2008 AMC 10B Problem 19

The submerged cross-section is a circular segment. The chord is $4-2=2$ feet below the center, and $\cos\theta=\tfrac{2}{4}=\tfrac12,$ so the half-angle is $60^\circ$ and the central angle is $120^\circ.$ The sector area is $\tfrac{120}{360}\pi(4)^2=\tfrac{16\pi}{3},$ and the triangle formed by the two radii has area $\tfrac12(4)^2\sin 120^\circ=4\sqrt3.$ The segment area is $\tfrac{16\pi}{3}-4\sqrt3.$ Multiplying by the length $9$ gives $9\left(\tfrac{16\pi}{3}-4\sqrt3\right)=48\pi-36\sqrt3.$ Thus, the correct answer is E.