题目

Problem

The diagram shows part of the curve with equation $y=\frac{5}{2x-1}$ and the line $y=7-2x$ .

Find the exact volume formed when the shaded region is rotated through $360^\circ$ about the $x$ -axis.

(8)

题目中文翻译

图中所示的是曲线 $y=\frac{5}{2x-1}$ 的一部分及直线 $y=7-2x$ 。

求将阴影区域绕 $x$ 轴旋转 $360^\circ$ 所形成旋转体体积的精确值。

(8)

原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

解法一

思路

展开

先求图中直线和曲线的交点。绕 $x$ 轴旋转时，外半径是直线高度，内半径是曲线高度。

答题过程

展开

At an intersection,

5=(7-2x)(2x-1) \quad\Longrightarrow\quad x^2-4x+3=(x-1)(x-3)=0.

Between $x=1$ and $x=3$ , the line lies above the positive curve. The volume is

\begin{align*} V=&\,\pi\int_1^3\Big((7-2x)^2-\frac{25}{(2x-1)^2}\Big)\,\mathrm{d}x\\ =&\,\pi\Big[-\frac{(7-2x)^3}{6}+\frac{25}{2(2x-1)}\Big]_1^3\\ =&\,\pi\Big[\Big(-\frac16+\frac52\Big)\\ &\,-\Big(-\frac{125}{6}+\frac{25}{2}\Big)\Big]\\ =&\,\frac{32\pi}{3}. \end{align*}

CIE 9709 2026 June Paper 11 Q8

题目

解答

解法一

思路

答题过程