Question

A tank containing oil is in the shape of a downward-pointing cone with its vertical axis perpendicular to ground level. (See a graph of the tank here.) In this exercise we will assume that the height of the tank is 10 ft , the circular top of the tank has radius 5 ft, and that the oil inside the tank weighs 56 lb per cubic ft.
How much work does it take to pump oil from the tank to an outlet that is 3 ft above the top of the tank if, prior to pumping, there is only a half-tank of oil? Note: "half-tank" means half the volume in the tank.

Answer 1

Do you know any physical laws, equations that could help? @skimo16

Answer 2

I thought this might be a pressure problem, but it looks like it's more of just a work problem. So I suppose the only equation of pertinence is: \[
\vec{W} = \int \vec{F} \cdot d\vec{r}
\]

Answer 3

So we need to come up with some equations of force here, with respect to distance.

Answer 4

"56 lb per cubic ft" is our Force/Volume, so we need an equation for volume now.

Answer 5

We'll integrate with respect to \(x\).

Answer 6

The volume at any \(x\) value will be:
\(\pi r(x)^2dx\) where \(r(x)\) is our radius in terms of \(x\) and \(dx\) is our thickness

Answer 7

The distance it must travel for any \(x\) is just \(x\), so our work for each is thus: \(x \times (56\pi r(x)^2dx)\)

Answer 8

All together we have: \[ \Large
\int_{x_i}^{x_f} 56xr(x)^2dx
\]
So what we need is \(r(x) \) and our limits of integration \(x_i \ x_f\).

Answer 9

\[
\Large
\int_{x_i}^{x_f} 56x\pi r(x)^2dx
\]

Answer 10

@skimo16 Is it starting to make sense?