📝 题目
例 1 计算第一型曲线积分 $\displaystyle{I = {\int }_{L}\sqrt{{x}^{2} + {y}^{2}}\mathrm{\;d}s,L : {x}^{2} + {y}^{2} = {ax}}$ .
💡 答案与解析
解法 1 写出曲线的参数方程:
$$ x = \frac{a}{2} + \frac{a}{2}\cos t,\;y = \frac{a}{2}\sin t\;\left( {0 \leq t \leq {2\pi }}\right) . $$
因为 $\mathrm{d}s = \sqrt{{\left( \frac{a}{2}\sin t\right) }^{2} + {\left( \frac{a}{2}\cos t\right) }^{2}}\mathrm{\;d}t = \frac{a}{2}\mathrm{\;d}t$ ,所以
$$ I = {\int }_{L}\sqrt{{x}^{2} + {y}^{2}}\mathrm{\;d}s = {\int }_{0}^{2\pi }\frac{a}{2}\sqrt{\frac{{a}^{2}\left( {1 + \cos t}\right) }{2}}\mathrm{\;d}t $$
$$ = \frac{{a}^{2}}{2}{\int }_{0}^{2\pi }\left| {\cos \frac{t}{2}}\right| \mathrm{d}t = 2{a}^{2}{\int }_{0}^{\frac{\pi }{2}}\cos t\mathrm{\;d}t = 2{a}^{2}. $$
解法 2 由对称性只需考虑沿上半圆周 ${L}_{1} : y = \sqrt{{ax} - {x}^{2}}(0 \leq$ $x \leq a)$ 的积分,这时 $\mathrm{d}s = \frac{a}{2}\frac{\mathrm{d}x}{\sqrt{x\left( {a - x}\right) }}$ ,所以
$$ I = 2{\int }_{{L}_{1}}\sqrt{{x}^{2} + {y}^{2}}\mathrm{\;d}s = 2{\int }_{0}^{a}\sqrt{ax} \cdot \frac{a}{2}\frac{\mathrm{d}x}{\sqrt{x\left( {a - x}\right) }} $$
$$ = a\sqrt{a}{\int }_{0}^{a}\frac{\mathrm{d}x}{\sqrt{a - x}} = 2{a}^{2}. $$