Finding volume of hyperboloid bounded by two planes I want to find the volume bounded by hyperboloid $\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}-\cfrac{z^2}{c^2} = 1$ and the planes $z=-c, z=c$.
I do not know whether should I use the cylindirical coordinates or spherical coordinates.  
At first, I am thinking to set $x =au, y=bv,z=cw$ and now we have $u^2+v^2 -w^2 = 1$. Jacobian of this transformation is $abc$. I guess that after this change of variables If I take a cross section then it will give me a circle on $uv-$plane but I do not know the reason.
If we consider the part of the volume that is inside the first octant, my intuition says that the desired volume is $\displaystyle \int_{0}^{1}\int_{0}^{1-v^2}\int_{0}^{u^2+v^2-1}g(u,v,w)abc\times dwdudv$
 and I am not sure about $g$...
To sum up, I appreciate if you could explain me what is going on exactly in a basic way. 
 A: I like your change of coordinates step.  So it's enough to show that the volume enclosed by the “standard” of one sheet 
$$
    x^2 + y^2 - z^2 = 1,\ \ -1 \leq z \leq 1
$$
is $\frac{8\pi}{3}$.  
For this, use cylindrical coordinates.  The curved surface has equation $r^2 - z^2 = 1$, and the enclosed solid can be represented as
$$
    \left\{(r,\theta,z) : -1 \leq z \leq 1,\ 0 \leq \theta \leq 2\pi,\ 0 \leq r \leq \sqrt{1+z^2}\right\}
$$
Therefore, the volume is
\begin{align*}
    \int_{-1}^1\int_0^{2\pi}\int_0^\sqrt{1+z^2} r\,dr \,d\theta\,dz = \frac{8\pi}{3}
\end{align*}
as desired.

Another user suggested, but unfortunately deleted, a “Calc II” solution where you realize this as a solid of revolution.  It's the curve $z^2-x^2 = 1$ revolved around the $z$-axis, between $z=-1$ and $z=1$.  A slice of the solid at a constant $z$-coordinate is a disk of radius $x^2$.  Since $x^2=1+z^2$, the area of that disk is $\pi(1+z^2)$.  Therefore the volume is
$$
    \int_{-1}^1 \pi(1+z^2)\,dz = \frac{8\pi}{3}
$$
In fact, $\int_0^{2\pi}\int_0^{\sqrt{1+z^2}}r\,dr\,d\theta = \pi(1+z^2)$, so this is no coincidence.
