Volume of a Torus Segment

I am calling a segment analogous to a circle's, that is a splitting a torus in two by a chord. The resulting piece has an ellipse formed by the chord. The chords distance from the torus centre lies between the torus's two radii.

Out of interest I'm trying to work out the volume of a liquid that partially fills a torus as in a tyres inner tube. The liquid does not come high enough to touch the inner radius. So there are just 3 variables, Torus radii q & r and height of liquid h.

tl; dr: For the stated dimensions, the quadratic approximation in the comments over-estimates the volume to be $(4\pi/3)h^{2}\sqrt{25 \cdot 360} \approx 400h^{2}\, \text{mm}^{3}$, or about $0.4h^{2}$ ml at a water depth of $h$ mm.

This estimate is only reasonably accurate if the depth is smaller than the "minor radius" of the tyre, i.e., half the difference of the outer and inner radii.

Let's denote by $R_{0}$ the outer radius (say $360$ mm) and by $R_{1}$ the inner radius (say, $310$ mm).

Mathematicians generally specify a torus in terms of its major radius $R = \frac{1}{2}(R_{0} + R_{1})$ (the radius of the core circle) and its minor radius $r = \frac{1}{2}(R_{0} - R_{1})$.

Suppose $x$ is the axis of the tyre, and the depth of water is measured along the $z$-axis. Near the bottom, the tyre surface is a graph of $z$ as a function of $(x, y)$. Over the $x$-axis this graph is the circle of radius $r$ with equation $x^{2} + (z + R)^{2} = r^{2}$, or $$ z = -R - \sqrt{r^{2} - x^{2}} \approx -(R + r) + \frac{x^{2}}{2r}. $$ Similarly, over the $y$-axis the graph is the circle of radius $R_{0} = (R + r)$ with equation $y^{2} + z^{2} = (R + r)^{2}$, or $$ z = -\sqrt{(R + r)^{2} - y^{2}} \approx -(R + r) + \frac{y^{2}}{2(R + r)}. $$ The desired quadratic approximation is $$ z \approx -(R + r) + \frac{x^{2}}{2r} + \frac{y^{2}}{2(R + r)}. $$ We seek the volume of the paraboloid $$ \frac{x^{2}}{2r} + \frac{y^{2}}{2(R + r)} \leq h. $$ If the top surface of the paraboloid is the ellipse $(x/a)^{2} + (y/b)^{2} = 1$, the volume is $(2\pi/3)abh$, i.e., two-thirds the volume of the circumscribed elliptical cylinder. Here, $a = \sqrt{2rh}$ and $b = \sqrt{2(R + r)h}$, so $$ \text{Volume} = \frac{4\pi}{3} \sqrt{r(R + r)} h^{2}. $$ If $h$ is small compared to $r = \frac{1}{2}(R_{0} - R_{1})$ this is a reasonable approximation of the volume of water in the tyre. For the stated dimensions $r = 50$, $R + r = 360$, this is $(4\pi/3)\sqrt{25\cdot360}h^{2} \approx 400h^{2}\, \text{mm}^{3}$, or $0.4h^{2}$ ml at a water depth of $h$ mm.

The diagram shows the situation $h = 25$, i.e., the tyre is filled to a depth equal to its minor radius. The green surface bounds the actual volume of water, while the gold ellipse shows the bound given by the quadratic approximation, which is visibly larger, but looks correct to within ten percent or so.