2008 AMC 10B Problem 19

The tank lies on its side, so every slice across it looks the same: a circle with water in the bottom. The volume is the area of the wet part of that circle times the tank's length, $9$.

The wet part of one circle.

1. The circle has radius $4$ and center $O$. The water is $2$ feet deep, so the water surface is a straight line $4 - 2 = 2$ feet below the center.
2. Let the surface line meet the circle at $P$ and $Q$, and let $M$ be the point on it straight below $O$. Then $OM = 2$ and $OP = 4$ (a radius).
3. Triangle $OMP$ has a right angle at $M$, and its short leg $OM$ is half its longest side $OP$. That makes it a $30$-$60$-$90$ triangle, with the $60^\circ$ angle at $O$. So $MP = 2\sqrt3$.
4. The same holds on the $Q$ side, so the angle $POQ$ is $60^\circ + 60^\circ = 120^\circ$, and $PQ = 4\sqrt3$.
5. The wet region is the pie slice $POQ$ with the triangle $POQ$ removed.
- The slice is $\tfrac{120}{360} = \tfrac13$ of the circle: $\tfrac13 \times \pi \times 4^2 = \tfrac{16\pi}{3}$.
- The triangle has base $PQ = 4\sqrt3$ and height $OM = 2$: $\tfrac12 \times 4\sqrt3 \times 2 = 4\sqrt3$.
- Wet area: $\tfrac{16\pi}{3} - 4\sqrt3$.

The volume.

6. Multiply by the length $9$:
$$9\left(\frac{16\pi}{3} - 4\sqrt3\right) = 48\pi - 36\sqrt3.$$

The answer is $\boxed{\textbf{(E)}\ 48\pi - 36\sqrt{3}}$.

A tank lying on its side is a long shape with the same slice all the way along, so its volume is slice area times length. The slice is a piece of a circle cut off by a straight line, and that is always found as "pie slice minus triangle". The depth $2$ being half the radius $4$ is what creates the $30$-$60$-$90$ triangle, and that is where the $\sqrt3$ in the answer comes from.