题目

Problem

The shaded region $R$ is bounded by the $x$ -axis, the line with equation $x=1$ , the curve with equation $y=1+\sqrt{x}$ and the $y$ -axis, as shown in Figure 1. The unit of length on both of the axes is $1\,\mathrm{m}$ .

The region $R$ is rotated through $2\pi$ radians about the $x$ -axis to form a solid of revolution which is used to model a uniform solid $S$ .

Show, using the model and algebraic integration, that

(a) the volume of $S$ is $\dfrac{17\pi}6\,\mathrm{m^3}$

(3)

(b) the centre of mass of $S$ is $\dfrac{49}{85}\,\mathrm{m}$ from $O$ .

(5)

题目中文翻译

如图 1 所示，阴影区域 $R$ 由 $x$ 轴、直线 $x=1$ 、曲线 $y=1+\sqrt{x}$ 及 $y$ 轴围成。两个坐标轴上的长度单位均为 $1\,\mathrm{m}$ 。

将区域 $R$ 绕 $x$ 轴旋转 $2\pi$ 弧度，形成一个旋转体，用来模拟均匀立体 $S$ 。

利用此模型及代数积分证明

(a) $S$ 的体积为 $\dfrac{17\pi}6\,\mathrm{m^3}$ ；

(3)

(b) $S$ 的质心到 $O$ 的距离为 $\dfrac{49}{85}\,\mathrm{m}$ 。

(5)

解答

(a)

解法一

思路

展开

绕 $x$ 轴旋转后，用垂直于 $x$ 轴的薄圆盘求体积。先展开半径的平方，再逐项积分，保留上下限和体积单位。

答题过程

展开

A cross-section perpendicular to the $x$ -axis has radius $y=1+\sqrt{x}$ . Its area is

\pi y^2=\pi(1+\sqrt{x})^2=\pi(1+2x^{1/2}+x).

The solid extends from $x=0$ to $x=1$ . Hence its volume is

\begin{align*} V=&\,\pi\int_0^1(1+2x^{1/2}+x)\,\mathrm{d}x\\ =&\,\pi\left[x+\frac43x^{3/2}+\frac12x^2\right]_0^1\\ =&\,\pi\left(1+\frac43+\frac12\right)\\ =&\,\boxed{\frac{17\pi}{6}\,\mathrm{m^3}}. \end{align*}

(b)

解法一

思路

展开

均匀旋转体的质心位于对称轴上。每片圆盘的质量正比于体积，以 $x=0$ 平面取质量矩后，密度可约去。

答题过程

展开

By rotational symmetry, the centre of mass lies on the $x$ -axis. Let its coordinate be $\bar x$ and let the uniform density be $\rho$ . A thin disc has mass

\mathrm{d}m=\rho\pi(1+\sqrt{x})^2\,\mathrm{d}x.

Taking first moments about the plane $x=0$ and cancelling $\rho$ ,

V\bar x=\pi\int_0^1x(1+\sqrt{x})^2\,\mathrm{d}x.

The integral is

\begin{align*} \pi\int_0^1(x+2x^{3/2}+x^2)\,\mathrm{d}x =&\,\pi\left[\frac12x^2+\frac45x^{5/2}+\frac13x^3\right]_0^1\\ =&\,\pi\left(\frac12+\frac45+\frac13\right)\\ =&\,\frac{49\pi}{30}. \end{align*}

Using part (a),

\bar x=\frac{49\pi/30}{17\pi/6} =\frac{49}{30}\cdot\frac6{17} =\frac{49}{85}\,\mathrm{m}.

Since the centre of mass lies on the positive $x$ -axis, its distance from $O$ is

\boxed{\frac{49}{85}\,\mathrm{m}.}

M3 2023 Jan Q1

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程