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 byR2for any tank, whereRis the tank's radius.

The equation for a unit circle isx2+y2=1whereR2=1We will be integrating over the change iny(change in fluid height), therefore we needxin terms ofyx2+y2x2x=1=1y2=1y2length of chord will be=2|x|therefore the length of the chord is:lchord=2(1y2)1/2The area of the circle in contact with the fluid isA=h12(1y2)1/2dyThe volume of fluid in the tank isV=h12(1y2)1/2dyLIntegrating:V=Lh12(1y2)1/2dy*(1y2)is a trig identity form ofa2y2;a=1sety=asin()ory=asin(u),dydu=acos(u)dy=acos(u)du=Lh22(1sin2(u))1/2cos(u)du=Lh22(cos2(u))1/2cos(u)du*wheresin2()+cos2()=1, therefore:1sin2()=cos2()=Lh22cos(u)cos(u)du=Lh22cos2(u)du*A double angle identity forcos(2u)=2cos2(u)1, therefore:2cos2(u)=1+cos(2u)=Lh2(1+cos(2u))dulooking at the underbraced portion:set(2u)=s, therefore:dsdu=2ds=2dudu=12dstherefore the underbraced portion can be written as:cos(2u)du=cos(s)12ds=12cos(s)ds=12sin(s)+cback-sub the substitution=12sin(2u)+cwe can now do our integration by parts:=L1du+Lcos(2u)duL(u+c)h2+L[12sin(2u)+c]h2*thec's are the constants of integration and will cancel out when the integral is evaluated.=L[u+12sin(2u)]h2*A double angle identity forsin(2u):sin(2u)=2sin(u)cos(u)=2sin(u)(1sin2(u))12therefore=L[u+122sin(u)(1sin2(u))12]h2*Recally=sin(u)u=arcsin(y)=L[arcsin(y)+y(1y2)12]h1=L[(arcsin(h)+h(1h2)12)(arcsin(1)+(1)(1(1)2)12)]=L[2+(arcsin(h)+h(1h2)12)]*For any tank with radiusRV=R2L[/2+(arcsin(h)+h (1h2))]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.