题目

Problem

In this question you must show all stages in your working.

Solutions relying on calculator technology are not acceptable.

The finite region $R$ , shown shaded in Figure 1, is bounded by the $x$ -axis, the line with equation $x=3$ , the curve with equation $y=\sqrt{x+1}$ and the $y$ -axis.

Find the $y$ coordinate of the centre of mass of a uniform lamina in the shape of $R$ .

(5)

题目中文翻译

本题必须展示所有计算步骤。

不接受依赖计算器技术的解法。

图 1 阴影所示的有限区域 $R$ 由 $x$ 轴、直线 $x=3$ 、曲线 $y=\sqrt{x+1}$ 及 $y$ 轴围成。

求形状为 $R$ 的均匀薄片的质心的 $y$ 坐标。

(5)

解答

解法一

思路

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这是平面薄片的 $y$ 坐标。取竖直窄条，其面积为 $y\,\mathrm{d}x$ ，条的质心高度是 $y/2$ ；用对 $x$ 轴的面积矩除以总面积。

答题过程

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Write $y=\sqrt{x+1}$ . A thin vertical strip has area $y\,\mathrm{d}x$ and its centre is at height $y/2$ above the $x$ -axis.

The total area is

\begin{align*} A=&\,\int_0^3(x+1)^{1/2}\,\mathrm{d}x\\ =&\,\left[\frac23(x+1)^{3/2}\right]_0^3\\ =&\,\frac23(8-1)=\frac{14}{3}. \end{align*}

For a uniform lamina, the mass per unit area cancels in the first-moment equation. Hence

\begin{align*} A\bar y=&\,\int_0^3\frac y2\,y\,\mathrm{d}x\\ =&\,\frac12\int_0^3(x+1)\,\mathrm{d}x\\ =&\,\frac12\left[\frac{x^2}{2}+x\right]_0^3\\ =&\,\frac12\left(\frac92+3\right)=\frac{15}{4}. \end{align*}

Therefore the required $y$ coordinate is

\boxed{\bar y=\frac{15/4}{14/3}=\frac{45}{56}.}

M3 2023 June Q1

题目

解答

解法一

思路

答题过程