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M3 2026 Jan A Q5

A Level / Edexcel / M3

IAL 2026 Jan A Paper · Question 5

题目

Problem

5.

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

Solutions relying on calculator technology are not acceptable.

Figure 1 shows the finite region RR which is bounded by part of the curve with equation y=sin⁡xy=\sin x, the xx-axis and the line with equation x=π2x=\dfrac\pi2

A uniform solid SS is formed by rotating RR through 2π2\pi radians about the xx-axis.

Using algebraic integration,

(a) show that the volume of SS is π24\dfrac{\pi^2}4

(4)

(b) find, in terms of π\pi, the xx coordinate of the centre of mass of SS.

(7)
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 所示的有限区域 RR 由曲线 y=sin⁡xy=\sin x 的一部分、xx 轴及直线 x=π2x=\dfrac\pi2 围成。

将 RR 绕 xx 轴旋转 2π2\pi 弧度，形成均匀立体 SS。

用代数积分

(a) 证明 SS 的体积为 π24\dfrac{\pi^2}4；

(4)

(b) 用 π\pi 表示 SS 的质心的 xx 坐标。

(7)

解答

(a)

解法一

思路

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旋转体用圆盘法，截面积为 πsin⁡2x\pi\sin^2x。使用二倍角公式手工积分，并明确代入上下限。

答题过程

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A thin disc perpendicular to the xx-axis has radius sin⁡x\sin x, so

V=π∫0π/2sin⁡2x dx=π2∫0π/2(1−cos⁡2x) dx.V=\pi\int_0^{\pi/2}\sin^2x\,\mathrm dx =\frac\pi2\int_0^{\pi/2}(1-\cos2x)\,\mathrm dx.

Hence

V=π[x2−sin⁡2x4]0π/2=π(π4−0)=π24.V=\pi\left[\frac x2-\frac{\sin2x}{4}\right]_0^{\pi/2} =\pi\left(\frac\pi4-0\right) =\boxed{\frac{\pi^2}{4}}.

(b)

解法一

思路

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均匀旋转体的质量矩可用 π∫xsin⁡2x dx\pi\int x\sin^2x\,\mathrm dx 计算。先展开二倍角，再对 xcos⁡2xx\cos2x 分部积分，最后除以体积。

答题过程

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Let the constant density be ρ\rho. The mass is ρV\rho V and the moment about the plane x=0x=0 is

ρπ∫0π/2xsin⁡2x dx.\rho\pi\int_0^{\pi/2}x\sin^2x\,\mathrm dx.

Write I=∫0π/2xsin⁡2x dxI=\int_0^{\pi/2}x\sin^2x\,\mathrm dx. Using sin⁡2x=(1−cos⁡2x)/2\sin^2x=(1-\cos2x)/2,

I=∫0π/2x2 dx−12∫0π/2xcos⁡2x dx.I=\int_0^{\pi/2}\frac x2\,\mathrm dx -\frac12\int_0^{\pi/2}x\cos2x\,\mathrm dx.

Integration by parts gives

∫xcos⁡2x dx=xsin⁡2x2−12∫sin⁡2x dx=xsin⁡2x2+cos⁡2x4.\int x\cos2x\,\mathrm dx =\frac{x\sin2x}{2}-\frac12\int\sin2x\,\mathrm dx =\frac{x\sin2x}{2}+\frac{\cos2x}{4}.

Therefore

I=[x24−xsin⁡2x4−cos⁡2x8]0π/2=(π216+18)−(−18)=π216+14.I=\left[\frac{x^2}{4}-\frac{x\sin2x}{4}-\frac{\cos2x}{8}\right]_0^{\pi/2} =\left(\frac{\pi^2}{16}+\frac18\right)-\left(-\frac18\right) =\frac{\pi^2}{16}+\frac14.

Dividing the mass moment by the mass,

xˉ=ρπIρV=π(π2/16+1/4)π2/4=π2+44π.\bar x=\frac{\rho\pi I}{\rho V} =\frac{\pi(\pi^2/16+1/4)}{\pi^2/4} =\boxed{\frac{\pi^2+4}{4\pi}}.

解法二

思路

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也可以直接使用第(a)问中 sin⁡2x\sin^2x 的原函数做分部积分。这种写法把上一问的积分结果用于质量矩，并同样完整展示边界项。

答题过程

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As in (a), an antiderivative of sin⁡2x\sin^2x is

F(x)=x2−sin⁡2x4.F(x)=\frac x2-\frac{\sin2x}{4}.

Integrate I=∫0π/2xsin⁡2x dxI=\int_0^{\pi/2}x\sin^2x\,\mathrm dx by parts, taking u=xu=x and dv=sin⁡2x dx\mathrm dv=\sin^2x\,\mathrm dx:

I=[xF(x)]0π/2−∫0π/2F(x) dx.I=\left[xF(x)\right]_0^{\pi/2}-\int_0^{\pi/2}F(x)\,\mathrm dx.

The first term is π2/8\pi^2/8, while

∫0π/2F(x) dx=[x24+cos⁡2x8]0π/2=π216−14.\int_0^{\pi/2}F(x)\,\mathrm dx =\left[\frac{x^2}{4}+\frac{\cos2x}{8}\right]_0^{\pi/2} =\frac{\pi^2}{16}-\frac14.

It follows that

I=π28−(π216−14)=π216+14.I=\frac{\pi^2}{8}-\left(\frac{\pi^2}{16}-\frac14\right) =\frac{\pi^2}{16}+\frac14.

Using the disc volume V=π2/4V=\pi^2/4, the centre of mass has coordinate

xˉ=πIV=π2+44π.\boxed{\bar x=\frac{\pi I}{V}=\frac{\pi^2+4}{4\pi}}.