Height of liquid inside a cylindrical container. Please refer to details before answering!?

Question

A cylinder 15m by 4m lies on its 'side', not upright. Assume the volume is 42,000 gallons (Imperial gallons, not US). The tank is 40,000 gallons full. What is the height of the liquid present inside?

Answer 1

The volume of the tank is:
V = A · L
and the volume of the liquid with in the tank
V' = A' · L

Thus the difference in volume is represented by difference cross sectional area:
ΔA = A - A' = ΔV / L
= (42000gal - 4000gal) · 0.00454609 m³/gal / 15m
= 0.606m²

ΔA is the cross sectional area of the empty space above the liquid. This area is a circular segment of the area
ΔA = R/2 · ( θ - sin θ)
where R is radius of the circle (here R = 2m
and θ the opening angle of the segment
(follow the link for a sketch)

The equation is only numerically solvable.
e.g. by fix point iteration:
Rearrange the formula above to:
θ = 2ΔA/ R + sin θ
Therefore iterate by
θ_(n+1) = 2ΔA/ R + sin θ_n
(subscripts n+1 and n denote the number of the iteration step)
Starting with an intermediate initial value
θ_0 = π /2
leads to
θ = 1.252999 (= 71.8°)
with a precision of 6 digits in twelve iteration steps.

The height of segment (distance from top to liquid surface level) is
h = R (1 - cos(θ/2))

The height of the liquid surface measured from the bottom of the tank is
H = 2R - h = R (1 + cos(θ/2)) = 3.62m

Answer 2

Imagine looking down the cross section, you can integrate the area based on the angle from the horizontal, e.g. x=Rcos(theta) y=Rsin(theta) where R is the radius of the cylinder. Let the length of the cylinder be L Then the volume is:

R^2*L((theta)+pi/2+cos( theta ) * sin( theta ))= V

I don't think you can analytically solve this prob for the angle, but find it graphically, then you will also know the height for a known volume.