题目

Problem

The diagram shows part of the curve $y=x^2-\frac{1}{x^2}$ . The shaded region is bounded by the curve, the line $x=2$ and the $x$ -axis.

Find the volume formed when the shaded region is rotated through $360^\circ$ about the $x$ -axis, giving your answer correct to 2 decimal places.

[5]

题目中文翻译

图中显示曲线 $y=x^2-\frac{1}{x^2}$ 的一部分。阴影区域由该曲线、直线 $x=2$ 和 $x$ 轴围成。

将阴影区域绕 $x$ 轴旋转 $360^\circ$ ，求所得立体的体积，答案精确到 2 位小数。

解答

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先求图中正半轴上的交点。绕 $x$ 轴旋转后，截面积是 $\pi y^2$ ，平方时要保留交叉项。

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The positive $x$ -intercept satisfies $x^4=1$ , so it is at $x=1$ . Between $1$ and $2$ , the disc radius is $x^2-x^{-2}$ . Thus

\begin{align*} V=&\,\pi\int_1^2(x^2-x^{-2})^2\,\mathrm{d}x\\ =&\,\pi\int_1^2(x^4-2+x^{-4})\,\mathrm{d}x\\ =&\,\pi\Big[\frac{x^5}{5}-2x-\frac{1}{3x^3}\Big]_1^2\\ =&\,\frac{539\pi}{120}. \end{align*}

To two decimal places, the volume is $14.11$ cubic units.