1. ## Work Problem

A tank has the shape of an inverted circular cone (point at the bottom) with height 8 feet and radius 4 feet. The tank is full of water. We pump out water (to a pipe at the top of the tank) until the water level is 4 feet from the bottom. What is the work required to do?

2. The first drop is pumped 0 ft.
The last drop is pumped 4 ft.

Okay, let's define 't' as the distance from the top of the tank. This gives us a nice vertical distance function, d(t) = t

The first drop has enough friends to cover $\pi (4\;ft)^{2}$

The last drop has enough friends to cover $\pi (2\;ft)^{2}$

No vertical relationships are allowed in this inverted cone!

We need a radius function, r(t), such that r(0) = 4 and r(4) = 2. It's linear, so we're done, r(t) = 4 - t/2

Thus, $\int_{0}^{4}\;WaterWeight \;\cdot d(t) \cdot \pi (r(t))^{2}\;dt$

You just have to think it through. It isn't magic.