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