how to find the work done in pumping fluid out of a tank using integration and formula?

For example, a half-full cylindrical tank that holds a liquid that is 12 pounds per cubic foot(4ft radius by 8 ft tall). How do we find the work done in pumping the fluid out of the tank from the top outlet?

I tried using Work = Force * distance:

• determined the force is the volume and weight of the liquid, $$\pi*(4)^2 * 8/2 * 12 = 768\pi$$
• Because work is the integral of force: Work = $$\int_{4}^{8}768 \pi dx = 3072\pi$$ , which is incorrect (Ans: 4608ft-lb)

Where have I gone wrong here, did I get the value of the force wrong, where is my knowledge gap?

• What does $x$ signify in your integral? Commented Sep 30, 2021 at 12:48
• Change in distance? Commented Oct 1, 2021 at 7:19
• A change in distance could work, but you have to be much, much more precise than that in how you think about it. First, what object undergoes a change in distance? Second, how is this distance measured? Distance is from point A to point B; what is point A, what is point B? Is it straight line distance, distance along a curve, only the horizontal component of distance, only the vertical component? Commented Oct 1, 2021 at 11:10

The work done to pump out liquid from the top outlet is different for the liquid at different depths.

Volume of infinitely thin layer of liquid ($$dx$$) at depth $$x$$ from top is,

$$dV = \pi \cdot 4^2 \cdot dx = 16 \pi ~ dx ~$$ and this liquid needs to be taken $$x$$ distance against gravity.

So work done in pumping out the liquid in unit ft-lb is,

$$\displaystyle \int_4^8 \rho~ x ~ dV$$

where $$\rho = 12$$ lb / ft$$^3$$

• Incorrect though, the correct answer was 4608 ft-lb. $\displaystyle \int_4^8 \rho~ x ~ dV \implies \displaystyle \int_4^8 12 x ~ dV= 288$ ? Commented Oct 1, 2021 at 7:24
• $dV = 16 \pi dx$, right? Commented Oct 1, 2021 at 7:31
• So it is $\int_4^8 (12 \cdot 16 \pi) x ~ dx = 4608 \pi$ Commented Oct 1, 2021 at 7:34