📝 题目
例 4 计算第二型曲线积分
$$ I = {\oint }_{L}\frac{x\mathrm{\;d}y - y\mathrm{\;d}x}{{x}^{2} + {y}^{2}}. $$
(1) $L : {x}^{2} + {y}^{2} = {a}^{2}$ 沿逆时针方向;
(2) $L : \left| x\right| \leq 1,\left| y\right| \leq 1$ 的边界,沿逆时针方向.
💡 答案与解析
解 (1) $L$ 的参数方程为 $x = a\cos t,y = a\sin t\left( {0 \leq t \leq {2\pi }}\right)$ ,所以
$$ I = \frac{1}{{a}^{2}}{\oint }_{L}x\mathrm{\;d}y - y\mathrm{\;d}x $$
$$ = \frac{1}{{a}^{2}}{\int }_{0}^{2\pi }\left( {{a}^{2}{\cos }^{2}t + {a}^{2}{\sin }^{2}t}\right) \mathrm{d}t = {2\pi }. $$
(2) $\displaystyle{I = {\oint }_{L}\frac{x\mathrm{\;d}y - y\mathrm{\;d}x}{{x}^{2} + {y}^{2}}}$
$$ = {\int }_{-1}^{1}\frac{\mathrm{d}x}{1 + {x}^{2}} + {\int }_{-1}^{1}\frac{\mathrm{d}y}{1 + {y}^{2}} + {\int }_{1}^{-1}\frac{-1 \cdot \mathrm{d}x}{1 + {x}^{2}} - {\int }_{1}^{-1}\frac{-1 \cdot \mathrm{d}y}{1 + {y}^{2}} $$
$$ = 4{\int }_{-1}^{1}\frac{\mathrm{d}x}{1 + {x}^{2}} = {2\pi }. $$
评注 (1) 因 $\theta = \arctan \frac{y}{x}$ ,它在 ${x}^{2} + {y}^{2} > 0$ 上是多值函数. 注意 $\mathrm{d}\theta = \frac{x\mathrm{\;d}y - y\mathrm{\;d}x}{{x}^{2} + {y}^{2}}$ ,所以 $\displaystyle{I = {\int }_{L}\mathrm{\;d}\theta = {\left. \theta \right| }_{0}^{2\pi } = {2\pi }}$ .
(2)除利用参数方程或原函数求积外, 也可化为第一型曲线积分. 令 $\mathbf{r} = x\mathbf{i} + y\mathbf{j},\mathbf{r} = \left| \mathbf{r}\right|$ . 则
$$ I = {\oint }_{L}\frac{x\cos \langle \tau ,y\rangle - y\cos \langle \tau ,x\rangle }{{x}^{2} + {y}^{2}}\mathrm{\;d}s $$
$$ = {\oint }_{L}\frac{x\cos \langle n,x\rangle + y\cos \langle n,y\rangle }{{x}^{2} + {y}^{2}}\mathrm{\;d}s $$
$$ = {\oint }_{L}\frac{\mathbf{r} \cdot \mathbf{n}}{{r}^{2}}\mathrm{\;d}s\overset{\mathbf{r}\text{ 与 }\mathbf{n}\text{ 方向一致 }}{ = }\frac{1}{a}{\oint }_{L}\mathrm{\;d}s = {2\pi }. $$