2021年考研数二第19题解析：曲线的弧长、旋转曲面的面积

一、题目

难度评级：

二、解析

首先，对 $\int \frac{f\left(x\right)}{\sqrt{x}}\mathrm{~d}x=\frac{1}{6}x^{2}-x+C$ 求导，可得函数 $f(x)$ 的表达式为：

$$
\begin{aligned}
& \ \frac{f\left(x\right)}{\sqrt{x}} = \frac{1}{3}x-1 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{lightgreen}{ f\left(x\right) = \frac{1}{3}x^{\frac{3}{2}}-x^{\frac{1}{2}} } = y
\end{aligned}
$$

于是：

$$
\begin{aligned}
y ^{\prime} = \frac{1}{3} \cdot \frac{3}{2} x^{\frac{1}{2}} – \frac{1}{2} x^{\frac{-1}{2}} = \frac{1}{2} \left( \sqrt{x} – \frac{1}{\sqrt{x}} \right)
\end{aligned}
$$

进而：

$$
\textcolor{lightgreen}{ {y ^{\prime}}^{2} } = \frac{1}{4} \left( x + \frac{1}{x} – 2 \right) = \textcolor{lightgreen}{ \frac{x}{4} + \frac{1}{4x} – \frac{1}{2} }
$$

根据基于普通方程计算平面曲线弧长的公式可知，该曲线的弧长 $s$ 为：

$$
\begin{aligned}
\textcolor{lightgreen}{s} & = \int_{4}^{9}\sqrt{1+{y ^{\prime}}^{2}}\mathrm{~d}x \\ \\
& = \int_{4}^{9}\sqrt{\frac{1}{2}+\frac{x}{4}+\frac{1}{4x}}\mathrm{~d}x \\ \\
& = \int_{4}^{9}\sqrt{\frac{2x}{4x}+\frac{x^2}{4x}+\frac{1}{4x}}\mathrm{~d}x \\ \\
& = \int_{4}^{9}\sqrt{\frac{x^2+2x+1}{4x}}\mathrm{~d}x \\ \\
& = \int_{4}^{9}\sqrt{\frac{(x+1)^2}{4x}}\mathrm{~d}x \\ \\
& = \int_{4}^{9}\frac{x+1}{2\sqrt{x}}\mathrm{~d}x \\ \\
& = \frac{1}{2}\int_{4}^{9}\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)\mathrm{~d}x \\ \\
& = \frac{1}{2}\int_{4}^{9}\left(x^{\frac{1}{2}}+x^{-\frac{1}{2}}\right)\mathrm{d}x \\ \\
& = \frac{1}{2}\left[\frac{2}{3}x^{\frac{3}{2}}+2x^{\frac{1}{2}}\right]_{4}^{9} \\ \\
& = \left[\frac{1}{3}x^{\frac{3}{2}}+x^{\frac{1}{2}}\right]_{4}^{9} \\ \\
& = \left(\frac{1}{3}\cdot 9^{\frac{3}{2}}+9^{\frac{1}{2}}\right)
-\left(\frac{1}{3}\cdot 4^{\frac{3}{2}}+4^{\frac{1}{2}}\right) \\ \\
& = \left(\frac{1}{3}\cdot 27+3\right)
-\left(\frac{1}{3}\cdot 8+2\right) \\ \\
& = 12-\frac{14}{3} \\ \\
& = \textcolor{lightgreen}{ \frac{22}{3} }
\end{aligned}
$$

由于 $L$ 绕 $X$ 轴旋转一周所形成的曲面，实际上就是该旋转体的侧面积，所以，根据旋转体侧面积的计算公式可知，该曲面的侧面积 $A$ 为：

$$
\begin{aligned}
\textcolor{lightgreen}{A} & = 2\pi\int_{4}^{9}y\sqrt{1+{y ^{\prime}}^{2}}\mathrm{~d}x \\ \\
& = 2\pi\int_{4}^{9}\left(\frac{1}{3}x^{\frac{3}{2}}-x^{\frac{1}{2}}\right)\sqrt{\frac{1}{2}+\frac{x}{4}+\frac{1}{4x}}\mathrm{~d}x \\ \\
& = 2\pi\int_{4}^{9}\left(\frac{1}{3}x^{\frac{3}{2}}-x^{\frac{1}{2}}\right)\sqrt{\frac{x^{2}+2x+1}{4x}}\mathrm{~d}x \\ \\
& = 2\pi\int_{4}^{9}\left(\frac{1}{3}x^{\frac{3}{2}}-x^{\frac{1}{2}}\right)\sqrt{\frac{(x+1)^{2}}{4x}}\mathrm{~d}x \\ \\
& = 2\pi\int_{4}^{9}\left(\frac{1}{3}x^{\frac{3}{2}}-x^{\frac{1}{2}}\right)\frac{x+1}{2\sqrt{x}}\mathrm{~d}x \\ \\
& = \pi\int_{4}^{9}\left(\frac{1}{3}x^{\frac{3}{2}}-x^{\frac{1}{2}}\right)\frac{x+1}{\sqrt{x}}\mathrm{~d}x \\ \\
& = \pi\int_{4}^{9}\sqrt{x}\left(\frac{x}{3}-1\right)\frac{x+1}{\sqrt{x}}\mathrm{~d}x \\ \\
& = \pi\int_{4}^{9}\left(\frac{x}{3}-1\right)(x+1)\mathrm{~d}x \\ \\
& = \pi\int_{4}^{9}\left(\frac{x^{2}}{3}+\frac{x}{3}-x-1\right)\mathrm{~d}x \\ \\
& = \pi\int_{4}^{9}\left(\frac{x^{2}}{3}-\frac{2x}{3}-1\right)\mathrm{~d}x \\ \\
& = \pi\left[\frac{x^{3}}{9}-\frac{x^{2}}{3}-x\right]_{4}^{9} \\ \\
& = \pi\left[\left(\frac{9^{3}}{9}-\frac{9^{2}}{3}-9\right)-\left(\frac{4^{3}}{9}-\frac{4^{2}}{3}-4\right)\right] \\ \\
& = \pi\left[\left(81-27-9\right)-\left(\frac{64}{9}-\frac{16}{3}-4\right)\right] \\ \\
& = \pi\left[45-\left(\frac{64}{9}-\frac{48}{9}-\frac{36}{9}\right)\right] \\ \\
& = \pi\left(45+\frac{20}{9}\right) \\ \\
& = \pi\cdot\frac{425}{9} \\ \\
& = \textcolor{lightgreen}{\frac{425\pi}{9}}
\end{aligned}
$$

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