If the cut is parallel to the axis of the cone then you'll get parabolas as cross-sections.
Have a look here: Conic section - Wikipedia, the free encyclopedia
It is a lot easier to do it the other way but here it goes.
The equation of a cone is $\displaystyle x^2+y^2=R^2z^2$.
Suppose the radius of the base at $\displaystyle z=h$ is $\displaystyle r$. Then we have $\displaystyle Rh=r$ or $\displaystyle R=\dfrac{r}{h}$
Now we let $\displaystyle z\in\left[0,h\right]$
Next we have the equation of a general plane $\displaystyle Ax+By+Cz=D$. We want to construct a plane $\displaystyle \mathcal{P}||(\hat{\textbf{j}},\hat{\textbf{k}})$
This would mean that $\displaystyle x=const$ or $\displaystyle \mathcal{P}\rightarrow x=x_0$.
Now lets intersect this plane with our cone. This gives us the curve
$\displaystyle R^2z^2-y^2=x_0^2$ or $\displaystyle y=\pm\sqrt{R^2z^2-x_0^2}$
Now at $\displaystyle z=h$ $\displaystyle y=\pm\sqrt{r^2-x_0^2}$ which will be our limit of integration. Keep in mind that the vertex of the cone is at $\displaystyle z=0$. We know that $\displaystyle z>0$ therefore $\displaystyle z=\dfrac{\sqrt{y^2+x_0^2}}{R}$
Next we find the area $\displaystyle A(x_0)=2hr-\int_{-\sqrt{r^2-x_0^2}}^{\sqrt{r^2-x_0^2}}\dfrac{\sqrt{y^2+x_0^2}}{R}\,dy=$ and I'm going to stop here as it is getting quite messy. I guess that helps to a point but I don't see any point in continuing.
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)