Fluid force/calculus

Question

A certain open top container is made by rotating the curve y = x^2/a about the y-axis from x = 0 meters to x = 3 meters, where a = 1 meter. Initially the container is filled to a height of H = 4 meters with water of density ρ = 1000 kg/m^3 . How much work does it take to pump all of the water out of the container? Take g = 10 m/s^2

Answer 1

Rotating the curve y=x²/a about the y-axis creates a paraboloid with a radius of x that increases with your y-value. Solve for the radius in terms of y:

r = x = √(y/a)

Now use the bounds of the container in terms of x to solve for the bounds of the container in terms of y:

x=0: y(0) = x²/a = (0)²/1 = 0

x=3: y(3) = x²/a = (3)²/1 = 9

Use the radius and height to solve for the volume of water:

V = (π r²)(h) = π (√(y/a))² (Δy) = π (y/a) (Δy)

Use the volume and density to solve for the mass of water:

m = ρ * V = (1000 kg/m³) * [π (y/a) (Δy)] = (1000π/a) * y * Δy

Use mass and acceleration to solve for the weight of the water, which is the force that the pump has to overcome:

F = m * a = m * g = [(1000π/a) * y * Δy] * (10 m/s²) = (10,000π/a) * y * Δy

Use force of the water and distance it has to be moved to solve for work (the distance each layer of water would have to move would be the height of the top of the container subtracted by the height of each layer of water):

W = F * d = [(10,000π/a) * y * Δy] * (9 - y) = (10,000π/a) * y * (9 - y) * Δy

= (10,000π/a) * (9y - y²) * Δy

Create an integral to determine the sum of the work to pump all of the water from the tank (note that the integral is from y=0 to y=4 since the water starts at a depth of 4m and ends at a depth of 0m):

W = ∫(10,000π/a) * (9y - y²) dy = (10,000π/a) ∫(9y - y²) dy = (10,000π/a) [4.5y² - (y³/3)]

= (10,000π/1) [(4.5*4² - (4³/3)) - (4.5*0² - (0³/3))] = (10,000π) [(72 - 64/3) - (0 - 0)] = (10,000π)(152/3)

= 1,520,000π/3 ≈ 1,591,738.93 N*m ≈ 1,591,738.93 J ≈ 1,592 kJ