Find the volume bounded by the hyperboloid of one sheet

Find the volume bounded by the hyperboloid of one sheet $\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1$ and the planes $z = 0$ and $z = h$.

We slice the hyperboloid into elliptical cross-sections perpendicular to its axis.

Step 1: Cross-section at height $z$

\[\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1+\frac{z^2}{c^2}\]

an ellipse with semi-axes $a_z=a\sqrt{1+z^2/c^2}$ and $b_z=b\sqrt{1+z^2/c^2}$.

Step 2: Area of the cross-section

\[A(z) = \pi a_zb_z = \pi ab\left(1+\frac{z^2}{c^2}\right)\]

Step 3: Integrate from $z=0$ to $z=h$

\[V = \pi ab\int_0^h\left(1+\frac{z^2}{c^2}\right)dz = \pi ab\left[z+\frac{z^3}{3c^2}\right]_0^h = \pi abh\left(1+\frac{h^2}{3c^2}\right)\]

\[\boxed{V = \pi abh\left(1+\frac{h^2}{3c^2}\right) = \frac{\pi abh(3c^2+h^2)}{3c^2} \text{ cubic units}}\]

Step 4: Special case check

At $h=0$: $V=0$ ✓, as expected for a degenerate slice.

Step 5: Interpretation

\[V = \underbrace{\pi abh}_{\text{cylinder of base area }\pi ab} + \underbrace{\frac{\pi abh^3}{3c^2}}_{\text{extra volume from flaring outward}}\]

Summary

\[\boxed{V = \pi abh\left(1+\frac{h^2}{3c^2}\right) \text{ cubic units}}\]

Key insight: Near its narrowest point ($z=0$, the "waist"), the hyperboloid closely resembles a plain elliptical cylinder of cross-section $\pi ab$; the correction term $\frac{\pi abh^3}{3c^2}$ captures the extra volume from the hyperboloid's outward flare as $z$ increases, with the flare rate controlled by $c$.