第二型曲面积分、Gauss公式、Stokes公式

多变量函数的积分学

张瑞
中国科学技术大学数学科学学院

第二型曲面积分

物理背景： $V$中流体每一点有速度$\vec v$，若有曲面$S\subset V$，则流过$S$的流体有多少？

曲面的定向

  • 通常的曲面都有正侧和反侧，或内侧和外侧。如果一个油漆匠油漆曲面的某一侧，只要不超过边界， 是不可能油漆到另一侧的。
  • Mobius带只有一侧。

    fig-mobius

  • 有光滑曲面$S$，曲面上每一点$M$都有非零法向量$\vec n(M)$， 则$-\vec n(M)$也是$S$在$M$点的法向量。
  • 对点$M_0\in S$，取定$M_0$处法向量的一个，记为$\vec n(M_0)$。

fig-dir-surface-1
  • 作任意$S$上过$M_0$的闭曲线$L$
  • 让点$M$从$M_0$沿$L$运动， 并取法向量$\vec n(M)$使其连续变化。
  • 若$M$回到$M_0$时，取到的法向量还是$\vec n(M_0)$， 则称$M_0$是$S$上的双侧点。
  • 如果曲面上所有的点都是双侧点，则称曲面是双侧曲面。否则，称为单侧曲面。
  • 只要曲面上有一个点是双侧点，则曲面就是一个双侧曲面。
  • 双侧曲面也称为可定向曲面。
  • 只讨论双侧曲面。

对于可定向曲面，指定一个连续的单位法向量场$\vec n$为正向，称曲面为定向曲面， $\vec n$称为定向曲面的正向。

  • 若可定向曲面$S$有参数表示

    \[\vec r= x(u,v)\vec i+ y(u,v)\vec j+ z(u,v)\vec k, \quad (u,v)\in D \]

    则它有法向量

    \[\pm\vec r'_u\times\vec r'_v=\pm\left(\frac{\partial (y,z)}{\partial(u,v)}\vec i +\frac{\partial (z,x)}{\partial(u,v)}\vec j +\frac{\partial (x,y)}{\partial(u,v)}\vec k\right) \]

    若$\vec r'_u\times\vec r'_v$与曲面的正向$\vec n$指向相同， 则称$(u,v)$是定向曲面$S$的正向参数。此时，有

    \[\vec n=\frac{\vec r'_u\times\vec r'_v}{|\vec r'_u\times\vec r'_v|} \]

    否则，称$(u,v)$是反向参数。

  • 可以验证，若$(u,v)$是反向参数， 则$(v,u)$是正向参数。

tangent-plane

对于一般的参数曲面，不做特别声明的话，取$(u,v)$是定向曲面的正向参数。 即一般的参数曲面的正向是

\[\vec n=\frac{\vec r'_u\times\vec r'_v}{|\vec r'_u\times\vec r'_v|} \]
  • 特别地，$S$可以显式表达为 $z=f(x,y)$，$(x,y)\in D$。则参数正向
    \[\vec n=\frac{-\frac{\partial f}{\partial x}\vec i -\frac{\partial f}{\partial y}\vec j+\vec k}{\sqrt{1+(\frac{\partial f}{\partial x})^2+(\frac{\partial f}{\partial y})^2}} \]
    与$z$轴的夹角$\alpha$的余弦为$\vec n\cdot \vec k>0$，因而夹角为锐角。 即显式表达的曲面的参数正向为通常所说的曲面的上侧。另一侧就是曲面的下侧。

由二元函数表达的曲面的切平面的法向量

\[z=f(x,y), (x,y)\in D \]

写成参数方程

\[\vec r=\vec r(x,y)=(x,y,f(x,y)) \]

所以

\[\begin{aligned} \vec r'_x=(1,0,f'_x) \\ \vec r'_y=(0,1,f'_y) \end{aligned} \]

可以得到切平面的法向量

\[\begin{aligned} \left| \begin{array}{ccc} \vec i & \vec j & \vec k \\ 1 & 0 & f'_x \\ 0 & 1 & f'_y \end{array} \right|=(-f'_x,-f'_y,1) \end{aligned} \]

单位法向量$\vec n$与$z$轴的夹角的余弦为

\[\cos\alpha=\vec n\cdot(0,0,1)=\frac1{\sqrt{1+(\frac{\partial f}{\partial x})^2+(\frac{\partial f}{\partial y})^2}}>0 \]

因而，夹角为锐角($<90^\circ$)。

例 1. 球面方程

\[\begin{cases} x(\theta, \phi)=R\sin\theta\cos\phi , \\ y(\theta, \phi)=R\sin\theta\sin\phi , \\ z(\theta, \phi)=R\cos\theta \end{cases} \]

$(\theta,\phi)\in[0,\pi]\times[0,2\pi]$

参数正向为球面的外侧。

% sphere coordinate， 球坐标 \begin{tikzpicture} [global scale=0.8] \begin{axis} [ view={110}{20}, axis lines=middle, width=7cm, ztick=\empty,ytick=\empty,xtick=\empty, %xlabel=$x$, %zlabel={$z$} ] \addplot3[ domain=0:90, samples = 60, samples y=0, %thick, ] ({cos(x)}, {sin(x)}, 0); \addplot3[ domain=0:90, samples = 60, samples y=0, %thick, ] (0, {cos(x)}, {sin(x)}); \addplot3[ domain=0:90, samples = 60, samples y=0, %thick, ] ({cos(x)}, 0, {sin(x)}); \addplot3[line legend] coordinates { (-0.1,0,0) (1.3,0,0) } node[left] {$x$}; \addplot3[->] coordinates { (0,-0.1,0) (0,1.3,0) } node[right] {$y$}; \addplot3[->] coordinates { (0,0,-0.1) (0,0,1.3) } node[above] {$z$}; % \addplot3[dashed, blue] coordinates { (0,0,0) ({sin(45)},{cos(45)},0) }; \addplot3[ domain=0:90, samples = 60, samples y=0, dashed, blue, ] ({sin(x)*cos(45)}, {sin(x)*sin(45)}, {cos(x)}); \addplot3[ domain=0:90, samples = 60, samples y=0, blue, ] ({sin(30)*cos(x)}, {sin(30)*sin(x)}, {cos(30)}); \addplot3[-latex] coordinates { (0,0,0) ({sin(30)*cos(45)}, {sin(30)*sin(45)}, {cos(30)}) } node[above] {$P$} node[pos=0.5, below] {$\rho$}; \addplot3[-latex] coordinates { ({sin(30)*cos(45)}, {sin(30)*sin(45)+0.1}, {cos(30)}) ({sin(30)*cos(45)}, {sin(30)*sin(45)}, {cos(30)}) } node[pos=0, right] {$(\rho,\theta,\phi)$}; \addplot3[ domain=0:30, samples = 5, samples y=0, blue, -latex, ] ({0.2*sin(x)*cos(45)}, {0.2*sin(x)*sin(45)}, {0.2*cos(x)}) node[pos=0.5, above] {$\theta$}; \addplot3[ domain=0:45, samples = 5, samples y=0, blue, -latex, ] ({0.2*sin(90)*cos(x)}, {0.2*sin(90)*sin(x)}, {0.2*cos(90)}) node[pos=0.5, below] {$\phi$}; \addplot3[red] coordinates { ({sin(30)*cos(45)}, {sin(30)*sin(45)}, {cos(30)}) ({sin(30)*cos(45)}, {sin(30)*sin(45)}, 0) } node[pos=1, below] {$P_1$} node[pos=0.5, right] {$z$}; \addplot3[red] coordinates { ({sin(30)*cos(45)}, {sin(30)*sin(45)}, 0) (0, {sin(30)*sin(45)}, 0) } %({sin(30)*cos(45)}, 0, 0) } node[pos=1, above] {$P_2$} node[pos=0.5, below] {$x$}; \addplot3[red] coordinates { (0, {sin(30)*sin(45)}, 0) (0, 0, 0) } node[pos=1, above left, black] {$O$} node[pos=0.5, above] {$y$}; \end{axis} \end{tikzpicture}

问题. 若将下半球面写成$z=-\sqrt{R^2-x^2-y^2}$，则参数正向为？

  • 对于封闭曲面，通常把两侧分为内侧与外侧。如，对于球面
    \[x^2+y^2+z^2=R^2 \]
    的外侧法向量为
    \[\vec n=\frac{x\vec i+y\vec j+z\vec k}{R}=\frac{\vec r}{R} \]
  • 需要将曲面的取向与其边界曲线的方向相协调： 曲面的正向与边界曲线的正向构成右手系。 或者说： 在正侧沿边界正向走，曲面在左手。

    fig-dir-surface-consist

  • 对于拼接的曲面$S$（由有限多块曲面拼接，任意两块至多只相交于边界上的一段曲线，任意三块或更多曲面至多只能相交于一点）。
    1. 每一块都应该是可定向的，当其中一块定义了正方向后，它的边界也定义好了协调的正方向。
    2. 如果有两个曲面片有一段公共的边界，它们各自的定向使得在它们公共边界上的定向正好相反。

fig-dir-surface-consist-multi

第二型曲面积分

设$\vec v$是一个不可压流体的速度场， $S$是一张定向曲面。

  • $S$的外法向量为$\vec n$。
  • 取$S$上的小块面积元$dS$，则有有向面积元$d\vec S=\vec n dS$。
  • 在$dS$上取一点$M$，则流体在$dS$的速度可以近似为$\vec v(M)$。
  • 则单位时间渡过$dS$的流量为
    \[dN=\vec v\cdot d\vec S=\vec v\cdot\vec n dS \]
%第2型曲面积分 \begin{tikzpicture}[x=(215:3em/sqrt 2), y=(0:3em), z=(90:3em), global scale=0.6, declare function={ f(\x,\y)=((\x-3)^2+(-\y+3)^3)/8+3; fx(\x,\y)=(\x-3)/4; fy(\x,\y)=-3*(3-\y)^2/8; nlength(\x,\y)=sqrt(fx(\x,\y)*fx(\x,\y)+fy(\x,\y)*fy(\x,\y)+1); nx(\x,\y)= -fx(\x,\y)/nlength(\x,\y); ny(\x,\y)= -fy(\x,\y)/nlength(\x,\y); nz(\x,\y)= 1/nlength(\x,\y); vx(\x,\y,\z)=1/sqrt(1+(\x*\x+\y*\y-1)^2+(\z*\z+\y*\y+1)^2); vy(\x,\y,\z)=(\x*\x+\y*\y-1)/sqrt(1+(\x*\x+\y*\y-1)^2+(\z*\z+\y*\y+1)^2); vz(\x,\y,\z)=(\z*\z+\y*\y+1)/sqrt(1+(\x*\x+\y*\y-1)^2+(\z*\z+\y*\y+1)^2); },line join=round,thick] % surface \draw[fill=black!50,thin,fill opacity=0.125,domain=0:4,smooth,variable=\t] plot (1+\t, 1, {f(1+\t,1)}) -- plot (5, 1+\t, {f(5,1+\t)}) -- plot (5-\t, 5, {f(5-\t,5)}) -- plot (1, 5-\t, {f(1,5-\t)}) -- cycle; \node[left] at (2.2,0.6,{f(2.6,1)}) {$S$}; % one piece \def\a{2} \def\b{2} \pgfmathsetmacro{\c}{f(\a,\b)} \draw[domain=0:1,variable=\t,smooth,fill=black,fill opacity=0.125] plot (\a+\t, \b, {f(\a+\t,\b)}) -- plot (\a+1, \b+\t, {f(\a+1,\b+\t)}) -- plot (\a+1-\t, \b+1, {f(\a+1-\t,\b+1)}) -- plot (\a, \b+1-\t, {f(\a,\b+1-\t)}) -- cycle; \node[below,inner sep=0.04em,font=\small] at (3,2.5,{f(3,2.5)}) {$\mathrm{d}S$}; % velocity vector \coordinate (p) at ({\a+0.5},{\b+0.5},{\c-0.2}); \coordinate (v) at (-1,1,2); \draw[-stealth,color=blue] (p)--++(v) coordinate(vv) node[right,inner sep=0pt]{$\vec v$}; % amount of fluid \def\vx{-1} \def\vy{1} \def\vz{2} \draw[domain=0:1,variable=\t,smooth] plot (2+\t+\vx,2+\vy,{f(2+\t,2)+\vz}) -- plot (3+\vx,2+\t+\vy,{f(3,2+\t)+\vz}) -- plot (3-\t+\vx,3+\vy,{f(3-\t,3)+\vz}) -- plot (2+\vx,3-\t+\vy,{f(2,3-\t)+\vz}) -- cycle; \draw[thin] (2,2,{f(2,2)}) -- ++(v) (2,3,{f(2,3)}) -- ++(v) (3,2,{f(3,2)}) -- ++(v) (3,3,{f(3,3)}) -- ++(v); % normal vector \draw[-stealth,red] (p) --++({nx(2.5,2.5)*3},{ny(2.5,2.5)*3},{nz(2.5,2.5)*3}) coordinate(vn) node[right]{$\vec n$}; \draw[-stealth, densely dashed, thin] (0,0,2) node[left] {$M$} -- (p); % projection \draw[densely dashed] (vv) -- ($(p)!(vv)!(vn)$); \end{tikzpicture}

定义 1.
$\vec v(M)$定义在$\mathbb{R}^3$中区域$V$的一个向量场，$S$是$V$中一张光滑的指定了正侧的双侧曲面，$\vec n(M)$为$S$上指向正侧的单位法向量。则积分

\[\iint_S \vec v\cdot d\vec S=\iint_S \vec v\cdot \vec ndS \]

称为向量场$\vec v$在有向曲面$S$上的第二型曲面积分。 也就是说， 向量场在曲面上的积分是通过数量场$\vec v\cdot\vec n$在曲面上的第一型积分给出的。

当曲面$S$是一个封闭曲面时，称积分为向量场通过封闭曲面的通量，也记为

\[%\mathop{{\int\!\!\!\!\!\int}\mkern-21mu \bigcirc}_S \vec v\cdot d\vec S % \oiint 需要 esint package \oiint_S \vec v\cdot d\vec S \]

用任意分法将曲面分为$S_1$, $S_2$, $\cdots$, $S_n$，并记面积为$\Delta S_i$。取$M_i\in S_i$，做Riemann和

\[\sigma=\sum_{i=1}^n \vec v(M_i)\cdot\vec n(M_i)\Delta S_i \]

记$\lambda$为曲面块$S_i$的最大直径。 若$\lim\limits_{\lambda\to0}\sigma$存在，且与$M_i$, $S_i$的取法无关，则称这个极限为$\vec v$在$S$上的第二型曲面积分，记为

\[\iint_S \vec v \cdot \vec n dS \]

第二型曲面积分的特性

（1） 场的线性

\[\iint_S(c_1\vec v_1+c_2\vec v_2)\cdot\vec ndS=c_1\iint_S \vec v_1 \cdot \vec n dS+c_2\iint_S \vec v_2 \cdot \vec n dS \]

（2） 积分曲面的可加性， $S$由曲面$S_1$和$S_2$拼接而成，则

\[\iint_S\vec v\cdot\vec ndS=\iint_{S_1} \vec v \cdot \vec n dS+\iint_{S_2} \vec v \cdot \vec n dS \]

（3） 曲面的方向性。若$S^-$和$S^+$表示曲面的两侧，则

\[\iint_{S^-} \vec v \cdot \vec n dS=\iint_{S^+} \vec v \cdot \vec n dS \]
  • 设$\vec v=(P,Q,R)$，正法向量
    \[\vec n=(\cos\alpha, \cos\beta, \cos\gamma) \]
    则第二型积分可以写成
    \[\iint_{S} \vec v \cdot \vec n dS=\iint_S(P\cos\alpha+Q\cos\beta+R\cos\gamma)dS \]
    记
    \[dydz=\cos\alpha dS, dzdx=\cos\beta dS, dxdy=\cos\gamma dS \]
    则得到第二型曲面积分的一个普遍使用的记法
    \[\iint_{S} \vec v \cdot \vec n dS=\iint_S Pdydz+Qdzdx+Rdxdy \]
  • 若$S\perp z$轴，则
    \[\iint_{S} \vec v \cdot \vec n dS=\iint_S Rdxdy \]
  • 若$S$平行于$x$轴，则它在$Oyz$面上投影为线，面积为$0$，则有
    \[\iint_SPdydz=0 \]
    同样，若$S$平行于$y$轴，则
    \[\iint_SQdzdx=0 \]
    若$S$平行于$z$轴，则
    \[\iint_SRdxdy=0 \]

若光滑曲面有参数方程

\[\vec r(u,v)=(x(u,v), y(u,v), z(u,v)), (u,v)\in D \]

则$S$的单位法向量为

\[\vec n=\pm\frac{r'_u\times r'_v}{|r'_u\times r'_v|} \]

由

\[dS=|r'_u\times r'_v|dudv \]

则

\[\iint_S \vec v\cdot\vec ndS=\pm\iint_D \vec v\cdot(r'_u\times r'_v)dudv \]
\[\begin{aligned} &\iint_S \vec v\cdot\vec ndS \\ = & \pm\iint_D\left|{ \begin{matrix} P & Q & R \\ x'_u & y'_u & z'_u \\ x'_v & y'_v & z'_v \end{matrix} }\right|dudv \\ =& \pm\iint_D \left[P\frac{\partial (y,z)}{\partial (u,v)}+Q\frac{\partial (z,x)}{\partial (u,v)}+R\frac{\partial (x,y)}{\partial (u,v)}\right]dudv \end{aligned} \]

显式曲面

\[z=f(x,y), (x,y)\in D \]

有

\[\iint_S\vec v\cdot \vec n dS=\pm\iint_D(-Pf'_x-Qf'_y+R)dxdy \]

特别地，

\[\iint_S Rdxdy=\pm\iint_DR(x,y,f(x,y))dxdy \]

例 2. $S$是顶点为$(1,0,0)$, $(0,1,0)$, $(0,0,1)$的三角形的上侧，求

\[\iint_S xdydz+ydzdx+zdxdy \]
% x+y+z=1 平面 \begin{tikzpicture}[thick,scale=1.4,font=\small,inner sep=0pt] %coordinates \coordinate (O) at (0,0); \coordinate (A) at (-0.5,-0.7); \coordinate (B) at (1.05,0); \coordinate (C) at (0,1); \draw [-stealth, black!75] (A) -- (-0.7,-0.98) node [above=6pt, font={\footnotesize}] {$x$}; \draw [-stealth, black!75] (B) -- (1.3,0) node [above left=2pt and 2pt, font={\footnotesize}] {$y$}; \draw [-stealth, black!75] (C) -- (0,1.2) node [below left=2pt and 4pt, font={\footnotesize}] {$z$}; \draw [cyan,thick,densely dashed] (O) -- (A) (O) -- (B) (O) -- (C); \draw [cyan,thick] (A) -- (B) -- (C) -- cycle; \fill[fill=red] (O) node[left=1.5pt,font={\footnotesize}] {$O$} circle (0.6pt); \fill[fill=red] (A) node[color=blue,below right] {$A(1,0,0)$} circle (0.6pt); \fill[fill=red] (B) node[color=blue,below right=2pt and -3pt] at (B) {$B(0,1,0)$} circle (0.6pt); \fill[fill=red] (C) node[color=blue,right=5pt] at (C) {$C(0,0,1)$} circle (0.6pt); %projection %\visible<4->{ \fill[gray!30,opacity=0.7] (A) -- (B) -- (0,0); \node[font={\footnotesize}] at(0.18,-0.19) {$D_{xy}$}; %} \end{tikzpicture}

例 3. $S$为球$x^2+y^2+z^2=a^2$的外表面，求 $\displaystyle\iint_S z dxdy$

例 4. $S$为椭球面$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$， 计算

\[\iint_S \frac1x dydz+\frac1y dzdx+\frac 1z dxdy \]

例 5. $S$为锥$x^2+y^2=z^2$, $0\leq z\leq h$的外表面，求

\[\iint_S (y-z)dydz+(z-x)dxdz+(x-y)dxdy \]
% 圆柱面截锥面的交线 \tdplotsetmaincoords{70}{60} \begin{tikzpicture}[scale=1.5,tdplot_main_coords] %\begin{tikzpicture}[x={(215:2cm/sqrt 2)},y={(2cm,0cm)},z={(0cm,2cm)},samples=40,scale=1.5] \draw[densely dashed] (0,0,0)--(1,0,0); %,(2,0,0)}; \draw[->] (1,0,0) -- (2,0,0) node[left,below] {$x$}; \draw[densely dashed] (0,0,0)--(0,1,0); %,(2,0,0)}; \draw[->] (0,1,0) -- (0,1.5,0) node[right] {$y$}; \draw[densely dashed] (0,0,0)--(0,0,1); %,(2,0,0)}; \draw[->] (0,0,1) -- (0,0,1.5) node[above] {$z$}; % 锥面 \draw[thick] plot[domain=0:2*pi, samples=40] ({sin(\x r)},{cos(\x r)},1); %%% 涂色 \foreach \x in {0,1,...,5}{ \pgfmathsetmacro{\MyPgfMathResult}{100-\x * 10} \fill[cyan!\MyPgfMathResult!white, opacity=0.5] ({cos(\x*pi/20 r)},{sin(\x*pi/20 r)},1) -- (0,0,0) -- ({cos((\x+1)*pi/20 r)},{sin((\x+1)*pi/20 r)},1) -- cycle; } \foreach \x in {0,1,...,10}{ \pgfmathsetmacro{\MyPgfMathResult}{100-\x * 10} \fill[cyan!\MyPgfMathResult!white, opacity=0.5] ({cos(\x*pi/20 r)},{-sin(\x*pi/20 r)},1) -- (0,0,0) -- ({cos((\x+1)*pi/20 r)},{-sin((\x+1)*pi/20 r)},1) -- cycle; } \fill (0,0,1) circle(1pt) node[left] {$h$}; % % 两个曲面的交线 % \draw[color=red, densely dashed] plot[domain=0:pi, samples=50] % ({1/2*(1+cos(\x r))}, % {1/2*sin(\x r)}, % {1/2*sqrt(2+2*cos(\x r))}); % \draw[color=red] plot[domain=pi:2*pi, samples=50] % ({1/2*(1+cos(\x r))}, % {1/2*sin(\x r)}, % {1/2*sqrt(2+2*cos(\x r))}); \end{tikzpicture}

Gauss公式

如图所示的Y型区域:

\[V=\{(x,y,z)| y_1(z,x)\leq y\leq y_2(z,x), (z,x)\in D\} \]

fig-gauss-domain-y

$S$是$V$的外表面$S=\partial V$，方向指向$V$的外侧。

$Q(x,y,z)$在$V$上有连续的一阶偏导数，则

\[\begin{aligned} &\iint_S Q(x,y,z) dzdx \\ =&\iiint_V Q'_y dxdydz \end{aligned} \]

显然，$S$由上表面$S_2$(方向指向上侧), 下表面$S_1$(方向指向下侧)和侧面$S_3$(与$y$轴平行)组成。

$S_3$与$y$轴平行，因而$\displaystyle \iint_{S_3} Q dzdx=0$。这样

\[\begin{aligned} \iint_S Q(x,y,z)dzdx=&\iint_{S_1}Q dzdx+\iint_{S_2}Q dzdx \\ =&\iint_{D} Q(x,y_2(z,x),z)dxdz\\ &-\iint_{D} Q(x,y_1(z,x),z)dxdz \\ =&\iint_{D} [Q(x,y_2(z,x),z)- Q(x,y_1(z,x),z)]dxdz \\ \end{aligned} \]
\[\begin{aligned} \iint_S Q(x,y,z)dzdx =&\iint_{D} [Q(x,y_2(z,x),z)- Q(x,y_1(z,x),z)]dxdz \\ =&\iint_D \left[ \int_{y_1(z,x)}^{y_2(z,x)}\frac{\partial Q}{\partial y}dy\right] dxdz \\ =&\iiint_V \frac{\partial Q}{\partial y} dxdydz \end{aligned} \]

完全类似地，可以证明

对于Z型区域

\[V=\{(x,y,z)| z_1(x,y)\leq z\leq z_2(x,y), (x,y)\in D\} \]

取$S$为$V$的外表面，方向指向$V$的外侧，有

\[\iint_S R(x,y,z)dxdy=\iiint_V\frac{\partial R}{\partial z} dxdydz \]

对于X型区域

\[V=\{(x,y,z)| x_1(y,z)\leq x\leq x_2(y,z), (y,z)\in D\} \]

取$S$为$V$的外表面，方向指向$V$的外侧，有

\[\iint_S P(x,y,z)dydz=\iiint_V\frac{\partial P}{\partial x} dxdydz \]

事实上，Y型区域也可以看作是没有柱面的X型区域或Z型区域。

fig-gauss-domain-yx

定理 1. (Guss公式)
$V$由分片光滑的双侧封闭曲面$S$围成。 向量场函数

\[\vec v=P(x,y,z)\vec i+Q(x,y,z)\vec j+R(x,y,z)\vec k \]

在$V$中有一阶连续偏导数。 如果$V$可以同时分解成有限个互不重叠的X型、Y型、Z型区域的并，则成立

\[\begin{aligned} \oiint_S Pdydz+&Qdzdx+Rdxdy= \\ &\iiint_V(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z})dxdydz \end{aligned} \]

其中$S$方向指向$V$的外侧。

三维区域的体积，同样可以写成

\[\begin{aligned} V=&\iint_S xdydz=\iint_S ydzdx=\iint_S z dxdy \\ =&\frac13\iint_S xdydz+ydzdx+zdxdy \end{aligned} \]

例 6. $S$为球壳$(x-a)^2+(y-b)^2+(z-c)^2=R^2$，求

\[\iint_S x^2dydz+y^2dzdx+z^2dxdy \]

例 7. $S$为立方体$0\leq x\leq a$, $0\leq y\leq a$, $0\leq z\leq a$的外表面，求

\[\iint_S x^2dydz+y^2dzdx+z^2dxdy \]

例 8. $S$为球$x^2+y^2+z^2=a^2$外侧，求

\[\iint_S x^3dydz+y^3dzdx+z^3dxdy \]

16.

例 9. $S$为某个三维体$V$的外表面，求

\[\iint_S xydydz+yzdzdx+zxdxdy \]

例 10. $S$为某个三维体$V$的外表面，求

\[\iint_S (y-z)dydz+(z-x)dzdx+(x-y)dxdy \]
% Stoltz 公式 \begin{tikzpicture}[ scale=1.5, every node/.append style={scale=1.0}, y={(0:1cm)}, z={(90:1cm)}, x={(225:0.7cm)}] % coordinate system \coordinate (O) at (0, 0, 0); \draw[-latex] (O) -- +(2.1, 0, 0) node [left] {$x$}; \draw[-latex] (O) -- +(0, 2.1, 0) node [right] {$y$}; \draw[-latex] (O) -- +(0, 0, 2.1) node [above] {$z$}; %\draw[domain=0:90, color=red, thick, smooth]%, samples=101 % plot ({sin(\x)},{cos(\x)},{1}); \draw[color=red, thick] plot [smooth ] coordinates { (1,2,1.8) (1.5, 1.5, 2.35) (2,1,2.01) } plot [smooth ] coordinates { (1,1,1.79) (1.5, 1.5, 2.35) (2,2,1.95) }; \draw[red, -latex, thick] (1.5,1.5,2.35) -- +(0.1, -0.1, 0.5) node[above] {$\vec n$}; \draw[color=blue, thick, tension=0.8, postaction={decorate}, decoration={markings, mark=at position 0.35 with {\arrow[blue]{stealth}}}, decoration={markings, mark=at position 0.85 with {\arrow[blue]{stealth}}}] plot [smooth cycle] coordinates { (1,1,0) (2,1,0) (2,2,0) (1,2,0) }; \draw[color=blue, thick, tension=0.8, postaction={decorate}, decoration={markings, mark=at position 0.35 with {\arrow[blue]{stealth}}}, decoration={markings, mark=at position 0.85 with {\arrow[blue]{stealth}}}] plot [smooth cycle] coordinates { (1,1,1.79) (2,1,2.01) (2,2,1.95) (1,2,1.8) }; \draw[color=blue, dashed] (1,1,0) -- (1,1,1.79) (1,2,0) -- (1,2,1.8) (2,2,0) -- (2,2,1.95) (2,1,0) -- (2,1,2.01); \draw (1.5,1.5,0) node {$D$}; \draw (2,2,0) node[below] {$L_1$}; \draw (2,2,1.95) node[below right] {$L$}; \draw (1,2,1.9) node[red, above left] {$S$}; \end{tikzpicture}

曲面$S$以$L$为边界，且方向相协调。 $L$在$Oxy$平面上的投影为$L_1$， 若$S$有显式表达

\[z=z(x,y), (x,y)\in D \]

则有

\[\begin{aligned} &\oint_L P(x,y,z)dx \\ =&\iint_S \frac{\partial P}{\partial z} dzdx-\frac{\partial P}{\partial y} dxdy \\ =&\oint_{L_1} P(x,y,z(x,y)) dx \end{aligned} \]
%Stokes定理， 两个$z=z(x,y)$拼接 \tdplotsetmaincoords{80}{60} \begin{tikzpicture}[scale=1.5, every node/.append style={scale=0.8}, tdplot_main_coords] %\begin{tikzpicture}[x={(215:2cm/sqrt 2)},y={(2cm,0cm)},z={(0cm,2cm)},samples=40,scale=1.5] \draw[densely dashed] (0,0,0)--(1,0,0); %,(2,0,0)}; \draw[->] (1,0,0) -- (2,0,0) node[left,below] {$x$}; \draw[densely dashed] (0,0,0)--(0,1,0); %,(2,0,0)}; \draw[->] (0,1,0) -- (0,1.5,0) node[right] {$y$}; \draw[densely dashed] (0,0,0)--(0,0,1); %,(2,0,0)}; \draw[->] (0,0,1) -- (0,0,1.5) node[above] {$z$}; % 柱面 \draw[color=cyan, densely dashed] plot[domain=0:2*pi] ({cos(\x r)}, {sin(\x r)}, 0); %\draw[ color=cyan] % plot[domain=0:2*pi] ({cos(\x r)}, {sin(\x r)}, 1); \draw[densely dashed, color=cyan] (0,-1,0) -- (0,-1,1) (0,1,0) -- (0,1,1) (-1,0,0) -- (-1,0,1) (1,0,0) -- (1,0,1) ({sin(30)}, {cos(30)}, 0) -- ({sin(30)}, {cos(30)}, 1) ({-cos(60)}, {-sin(60)}, 0) -- ({-cos(60)}, {-sin(60)}, 1); % 两个曲面 \draw[color=red] plot[domain=-1:1, samples=50] ({\x*sin(30)}, {\x*cos(30)}, {sqrt(2-\x*\x)}); \draw[color=red] plot[domain=-1:-0.7, samples=50] ({\x*sin(30)}, {\x*cos(30)}, {\x*\x}); \draw[color=red] plot[domain=0.7:1, samples=50] ({\x*sin(30)}, {\x*cos(30)}, {\x*\x}); \draw[color=red, densely dashed] plot[domain=pi/3:4*pi/3] ({cos(\x r)}, {sin(\x r)}, 1); \draw[color=red] plot[domain=4*pi/3:7*pi/3] ({cos(\x r)}, {sin(\x r)}, 1); \draw[color=red, densely dashed] plot[domain=pi/3:4*pi/3] ({0.7*cos(\x r)}, {0.7*sin(\x r)}, 0.49); \draw[color=red] plot[domain=4*pi/3:7*pi/3] ({0.7*cos(\x r)}, {0.7*sin(\x r)}, 0.49); \draw[red] ({0.7*cos(0)}, {0.7*sin(0)}, 0.49) node[below] {$L$}; \draw[red] ({cos(0)}, {sin(0)}, 1) node[above] {$L'$}; \draw[red] ({0.8*cos(-90)}, {0.8*sin(-90)}, 0.49) node[above left] {$S_2$}; \draw[red] ({cos(-90)}, {sin(-90)}, 1.2) node[above] {$S_1$}; \end{tikzpicture}

若曲面有两个部分，不能简单地表示为$z=z(x,y)$。

则利用曲面拼接时的协调性要求，同样可以得到

\[\begin{aligned} &\oint_{L}P(x,y,z) dx \\ =&\iint_{S_1+S_2} P'_z dzdx-P'_y dxdy \end{aligned} \]

类似地有

  • 若$S$可以表示为 $x=x(y,z)$，则
    \[\oint_L Q(x,y,z) dy = \iint_S \frac{\partial Q}{\partial x} dxdy - \frac{\partial Q}{\partial z} dydz \]
  • 若$S$可以表示为 $y=x(z,x)$，则
    \[\oint_L R(x,y,z) dy = \iint_S \frac{\partial R}{\partial y} dydz - \frac{\partial R}{\partial x} dzdx \]

Stokes 公式

定理 2.
$S$是以曲线$L$为边界的分片光滑的定向曲面。若$P(x,y,z)$, $Q(x,y,z)$, $R(x,y,z)$定义在某个含$S$的空间区域上，且有一阶连续偏导数，则

\[\begin{aligned} \oint_L Pdx+Qdy+Rdz & \\ =\iint\limits_S(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z})dydz&+(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x})dzdx \\ &+(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dxdy \\ \end{aligned} \]

其中$L$的正方向与$S$的正向符合右手法则

写成矩阵形态，

\[\oint_L Pdx+Qdy+Rdz=\iint\limits_S \left|{\begin{matrix} \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ P & Q & R \\ dydz & dzdx & dxdy \end{matrix}}\right| \]

证明.

例 11. $C$为圆周 $\displaystyle\begin{cases}x^2+y^2+z^2=a^2 \\x+y+z=0\end{cases}$， $Oz$轴正向看，逆时针。求

\[\oint_C ydx+zdy+xdz \]
% 球面和平面的交线 \begin{tikzpicture}[x={(215:4cm/sqrt 2)},y={(4cm,0cm)},z={(0cm,4cm)}, scale=0.6, samples=40, global scale=0.5] \draw[densely dashed] (-1,0,0)--(0,0,0); %,(2,0,0)}; \draw (0,0,0)--(1,0,0); %,(2,0,0)}; \draw[->] (1,0,0) -- (2,0,0) node[left,below] {$x$}; \draw (0,0,0)--(0,1,0); %,(2,0,0)}; \draw[->] (0,1,0) -- (0,1.5,0) node[right] {$y$}; \draw (0,0,0)--(0,0,1); %[densely dashed] \draw[->] (0,0,1) -- (0,0,1.5) node[above] {$z$}; % 平面 x+y+z=0 \draw[densely dashed, cyan] ({0.5*sqrt(2)},{-0.5*sqrt(2)},0) node[left] {$(-\frac{\sqrt2}2,\frac{\sqrt2}2,0)$} --({-0.5*sqrt(2)},{0.5*sqrt(2)},0); \coordinate (p1) at ($ 0.5*({sqrt(2)},{-sqrt(2)},0)+0.6*(1,1,-1) $); \coordinate (p2) at ($ 0.5*({sqrt(2)},{-sqrt(2)},0)-0.6*(1,1,-1) $); \coordinate (p4) at ($ 0.5*({-sqrt(2)},{sqrt(2)},0)+0.6*(1,1,-1) $); \coordinate (p3) at ($ 0.5*({-sqrt(2)},{sqrt(2)},0)-0.6*(1,1,-1) $); \draw[cyan, thick] (p1)--(p2)--(p3)--(p4)--cycle; \draw[cyan, thick, -latex] (0,0,0)--(1,1,1) node[above] {$\vec n$}; % 球面 \draw[thick] plot[domain=0:pi] ({cos(\x r)/sqrt(2)+sin(\x r)/sqrt(3)}, {-cos(\x r)/sqrt(2)+sin(\x r)/sqrt(3)}, {sin(\x r)/sqrt(3)} ); \draw[thick, densely dashed] plot[domain=pi:2*pi] ({cos(\x r)/sqrt(2)+sin(\x r)/sqrt(3)}, {-cos(\x r)/sqrt(2)+sin(\x r)/sqrt(3)}, {sin(\x r)/sqrt(3)} ); %\draw[thick] plot[domain=0:pi] ({sin(\x r)},{cos(\x r)},0); %\draw[thick, densely dashed] plot[domain=pi:2*pi] ({sin(\x r)},{cos(\x r)},0); %\draw[thick] plot[domain=0:2*pi] (0,{cos(\x r)},{sin(\x r)}); % 两个曲面的交线 \draw[thick,color=red] plot[domain=0:2*pi] ({cos(\x r)/sqrt(2)-sin(\x r)/sqrt(3)}, {-cos(\x r)/sqrt(2)-sin(\x r)/sqrt(3)}, {sin(\x r)/sqrt(3)} ); \end{tikzpicture}

16.

例 12. $C$为平面$x+y+z=\frac32a$切立方体$0\leq x\leq a$, $0\leq y\leq a$, $0\leq z\leq a$得到的曲线，求

\[\oint_C(y^2-z^2)dx+(z^2-x^2)dy+(x^2-y^2)dz \]
%面$x+y+z=1.5$与单位cube相交的部分 \begin{tikzpicture}[x=(215:2em/sqrt 2), y=(0:2em), z=(90:2em), inner sep=3pt,thick,font=\small,scale=2] %coordinates \draw [-stealth,black!75] (0,0,0) -- (0,0,1.8) node [below right, font={\footnotesize}] {$z$}; \draw [-stealth,black!75] (0,0,0) -- (0,1.8,0) node [above left, font={\footnotesize}] {$y$}; \draw [-stealth,black!75] (0,0,0) -- (1.8,0,0) node [above=3pt, font={\footnotesize}] {$x$}; %curve \draw[-latex,blue,thick,postaction={decorate}, decoration={markings, mark=at position 0.16664 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.5 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.866664 with {\arrow{stealth}}}] (1.5,0,0)--(0,1.5,0)--(0,0,1.5)--cycle; \node[below,blue] at (1.5,0,0) {$1.5a$}; \node[below,blue] at (0,1.5,0) {$1.5a$}; \node[left,blue] at (0,0,1.5) {$1.5a$}; \node[below=3pt,blue] at (0.5,0.5,0) {$\Gamma$}; %surface \fill[gray,opacity=0.6] (1.,0,0.5)--(1, 0.5, 0) -- (0.5,1,0) --(0,1,0.5)--(0,0.5,1)--(0.5,0,1) --cycle; %normal \draw[-latex,red,thick] ({1/3},{1/3},{1/3})--++(1,1,1) node[right] {$\vec{n}$}; % cube \draw (1,0,0) -- (1,1,0) -- (0,1,0) -- (0,1,1) -- (0,0,1) -- (1,0,1) -- (1,0,0) (1,0,1) -- (1,1,1) -- (1,1,0) (1,1,1) -- (0,1,1); \end{tikzpicture}

例 13. 椭圆抛物面$z=3x^2+4y^2$与柱面$4x^2+y^2=4y$相交得到$L$， 从$z$轴的正方向看，$L$的方向是逆时针。求

\[\oint_L y(z+1)dx+zx dy+(xy-z)dz \]
%抛物面与柱面的交线 \tdplotsetmaincoords{80}{120} \begin{tikzpicture}[scale=0.8, every node/.append style={scale=0.8}, tdplot_main_coords] %\begin{tikzpicture}[x={(215:2cm/sqrt 2)},y={(2cm,0cm)},z={(0cm,2cm)},samples=40,scale=1.5] \draw[densely dashed] (0,0,0)--(1,0,0); %,(2,0,0)}; \draw[->] (1,0,0) -- (2,0,0) node[left,below] {$x$}; \draw[densely dashed] (0,0,0)--(0,1,0); %,(2,0,0)}; \draw[->] (0,1,0) -- (0,2.5,0) node[right] {$y$}; \draw[densely dashed] (0,0,0)--(0,0,4); %,(2,0,0)}; \draw[->] (0,0,4) -- (0,0,4.35) node[above] {$z$}; % 柱面 \draw[color=blue, densely dashed] plot[domain=0:2*pi] ({0.5*cos(\x r)}, {(1+sin(\x r))}, 0); \draw[color=blue, densely dashed] plot[domain=0:2*pi] ({0.5*cos(\x r)}, {(1+sin(\x r))}, 4); %\draw[ color=cyan] % plot[domain=0:2*pi] ({cos(\x r)}, {sin(\x r)}, 1); \draw[densely dashed, color=blue] %(0.5,1,0) -- (0.5,1,4) %(-0.5,1,0) -- (-0.5,1,4) (0,2,0) -- (0,2,4); % - 抛物面 \draw[color=red] plot[domain=-1:1, samples=50] ({2*\x*cos(120)}, {2*\x*sin(120)}, {4*\x*\x}); \draw[color=red] plot[domain=0:2*pi] ({2*cos(\x r)}, {2*sin(\x r)}, 4); % - 交线 \draw[color=blue] plot[domain=0:2*pi] ({0.5*cos(\x r)}, {1+sin(\x r)}, {0.25*cos(\x r)^2+(1+sin(\x r))^2}); % 虚线，看着像3D \draw[color=cyan, densely dashed] ({0.5*cos(0)}, {1+sin(0)}, {0.25*cos(0)^2+(1+sin(0))^2}) node[left,blue] {$a$} -- ({0.5*cos(0)}, {1+sin(0)}, 4); \draw[color=cyan, densely dashed] ({0.5*cos(20)}, {1+sin(20)}, {0.25*cos(20)^2+(1+sin(20))^2}) node[left, blue] {$b$} -- ({0.5*cos(20)}, {1+sin(20)}, 4); \draw[color=cyan, densely dashed] ({0.5*cos(160)}, {1+sin(160)}, {0.25*cos(160)^2+(1+sin(160))^2}) node[left,blue] {$c$} -- ({0.5*cos(160)}, {1+sin(160)}, 4); \fill ({0.5*cos(160)}, {1+sin(160)}, {0.25*cos(160)^2+(1+sin(160))^2}) circle(1pt); \fill ({0.5*cos(20)}, {1+sin(20)}, {0.25*cos(20)^2+(1+sin(20))^2}) circle(1pt); \fill ({0.5*cos(0)}, {1+sin(0)}, {0.25*cos(0)^2+(1+sin(0))^2}) circle(1pt); \end{tikzpicture}

例题

例 14. $S$是上半球面$x^2+y^2+z^2=a^2(z\leq 0)$的上侧。求

\[\iint_S (y^2+z^2)dydz+(z^2+x^2)dzdx+(x^2+y^2)dxdy \]

例 15. $L$是柱面$x^2+y^2=a^2$与平面$\frac{x}a+\frac{z}h=1$的交线，从$x$轴正向看$L$沿逆时针方向。 求

\[\oint_L (y-z)dx+(z-x)dy+(x-y)dz \]

目录

谢谢

例 16. 本节读完

16.