My case is the third figure in your link, which is a hyperbola. Is the cross-section that I have taken in my original attachment a parabola or a hyperbola?

Of course you are right and I apologize for the confusion.

Obviously a hyperbolic cross-section is hard to handle as fobos3 has demonstrated. So I suggest to use parabolic cross-sections.

According to Archimedes a segment of a parabola has an area

\(\displaystyle A = \frac43 \cdot \underbrace{\frac12 \cdot 2 b \cdot h}_{area\ of\ triangle} = \frac43 b h\)

Use the proportion \(\displaystyle \dfrac hs = \dfrac{2R-x}{2R}~\implies~h=\dfrac{s(2R-x)}{2R}\)

Use Pthagorian theorem to calculate the length of \(\displaystyle b = \sqrt{R^2-(R-x)^2}\)

**Caution: This step probably contains a mistake which causes a wrong final result! I wasn't able to spot my error.**
Now the colume of the cone should be:

\(\displaystyle V=\displaystyle{\int}_0^{2R}\left(\frac43 \cdot \dfrac{s(2R-x)}{2R} \cdot \sqrt{R^2-(R-x)^2}\right) dx\)

My computer tells me that \(\displaystyle V = \dfrac \pi3 \cdot R^2 \cdot s\) which is definitely too large.

(... but it looks like a very near miss)