题目

Problem

A curve has equation

y = \sqrt{9 - x^2} \qquad 0 \le x \le 3.

(a) Using calculus, show that the length of the curve is

\frac{3\pi}{2}.

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

(b) Using calculus, find the exact area of the surface generated.

(7)

题目中文翻译

曲线的方程为

y = \sqrt{9 - x^2} \qquad 0 \le x \le 3

(a) 用微积分证明该曲线的长度为

\frac{3\pi}{2}

将该曲线绕 $x$ 轴旋转 $2\pi$ 弧度。

(b) 用微积分求所生成曲面的精确面积。

解答

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先对曲线方程求导，再代入弧长公式。根式内的两项通分后会化为 $\frac{9}{9-x^2}$ ，因此弧长积分成为标准的反正弦积分；必须明确代入上下限，才能自然得到题设结果。

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Differentiating,

\frac{\mathrm{d}y}{\mathrm{d}x} =-\frac{x}{\sqrt{9-x^2}}.

Hence

\begin{align*} \sqrt{1+\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2} =&\,\sqrt{1+\frac{x^2}{9-x^2}} \\ =&\,\frac{3}{\sqrt{9-x^2}}. \end{align*}

Therefore, the length of the curve is

\begin{align*} L =&\,\int_0^3 \frac{3}{\sqrt{9-x^2}}\,\mathrm{d}x \\ =&\,3\left[ \arcsin\bigg(\frac{x}{3}\bigg) \right]_0^3 \\ =&\,3\bigg(\frac{\pi}{2}-0\bigg) \\ =&\,\boxed{\frac{3\pi}{2}}. \end{align*}

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绕 $x$ 轴旋转所得曲面的面积公式为 $\int 2\pi y\,\mathrm{d}s$ 。沿用 (a) 已求出的弧长因子后， $y=\sqrt{9-x^2}$ 与分母完全约去，被积函数变成常数。

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The area of the surface generated is

\begin{align*} S =&\,\int_0^3 2\pi y \sqrt{1+\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2} \,\mathrm{d}x \\ =&\,\int_0^3 2\pi\sqrt{9-x^2} \frac{3}{\sqrt{9-x^2}}\,\mathrm{d}x \\ =&\,\int_0^3 6\pi\,\mathrm{d}x \\ =&\,6\pi[x]_0^3 \\ =&\,\boxed{18\pi}. \end{align*}