题目

Problem

The curve $C$ , shown in Figure 3, has equation

y=\frac{x^{-\frac14}}{\sqrt{1+x}\left(\arctan\sqrt{x}\right)}

The region $R$ , shown shaded in Figure 3, is bounded by $C$ , the line with equation $x=3$ , the $x$ -axis and the line with equation $x=\dfrac13$ .

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

Using the substitution $\tan u=\sqrt{x}$

(a) show that the volume $V$ of the solid formed is given by

k\int_a^b \frac{1}{u^2}\,\mathrm{d}u

where $k,a$ and $b$ are constants to be found.

(6)

(b) Hence, using algebraic integration, find the value of $V$ in simplest form.

(3)

题目中文翻译

曲线 $C$ 如图 3 所示，其方程为

y=\frac{x^{-\frac14}}{\sqrt{1+x}\left(\arctan\sqrt{x}\right)}

阴影区域 $R$ 由 $C$ 、直线 $x=3$ 、 $x$ 轴以及直线 $x=\dfrac13$ 围成。

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

使用代换 $\tan u=\sqrt{x}$ ，

(a) 证明所形成立体的体积 $V$ 为

k\int_a^b \frac{1}{u^2}\,\mathrm{d}u

其中 $k,a,b$ 为待求常数。

(b) 进而使用代数积分，求 $V$ 的最简值。

解答

(a)

解法一

思路

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绕 $x$ 轴旋转的体积公式是

V=\pi\int y^2\,\mathrm{d}x

本题给出代换 $\tan u=\sqrt{x}$ 。关键是把 $x$ 、 $1+x$ 、 $\arctan\sqrt{x}$ 和 $\mathrm{d}x$ 全部换成 $u$ ，同时把上下限 $x=\dfrac13,3$ 换成对应的 $u$ 值。

答题过程

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The volume of revolution about the $x$ -axis is

V=\pi\int_{\frac13}^{3} y^2\,\mathrm{d}x

Since

y=\frac{x^{-\frac14}}{\sqrt{1+x}\left(\arctan\sqrt{x}\right)}

we have

y^2=\frac{x^{-\frac12}}{(1+x)\left(\arctan\sqrt{x}\right)^2}

Therefore

V=\pi\int_{\frac13}^{3} \frac{x^{-\frac12}}{(1+x)\left(\arctan\sqrt{x}\right)^2}\,\mathrm{d}x

Use the substitution

\tan u=\sqrt{x}

Then

x=\tan^2u

and

\frac{\mathrm{d}x}{\mathrm{d}u}=2\tan u\sec^2u

\mathrm{d}x=2\tan u\sec^2u\,\mathrm{d}u

Also,

x^{-\frac12}=\frac{1}{\sqrt{x}}=\frac1{\tan u}

and

1+x=1+\tan^2u=\sec^2u

Also,

\arctan\sqrt{x}=u

Substitute into the integral:

\begin{aligned} V =&\,\pi\int \frac{\frac1{\tan u}}{\sec^2u\cdot u^2} \left(2\tan u\sec^2u\right)\,\mathrm{d}u \\ =&\,2\pi\int \frac1{u^2}\,\mathrm{d}u \end{aligned}

Now find the new limits.

When

x=\frac13

we have

\tan u=\frac1{\sqrt3}

u=\frac{\pi}{6}

When

x=3

we have

\tan u=\sqrt3

u=\frac{\pi}{3}

Therefore

V=2\pi\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac1{u^2}\,\mathrm{d}u

Hence

\boxed{k=2\pi,\qquad a=\frac{\pi}{6},\qquad b=\frac{\pi}{3}}

(b)

解法一

思路

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承接 (a)，只需要对 $u^{-2}$ 做代数积分：

\int u^{-2}\,\mathrm{d}u=-\frac1u

然后代入上下限 $\dfrac{\pi}{6}$ 和 $\dfrac{\pi}{3}$ 。

答题过程

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From part (a),

V=2\pi\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}u^{-2}\,\mathrm{d}u

Integrate:

\int u^{-2}\,\mathrm{d}u=-u^{-1}=-\frac1u

\begin{aligned} V =&\,2\pi\left[-\frac1u\right]_{\frac{\pi}{6}}^{\frac{\pi}{3}} \\ =&\,2\pi\left( -\frac{1}{\frac{\pi}{3}} +\frac{1}{\frac{\pi}{6}} \right) \\ =&\,2\pi\left(-\frac3\pi+\frac6\pi\right) \\ =&\,2\pi\cdot\frac3\pi \\ =&\,6 \end{aligned}

Therefore

\boxed{V=6}

IAL 2024 May Q9

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程