下册 9.2 第二型曲线积分 第35题

\section*{解题过程：} （1）$\displaystyle \oint_{L} \frac{-y \mathrm{~d} x+x \mathrm{~d} y}{\left((\alpha x+\beta y)^{2}+(r x+\delta y)^{2}\right)^{a}}=\oint_{L}-y \mathrm{~d} x+x \mathrm{~d} y=2 \iint_{D} \mathrm{~d} x \mathrm{~d} y$ ，其中 $D:(\alpha x+\beta y)^{2}+(r x+\delta y)^{2} \leqslant 1$ ． 下求椭圆 $D:(\alpha x+\beta y)^{2}+(r x+\delta y)^{2} \leqslant 1$ 的面积． 令 $s=\alpha x+\beta y, t=r x+\delta y$ ，则 $D$ 变成 $D^{\prime}: s^{2}+t^{2} \leqslant 1, J=\alpha \delta-\beta r$ 。于是 $$ \iint_{D} \mathrm{~d} x \mathrm{~d} y=\frac{1}{|J|} \iint_{D^{\prime}} \mathrm{d} x \mathrm{~d} y=\frac{\pi}{|\alpha \delta-\beta r|} $$ 所以 $$ \oint_{L} \frac{-y \mathrm{~d} x+x \mathrm{~d} y}{\left((\alpha x+\beta y)^{2}+(r x+\delta y)^{2}\right)^{a}}=\frac{2 \pi}{|\alpha \delta-\beta r|} $$ （2）记 $\displaystyle P=\frac{-y}{A x^{2}+2 B x y+C y^{2}}, Q=\frac{x}{A x^{2}+2 B x y+C y^{2}}$ ，则当 $x^{2}+y^{2} \neq 0$ 时， $$ P_{y}=\frac{C y^{2}-A x^{2}}{\left(A x^{2}+2 B x y+C y^{2}\right)^{2}}=Q_{x} $$ 取一条特殊的有向曲线 $L_{1}: A x^{2}+2 B x y+C y^{2}=\varepsilon^{2}$（ $\varepsilon>0$ 充分小），让 $L_{1}$ 完全落在 $L$ 内，并规定 $L_{1}$ 的方向为逆时针。设 $L+L_{1}^{-}$所围城的区域为 $D$ 。在 $L+L_{1}^{-}$上应用 Green 公式得 $$ \oint_{L+L_{1}} \frac{x \mathrm{~d} y-y \mathrm{~d} x}{A x^{2}+2 B x y+C y^{2}}=0 $$ 于是 $$ \oint_{L} \frac{x \mathrm{~d} y-y \mathrm{~d} x}{A x^{2}+2 B x y+C y^{2}}=\oint_{L_{4}} \frac{x \mathrm{~d} y-y \mathrm{~d} x}{A x^{2}+2 B x y+C y^{2}}=\frac{1}{\varepsilon^{2}} \oint_{L_{4}} x \mathrm{~d} y-y \mathrm{~d} x=\frac{2}{\varepsilon^{2}} \iint_{D} \mathrm{~d} x \mathrm{~d} y $$ 下面求椭圆区域 $D: A x^{2}+2 B x y+C y^{2} \leqslant \varepsilon^{2}$ 的面积． 化椭圆 $L_{1}: A x^{2}+2 B x y+C y^{2}=\varepsilon^{2}$ 为 $$ \left[\frac{\sqrt{A}}{\varepsilon}\left(x+\frac{B}{A} y\right)\right]^{2}+\left(\frac{1}{\varepsilon} \sqrt{C-\frac{B^{2}}{A}} y\right)^{2}=1 $$ 令 $\displaystyle s=\frac{\sqrt{A}}{\varepsilon}\left(x+\frac{B}{A} y\right), t=\frac{1}{\varepsilon} \sqrt{C-\frac{B^{2}}{A}} y$ ，则 $D$ 变成 $\displaystyle D^{\prime}: s^{2}+t^{2} \leqslant 1, J=\frac{1}{\varepsilon^{2}} \sqrt{A C-B^{2}}$ ．于是 $$ \begin{aligned} & \iint_{D} \mathrm{~d} x \mathrm{~d} y=\frac{1}{|J|} \iint_{D^{\prime}} \mathrm{d} x \mathrm{~d} y=\frac{\varepsilon^{2} \pi}{\sqrt{A C-B^{2}}} \\ & \oint_{L} \frac{x \mathrm{~d} y-y \mathrm{~d} x}{A x^{2}+2 B x y+C y^{2}}=\frac{2 \pi}{\sqrt{A C-B^{2}}} \end{aligned} $$ 所以 特别地，$\displaystyle \oint_{L} \frac{x \mathrm{~d} y-y \mathrm{~d} x}{x^{2}+2 B x y+C y^{2}}=\frac{2 \pi}{\sqrt{C-B^{2}}}$ ． （3）由（2）得 $\displaystyle \int_{L} \frac{x \mathrm{~d} y-y \mathrm{~d} x}{(x+y)^{2}+y^{2}}=\int_{L} \frac{x \mathrm{~d} y-y \mathrm{~d} x}{x^{2}+2 x y+2 y^{2}}=\frac{2 \pi}{\sqrt{2-1^{2}}}=2 \pi$ ． （4）由（2）得 $\displaystyle \int_{L} \frac{x \mathrm{~d} y-y \mathrm{~d} x}{x^{2}+x y+y^{2}}=\frac{2 \pi}{\sqrt{1-\left(2^{-1}\right)^{2}}}=\frac{4 \pi}{\sqrt{3}}$ ． （5）$\displaystyle I=\frac{1}{2 \pi} \int_{L} \frac{X \mathrm{~d} Y-Y \mathrm{~d} X}{X^{2}+Y^{2}}=\frac{1}{2 \pi} \int_{L} \frac{(a x+b y) d(c x+e y)-(c x+e y) d(a x+b y)}{(a x+b y)^{2}+(c x+e y)^{2}}$ $$ =\frac{1}{2 \pi}(a e-b c) \int_{L} \frac{x \mathrm{~d} y-y \mathrm{~d} x}{(a x+b y)^{2}+(c x+e y)^{2}}=\frac{1}{2 \pi}(a e-b c) \frac{2 \pi}{|a e-b c|}=\frac{a e-b c}{|a e-b c|} . $$