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IAL 2022 Oct Q5

A Level / Edexcel / P4

IAL 2022 Oct Paper · Question 5

题目

Problem

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

Solutions relying entirely on calculator technology are not acceptable.

Figure 2 shows a sketch of part of the curve with equation

y=12x(2x2+3)1.5y=\frac{12\sqrt{x}}{(2x^2+3)^{1.5}}

The region RR, shown shaded in Figure 2, is bounded by the curve, the line with equation x=12x=\dfrac{1}{\sqrt{2}}, the xx-axis and the line with equation x=kx=k.

This region is rotated through 360360^\circ about the xx-axis to form a solid of revolution.

Given that the volume of this solid is 713648π\dfrac{713}{648}\pi, use algebraic integration to find the exact value of the constant kk.

(6)
题目中文翻译

本题中你必须写出解题过程的所有步骤。

完全依赖计算器技术的解法不被接受。

图 2 给出了部分曲线的示意图,其方程为

y=12x(2x2+3)1.5y=\frac{12\sqrt{x}}{(2x^2+3)^{1.5}}

图 2 中阴影区域 RR 由该曲线、直线 x=12x=\dfrac{1}{\sqrt{2}}xx 轴以及直线 x=kx=k 围成。

该区域绕 xx 轴旋转 360360^\circ,形成一个旋转体。

已知该旋转体的体积为 713648π\dfrac{713}{648}\pi,用代数积分法求常数 kk 的精确值。

解答

解法一

思路

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旋转体绕 xx 轴生成,使用圆盘法 V=πy2dxV=\pi\int y^2\,\mathrm{d}x。平方后,被积函数为 144x(2x2+3)3\frac{144x}{(2x^2+3)^3},分母内部的导数与分子成比例,可以直接积分。代入上下限并与已知体积相等,最后利用图中 k>0k>0 选取正确的根。

答题过程

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The volume of the solid of revolution is

V=π12ky2dx.V=\pi\int_{\frac{1}{\sqrt2}}^k y^2\,\mathrm{d}x.

Since

y2=144x(2x2+3)3,y^2=\frac{144x}{(2x^2+3)^3},

we have

144x(2x2+3)3dx=18(2x2+3)2.\int\frac{144x}{(2x^2+3)^3}\,\mathrm{d}x =-18(2x^2+3)^{-2}.

Therefore,

V=π[18(2x2+3)2]12k=π(18(2k2+3)2+1842)=π(18(2k2+3)2+98).\begin{align*} V =&\,\pi\bigg[ -\frac{18}{(2x^2+3)^2} \bigg]_{\frac{1}{\sqrt2}}^k\\ =&\,\pi\bigg( -\frac{18}{(2k^2+3)^2} +\frac{18}{4^2} \bigg)\\ =&\,\pi\bigg( -\frac{18}{(2k^2+3)^2} +\frac98 \bigg). \end{align*}

Using the given volume,

18(2k2+3)2+98=713648.-\frac{18}{(2k^2+3)^2} +\frac98 =\frac{713}{648}.

Since 98=729648\frac98=\frac{729}{648},

18(2k2+3)2=16648=281.\begin{align*} \frac{18}{(2k^2+3)^2} =&\,\frac{16}{648}\\ =&\,\frac{2}{81}. \end{align*}

Hence

(2k2+3)2=729.(2k^2+3)^2=729.

As 2k2+3>02k^2+3>0,

2k2+3=27,k2=12.\begin{align*} 2k^2+3=&\,27,\\ k^2=&\,12. \end{align*}

The diagram shows that kk is positive, so

k=23.\boxed{k=2\sqrt3}.