ which is equivalent to $z \in (0, 2)$. So in our new coordinates, $A$ becomes
$$
A = \lbrace (r, \theta, u) \in (0, \infty) \times (0, 2\pi) \times (0, 2): u^2 < r^2 < 2u \rbrace
$$
which means that the integral transforms to
$$
\int_A~\mathrm{d}\lambda_3 = \int_0^{2\pi} \int^2_0  \int^{\sqrt{2u}}_{u} 6r~\mathrm{d}r  ~\mathrm{d}u ~\mathrm{d}\theta = \int^{2\pi}_0 \int^2_0 -3\left(u-2\right)u~\mathrm{d}u = \int^{2\pi}_0 4~\mathrm{d}\theta = 8\pi.
$$
Correct me if I made a computational error.
A: The area of the ellipsis $\displaystyle\frac{x^2}{4}+\frac{y^2}{9}\le r^2$ is $6\pi r^2$, hence the integral has value
\begin{equation}
I = 6\pi \int_0^2 (2 z - z^2) d z = 6\pi\left(4-\frac{8}{3}\right)=8\pi
\end{equation}
