As usual the teachers solution sheet takes leaps and bounds over steps in the solution that I need to understand it.
Q:
Determine the volume of the body limited by $x^2+y^2-3z^2=1$, $z=1$ and $z=-1$.
So that's a hyperboloid, something like this (but also limited by $z=1$ and $z=-1$).
The teacher jumps over the part where I always struggle, where you determine the limits of integration based on the given curves. I never get it right.
So, the first step of the solution here:
$$V=\int_{-1}^1\iint\limits_{x^2+y^2\le1+3z^2}dxdydz=...$$
I get what is happening, but what's the reasoning behind rewriting the limit to $x^2+y^2\le1+3z^2$? As I interpret it you get a circle with a radius from $0$ to $1+3z^2$. Am I mistaken?
In the next step the teacher jumps straight into this:
$$...=\int_{-1}^1\pi(1+3z^2)dz=...$$
What happened there? Usually in these cases I have to turn to polar coordinates, because they are easier to understand.
It ends straight up with
$$...=\pi[z+z^3]_{-1}^1=4\pi$$
Except my questions above, I would really like to get some guidance in how to think when it comes to the limits of integration in cases like these. What am I aiming to achieve?
Thank you for your time, it's invaluable to me.
It is natural to slice the region by planes parallel to the $xy$-plane going from $z=-1$ to $z=1$. The intersection of each such plane with the region is a disk (filled in circle) of radius $\sqrt{1+3z^2}$. So the volume of the region is given by the single integral $$\int_{-1}^1 A(z)\,dz\,,$$ where $A(z)$ is the cross-sectional area of the slice at height $z$. This is, in turn, $\pi(1+3z^2)$, since the area of a circle of radius $r$ is $\pi r^2$.
It's really a mess to write the integral $A(z)$ as a double integral in Cartesian coordinates, so your teacher skipped that and just wrote the region as $x^2+y^2\le 1+3z^2$.
Your instinct to use cylindrical coordinates isn't bad. Because of the shape of the region, we want $dz$ on the outside and then we can represent $A(z)$ as a double integral in polar coordinates:
$$\int_{-1}^1 \underbrace{\int_0^{2\pi}\int_0^{\sqrt{1+3z^2}} r \,dr\,d\theta}_{A(z)}\ dz$$