第六章 多元函数积分学 · 第2题

例题

📝 题目

解 由对称性知 $\displaystyle{\int }_{L}{x}^{2}\mathrm{\;d}s = {\int }_{L}{y}^{2}\mathrm{\;d}s = {\int }_{L}{z}^{2}\mathrm{\;d}s}$ ,所以

$$ I = {\int }_{L}{x}^{2}\mathrm{\;d}s = \frac{1}{3}{\int }_{L}\left( {{x}^{2} + {y}^{2} + {z}^{2}}\right) \mathrm{d}s $$

$$ = \frac{{a}^{2}}{3}{\int }_{L}\mathrm{\;d}s = \frac{{a}^{2}}{3} \cdot {2\pi a} = \frac{2}{3}\pi {a}^{3}. $$

💡 答案与解析

解 由对称性知 $\displaystyle{\int }_{L}{x}^{2}\mathrm{\;d}s = {\int }_{L}{y}^{2}\mathrm{\;d}s = {\int }_{L}{z}^{2}\mathrm{\;d}s}$ ,所以

$$ I = {\int }_{L}{x}^{2}\mathrm{\;d}s = \frac{1}{3}{\int }_{L}\left( {{x}^{2} + {y}^{2} + {z}^{2}}\right) \mathrm{d}s $$

$$ = \frac{{a}^{2}}{3}{\int }_{L}\mathrm{\;d}s = \frac{{a}^{2}}{3} \cdot {2\pi a} = \frac{2}{3}\pi {a}^{3}. $$