This is a standard problem in Calculus I or II for Work, Power, and Energy.
[Prologue] Work done on a body is fundamentally given by: W = Force*(distance the body is moved in the direction of the Force).
When considering a tank full of water, we need to consider given tiny layer of water in the tank, at a certain depth ‘h’ in the tank, and the work done to raise that tiny layer of water to the top of the tank. The layer of water is infinitesimal, so we need to Integrate Force with respect to depth.
In our case, the “Force” is the “weight” of the water layer, i.e. mass*gravity, and distance incurred during the Work is height “h” of some water to the top.
For “Force” = “Weight”, we need “mass”, but we are given water “density”, and we can figure out “volume” to produce “mass”, then “weight” = “force”.
Infinitesimal volume: Δvᵢ = πr²Δhᵢ
Infinitesimal mass: Δmᵢ = (πr²Δhᵢ)(density 'ρ' provided) ==> Δmᵢ = (πr²Δhᵢ)(1000 kg/m³)
Infinitesimal weight = force: ΔFᵢ = (Δmᵢ)(9.8 m/s² for gravity 'g') = (πr²Δhᵢ)(ρ = 1000 kg/m³)(g = 9.8 m/s² for gravity)
Infinitesimal work = ΔWᵢ = ΔFᵢ*hᵢ = (πr²Δhᵢ)(1000)(9.8 m/s²)*hᵢ = (πr²)(1000)(9.8 m/s²)*hᵢΔhᵢ
Approximate total work = Σᵢ ΔWᵢ
Total Work W = ∫₀ᴴ (πr²)(1000)(9.8 m/s2)*hᵢΔhᵢ = (πr²)(1000)(9.8 m/s2) ∫₀ᴴ hdh = (πr²)(1000)(9.8 m/s2) (h²/2) |₀ᴴ ==>
W = (πr²)(1000 kg/m³)(9.8 m/s²) (1/2)H² [ m²*kg/m³*m/s²*m² = kg*m²/s²*m = Newton-meter = Joules ]
Initial details: Problem description gives us a = 4 m, b = 4 m, c = 9 m, and d = 1 m)...
No idea what are a, b, c, d... we only require radius 'r' and height 'H' of the drum, as follows:
W = (πr²)(1000 kg/m³)(9.8 m/s²)(1/2) H²
Let's take the dimensions of a standard steel drum:
Internal Diameter: 572 mm ==> Internal Radius = 286 mm ==> r = 0.286 m
Internal Height: H = 851 mm = 0.851 m
W = π(0.286)²(1000)(9.8)(1/2)(0.851)² = 911 Joules
