📝 题目
4．求匀质抛物面壳 $z=x^{2}+y^{2}\left(0 \leqslant z \leqslant \frac{1}{4}\right)$ 的质心。
💡 答案与解析
[AI解答]
[AI解答] 设匀质抛物面壳的面密度为常数 $\rho$，其方程为 $$ z = x^2 + y^2, \quad 0 \le z \le \frac14. $$ 由于曲面关于 $z$ 轴对称，质心坐标 $(\bar{x}, \bar{y}, \bar{z})$ 中显然有 $$ \bar{x} = 0,\quad \bar{y} = 0. $$ 只需计算 $\bar{z}$。
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**第一步：曲面面积元素** 曲面 $z = f(x,y) = x^2 + y^2$，则 $$ \frac{\partial z}{\partial x} = 2x,\quad \frac{\partial z}{\partial y} = 2y, $$ 面积元素 $$ \mathrm{d}S = \sqrt{1 + (2x)^2 + (2y)^2}\,\mathrm{d}x\mathrm{d}y = \sqrt{1 + 4(x^2 + y^2)}\,\mathrm{d}x\mathrm{d}y. $$ 在极坐标下，$x = r\cos\theta,\; y = r\sin\theta$，且 $z = r^2$，条件 $0 \le z \le \frac14$ 给出 $0 \le r \le \frac12$。 于是 $$ \mathrm{d}S = \sqrt{1 + 4r^2}\; r\,\mathrm{d}r\,\mathrm{d}\theta. $$
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**第二步：总质量** $$ M = \iint_S \rho\,\mathrm{d}S = \rho \int_{0}^{2\pi} \mathrm{d}\theta \int_{0}^{1/2} r\sqrt{1+4r^2}\,\mathrm{d}r. $$ 先计算内层积分：令 $u = 1+4r^2$，则 $\mathrm{d}u = 8r\,\mathrm{d}r$，当 $r=0$ 时 $u=1$，$r=1/2$ 时 $u=2$， $$ \int_{0}^{1/2} r\sqrt{1+4r^2}\,\mathrm{d}r = \int_{1}^{2} \sqrt{u}\,\frac{\mathrm{d}u}{8} = \frac18 \cdot \frac{2}{3} u^{3/2}\Big|_{1}^{2} = \frac{1}{12}(2^{3/2} - 1). $$ 因此 $$ M = \rho \cdot 2\pi \cdot \frac{1}{12}(2\sqrt{2} - 1) = \frac{\pi\rho}{6}(2\sqrt{2} - 1). $$
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**第三步：对 $z$ 的静矩** $$ M_{xy} = \iint_S z\,\rho\,\mathrm{d}S = \rho \iint_S (x^2 + y^2)\,\mathrm{d}S = \rho \int_{0}^{2\pi} \mathrm{d}\theta \int_{0}^{1/2} r^2 \cdot r\sqrt{1+4r^2}\,\mathrm{d}r = 2\pi\rho \int_{0}^{1/2} r^3\sqrt{1+4r^2}\,\mathrm{d}r. $$ 计算内层积分：令 $u = 1+4r^2$，则 $r^2 = \frac{u-1}{4}$，$r\,\mathrm{d}r = \frac{\mathrm{d}u}{8}$，于是 $$ r^3\sqrt{1+4r^2}\,\mathrm{d}r = r^2 \cdot \sqrt{u} \cdot r\,\mathrm{d}r = \frac{u-1}{4} \cdot \sqrt{u} \cdot \frac{\mathrm{d}u}{8} = \frac{1}{32}(u^{3/2} - u^{1/2})\,\mathrm{d}u. $$ 积分限：$r=0 \Rightarrow u=1$，$r=1/2 \Rightarrow u=2$， $$ \int_{0}^{1/2} r^3\sqrt{1+4r^2}\,\mathrm{d}r = \frac{1}{32} \int_{1}^{2} (u^{3/2} - u^{1/2})\,\mathrm{d}u = \frac{1}{32} \left[ \frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2} \right]_{1}^{2}. $$ 计算： 当 $u=2$： $$ \frac{2}{5} \cdot 2^{5/2} - \frac{2}{3} \cdot 2^{3/2} = \frac{2}{5} \cdot 4\sqrt{2} - \frac{2}{3} \cdot 2\sqrt{2} = \frac{8\sqrt{2}}{5} - \frac{4\sqrt{2}}{3} = \sqrt{2}\left( \frac{24}{15} - \frac{20}{15} \right) = \frac{4\sqrt{2}}{15}. $$ 当 $u=1$： $$ \frac{2}{5} \cdot 1 - \frac{2}{3} \cdot 1 = \frac{2}{5} - \frac{2}{3} = \frac{6}{15} - \frac{10}{15} = -\frac{4}{15}. $$ 相减得 $$ \frac{4\sqrt{2}}{15} - \left(-\frac{4}{15}\right) = \frac{4(\sqrt{2}+1)}{15}. $$ 于是 $$ \int_{0}^{1/2} r^3\sqrt{1+4r^2}\,\mathrm{d}r = \frac{1}{32} \cdot \frac{4(\sqrt{2}+1)}{15} = \frac{\sqrt{2}+1}{120}. $$ 因此 $$ M_{xy} = 2\pi\rho \cdot \frac{\sqrt{2}+1}{120} = \frac{\pi\rho(\sqrt{2}+1)}{60}. $$
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**第四步：质心竖坐标** $$ \bar{z} = \frac{M_{xy}}{M} = \frac{\displaystyle \frac{\pi\rho(\sqrt{2}+1)}{60}} {\displaystyle \frac{\pi\rho}{6}(2\sqrt{2} - 1)} = \frac{\sqrt{2}+1}{60} \cdot \frac{6}{2\sqrt{2} - 1} = \frac{\sqrt{2}+1}{10(2\sqrt{2} - 1)}. $$ 有理化分母： $$ 2\sqrt{2} - 1 \quad\text{乘共轭}\quad 2\sqrt{2} + 1, $$ $$ \bar{z} = \frac{(\sqrt{2}+1)(2\sqrt{2}+1)}{10[(2\sqrt{2})^2 - 1^2]} = \frac{(\sqrt{2}+1)(2\sqrt{2}+1)}{10(8 - 1)} = \frac{(\sqrt{2}+1)(2\sqrt{2}+1)}{70}. $$ 展开分子： $$ (\sqrt{2}+1)(2\sqrt{2}+1) = 2\cdot 2 + \sqrt{2} + 2\sqrt{2} + 1 = 4 + 3\sqrt{2} + 1 = 5 + 3\sqrt{2}. $$ 所以 $$ \bar{z} = \frac{5 + 3\sqrt{2}}{70}. $$
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**最终结果** $$ \boxed{(0,\;0,\;\frac{5+3\sqrt{2}}{70})}. $$
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难度：★★★☆☆ （涉及曲面面积元素、极坐标变换、换元积分及有理化化简，计算量中等，思路常规。）