设 $L$ 为椭圆 $\frac{x^2}{9}+\frac{y^2}{4}=1$, 取顺时针方向, 求下面第二型曲线积分的值
\[
\oint_{L}\frac{(x+y)\mathrm{d}x-(x-y)\mathrm{d}y}{4x^2+9y^2}.
\]
1
记 $D$ 为 $\{(x,y)\mid\dfrac{x^2}{9}+\dfrac{y^2}{4}\leqslant 1\}$, 即 $L$ 所围的有界平面区域. 注意到 $(0,0)\in D$, 但是 $(0,0)$ 不在被积表达式 $\dfrac{(x+y)\mathrm{d}x-(x-y)\mathrm{d}y}{4x^2+9y^2}$ 的定义域内.
令 $P(x,y)=\dfrac{x+y}{4x^2+9y^2}$, $Q(x,y)=\dfrac{-(x-y)}{4x^2+9y^2}$, 则
\[
\frac{\partial Q}{\partial x}=\frac{(-1)(4x^2+9y^2)+(x-y)\cdot 8x}{(4x^2+9y^2)^2}=\frac{4x^2-9y^2-8xy}{(4x^2+9y^2)^2},
\]
\[
\frac{\partial P}{\partial y}=\frac{1\cdot(4x^2+9y^2)-(x+y)\cdot 18y}{(4x^2+9y^2)^2}=\frac{4x^2-9y^2-18xy}{(4x^2+9y^2)^2}.
\]
在 $D$ 内并不总有 $\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}$. 故曲线的积分与路径相关.
令 $x=3\cos\theta$, $y=2\sin\theta$, $\theta$ 从 $2\pi$ 变化到 $0$. 则 $\mathrm{d}x=-3\sin\theta\mathrm{d}\theta$, $\mathrm{d}y=2\cos\theta\mathrm{d}\theta$,
原积分化为
\[
\begin{split}
&\int_{2\pi}^{0}\frac{(3\cos\theta+2\sin\theta)(-3\sin\theta)\mathrm{d}\theta-(3\cos\theta-2\sin\theta)2\cos\theta\mathrm{d}\theta}{4(3\cos\theta)^2+9(2\sin\theta)^2}\\
=&\int_{2\pi}^{0}\frac{(-9\sin\theta\cos\theta-6\sin^2\theta)-(6\cos^2\theta-4\sin\theta\cos\theta)}{36}\mathrm{d}\theta\\
=&\frac{1}{36}\int_{2\pi}^{0}(-6-5\sin\theta\cos\theta)\cos\theta\mathrm{d}\theta\\
=&\frac{1}{36}\int_{0}^{2\pi}(6+5\sin\theta\cos\theta)\cos\theta\mathrm{d}\theta\\
=&\frac{1}{36}\biggl[12\pi+\frac{5}{2}\int_{0}^{2\pi}\sin 2\theta\mathrm{d}\theta\biggr]\\
=&\frac{1}{3}\pi+\frac{5}{72}\cdot\frac{-1}{2}\cos 2\theta\biggr|_{0}^{2\pi}\\
=&\frac{1}{3}\pi.
\end{split}
\]