Question

一个圆柱形水箱半径为 $4$ 英尺，高为 $9$ 英尺，横放在地上。水箱中水深为 $2$ 英尺。水的体积是多少立方英尺？

A cylindrical tank with radius $4$ feet and height $9$ feet is lying on its side. The tank is filled with water to a depth of $2$ feet. What is the volume of the water, in cubic feet?

$24\pi-36\sqrt{2}$

$24\pi-24\sqrt{3}$

$36\pi-36\sqrt{3}$

$36\pi-24\sqrt{2}$

$48\pi-36\sqrt{3}$

答案：E

知识点：圆柱扇形体积

难度评级：1680

解答：

被水浸没的横截面是圆弓形。弦在圆心下方 $4-2=2$ 英尺，且 $\cos\theta=\tfrac{2}{4}=\tfrac12$ ，所以半角为 $60^\circ$ ，圆心角为 $120^\circ$ 。

扇形面积为 $\tfrac{120}{360}\pi(4)^2=\tfrac{16\pi}{3}$ ，两条半径形成的三角形面积为 $\tfrac12(4)^2\sin 120^\circ=4\sqrt3$ ，所以圆弓形面积为 $\tfrac{16\pi}{3}-4\sqrt3$ 。

乘以长度 $9$ ，体积为 $9\left(\tfrac{16\pi}{3}-4\sqrt3\right)=48\pi-36\sqrt3$ 。

所以正确答案是 E。

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.

2008 AMC 10B 第 19 题

答案：E

难度评级：1680

解答：

其他年份的第 19 题