题目

Problem

Figure 1 shows a sketch of part of the curve with equation $y=e^{0.5x}-2$ .

The region $R$ , shown shaded in Figure 1, is bounded by the curve, the $x$ -axis and the $y$ -axis.

The region $R$ is rotated $360^\circ$ about the $x$ -axis to form a solid of revolution.

Show that the volume of this solid can be written in the form $a\ln 2+b$ , where $a$ and $b$ are constants to be found.

(6)

题目中文翻译

图 1 给出了曲线 $y=e^{0.5x}-2$ 的一部分草图。

图中阴影区域 $R$ 由该曲线、 $x$ 轴和 $y$ 轴围成。

将区域 $R$ 绕 $x$ 轴旋转 $360^\circ$ ，形成一个旋转体。

证明该旋转体的体积可写成 $a\ln 2+b$ 的形式，其中 $a,b$ 为待求常数。

解答

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先令 $y=0$ 求出区域的右边界。区域绕 $x$ 轴旋转，体积为 $\pi\int y^2\,\mathrm{d}x$ ；把 $(e^{0.5x}-2)^2$ 展开后逐项积分，再代入上下限并整理为指定形式。

展开

The curve meets the $x$ -axis when

e^{0.5x}-2=0.

Therefore,

\begin{align*} e^{0.5x} =&\,2,\\ x =&\,2\ln2. \end{align*}

The required volume is

V=\pi\int_0^{2\ln2} \big(e^{0.5x}-2\big)^2\,\mathrm{d}x.

Expanding and integrating,

\begin{align*} \int\big(e^x-4e^{0.5x}+4\big)\,\mathrm{d}x =&\,e^x-8e^{0.5x}+4x. \end{align*}

At the upper limit,

e^{2\ln2}-8e^{\ln2}+8\ln2 =8\ln2-12.

At the lower limit,

e^0-8e^0=-7.

Therefore,

\begin{align*} V =&\,\pi\big[(8\ln2-12)-(-7)\big]\\ =&\,\boxed{8\pi\ln2-5\pi}. \end{align*} Thus $a=8\pi$ and $b=-5\pi$. </details>