I am stuck on a triple integral problem that reads:
Find the volume of the compact $$ A = \left\{(x, y, z)\in\mathbb{R}^3; x^2 + 4y^2 \leq 2, x^2 + 4y^2 \leq z^2+1, 0 \leq z \leq 3\right\}$$
So I have to calculate $$\int_A \text{d}x\ \text{d}y\ \text{d}z$$
Now this is what I though: using elliptic coordinates like $x = R\cos\theta$ $y = \dfrac{R}{2}\sin\theta$ gives me
$$R^2\leq 2$$ but also
$$R^2 \leq z^2+1$$
Since I already have the range for $z$ ($z\in [0, 3]$), and since $\theta\in[0, 2\pi]$ how do I interpret the two results above, for $R$?
I don't understand this part, basically.
Thank you in advance!
Your region can be split into two simple subregions, according to whether $0 \leq z \leq 1$ or $1 \leq z \leq 3$. For the former, you will integrate $R$ from $0$ to $\sqrt{z^2+1}$, since the constraint $R^2 \leq z^2+1$ implies the constraint $R^2 \leq 2$. For the latter it's the opposite: the constraint $R^2 \leq 2$ implies the constraint $R^2 \leq z^2+1$, so you will integrate $R$ from $0$ to $\sqrt 2$ (this subregion is just a filled up elliptic cylinder).
EDIT: More clarification for the second part: when $1 \leq z \leq 3$, automatically $\sqrt{z^2+1} \geq \sqrt 2$, so $$R^2 \leq 2 \implies R^2 \leq 2 \leq z^2 + 1.$$ Thus there is an equality of sets $$ \{(R,z) : R^2 \leq 2, \ R^2 \leq z^2 +1, \ 1 \leq z \leq 3\} = \{(R,z) : R^2 \leq 2, \ 1 \leq z \leq 3\},$$ i.e., you can drop the second constraint, and then your second integral will look like $$ \int_0^{2\pi}\int_0^\sqrt2 \int_1^3 \underbrace{\frac R2 \, dz \, dR \, d\theta}_{dV}.$$