I would like to share the mathematical steps I took to solve this problem. I solved for a horizontal tank without the hemispherical caps. So essentially the problem is, "How does the cross-sectional area of a circle change with respect to height from the bottom of the tank?"
I know that the length of the tank does not change, therefore I only need to solve for the change in area of the circle in contact with the fluid and then multiply by the tank's length to get the fluid's volume. I can first solve the problem for a Unit Circle and then scale by $R^2$ for any tank, where $R$ is the tank's radius.

The equation for a unit circle is $x^2+y^2=1$ where $R^2=1$
We will be integrating over the change in $y$ (change in fluid height), therefore we need $x$ in terms of $y$
\begin{align} x^2+y^2 &=1 \\ x^2 &= 1-y^2\\ x &=\pm \sqrt{1-y^2} \ \text{length of chord will be} =2 \times |x| \end{align}
therefore the length of the chord is: $l_{\text{chord}}=2(1-y^2)^{1/2}$
The area of the circle in contact with the fluid is
$A=\int_{-1}^h 2(1-y^2)^{1/2} \ dy$
The volume of fluid in the tank is
$V=\int_{-1}^h 2(1-y^2)^{1/2} \ dy \times L$
Integrating:
$V=L \int_{-1}^h 2(1-y^2)^{1/2} \ dy $
*$(1-y^2)$ is a trig identity form of $a^2-y^2$; $a=1$
set $y=a\sin(\theta)$ or $y=a\sin(u)$, $\frac{dy}{du}=a\cos(u) \Rightarrow dy=a\cos(u) \ du$
\begin{align} &=L \int_{\frac{-\pi}{2}}^{h'}2(1-\sin^2(u))^{1/2}\cos(u) \ du \\ \nonumber\\ &=L \int_{\frac{-\pi}{2}}^{h'}2(\cos^2(u))^{1/2} \cos(u) \ du \end{align}
*where $\sin^2(\theta)+\cos^2(\theta)=1$, therefore: $1-\sin^2(\theta)=\cos^2(\theta)$
\begin{align} &=L \int_{\frac{-\pi}{2}}^{h'}2\cos(u)\cos(u) \ du \\ \nonumber\\ &=L \int_{\frac{-\pi}{2}}^{h'}2\cos^2(u) \ du \end{align}
*A double angle identity for $\cos(2u)=2\cos^2(u)-1$, therefore: $2\cos^2(u)=1+\cos(2u)$
$=L \int_{\frac{-\pi}{2}}^{h'}(1+\underbrace{\cos(2u)}) \ du$
looking at the underbraced portion:
set $(2u)=s$, therefore: $\frac{ds}{du}=2$
$ds=2 \ du \Rightarrow du=\frac{1}{2} \ ds$
therefore the underbraced portion can be written as:
\begin{align} \int \cos(2u) \ du &=\int \cos(s) \times \frac{1}{2} \ ds \\ &=\frac{1}{2} \int \cos(s) \ ds \\ &=\frac{1}{2} \sin(s)+c \ \ \text{back-sub the substitution}\\ &=\frac{1}{2} \sin(2u)+c \end{align}
we can now do our integration by parts:
$=L \int 1 \ du + L \int \cos(2u) \ du$
$L(u+c)\Biggr|_{\frac{\pi}{2}}^{h'}+L\left[\frac{1}{2}\sin(2u)+c\right]_{\frac{\pi}{2}}^{h'}$
*the $c$'s are the constants of integration and will cancel out when the integral is evaluated.
$=L\left[u+\frac{1}{2}\sin(2u)\right]_{\frac{\pi}{2}}^{h'}$
*A double angle identity for $\sin(2u)$:
\begin{align} \sin(2u) &=2\sin(u)\cos(u)\\ &=2\sin(u)(1-\sin^2(u))^{\frac{1}{2}} \end{align}
therefore
$=L\left[u+\frac{1}{2} \times 2\sin(u)(1-\sin^2(u))^{\frac{1}{2}}\right]_{\frac{\pi}{2}}^{h'}$
*Recall $y=\sin(u) \Rightarrow u=\arcsin(y)$
\begin{align} &=L\left[\arcsin(y)+y \times (1-y^2)^{\frac{1}{2}}\right]_{-1}^{h} \\ \\ &=L\left[\left(\arcsin(h)+h \times (1-h^2)^{\frac{1}{2}}\right)-\left(\arcsin(-1)+(-1)(1-(-1)^2)^{\frac{1}{2}}\right)\right] \\ \\ &=L\left[\frac{\pi}{2}+\left(\arcsin(h)+h(1-h^2)^{\frac{1}{2}}\right)\right] \end{align}
*For any tank with radius $R$
$V=R^2 \times L\left[\frac{\pi}{2}+\left(\arcsin(h)+h\sqrt{(1-h^2)}\right)\right]$
Interestingly, I ran into this problem while out on a frac job and we had missplaced our dipstick to our acid transport. We needed to keep tabs on how much acid we were pumping into the well.