证明 由题得$$\begin{array}{}\text{(2)}& I(\delta ):={\int }_{D_{\delta }}\frac{x{f}_x(x,y)+y{f}_y(x,y)}{x^2+y^2}\mathrm{d}x\mathrm{d}y={\int }_{D_{\delta }}\mathrm{\nabla }f\cdot \frac{\mathit{r}}{r^2}\mathrm{d}S={\int }_{D_{\delta }}\mathrm{\nabla }f\cdot \mathrm{\nabla }\ln r\mathrm{d}S\end{array}$$ 注意到²$$\begin{array}{}\text{(3)}& \mathrm{\nabla }\cdot (f\mathrm{\nabla }\ln r)=\mathrm{\nabla }f\cdot \mathrm{\nabla }\ln r+f{\mathrm{\nabla }}^2\ln r,{\mathrm{\nabla }}^2\ln r=\mathrm{\Delta }\ln r=0\end{array}$$ 记$C_r:=\{(x,y)\mid x^2+y^2=r^2\}$（自然正向），则有$$\begin{array}{}\text{(4)}& \begin{array}{rl}I(\delta )& ={\int }_{D_{\delta }}\mathrm{\nabla }\cdot (f\mathrm{\nabla }\ln r)\mathrm{d}S={\oint }_{\partial D_{\delta }^{+}}f\mathrm{\nabla }\ln r\cdot \mathit{n}\mathrm{d}l\\ & ={\int }_{C_1}f\frac{\mathit{r}}{r^2}\cdot \frac{\mathit{r}}{r}\mathrm{d}l-{\int }_{C_{\delta }}f\frac{\mathit{r}}{r^2}\cdot \frac{\mathit{r}}{r}\mathrm{d}l=-{\int }_{C_{\delta }}f\frac{\mathrm{d}l}{r}\end{array}\end{array}$$ 由积分中值定理可得$\mathrm{\exists }{\theta }_{\delta }$使得$$\begin{array}{}\text{(5)}& I(\delta )=-f(\delta \cos {\theta }_{\delta },\delta \sin {\theta }_{\delta }){\int }_{-C_{\delta }}\frac{\mathrm{d}l}{r}=-2\pi f(\delta \cos {\theta }_{\delta },\delta \sin {\theta }_{\delta })\stackrel{\delta \to 0^{+}}{\to }-2\pi f(0,0)\end{array}$$ $◻$

证明 (1) 注意到$$\begin{array}{}\text{(8)}& u\frac{\partial \ln r}{\partial \mathit{n}}-\ln r\frac{\partial u}{\partial \mathit{n}}=(u\mathrm{\nabla }\ln r-\ln r\mathrm{\nabla }u)\cdot \mathit{n}\end{array}$$ 原积分在$(x_0,y_0)$处有瑕点。记$C_{\delta }:=\{(x,y)\mid (x-x_0{)}^2+(y-y_0{)}^2={\delta }^2\}$（自然正向），$D_{\delta }=D\setminus \{(x,y)\mid (x-x_0{)}^2+(y-y_0{)}^2\le {\delta }^2\}$，由Green公式可得$$\begin{array}{}\text{(9)}& \begin{array}{rl}\mathrm{RHS}& =\frac{1}{2\pi }{\oint }_{\partial D^{+}-C_{\delta }+C_{\delta }}(u\mathrm{\nabla }\ln r-\ln r\mathrm{\nabla }u)\cdot \mathit{n}\mathrm{d}l,(\partial D^{+}-C_{\delta }=\partial D_{\delta }^{+})\\ & =\frac{1}{2\pi }{\int }_{D_{\delta }}\mathrm{\nabla }\cdot (u\mathrm{\nabla }\ln r-\ln r\mathrm{\nabla }u)\mathrm{d}S+\frac{1}{2\pi }{\int }_{C_{\delta }}(u\mathrm{\nabla }\ln r-\ln r\mathrm{\nabla }u)\cdot \mathit{n}\mathrm{d}l\\ & =\frac{1}{2\pi }{\int }_{D_{\delta }}(u{\mathrm{\nabla }}^2\ln r-\ln r{\mathrm{\nabla }}^{2}u)\mathrm{d}S+\frac{1}{2\pi }{\int }_{C_{\delta }}(u\frac{\mathit{r}}{r^2}-\ln r\mathrm{\nabla }u)\cdot \frac{\mathit{r}}{r}\mathrm{d}l\end{array}\end{array}$$ 注意到${\mathrm{\nabla }}^2\ln r={\mathrm{\nabla }}^{2}u=0$，计算可得$$\begin{array}{}\text{(10)}& \begin{array}{r}\mathrm{RHS}=0+\frac{1}{2\pi }{\int }_{C_{\delta }}u\frac{\mathrm{d}l}{r}-\frac{1}{2\pi }{\int }_{C_{\delta }}\frac{\partial u}{\partial \mathit{n}}\ln r\mathrm{d}l\end{array}\end{array}$$ 由积分中值定理可得$\mathrm{\exists }{\theta }_{\delta }$使得$$\begin{array}{}\text{(11)}& {\int }_{C_{\delta }}u\frac{\mathrm{d}l}{r}=u(x_0+\delta \cos {\theta }_{\delta },y_0+\delta \sin {\theta }_{\delta }){\int }_{C_{\delta }}\frac{\mathrm{d}l}{r}\stackrel{\delta \to 0^{+}}{\to }2\pi u(x_0,y_0)\end{array}$$ 由积分的三角不等式可得$$\begin{array}{}\text{(12)}& |{\int }_{C_{\delta }}\frac{\partial u}{\partial \mathit{n}}\ln r\mathrm{d}l|\le {\int }_{C_{\delta }}\left|\frac{\partial u}{\partial \mathit{n}}\right||\ln r|\mathrm{d}l\le 2\pi \delta |\ln \delta |\underset{C_{\delta }}{max}\parallel \mathrm{\nabla }u\parallel \stackrel{\delta \to 0^{+}}{\to }0\end{array}$$ 因此$$\begin{array}{}\text{(13)}& \mathrm{RHS}=u(x_0,y_0)=\mathrm{LHS}\end{array}$$

(2) 由(1)的结论可得$$\begin{array}{}\text{(14)}& \begin{array}{rl}u(x_0,y_0)& =\frac{1}{2\pi }{\oint }_{L^{+}}(u\frac{\partial \ln r}{\partial \mathit{n}}-\ln r\frac{\partial u}{\partial \mathit{n}})\mathrm{d}l\\ & =\frac{1}{2\pi }{\oint }_{L^{+}}u\frac{\mathit{r}}{r^2}\cdot \frac{\mathit{r}}{r}\mathrm{d}l-\frac{\ln R}{2\pi }{\oint }_{L^{+}}\mathrm{\nabla }u\cdot \mathit{n}\mathrm{d}l\\ & =\frac{1}{2\pi R}{\oint }_{L^{+}}u(x,y)\mathrm{d}l-\frac{\ln R}{2\pi }{\int }_{D_R}{\mathrm{\nabla }}^{2}u\mathrm{d}S=\frac{1}{2\pi R}{\oint }_{L^{+}}u(x,y)\mathrm{d}l\end{array}\end{array}$$ $◻$

证明 首先需要利用对称性。观察到曲线$L$关于$Oxz$平面对称，在对称点上$y^2\mathrm{d}x$（以及$x^2\mathrm{d}z$）大小相等、符号相反（见图 8.4.1），故有$$\begin{array}{}\text{(16)}& {\int }_{L^{+}}{y}^2\mathrm{d}x={\int }_{L^{+}}{x}^2\mathrm{d}z=0\end{array}$$

其次，本题的关键在于选择合适的曲线参数化方式。一种自然的想法是利用$x^2+y^2=ax\Rightarrow {(x-\frac{a}{2})}^2+y^2={\left(\frac{a}{2}\right)}^2$，故可取$$\begin{array}{}\text{(17)}& x=\frac{a}{2}(1+\cos \theta )=a{\cos }^2\frac{\theta }{2},y=\frac{a}{2}\sin \theta ,z=\sqrt{a^2-ax}=a|\sin \frac{\theta }{2}|,\theta \in [0,2\pi ]\end{array}$$ 计算可得$$\begin{array}{}\text{(18)}& I={\int }_{L^{+}}{z}^2\mathrm{d}y={\int }_0^{2\pi }{a}^2{\sin }^2\frac{\theta }{2}\cdot \frac{a}{2}\cos \theta \mathrm{d}\theta =-\frac{\pi }{4}{a}^3\end{array}$$

另一种想法是利用球坐标，即$$\begin{array}{}\text{(19)}& \{\begin{array}{l}x=a\sin \theta \cos \phi \\ y=a\sin \theta \sin \phi \\ z=a\cos \theta \end{array},x^2+y^2=ax\iff \sin \theta =\cos \phi ,-\frac{\pi }{2}\le \phi \le \frac{\pi }{2}\end{array}$$ 因此曲线可参数化为$$\begin{array}{}\text{(20)}& \{\begin{array}{l}x=a{\cos }^2\phi \\ y=a\cos \phi \sin \phi \\ z=a|\sin \phi |\end{array},-\frac{\pi }{2}\le \phi \le \frac{\pi }{2}\end{array}$$ 其余计算过程与前述相同。 $◻$

证明 不妨设$L$为自然正向，记$L$围成的区域为$\mathrm{\Omega }$。由于$L$在上半平面内，故有$$\begin{array}{}\text{(25)}& f(x,y)=f(\frac{x}{y}\cdot y,y)=\frac{1}{y^2}f(\frac{x}{y},1)\end{array}$$ 因此$$\begin{array}{}\text{(26)}& I={\oint }_L\frac{1}{y}f(\frac{x}{y},1)\mathrm{d}x-\frac{x}{y^2}f(\frac{x}{y},1)\mathrm{d}y={\oint }_{L}f(\frac{x}{y},1)\mathrm{d}\frac{x}{y}\end{array}$$ 设$f(\cdot ,1)$的原函数为$F$，则有$$\begin{array}{}\text{(27)}& I={F\left(\frac{x}{y}\right)|}_A^{B=A}=0\end{array}$$ $◻$

证明 首先，我们需要证明以下引理：

引理证明 我们首先需要给“曲线正向”一个数学表述。

对于平面闭合曲线$L$，其前向单位切向量为$\hat{\tau }$、外向单位法向量$\hat{n}$，$z$轴的（正向）单位向量为$\hat{k}$，则$\hat{k}\cdot (\hat{n}\times \hat{\tau })$的符号反映了$L$的定向。设$\overrightarrow{\tau },\overrightarrow{n}$分别为与$\hat{\tau },\hat{n}$同向的切向量、法向量，则$k=n_1{\tau }_2-n_2{\tau }_1$的符号反映了$L$的定向。

设$D_0$由$f(u,v)\le 0$确定（否则考虑$-f$），则$\overrightarrow{n}=\mathrm{\nabla }f$为外法向量（指向$f$增大的方向）。设$\partial D_0^{+}$可以参数化为$t\mapsto (\genfrac{}{}{0ex}{}{u}{v})$，则$\overrightarrow{\tau }=(\genfrac{}{}{0ex}{}{u^{\prime }(t)}{v^{\prime }(t)})$为前切向量。

现考虑变换$\phi :D_0\to D_1$、$(u,v)\mapsto (x,y)$，令$\tilde{f}=f\circ {\phi }^{-1}$，则$D_1$由$\tilde{f}(x.y)\le 0$确定，${\overrightarrow{n}}^{\prime }={\mathrm{\nabla }}_{(x,y)}\tilde{f}$；$\partial D_1^{+}$可以参数化为$t\mapsto (\genfrac{}{}{0ex}{}{u}{v})\stackrel{\phi }{\mapsto }(\genfrac{}{}{0ex}{}{x}{y})$，则${\overrightarrow{\tau }}^{\prime }=(\genfrac{}{}{0ex}{}{x^{\prime }(t)}{y^{\prime }(t)})$。记$J=\mathrm{J}\phi =\frac{\partial (x,y)}{\partial (u,v)}$，由链式法则可得$$\begin{array}{}\text{(29)}& (\genfrac{}{}{0ex}{}{x^{\prime }}{y^{\prime }})=(\mathrm{J}\phi )(\genfrac{}{}{0ex}{}{u^{\prime }}{v^{\prime }}),{\mathrm{\nabla }}_{(x,y)}\tilde{f}={\mathrm{\nabla }}_{(x,y)}(f\circ {\phi }^{-1})=(\mathrm{J}{\phi }^{-1}{)}^{\mathrm{T}}{\mathrm{\nabla }}_{(u,v)}f\end{array}$$ 因此${\overrightarrow{\tau }}^{\prime }=J\overrightarrow{\tau }$、${\overrightarrow{n}}^{\prime }=J^{-\mathrm{T}}\overrightarrow{n}$。

设$k=n_1{\tau }_2-n_2{\tau }_1$，等式两端分别同乘${\tau }_1,{\tau }_2$，结合$\overrightarrow{n}\cdot \overrightarrow{\tau }=n_1{\tau }_1+n_2{\tau }_2=0$可得$$\begin{array}{}\text{(30)}& \{\begin{array}{l}k{\tau }_1=n_1{\tau }_1{\tau }_2-n_2{\tau }_1^2=-n_2({\tau }_1^2+{\tau }_2^2)\\ k{\tau }_2=n_1{\tau }_2^2-n_2{\tau }_2{\tau }_1=n_1({\tau }_1^2+{\tau }_2^2)\end{array}\Rightarrow \{\begin{array}{l}{n}_1=\frac{k{\tau }_2}{{\tau }_1^2+{\tau }_2^2}=\tilde{k}{\tau }_2\\ n_2=-\frac{k{\tau }_1}{{\tau }_1^2+{\tau }_2^2}=-\tilde{k}{\tau }_1\end{array}\end{array}$$ 记$$\begin{array}{}\text{(31)}& J=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\Rightarrow J^{-\mathrm{T}}=\frac{1}{detJ}\left(\begin{array}{cc}d& -c\\ -b& a\end{array}\right)\end{array}$$ 则有$$\begin{array}{}\text{(32)}& \begin{array}{rl}{k}^{\prime }& =n_1^{\prime }{\tau }_2^{\prime }-n_2^{\prime }{\tau }_1^{\prime }\\ & =\frac{1}{detJ}[(d{n}_1-c{n}_2)(c{\tau }_1+d{\tau }_2)-(-b{n}_1+a{n}_2)(a{\tau }_1+b{\tau }_2)]\\ & =\frac{\tilde{k}}{detJ}[(d{\tau }_2+c{\tau }_1)(c{\tau }_1+d{\tau }_2)-(-b{\tau }_2-a{\tau }_1)(a{\tau }_1+b{\tau }_2)]\\ & =\frac{\tilde{k}}{detJ}({\tau }_1^{\prime 2}+{\tau }_2^{\prime 2})=\frac{k}{detJ}\frac{{\tau }_1^{\prime 2}+{\tau }_2^{\prime 2}}{{\tau }_1^2+{\tau }_2^2}\end{array}\end{array}$$ 故$k^{\prime }$的符号与$\frac{k}{detJ}$相同。 $◻$

引理证明结束，我们回到原命题。注意到$$\begin{array}{}\text{(33)}& \begin{array}{rl}{I}_1& ={\oint }_{\partial D_1}x\mathrm{d}y={\int }_a^{b}x(t)y^{\prime }(t)\mathrm{d}t={\int }_a^{b}x(t)[y_u{u}^{\prime }(t)+y_v{v}^{\prime }(t)]\mathrm{d}t\\ & ={\oint }_{\partial D_0}x{y}_u\mathrm{d}u+x{y}_v\mathrm{d}v\stackrel{\mathrm{Green}}{=}{\int }_{D_0}[\frac{\partial }{\partial u}(x{u}_v)-\frac{\partial }{\partial v}(x{u}_u)]\mathrm{d}u\mathrm{d}v\\ & ={\int }_{D_0}(x_u{y}_v-x_v{y}_u)\mathrm{d}u\mathrm{d}v={\int }_{D_0}det\frac{\partial (x,y)}{\partial (u,v)}\mathrm{d}u\mathrm{d}v\end{array}\end{array}$$ 若$detJ>0$，则映射前后曲线正向不变，故有$$\begin{array}{}\text{(34)}& |D_1|=I_1={\int }_{D_0}|det\frac{\partial (x,y)}{\partial (u,v)}|\mathrm{d}u\mathrm{d}v\end{array}$$ 若$detJ<0$，则映射前后曲线正向改变，故有$$\begin{array}{}\text{(35)}& |D_1|=-I_1={\int }_{D_0}|det\frac{\partial (x,y)}{\partial (u,v)}|\mathrm{d}u\mathrm{d}v\end{array}$$ 综上所述，原命题得证。 $◻$