📝 题目
例 8 已知抛物叶形线 ${y}^{2} = \frac{x}{9}{\left( 3 - x\right) }^{2}$ ,如图 3.11 所示,其中当 $0 \leq x \leq 3$ 时的叶形部分记作 $M$ . 求
(1) $M$ 的面积；
(2) $M$ 的周长；
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图 3.11
(3) $M$ 绕 $x$ 轴旋转所得旋转体的体积 ${V}_{x}$ ；
(4) $M$ 绕 $x$ 轴旋转所得旋转体的侧面积 ${P}_{x}$ ；
(5) $M$ 的重心.
💡 答案与解析
解 (1) 由对称性,只要求出 $y = \frac{1}{3}\sqrt{x}\left( {3 - x}\right)$ 与 $x$ 轴所围成的面积, 两倍即得结果, 即
$$ A = \frac{2}{3}{\int }_{0}^{3}\left( {3 - x}\right) \sqrt{x}\mathrm{\;d}x = \frac{2}{3}{\int }_{0}^{3}3\sqrt{x}\mathrm{\;d}x - \frac{2}{3}{\int }_{0}^{3}x\sqrt{x}\mathrm{\;d}x $$
$$ = 4\sqrt{3} - \frac{12}{5}\sqrt{3} = \frac{8}{5}\sqrt{3}. $$
(2) ${y}^{\prime } = \frac{1}{6\sqrt{x}}\left( {3 - x}\right) - \frac{1}{3}\sqrt{x} = \frac{1 - x}{2\sqrt{x}}$
$$ \Rightarrow 1 + {\left( \frac{1 - x}{2\sqrt{x}}\right) }^{2} = \frac{1}{4x}{\left( 1 - x\right) }^{2} + 1 = \frac{{\left( x + 1\right) }^{2}}{4x}\text{ , } $$
由此即得
$$ s = 2{\int }_{0}^{3}\sqrt{1 + {y}^{\prime 2}}\mathrm{\;d}x = {\int }_{0}^{3}\frac{x + 1}{\sqrt{x}}\mathrm{\;d}x = 4\sqrt{3}. $$
(3) ${V}_{x} = \pi {\int }_{0}^{3}\frac{x}{9}{\left( 3 - x\right) }^{2}\mathrm{\;d}x = \frac{3}{4}\pi$ .
(4) ${P}_{x} = {2\pi }{\int }_{0}^{3}\frac{\sqrt{x}}{3}\left( {3 - x}\right) \sqrt{1 + {\left( \frac{1 - x}{2\sqrt{x}}\right) }^{2}}\mathrm{\;d}x = {3\pi }$ .
(5) 由对称性, $\bar{y} = 0$ ,
$$ \bar{x} = \frac{2{\int }_{0}^{3}{xy}\mathrm{\;d}x}{A} = \frac{\frac{2}{3}{\int }_{0}^{3}x\left( {3 - x}\right) \sqrt{x}\mathrm{\;d}x}{\frac{8}{5}\sqrt{3}} = \frac{9}{7}. $$