I am told to solve the following problem using Riemann Sums and Definite Integrals:

A tank has the shape of an inverted cone with height $10$m and base radius $4$m. It is filled with water to a height of $8$m. It is required to pump all the water out of the tank by inserting a hose through the top to which a pump is attached. Calculate the work done against gravity to achieve this. (Assuming that the gravity constant is $g = 9.8$m/s$^2$ and the density (mass per unit volume) of the water is $1000$kg/m$^3$).

I have learned about Riemann sums $\sum_{i = 1}^{n} f(c_j)\Delta x_j$ and the basics of solids of revolution. Looking at this problem, given that we are told that the tank is an inverted cone, I understand that we should use the area of a circle ($\pi r^2$) to give us the volume for each interval of the Riemann sum. Excluding this insight, I am thoroughly confused with how to proceed.

I would greatly appreciate it if people could please take the time to write out a step-by-step solution of how to solve this problem and the reasoning behind each step. I'm eager to study any step-by-step solutions offered by people on this site, so that I may be able to independently solve these types of problems in the future.


The work done equals the weight of the water times the distance you need to lift it.

At any height (h) the volume of water is equals the surface area of the cross section at that height $\pi (r^2)$

Since the tank is cone shaped. $r = 0.4h$

and the weight of the water $= 9.8$ volume.

The distance you need to lift it $10 - h$

work $\int_0^8 (9.8)(\pi)(0.4 h)^2 (10-h) dh$

  • $\begingroup$ Thanks for the response. Why do we use $0$ to $8$ for the limits of integration instead of $0$ to $10$? Doesn't the water have the travel all $10$m to actually leave the tank, despite there only being $8$m of water? $\endgroup$ – The Pointer Mar 15 '17 at 0:02
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    $\begingroup$ @ThePointer The 10m height appears in the $(10-h)$ term. There’s only 8m of water to lift. $\endgroup$ – amd Mar 15 '17 at 0:08
  • $\begingroup$ @amd Ahh, because the limits of integration come from the Riemann sum, which models the volume of water rather than the height? $\endgroup$ – The Pointer Mar 15 '17 at 0:12
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    $\begingroup$ @ThePointer That’s it exactly. $\endgroup$ – amd Mar 15 '17 at 0:13
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    $\begingroup$ Limits of integration are 0-8 because that is the height that the water goes up the tank (10-h) because the water needs to be lifted to the top of the tank. 9.8 is the gravitational constant. I neglected the density factor. So, multiply by 1000 $\endgroup$ – Doug M Mar 15 '17 at 7:38

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