📝 题目
例 6 求第二型曲线积分
$$ I = {\oint }_{L}\frac{x\mathrm{\;d}y - y\mathrm{\;d}x}{{x}^{2} + {y}^{2}}. $$
(1) $L$ 是圆周: ${x}^{2} + {y}^{2} = {\varepsilon }^{2}$ ;
(2) $L$ 是不过原点的简单、可求长闭曲线,且 $L$ 所围区域 $D$ 不含原点;
(3) $L$ 是环绕原点的简单、可求长闭曲线；
(4) $L$ 是环绕原点两圈的可求长闭曲线.
💡 答案与解析
解 (1) $\displaystyle{I = {\oint }_{L}\frac{x\mathrm{\;d}y - y\mathrm{\;d}x}{{x}^{2} + {y}^{2}} = \frac{1}{{\varepsilon }^{2}}{\oint }_{L}x\mathrm{\;d}y - y\mathrm{\;d}x}$
$$ = \frac{1}{{\varepsilon }^{2}}{\iint }_{D}2\mathrm{\;d}x\mathrm{\;d}y = {2\pi }. $$
(2)令 $P = - \frac{y}{{x}^{2} + {y}^{2}},Q = \frac{x}{{x}^{2} + {y}^{2}}$ . 因 $\frac{\partial Q}{\partial x} = \frac{{y}^{2} - {x}^{2}}{{\left( {x}^{2} + {y}^{2}\right) }^{2}} = \frac{\partial P}{\partial y} \in$ ${C}^{1}\left( \bar{D}\right)$ ,所以由格林公式得
$$ I = {\oint }_{L}\frac{x\mathrm{\;d}y - y\mathrm{\;d}x}{{x}^{2} + {y}^{2}} = {\iint }_{D}\left( {\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}}\right) \mathrm{d}x\mathrm{\;d}y = 0. $$
(3)以原点为心,以 $\varepsilon$ 为半径作圆 ${C}_{\varepsilon } : {x}^{2} + {y}^{2} = {\varepsilon }^{2}$ ,其中 $\varepsilon$ 小于原点到集合 $L$ 的距离. 记 $L$ 与 ${C}_{\varepsilon }$ 所围的区域为 $D.{C}_{\varepsilon }^{ - }$ 表示顺时针方向的圆周. 则由格林公式得
$$ {\int }_{L}\frac{x\mathrm{\;d}y - y\mathrm{\;d}x}{{x}^{2} + {y}^{2}} + {\int }_{{C}_{\varepsilon }}\frac{x\mathrm{\;d}y - y\mathrm{\;d}x}{{x}^{2} + {y}^{2}} = {\iint }_{D}\left( {\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}}\right) \mathrm{d}x\mathrm{\;d}y = 0, $$
由此推出
$$ I = {\int }_{L}\frac{x\mathrm{\;d}y - y\mathrm{\;d}x}{{x}^{2} + {y}^{2}} = - {\int }_{{C}_{\varepsilon }^{ - }}\frac{x\mathrm{\;d}y - y\mathrm{\;d}x}{{x}^{2} + {y}^{2}} $$
$$ = {\int }_{{C}_{\varepsilon }}\frac{x\mathrm{\;d}y - y\mathrm{\;d}x}{{x}^{2} + {y}^{2}} = {2\pi }. $$
(4)把绕原点两圈的曲线 $L$ 拆成两条绕原点的简单闭曲线的并集: $L = {C}_{1} + {C}_{2},{C}_{i}$ 为简单闭曲线,则
$$ I = {\int }_{L}\frac{x\mathrm{\;d}y - y\mathrm{\;d}x}{{x}^{2} + {y}^{2}} = {\int }_{{C}_{1}}\frac{x\mathrm{\;d}y - y\mathrm{\;d}x}{{x}^{2} + {y}^{2}} + {\int }_{{C}_{2}}\frac{x\mathrm{\;d}y - y\mathrm{\;d}x}{{x}^{2} + {y}^{2}} = {4\pi }. $$