We are trying to ascertain the correct formula discussed in this post.
The volume formula for a capsule (a cylinder with a hemisphere at both ends) is,
$$V_c = \pi r^2 H + \frac{4}{3}\pi r^3\tag1$$
while that of a tank (a cylinder with a hemi-oblate spheroid at both ends) is,
$$V_t = \pi r^2 H + \frac{4}{3}\pi c r^2\tag2$$
with the capsule being the special case $c = r$.
For example, in this tank volume calculator diagram, with $H = 192$, $c = 18$, $r=36$, (in inches) the total fill volume is
$$V_t = 879444.88\;\text{in}^3 = 3807.12\; \text{US gallons}$$
and easily computed from $(2)$.
Question: Using the same diagram, what is the volume of water $V_p$ in a partially filled tank with water depth $h$?
P.S. Using $h=48\;\text{in}$, the website author claims that,
$$V_p = 2710\; \text{US gallons}$$
However, one cannot arrive at this value using the formula given by the author in the linked post so it is either wrong, or has some missing assumptions.
Let $H$ be the height and $R$ be the radius of the cylinder, and assume that the tank is filled up to height $h$, $0\leq h\leq 2R$. In the case of a capsule the volume of fluid consists of a segment of the cylinder and a segment of of a sphere, the latter coming in two separated halves. The volume of a sphere segment is computed exactly by Kepler's formula. One obtains $$V_{\rm sph}(h)={\pi h^2\over3} (3R-h)\ .\tag{1}$$ The volume of the cylinder segment is $H$ times the area of a circle segment of height $h$. One obtains $$V_{\rm cyl}(h)=H\left(R^2\left({\pi\over2}+\arcsin{h-R\over R}\right)+(h-R)\sqrt{h(2R-h)}\right)\ .$$ Equivalently, $$V_{\rm cyl}(h)=H\left(R^2\left(\arccos{R-h\over R}\right)+(h-R)\sqrt{h(2R-h)}\right)\ .$$ When instead of a sphere we have an ellipsoid with semiaxes $R$, $R$, and $C$ then $V_{\rm sph}(h)$ is multiplied by a factor ${C\over R}$, so that $(1)$ has to be replaced by $$V'_{\rm sph}(h)={\pi C h^2\over3R} (3R-h)\ .$$ Responding to a comment: Given $h$, we now have $$V_{\rm tank}(h)=V_{\rm cyl}(h)+V'_{\rm sph}(h)\ .$$ The expression on the right hand side contains algebraic terms in the variable $h$, as well as an $\arcsin$-term. Therefore it is impossible to express the inverse function $V\mapsto h(V)$ in terms of elementary functions.