2008 AMC 10B Problem 19

Since the tank lies on its side, the water occupies the same shape in every cross-section perpendicular to the axis, so its volume is (area of the cross-section) $\times 9$.

The cross-section is a circle of radius $4$. Water fills the bottom $2$ feet, so its surface is a chord at distance $4 - 2 = 2$ from the center $O$. If the chord has endpoints $P$ and $Q$ and $M$ is its midpoint, then in right triangle $OMP$, $\cos\angle MOP = \tfrac{2}{4}$, so $\angle MOP = 60^\circ$ and the central angle $\angle POQ = 120^\circ$.

The water region is the $120^\circ$ sector minus triangle $OPQ$:
$$\text{sector} = \frac{120}{360}\cdot \pi \cdot 4^2 = \frac{16\pi}{3}, \qquad \text{triangle} = \frac12 \cdot 4 \cdot 4 \cdot \sin 120^\circ = 4\sqrt{3}.$$
So the cross-sectional area is $\tfrac{16\pi}{3} - 4\sqrt3$, and the volume is
$$9\left(\frac{16\pi}{3} - 4\sqrt{3}\right) = 48\pi - 36\sqrt{3}.$$

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

A horizontal cylinder is a prism with a circular base, so partial filling is a 2-D problem: the area of a circular segment, always computed as sector minus isosceles triangle. The depth-to-radius ratio $2 : 4$ produced a $60^\circ$ half-angle, the hallmark of a $30$-$60$-$90$ triangle, which is why $\sqrt3$ appears in the answer.

Eliminate: the segment is less than a third of the circle's area, so the volume is under $\tfrac13 \cdot 16\pi \cdot 9 = 48\pi$, and it must involve $\sqrt3$ (from a $60^\circ$ angle), leaving (B), (C) or (E); only (E) has the correct $\pi$ coefficient $48$ from the sector.

• Using the full 9-foot height as if the tank were upright, or using a 60-degree sector instead of 120 degrees.
• Computing the triangle with base $PQ = 4\sqrt3$ and height $2$ but then adding it to the sector instead of subtracting.