Question

Laying on its side, in the basement of the math department is a large cylin- drical tank with a length of 8 m and a radius of 2 m. The tank is half full with the strongest coffee known to man having a density of rho = 1314 kg/m3. The department chair would like to serve the rest of the coffee to a conference of visiting mathematicians. Assuming g = 9.8 m/s2, set up but do not evaluate an integral to compute the work done to pump the coffee out through the top of the tank when the tank is half full.

Answer 1

To compute the work done to pump the coffee out of the tank, we need to consider the force required to move the coffee against gravity. The work done can be calculated using an integral that takes into account the density of the coffee, the height of the tank, and the area of the top surface of the tank.

Step-by-step explanation:

Work done to pump the coffee out of the tank:

To compute the work done to pump the coffee out, we need to consider the force required to move the coffee against gravity. The force can be calculated using the density of the coffee, the height of the tank, and the area of the top surface of the tank. The work done is then equal to the product of the force and the distance over which it is applied.

We can set up an integral to compute the work as follows:

1. First, let's consider a small element of height dh inside the tank.
2. The volume of this element is dV = (2πrh) * (2πr * dh) = 4π^2r^2h * dh.
3. The mass of this element is dm = dV * ρ = 4π^2r^2h * dh * ρ.
4. The force required to lift this element against gravity is dF = dm * g = 4π^2r^2h * dh * ρ * g.
5. The work done to lift this element is dW = dF * ds = 4π^2r^2h * dh * ρ * g * h, where ds is the distance over which the force is applied.
6. To find the total work done, we need to integrate this expression over the height of the tank. The limits of integration are from 0 to half the length of the tank (4m).
7. The integral becomes:
   Work = ∫ (0 to 4) 4π^2r^2h^2 * ρ * g * dh