A tank is full of oil weighing 20 lb/ft${^3}$. The tank is an inverted right rectangular pyramid (with the base at the top) with a width of 2 feet, a depth of 2 feet, and a height of 5 feet. Find the work required to pump the water to a height of 3 feet above the top of the tank.

I was thinking that the answer would be

Work = $\int_0^{8} (20)($'some equation I need help finding'$)(8-y)dy $

How would I find the missing equation and the work?

  • $\begingroup$ It is very difficult to tell what you are asking. You have a tank full of oil and a problem that asks for the work required to pump "the water". You, however, don't seem to be interested in that problem. Instead, you ask for the area of the tank. $\endgroup$
    – John Douma
    Commented Jan 15, 2019 at 21:01
  • $\begingroup$ What I am asking is if the equation I have posted to find the work is correct, and if so, how would I find what "Area" is equal to? Does this make sense? $\endgroup$ Commented Jan 15, 2019 at 21:07
  • $\begingroup$ @SomeGuy2312 You might be able to eliminate the confusion for readers by changing the subject line to something like "Help find work done to pump fluid". Also, what exactly do you mean by "area" of a 3D tank? $\endgroup$
    – Godfather
    Commented Jan 15, 2019 at 21:34
  • $\begingroup$ Thank you for the help, and what I mean by "area" is that I want to find the area of the cross-section in the shape. $\endgroup$ Commented Jan 15, 2019 at 21:44
  • $\begingroup$ @AdityaDua I tried editing to hopefully make more sense! $\endgroup$ Commented Jan 15, 2019 at 22:01

1 Answer 1


The work to move something with mass $m$ up a height $h$ against gravity $g$ is $mgh$.

Mass is density $\rho$ times volume, so the differential mass is $\rho\text{ }dV = \rho\text{ }dx\text{ }dy\text{ }dz$.

You're given the density, so to get the total work you calculate

$$W=\iiint_V \rho gh(z) \text{ }dV,$$

where $h(z)$ is the height moved up as a function of $z$. You already have $h(z)$ in your question if the tip of the pyramid is at the origin.

You can simplify things a bit by noting that the side of the cross section varies linearly from the tip to the base. It's zero at the origin, and $2$ feet at five feet above the origin. So the $x$ and $y$ integrations end up giving you the area of the cross section:

$$W=\int_0^5 \rho g (8-z) (2z/5)^2 \text{ } dz.$$


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