题目

Problem

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

Solutions relying entirely on calculator technology are not acceptable.

Figure 1

Figure 1 shows a sketch of the curve with equation

y = \cos 2x \qquad 0 \le x \le \frac{\pi}{4}

The curve is rotated through $2\pi$ radians about the x-axis.

(a) Show that the area of the curved surface generated is given by

S = 2\pi \int_0^{\pi/4} \cos 2x \sqrt{1 + 4\sin^2 2x}\,dx

(b) Hence, using the substitution $2\sin 2x = \sinh \theta$ , show that

S = \frac{\pi}{4}\left(\ln(a+\sqrt{b}) + a\sqrt{b}\right)

where $a$ and $b$ are integers to be determined.

(2)

(7)

题目中文翻译

在本题中，你必须写出所有解题步骤。

不能完全依赖计算器技术求解。

图 1

图 1 展示了曲线的草图，其方程为

y = \cos 2x \qquad 0 \le x \le \frac{\pi}{4}

该曲线绕 x 轴旋转 $2\pi$ 弧度。

(a) 证明所生成曲面的面积可表示为

S = 2\pi \int_0^{\pi/4} \cos 2x \sqrt{1 + 4\sin^2 2x}\,dx

(b) 由此，使用代换 $2\sin 2x = \sinh\theta$ ，证明

S = \frac{\pi}{4}\left(\ln(a+\sqrt{b}) + a\sqrt{b}\right)

其中 $a$ 和 $b$ 为待定整数。

解答

(a)

解法一

思路

展开

曲线绕 $x$ 轴旋转的曲面面积公式为 $2\pi\int y\sqrt{1+(\mathrm{d}y/\mathrm{d}x)^2}\,\mathrm{d}x$ 。先求导，再把 $y$ 、导数及原题给出的上下限完整代入。

答题过程

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For

y=\cos2x,

we have

\frac{\mathrm{d}y}{\mathrm{d}x}=-2\sin2x.

The area of the curved surface generated by rotation about the $x$ -axis is

S=2\pi\int_0^{\pi/4} y\sqrt{1+\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2} \,\mathrm{d}x.

Substituting $y=\cos2x$ and the derivative,

\begin{align*} S =&\,2\pi\int_0^{\pi/4} \cos2x\sqrt{1+(-2\sin2x)^2}\,\mathrm{d}x\\ =&\,2\pi\int_0^{\pi/4} \cos2x\sqrt{1+4\sin^22x}\,\mathrm{d}x, \end{align*}

as required.

(b)

解法一

思路

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按题目指定令 $2\sin2x=\sinh\theta$ 。求导后， $\cos2x\,\mathrm{d}x$ 会变成 $\frac14\cosh\theta\,\mathrm{d}\theta$ ，而根式也等于 $\cosh\theta$ ，所以积分化为 $\cosh^2\theta$ 。最后必须转换上下限，并把 $\operatorname{arsinh}2$ 写成自然对数。

答题过程

展开

Let

2\sin2x=\sinh\theta.

Differentiating,

4\cos2x\,\mathrm{d}x =\cosh\theta\,\mathrm{d}\theta,

\cos2x\,\mathrm{d}x =\frac14\cosh\theta\,\mathrm{d}\theta.

Also,

\begin{align*} \sqrt{1+4\sin^22x} =&\,\sqrt{1+\sinh^2\theta}\\ =&\,\cosh\theta. \end{align*}

The limits transform as follows:

\begin{array}{c|c} x&\theta\\ \hline 0&0\\ \dfrac{\pi}{4}&\operatorname{arsinh}2 \end{array}

Therefore, using part (a),

\begin{align*} S =&\,2\pi\int_0^{\operatorname{arsinh}2} \frac14\cosh^2\theta\,\mathrm{d}\theta\\ =&\,\frac{\pi}{2} \int_0^{\operatorname{arsinh}2} \cosh^2\theta\,\mathrm{d}\theta. \end{align*}

Using

\cosh^2\theta =\frac12(1+\cosh2\theta),

we obtain

\begin{align*} S =&\,\frac{\pi}{4} \bigg[\theta+\frac12\sinh2\theta\bigg] _0^{\operatorname{arsinh}2}. \end{align*}

Let $\alpha=\operatorname{arsinh}2$ . Then

\sinh\alpha=2

and

\cosh\alpha=\sqrt{1+\sinh^2\alpha}=\sqrt5.

Hence

\frac12\sinh2\alpha =\sinh\alpha\cosh\alpha =2\sqrt5.

Also,

\alpha=\operatorname{arsinh}2 =\ln(2+\sqrt5).

It follows that

\boxed{ S=\frac{\pi}{4} \big(\ln(2+\sqrt5)+2\sqrt5\big)}.

Therefore,

\boxed{a=2,\qquad b=5}.

IAL 2025 June FP3 Q7

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程