第二型曲线积分与Green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

第二型曲线积分与Green公式

第二型曲线与曲面积分与第一型的区别在于，曲线与曲面是有方向的。

问题. 空间域中力场$\vec F(M)$, $M\in V$，$L_{AB}\subset V$为光滑曲线。质点沿$AB$运动，求力$\vec F$对它做的功。

把$L_{AB}$分为$n$小段。 在第$i$段上任取一点$M_i$，则力$F$在第$i$段上近似为常值$\vec F(M_i)$。 力做功的近似为$\vec F(M_i)\cdot\Delta \vec r_i$，其中

\[\Delta \vec r_i=\overrightarrow{A_{i-1}A_{i}}=\vec r_{i}-\vec r_{i-1} \]

指向质点运动的方向。
% 第二型曲线积分 \begin{tikzpicture}[x=3cm, y=3cm, global scale=0.6,font=\small] \def\mya{260} \draw (0,0) to[out=118,in=\mya] (-0.1,0.5) to[out={\mya-180},in={\mya-25}] (0.23,1.1) to[out={\mya-25-180},in={\mya-70}] (1.0, 1.65) to[out={\mya-70-180},in={\mya-100}] coordinate(mi) (1.45, 1.55) to[out={\mya-100-180},in=175] (1.9,1.35); \draw[red] (0,0) node[left] {$P_0=A$} -- (-0.1,0.5) node[left] {$P_1$} -- (0.23,1.1) node[left] {$P_2$} (1.9,1.35) node[below] {$P_n=B$}; \draw[red, -latex] (1.0, 1.65) node[above] {$P_{i-1}$} -- (1.45, 1.55) node[below] {$\Delta \vec r_i$};%node[ right] {$P_i$} \draw[-latex,blue] (mi) node[below] {$M_i$} -- + (30:0.3) node[right] {$\vec F(M_i)$}; \fill[blue] (mi) circle(1pt); \end{tikzpicture}

把这些做功的近似值相加，并让分割的长度趋于0，就可以得到力场所做的总功。

  • 可以看到，即使质点在同一曲线上运动，但运动方向不同，力场所做的功也不同。

 

因此，有必要明确曲线的方向。

定义 1.
规定了方向的曲线就是定向曲线，它有起点和终点。

  • 若曲线$L$的一个端点为$A$，另一个端点为$B$，则 $L_{AB}$用来表示以$A$为起点，$B$为终点的定向曲线。
  • $L_{BA}$则是以$B$为起点，以$A$为终点的定向曲线。

定向曲线

若$L$有参数方程

\[\vec r=\vec r(t)=\begin{pmatrix} x(t) \\ y(t) \\ z(t) \end{pmatrix} , t\in[\alpha,\beta] \]

记$A=\vec r(\alpha)$, $B=\vec r(\beta)$。

  • 对于定向曲线$L_{AB}$，参数$t$增加的方向与$L_{AB}$的正向是一致的，称$t$为定向曲线$L_{AB}$的正向参数；
  • 对于定向曲线$L_{BA}$而言，参数$t$增加的方向与曲线的正向是相反的。
  • 若$L$为光滑曲线。则$L$的切向量$\vec r'(t)$指向$t$增加的方向。
    • 指定单位切向量为$\vec \tau=\dfrac{\vec r'(t)}{|\vec r'(t)|}$，就相当于选择定向曲线为$L_{AB}$
    • 指定单位切向量为$\vec \tau=-\dfrac{\vec r'(t)}{|\vec r'(t)|}$就是使用定向曲线$L_{BA}$

  • 若$L$是封闭曲线。 习惯上，称逆时针方向为正方向。

    此时，$L$围成的区域在$L$行进方向的左边。

第二型曲线积分

定义 2.
$L_{AB}$为$\mathbb{R}^3$中定向曲线，$r$为$L$上点的向量，$\vec F=\vec F(M)$为$L$上的向量场。

用分点$\{P_i\}$将$L_{AB}$分成$n$段，记第$i$段弧的长为$\Delta s_i$。 令$\Delta \vec r_i=\vec r_{i}-\vec r_{i-1}$， 任取第$i$段弧上的点$M_i$，作Riemann和

\[S=\sum_{i=1}^n\vec F(M_i)\cdot\Delta \vec r_i \]
% 第二型曲线积分 \begin{tikzpicture}[x=3cm, y=3cm, global scale=0.6,font=\small] \def\mya{260} \draw (0,0) to[out=118,in=\mya] (-0.1,0.5) to[out={\mya-180},in={\mya-25}] (0.23,1.1) to[out={\mya-25-180},in={\mya-70}] (1.0, 1.65) to[out={\mya-70-180},in={\mya-100}] (1.45, 1.55) to[out={\mya-100-180},in=175] (1.9,1.35); \draw[red] (0,0) node[left] {$P_0=A$} -- (-0.1,0.5) node[left] {$P_1$} -- (0.23,1.1) node[left] {$P_2$} (1.9,1.35) node[below] {$P_n=B$}; \draw[red, -latex] (1.0, 1.65) node[above] {$P_{i-1}$} -- (1.45, 1.55) node[below] {$\Delta \vec r_i$} ;%node[ right] {$P_i$} \coordinate (m) at ($ 0.45*(1.0, 1.65)+0.55*(1.45, 1.55) $); \draw[-latex,blue] (m) node[below] {$M_i$} -- + (30:0.3) node[right] {$\vec F(M_i)$}; \end{tikzpicture}

若$\displaystyle\lambda=\max_{1\leq i\leq n}\Delta s_i\to 0$时，$S$的极限存在，且与分点$\{P_i\}$及取点$\{M_i\}$都无关，则称这个极限为$F$在定向曲线$L_{AB}$上的第二型曲线积分，记为

\[\int_{L_{AB}}\vec F\cdot d\vec r=\lim_{\lambda\to 0}\sum_{i=1}^n\vec F(M_i)\cdot\Delta \vec r_i \]

定向曲线$L_{AB}$称为积分路径。

当$L$为封闭曲线时，$\displaystyle \int_L\vec F\cdot d\vec r$称为向量场沿回路$L$的环量，记为$\displaystyle\oint_L\vec F\cdot d\vec r$。

设$\vec F=(P,Q,R)$，则

\[\vec F\cdot d\vec r=Pdx+Qdy+Rdz \]

这样，第二型曲线积分又可以记为

\[\int_{L_{AB}}\vec F\cdot d\vec r=\int_{L_{AB}} Pdx+Qdy+Rdz \]

若$Q=R=0$，则记为$\displaystyle \int_{L_{AB}}Pdx$

  • 若$L$光滑，且有参数方程
    \[\vec r(t)=\begin{pmatrix} x(t) \\ y(t) \\ z(t) \end{pmatrix}, t\in[\alpha,\beta] \]
    则$d\vec r=(x'(t),y'(t),z'(t))dt$。这样，有
    \[\begin{aligned} \int_{L_{AB}}\vec F\cdot d\vec r =&\int_{\alpha}^{\beta}\vec F(x(t),y(t),z(t))\cdot \vec r'(t) dt \\ %=&\int_{\alpha}^{\beta} (Px'(t) +Q y'(t) +R z'(t)) dt \\ \end{aligned} \]
    • $\displaystyle \int_{L_{AB}} P dx = \int_\alpha^\beta P(x(t),y(t),z(t))x'(t) dt $
    • $\displaystyle \int_{L_{AB}} Q dy = \int_\alpha^\beta Q(x(t),y(t),z(t))y'(t) dt $

例 1. $C$的方程为

\[\begin{cases} & x^2+y^2+z^2=a^2 \\ & x^2+y^2=ax \end{cases} , z\geq0, a>0 \]

求曲线积分

\[\int_C y^2dx+z^2dy+x^2dz \]
\begin{tikzpicture}[x={(-.2cm,-.5cm)},y={(1cm,0cm)},z={(0cm,1cm)},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[densely dashed,color=cyan] plot[domain=0:2*pi] ({(1/2)*cos(\x r)+(1/2)},{(1/2)*sin(\x r)},0); \draw[densely dashed,color=cyan] plot[domain=0:2*pi] ({(1/2)*cos(\x r)+(1/2)},{(1/2)*sin(\x r)},.7); \draw[densely dashed,color=cyan] (.5,.5,0) -- (.5,.5,.7) (.5,-.5,0) -- (.5,-.5,.7); % 球面 \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:pi] (0,{cos(\x r)},{sin(\x r)}); % 两个曲面的交线 \draw[thick,color=red] plot[domain=0:2*pi] ({(1/2)*cos(\x r)+(1/2)},{(1/2)*sin(\x r)},{sqrt((1-cos(\x r))/2)}); \fill[red] (0,0,1) circle(1pt) node[above left] {$B$}; \fill[red] (1,0,0) circle(1pt) node[below right] {$A$}; \fill[red] (0.5,0.5,0.7) circle(1pt) node[above right] {$C$}; \fill[red] (0.5,-0.5,0.7) circle(1pt) node[left] {$D$}; \end{tikzpicture}

1.

例 2. (习题7.5(9-2)) 求曲线积分 $\displaystyle\int_C ydx + zdy+xdz$

(2) $C$为交线 $\begin{cases}z=xy \\x^2+y^2=1\end{cases}$，在$z$轴方向看是逆时针

(3) $C$为交线 $\begin{cases}x+y=2 \\x^2+y^2+z^2=2x+2y\end{cases}$，从原点看是逆时针

解. 需要将曲线$C$参数化，并确定参数是否为正向参数。

2

%$z=xy$, $x^2+y^2=1$ \begin{tikzpicture}[x={(-.2cm,-.5cm)},y={(1cm,0cm)},z={(0cm,1cm)},samples=40,scale=1.5] \draw[densely dashed] (0,0,0)--(1,0,0); \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[densely dashed,color=cyan] plot[domain=0:2*pi] ({cos(\x r)},{sin(\x r)},0); \draw[densely dashed,color=cyan] plot[domain=0:2*pi] ({cos(\x r)},{sin(\x r)},.7); \foreach \theta in {0,30,...,360}{ \draw[densely dashed, color=cyan] ({cos(\theta)},{sin(\theta)},0)--({cos(\theta)},{sin(\theta)},0.7); } %\draw[densely dashed,color=cyan] (.5,.5,0) -- (.5,.5,.7) (.5,-.5,0) -- (.5,-.5,.7); % 面$z=xy$ \foreach \theta in {0,45,...,360}{ \draw[color=blue] plot[domain=0:1.1] ({\x*cos(\theta)},{\x*sin(\theta)},{\x*\x*cos(\theta)*sin(\theta)}); } % 两个曲面的交线 \draw[thick,color=red] plot[domain=0:2*pi] ({cos(\x r)},{sin(\x r)},{cos(\x r)*sin(\x r)}); % \fill[red] (1,0,0) circle(1pt) node[below right] {$A$}; \end{tikzpicture}
%$z=xy$, $x^2+y^2=1$ \begin{tikzpicture}[x={(-.2cm,-.5cm)},y={(1cm,0cm)},z={(0cm,1cm)},samples=40,scale=1.5] \draw[densely dashed] (0,0,0)--(1,0,0); \draw[->] (1,0,0) -- (2.5,0,0) 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)--(0,0,1); %,(2,0,0)}; \draw[->] (0,0,1) -- (0,0,1.5) node[above] {$z$}; % 柱面 \draw[densely dashed,color=cyan] plot[domain=0:2*pi] ({1+sqrt(2)*cos(\x r)},{1+sqrt(2)*sin(\x r)},0); \draw[color=cyan] (2,0,-1.5)--(2,0,1.5)-- (0,2,1.5) -- (0,2,-1.5) --cycle; \draw (2,0,0) -- (0,2,0); \draw[blue] ({1+cos(30)},{1-cos(30)}, {sqrt(2)*sin(30)}) -- (1,1,0); \draw[blue] ({1+0.3*cos(30)},{1-0.3*cos(30)}, {sqrt(2)*0.3*sin(30)}) -- node[left] {$\theta$} (1.3,0.7,0); % 两个曲面的交线 \draw[thick,color=red] plot[domain=0:2*pi] ({1+cos(\x r)},{1-cos(\x r)}, {sqrt(2)*sin(\x r)}); \end{tikzpicture}
  • $L$上的单位切向量为 $\displaystyle\vec \tau=\frac{\vec r'(t)}{|\vec r'(t)|}$， 若与曲线$L_{AB}$的方向一致，则
    \[\vec \tau=\frac{\vec r'(t)}{|\vec r'(t)|} =\frac{\vec r'(t) dt}{|\vec r'(t)|dt} =\frac{d\vec r}{ds} \]
    其中$ds=|\vec r'(t)|dt$为$L$的弧长微元。 由$d\vec r=\vec \tau ds$，有
    \[\int_{L_{AB}}\vec F\cdot d\vec r=\int_{L_{AB}}\vec F\cdot\vec \tau ds \]
  • 第二型曲线积分就是向量场关于曲线有向弧长元素投影的积分， 即函数$\vec F\cdot\vec \tau$在$L$上的第一型曲线积分。
  • 记$d\vec s=\vec \tau ds$，称为有向弧长微元。($d\vec s=d\vec r$)

第二型曲线积分的性质

(1) 线性：

\[\int_{L_{AB}}(c_1\vec F_1+c_2\vec F_2)\cdot d\vec r =c_1\int_{L_{AB}}\vec F_1\cdot d\vec r+c_2\int_{L_{AB}}\vec F_2\cdot d\vec r \]

特别地，有

\[\int_{L_{AB}}\vec F\cdot d\vec r =\int_{L_{AB}}Pdx+\int_{L_{AB}}Qdy+\int_{L_{AB}}Rdz \]

(2) 积分曲线可加性： $L_{AC}$是由$L_{AB}$与$L_{BC}$连接而成，则

\[\int_{L_{AC}}\vec F\cdot d\vec r =\int_{L_{AB}}\vec F\cdot d\vec r+\int_{L_{BC}}\vec F\cdot d\vec r \]

(3) 方向性： $ \displaystyle\int_{L_{AB}}\vec F\cdot d\vec r=-\int_{L_{BA}}\vec F\cdot d\vec r$

  • 若 $L:\begin{cases}x=c \\y=y(t) \\z=z(t)\end{cases}$ 位于垂直于$x$轴的平面上，则
    \[\int_LPdx= \int_\alpha^\beta P(x(t),y(t),z(t)) 0 dt = 0 \]
    物理上来说：若力与运动方向是垂直的，则做功为$0$
  • 若$L$在$x$轴上，为$x$轴闭区间，则$L$可以参数化为
    \[L: \begin{cases} x= x \\ y= 0 \\ z= 0 \end{cases}, x\in[a,b] \]
    \[\int_{L_{AB}}\vec F\cdot d\vec r=\int_a^b P(x,0,0)dx \]
  • 若$L_{AB}: y=y(x), x\in[a,b]$，则$L$可以参数化为
    \[L: \begin{cases} x=x \\ y=y(x) \end{cases}, x\in[a,b] \]
    从而
    \[\begin{aligned} \int_{L_{AB}}Pdx=&\int_a^b P(x,y(x))dx \\ \int_{L_{AB}}Qdy=&\int_a^b Q(x,y(x))y'(x)dx \\ \end{aligned} \]

例 3. $C$为$ y=x^2 , x\in[-1,1]$，求

\[\int_C (x^2-2xy)dx+(y^2-2xy)dy \]

例 4. $O$为原点，$A=(1,2)$，求

\[\int_{OA} xdy-ydx \]

(1) $OA$为直线，$y=2x$

(2) $OA$为抛物线， $y=2x^2$

(3) $OA$为两段折线，$OB: y=0$和$BA: x=1$

3.

定义 3.
$D$为平面区域，如果$D$中任意一条简单闭曲线（自身不相交，首尾相连）的内部都包含在$D$内，则称$D$为单连通区域，否则称为多连通区域

% 单连通域 \begin{tikzpicture}[scale=1.2,font=\small] \draw[fill=gray!30] (0,0) circle (1); \node at(0,-1.3) {$D_1$}; %circle \draw[thick,blue,decoration={markings, mark=at position 0.1 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.6 with {\arrow{stealth}}},postaction={decorate}] (0.3,-0.3) to[out=0,in=-90] (0.7,0.3) to[out=90,in=0] (0.2,0.7) to[out=180,in=90] (-0.1,0.3) to[out=-90,in=180] (0.3,-0.3); \end{tikzpicture} % 平面上的多连通域 \begin{tikzpicture}[scale=1.8,font=\small] %body \def\mya{245} \draw[fill=gray!30] (0,0) to[out=-25,in=\mya] (2,0.5) to[out={\mya-180},in={\mya-180-90}] (1.8,1) to[out={\mya-90},in={-\mya+270}] (0.2,1.1) to[out={-\mya+90},in=165] (0,0); \node at(1,-0.4) {$D_2$}; %hole \draw[fill=white] (0.2,0.2) to[out=0,in=-90] (0.3,0.4) to[out=90,in=0] (0.2,0.5) to[out=180,in=90] (0.1,0.3) to[out=-90,in=180] (0.2,0.2); %hole \draw[fill=white] (1.0,0.6) to[out=0,in=-90] (1.3,0.8) to[out=90,in=0] (1.1,1.0) to[out=180,in=90] (0.8,0.75) to[out=-90,in=180] (1.0,0.6); %hole \draw[fill=white] (1.3,0.0) to[out=0,in=-90] (1.5,0.2) to[out=90,in=0] (1.2,0.3) to[out=180,in=90] (1.0,0.15) to[out=-90,in=180] (1.3,0.0); %hole \draw[fill=white] (0.7,0.3) circle (0.5pt); %circle %\visible<4->{ \draw[thick,red,decoration={markings, mark=at position 0.1 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.6 with {\arrow{stealth}}},postaction={decorate}] (1.0,0.5) to[out=0,in=-90] (1.4,0.9) to[out=90,in=0] (0.9,1.1) to[out=180,in=90] (0.7,0.8) to[out=-90,in=180] (1.0,0.5); %} %\visible<5->{ %circle \draw[thick,blue,decoration={markings, mark=at position 0.1 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.6 with {\arrow{stealth}}},postaction={decorate}] (0.7,0.15) to[out=0,in=-90] (0.9,0.3) to[out=90,in=0] (0.7,0.45) to[out=180,in=90] (0.5,0.3) to[out=-90,in=180] (0.7,0.15); %} \end{tikzpicture}

Green公式

\begin{tikzpicture}[x=2cm, y=2cm, font=\small] \draw[->] (-0.1,0) -- (1.1,0) node[right] {$x$}; \draw[->] (0,-0.1) -- (0,1.2) node[left] {$y$}; %circle \draw[ decoration={markings, mark=at position 0.1 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.6 with {\arrow{stealth}}}, postaction={decorate}] (0.7,0.1) to[out=0,in=-90] (0.9,0.4) to[out=90,in=0] (0.4,1.1) to[out=180,in=90] (0.1,0.7) to[out=-90,in=180] cycle; %(0.3,-0.3); \draw[densely dashed, blue] (0.1, 0) -- (0.1, 0.9); \draw[densely dashed, blue] (0.9, 0) -- (0.9, 0.9); \draw[densely dashed, red] (0, 1.1) -- (0.9, 1.1); \draw[densely dashed, red] (0, 0.1) -- (0.9, 0.1); \node at(0.5, 0.5) {$D$}; \node at(1, 0.5) {$L$}; \end{tikzpicture}

如左图所示的平面区域$D$的边界上 计算第二型积分

\[\oint_L Pdx+Qdy \]

$D$有两种表达形式，分别表示为

\[\begin{aligned} D_y=\{(x,y): y_1(x)\leq y\leq y_2(x), x\in[a,b]\} \\ D_x=\{(x,y): x_1(y)\leq x\leq x_2(y), y\in[c,d]\} \\ \end{aligned} \]
\begin{tikzpicture}[x=4cm, y=4cm, global scale=0.5, font=\small] \draw[->] (-0.1,0) -- (1.1,0) node[right] {$x$}; \draw[->] (0,-0.1) -- (0,1.2) node[left] {$y$}; %circle \draw[red, %decoration={markings, mark=at position 0.3 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.6 with {\arrow{stealth}}}, postaction={decorate}] (0.7,0.1) to[out=0,in=-90] (0.9,0.4) to[out=90,in=0] (0.4,1.1); %to[out=180,in=90] %(0.1,0.7) to[out=-90,in=180] %cycle; %(0.3,-0.3); \draw[blue, %decoration={markings, mark=at position 0.3 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.6 with {\arrow{stealth}}}, postaction={decorate}] (0.4,1.1) to[out=180,in=90] (0.1,0.7) to[out=-90,in=180] (0.7,0.1); %\draw[densely dashed, blue] (0.1, 0) -- (0.1, 0.9); %\draw[densely dashed, blue] (0.9, 0) -- (0.9, 0.9); \draw[densely dashed, red] (0, 1.1) node[left] {$c$} -- (0.9, 1.1); \draw[densely dashed, red] (0, 0.1) node[left] {$d$} -- (0.9, 0.1); \node at(0.5, 0.5) {$D$}; \node[red] at(1, 0.5) {$L_2$}; \node[red] at(1, 0.8) {$x=x_2(y)$}; \node[blue] at(0.1, 0.4) {$L_1$}; \node[blue] at(0.1, 0.3) {$x=x_1(y)$}; \end{tikzpicture}
\[\oint_{L}Qdy =\iint_{D}\frac{\partial Q}{\partial x}dxdy \]
\begin{tikzpicture}[x=4cm, y=4cm, global scale=0.5, font=\small] \draw[->] (-0.1,0) -- (1.1,0) node[right] {$x$}; \draw[->] (0,-0.1) -- (0,1.2) node[left] {$y$}; %circle \draw[red, %decoration={markings, mark=at position 0.3 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.6 with {\arrow{stealth}}}, postaction={decorate}] (0.9,0.4) to[out=90,in=0] (0.4,1.1) to[out=180,in=90] (0.1,0.7); \draw[blue, %decoration={markings, mark=at position 0.3 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.6 with {\arrow{stealth}}}, postaction={decorate}] (0.1,0.7) to[out=-90,in=180] (0.7,0.1) to[out=0,in=-90] (0.9,0.4); \draw[densely dashed, blue] (0.1, 0) node[below] {$a$} -- (0.1, 0.9); \draw[densely dashed, blue] (0.9, 0) node[below] {$b$} -- (0.9, 0.9); %\draw[densely dashed, red] (0, 1.1) -- (0.9, 1.1); %\draw[densely dashed, red] (0, 0.1) node[left] {$d$} -- (0.9, 0.1); \node at(0.5, 0.5) {$D$}; \node[red] at(0.8, 1) {$L_2$}; \node[red, above] at(0.5, 1.1) {$y=y_2(x)$}; \node[blue] at(0.2, 0.3) {$L_1$}; \node[blue, above] at(0.6, 0.2) {$y=y_1(x)$}; \end{tikzpicture}
\[\begin{aligned} \oint_L Pdx =\iint_{D}\left[-\frac{\partial P}{\partial y}\right]dxdy \end{aligned} \]
\[\begin{aligned} \oint_L Qdy =&\int_{L_1}Qdy+\int_{L_2}Qdy \\ =&-\int_c^d Q(x_1(y),y)dy +\int_c^d Q(x_2(y),y)dy \\ =&\int_c^d {\color{red}\left[Q(x_2(y),y)-Q(x_1(y),y)\right]}dy \\ =&\int_c^d {\color{red} \left[\int_{x_1(y)}^{x_2(y)}\frac{\partial Q}{\partial x}dx\right]} dy =\iint_{D}\frac{\partial Q}{\partial x}dxdy \end{aligned} \]
\[\begin{aligned} \oint_L Pdx =&\int_{L_1}Pdx+\int_{L_2}Pdx \\ =&\int_a^b P(x,y_1(x))dx - \int_a^b P(x, y_2(x))dx \\ =&\int_a^b {\color{red}\left[P(x,y_1(x))-P(x,y_2(x))\right]}dx \\ =&\int_a^b {\color{red} \left[-\int_{y_1(x)}^{y_2(x)}\frac{\partial P}{\partial y}dy\right]} dx =\iint_{D}\left[-\frac{\partial P}{\partial y}\right]dxdy \end{aligned} \]

因而，

\begin{tikzpicture}[x=2cm, y=2cm, font=\small] \draw[->] (-0.1,0) -- (1.1,0) node[right] {$x$}; \draw[->] (0,-0.1) -- (0,1.2) node[left] {$y$}; %circle \draw[ decoration={markings, mark=at position 0.1 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.6 with {\arrow{stealth}}}, postaction={decorate}] (0.7,0.1) to[out=0,in=-90] (0.9,0.4) to[out=90,in=0] (0.4,1.1) to[out=180,in=90] (0.1,0.7) to[out=-90,in=180] cycle; %(0.3,-0.3); \draw[densely dashed, blue] (0.1, 0) -- (0.1, 0.9); \draw[densely dashed, blue] (0.9, 0) -- (0.9, 0.9); \draw[densely dashed, red] (0, 1.1) -- (0.9, 1.1); \draw[densely dashed, red] (0, 0.1) -- (0.9, 0.1); \node at(0.5, 0.5) {$D$}; \node at(1, 0.5) {$L$}; \end{tikzpicture}
\[\begin{aligned} \oint_L Pdx+Qdy =&\iint_{D}\left[-\frac{\partial P}{\partial y}\right]dxdy \\ &+\iint_{D}\left[\frac{\partial Q}{\partial x}\right]dxdy \end{aligned} \]

即

\[\oint_L Pdx+Qdy=\iint\limits_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy \]
\begin{tikzpicture}[x=4cm, y=4cm, global scale=0.7, font=\small] \draw[->] (-0.1,0) -- (1.1,0) node[right] {$x$}; \draw[->] (0,-0.1) -- (0,1.2) node[left] {$y$}; %circle \draw[blue, %decoration={markings, mark=at position 0.3 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.6 with {\arrow{stealth}}}, postaction={decorate}] (0.9,0.6) to[out=110,in=0] (0.4,1.1) to[out=180,in=60] (0.1,0.9); \draw[blue, %decoration={markings, mark=at position 0.3 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.6 with {\arrow{stealth}}}, postaction={decorate}] (0.1,0.4) to[out=-70,in=180] (0.7,0.1) to[out=0,in=-100] (0.9,0.3); \draw[red] (0.9, 0.6) node[right] {$C$} -- node[right] {$L_3$} (0.9, 0.3) node[right] {$B$} (0.1, 0.9) node[left]{$D$} -- node[right] {$L_4$} (0.1, 0.4) node[left] {$A$}; \draw[densely dashed, blue] (0.1, 0) node[below] {$a$} -- (0.1, 0.4); \draw[densely dashed, blue] (0.9, 0) node[below] {$b$} -- (0.9, 0.3); \draw[densely dashed, red] (0, 1.1) node[left] {$c$} -- (0.9, 1.1); \draw[densely dashed, red] (0, 0.1) node[left] {$d$} -- (0.9, 0.1); \node at(0.5, 0.5) {$D$}; \node[red] at(0.8, 1) {$L_2$}; \node[red, above] at(0.5, 1.1) {$y=y_2(x)$}; \node[blue] at(0.2, 0.3) {$L_1$}; \node[blue, above] at(0.6, 0.2) {$y=y_1(x)$}; \end{tikzpicture}
\begin{tikzpicture}[x=4cm, y=4cm, global scale=0.7, font=\small] \draw[->] (-0.1,0) -- (1.1,0) node[right] {$x$}; \draw[->] (0,-0.1) -- (0,1.2) node[left] {$y$}; %circle \draw[blue, %decoration={markings, mark=at position 0.3 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.6 with {\arrow{stealth}}}, postaction={decorate}] (0.7,0.1) to[out=10,in=-90] (0.9,0.4) to[out=90,in=-10] (0.6,1.1); \draw[blue, %decoration={markings, mark=at position 0.3 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.6 with {\arrow{stealth}}}, postaction={decorate}] (0.4,1.1) to[out=200,in=90] (0.1,0.7) to[out=-90,in=170] (0.5,0.1); \draw[red] (0.6, 1.1) -- node[below] {$L_3$} (0.4, 1.1) (0.5, 0.1) -- node[below] {$L_4$} (0.7, 0.1); \draw[densely dashed, blue] (0.1, 0) node[below] {$a$} -- (0.1, 0.9); \draw[densely dashed, blue] (0.9, 0) node[below] {$b$} -- (0.9, 0.9); \draw[densely dashed, red] (0, 1.1) node[left] {$c$} -- (0.4, 1.1); \draw[densely dashed, red] (0, 0.1) node[left] {$d$} -- (0.5, 0.1); \node at(0.5, 0.5) {$D$}; \node[red] at(1, 0.5) {$L_2$}; \node[red] at(1, 0.8) {$x=x_2(y)$}; \node[red] at(0.1, 0.4) {$L_1$}; \node[red] at(0.1, 0.3) {$x=x_1(y)$}; \end{tikzpicture}

若区域如左图所示(称为I型)或右图(称为II型)，都同样可以得到

\[\begin{aligned} \oint_L Pdx+Qdy = \iint\limits_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy \end{aligned} \]

对于I型区域，

\[\begin{aligned} \oint_L Qdy = \iint_{D}\left[\frac{\partial Q}{\partial x}\right]dxdy \end{aligned} \]

而

\[\begin{aligned} \oint_L Pdx = \int_{L_1+L_2+L_3+L_4} Pdx \end{aligned} \]

注意到，在$L_3$和$L_4$上，$dx=0$，因而$\displaystyle \int_{L_3+L_4}Pdx=0$。

\[\oint_L Pdx=\int_{L_1+L_2} Pdx =\iint\limits_D\left(-\frac{\partial P}{\partial y}\right)dxdy \]
\begin{tikzpicture}[x=4cm ,y=4cm, font=\small] \draw[->] (0,0) -- (1.1,0) node[right] {$x$}; \draw[->] (0,0) -- (0,1.2) node[left] {$y$}; %circle \draw[ decoration={markings, mark=at position 0.1 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.6 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.8 with {\arrow{stealth}}}, postaction={decorate}] (0.5,0.1) to[out=0,in=-90] (0.9,0.4) to[out=90,in=0] (0.4,1.1) to[out=180,in=90] (0.1,0.8) to[out=-90,in=90] (0.4,0.5) to[out=-90,in=90] (0.1,0.3) to[out=-90,in=180] cycle; %(0.3,-0.3); %\draw[densely dashed, blue] (0.1, 0) -- (0.1, 0.9); %\draw[densely dashed, blue] (0.9, 0) -- (0.9, 0.9); %\draw[densely dashed, red] (0, 1.1) -- (0.9, 1.1); %\draw[densely dashed, red] (0, 0.1) -- (0.9, 0.1); \draw[densely dashed, red] (0.4, 0.5) -- (0.4, 1.1) (0.4, 0.5) -- (0.4, 0.1); \draw[->] (0.45, 0.4) -- (0.45, 0.2); \draw[->] (0.35, 0.2) -- (0.35, 0.4); \draw[->] (0.45, 1.0) -- (0.45, 0.7); \draw[->] (0.35, 0.7) -- (0.35, 1.0); \node[blue] at(0.7, 0.7) {$D_2$}; \node[blue] at(0.2, 0.3) {$D_3$}; \node[blue] at(0.2, 0.8) {$D_1$}; \end{tikzpicture}

对于由光滑曲线组成的一般的连通区域，可以用如图的处理方式。

定理 1. (Green公式)
$D$是由分段光滑闭曲线$L$围成的平面单连通区域，函数$P(x,y)$, $Q(x,y)$在$D$上有一阶连续偏导数，则有

\[\oint_L Pdx+Qdy=\iint\limits_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy \]

其中$L$的方向为：沿此方向行进时，区域$D$始终在左侧。

习惯上称这样的方向为正方向。

  • 在一定条件下，沿平面区域的边界的第二型曲线积分，可以转化成在这个区域上的二重积分

Green公式对多连通域同样成立。

仍然规定边界上的正方向为: 沿边界走，区域在左侧。 则有

\[\oint_{\partial D} Pdx+Qdy=\iint\limits_D(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dxdy \]
\begin{tikzpicture}[scale=2,font=\small] %body \def\mya{245} \draw[fill=gray!30] (0,0) to[out=-25,in=\mya] (2,0.5) to[out={\mya-180},in={\mya-180-90}] (1.8,1) to[out={\mya-90},in={-\mya+270}] (0.2,1.1) to[out={-\mya+90},in=165] (0,0); \draw[thick,blue,decoration={markings, mark=at position 0.3 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.6 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.9 with {\arrow{stealth}}},postaction={decorate}] (0,0) to[out=-25,in=\mya] (2,0.5) to[out={\mya-180},in={\mya-180-90}] (1.8,1) to[out={\mya-90},in={-\mya+270}] (0.2,1.1) to[out={-\mya+90},in=165] (0,0); \node at(0.5,0.2) {$D$}; %hole \draw[fill=white] (0.5,0.5) to[out=180,in=-90] (0.3,0.7) to[out=90,in=180] (0.6,1.0) to[out=0,in=90] (0.8,0.9) to[out=-90,in=0] (0.5,0.5); \draw[thick,red,decoration={markings, mark=at position 0.25 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.5 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.8 with {\arrow{stealth}}},postaction={decorate}] (0.5,0.5) to[out=180,in=-90] (0.3,0.7) to[out=90,in=180] (0.6,1.0) to[out=0,in=90] (0.8,0.9) to[out=-90,in=0] (0.5,0.5); %hole \draw[fill=white] (1.3,0.1) to[out=180,in=-90] (1.0,0.25) to[out=90,in=180] (1.2,0.5) to[out=0,in=90] (1.5,0.3) to[out=-90,in=0] (1.3,0.1); \draw[thick,red, decoration={markings, mark=at position 0.3 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.6 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.9 with {\arrow{stealth}}}, postaction={decorate}] (1.3,0.1) to[out=180,in=-90] (1.0,0.25) to[out=90,in=180] (1.2,0.5) to[out=0,in=90] (1.5,0.3) to[out=-90,in=0] (1.3,0.1); \end{tikzpicture}
\begin{tikzpicture}[scale=2,font=\small] %body \def\mya{245} \draw[fill=gray!30] (0,0) to[out=-25,in=\mya] (2,0.5) to[out={\mya-180},in={\mya-180-90}] (1.8,1) to[out={\mya-90},in={-\mya+270}] (0.2,1.1) to[out={-\mya+90},in=165] (0,0); \draw[thick,blue, name path=outer, decoration={markings, mark=at position 0.3 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.6 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.9 with {\arrow{stealth}}}, postaction={decorate}] (0,0) to[out=-25,in=\mya] (2,0.5) to[out={\mya-180},in={\mya-180-90}] (1.8,1) to[out={\mya-90},in={-\mya+270}] (0.2,1.1) to[out={-\mya+90},in=165] (0,0); \node at(0.5,0.2) {$D$}; %hole \draw[fill=white] (0.5,0.5) to[out=180,in=-90] (0.3,0.7) to[out=90,in=180] (0.6,1.0) to[out=0,in=90] (0.8,0.9) to[out=-90,in=0] (0.5,0.5); \draw[thick,red, decoration={markings, mark=at position 0.25 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.5 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.8 with {\arrow{stealth}}}, postaction={decorate}] (0.5,0.5) to[out=180,in=-90] (0.3,0.7) to[out=90,in=180] (0.6,1.0) to[out=0,in=90] (0.8,0.9) to[out=-90,in=0] (0.5,0.5); %hole \draw[fill=white] (1.3,0.1) to[out=180,in=-90] (1.0,0.25) to[out=90,in=180] (1.2,0.5) to[out=0,in=90] (1.5,0.3) to[out=-90,in=0] (1.3,0.1); \draw[thick,red, name path=inner, decoration={markings, mark=at position 0.3 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.6 with {\arrow{stealth}}}, decoration={markings, mark=at position 0.9 with {\arrow{stealth}}}, postaction={decorate}] (1.3,0.1) to[out=180,in=-90] (1.0,0.25) to[out=90,in=180] (1.2,0.5) to[out=0,in=90] (1.5,0.3) to[out=-90,in=0] (1.3,0.1); \draw[densely dashed, green] (0.6, -0.2) -- (0.6, 1.3); \draw[densely dashed, green] (1.1, -0.2) -- (1.1, 1.3); \draw[-stealth, thick, green] (1.05, -0.10) -- +(0, 0.2); \draw[-stealth, thick, green] (1.05, 0.5) -- +(0, 0.6); \draw[-stealth, thick, green] (1.15, 0.1) -- +(0, -0.2); \draw[-stealth, thick, green] (1.15, 1.1) -- +(0, -0.6); \end{tikzpicture}

例 5. 求$\displaystyle\oint_C (x+y)dx+(x-y)dy$，

  1. $C$为椭圆$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$沿逆时针方向，
  2. $C$为任意的简单闭曲线。

例 6. 曲线$AmO$为点$A=(a,0)$到点$O=(0,0)$的上半圆周$x^2+y^2=ax$，求

\[\int_{AmO}(e^x\sin y-my)dx+(e^x\cos y-m)dy \]
\begin{tikzpicture} [global scale=0.7] \draw [->] (-0.1,0)--(4.1,0) node[anchor=west] {$x$}; \draw [->] (0,-0.1)--(0,2.1) node[anchor=south] {$y$}; % \draw[red] (4,0) arc [start angle=0, end angle=180, radius=2]; \node[anchor=north] at (0,0) {$O$}; \node[anchor=north] at (4,0) {$A$}; \node[anchor=south] at (2,2) {$m$}; \node at (2,1) {$\Omega$}; \draw[blue,ultra thick] (0,0)--(4,0); \end{tikzpicture}

6.

例 7. 求$\displaystyle\oint_C \frac{(x+y)dx-(x-y)dy}{x^2+y^2}$，

  1. $C$为圆$x^2+y^2=a^2$沿逆时针方向，
  2. $C$为椭圆$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$沿逆时针方向，
  3. $C$为任意简单闭曲线，包含$(0,0)$在内，
  4. $C$为任意简单闭曲线，不包含$(0,0)$在内，

解. 注意到

\[\begin{aligned} \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} =&\frac{\partial}{\partial x}\left(\frac{y-x}{x^2+y^2}\right) -\frac{\partial}{\partial y}\left(\frac{x+y}{x^2+y^2}\right) =0 \end{aligned} \]

但$P(x,y)=\frac{x+y}{x^2+y^2}$在$(0,0)$处不连续。

7.

\[\begin{aligned} \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} =&\frac{\partial}{\partial x}\left(\frac{y-x}{x^2+y^2}\right) -\frac{\partial}{\partial y}\left(\frac{x+y}{x^2+y^2}\right) \\ =&\left[\frac{-2x}{(x^2+y^2)^2}(y-x)+\frac{-1}{x^2+y^2}\right] \\ &-\left[\frac{-2y}{(x^2+y^2)^2}(x+y)+\frac{1}{x^2+y^2}\right] \\ =&\frac{-2}{x^2+y^2}+\frac{2}{(x^2+y^2)^2}\left(y(x+y)-x(y-x)\right) \\ =&0 \end{aligned} \]
% Green 公式，区域无奇点 \begin{tikzpicture}[thick,scale=0.8,font=\small] %\path[use as bounding box] (-2.4,-2.3) rectangle (2.1,2.6); \draw[fill=gray!30,draw=blue,postaction={decorate}, decoration={markings, mark=at position 0.35 with {\arrow[blue]{stealth}}}, decoration={markings, mark=at position 0.85 with {\arrow[blue]{stealth}}}] (-1,-1) to[bend right=105] (1.5,2.2) to[bend left=-75] (-1,-1); \node at (0.2,0.5) {$D$}; \node[color=blue,above right] at (1.5,2) {$C$}; \draw[-stealth,thin] (-2.4,-2)--(2.1,-2) node[above left] {$x$}; \draw[-stealth,thin] (-2,-2.3)--(-2,2.6) node[below right] {$y$}; \end{tikzpicture}% % Green 公式，区域有奇点 \begin{tikzpicture}[thick,font=\small] \path[use as bounding box] (-1.4,-1.3) rectangle (2.3,2.6); \draw[fill=gray!30,draw=blue,postaction={decorate}, decoration={markings, mark=at position 0.35 with {\arrow[blue]{stealth}}}, decoration={markings, mark=at position 0.85 with {\arrow[blue]{stealth}}}] (-1,-1) to[bend right=105] (1.5,2.2) to[bend left=-75] (-1,-1); \draw[fill=white,draw=red,postaction={decorate}, decoration={markings, mark=at position 0.35 with {\arrow[red]{stealth}}}, decoration={markings, mark=at position 0.85 with {\arrow[red]{stealth}}}] (0.5,0) arc (0:-360:0.5); \node at (1.2,1.5) {$D_1$}; \node[color=blue,above right] at (1.5,2) {$C$}; \node[color=red,above right] at (0.4,0) {$c$}; \draw[-stealth,thin] (-1.4,0)--(1.9,0) node[below left] {$x$}; \draw[-stealth,thin] (0,-1.3)--(0,2.6) node[below left] {$y$}; \end{tikzpicture}%

例 8. (习题) $L$由$y^2=x-1$, $x=2$围成，沿逆时针方向。

\[\oint_L \sqrt{x^2+y^2}dx+y(xy+\ln(x+\sqrt{x^2+y^2}))dy \]
\begin{tikzpicture} [global scale=1] \draw [->] (-0.1,0)--(2.3,0) node[anchor=west] {$x$}; \draw [->] (0,-1.1)--(0,1.1) node[anchor=south] {$y$}; % \draw[red] plot[domain=-1.1:1.1] ({\x*\x+1}, \x); \draw[blue] (2, -1.1)--(2, 1.1); \end{tikzpicture}

$D$的面积，可以写成

\[A=\iint_D dxdy=\oint_Lxdy=\oint_L(-y)dx \]

这样，有

推论 1.
设$D$是满足Green公式的平面区域，则其面积$A$为

\[A=\oint_Lxdy=-\oint_L y dx=\frac12\oint_L xdy-ydx \]

若$L$的参数方程为$x=x(t)$, $y=y(t)$, $t\in[\alpha,\beta]$，则

\[A=\frac12\left|{ \int_{\alpha}^{\beta} [x(t)y'(t)-y(t)x'(t)]dt }\right| \]

例 9. 求面积$\displaystyle A=\frac12\oint_c xdy-ydx$

(1) 椭圆

\[\begin{cases} & x=a\cos t\\ & y=b\sin t \end{cases} , t\in[0,2\pi] \]

(2) 星形线

\[\begin{cases} & x=a\cos^3 t\\ & y=b\sin^3 t \end{cases} , t\in[0,2\pi] \]

(3) 双纽线 $(x^2+y^2)^2=a^2(x^2-y^2)$

9. (2)

%星形线 Star Curve \begin{tikzpicture}[x=2cm, y=2cm, global scale=0.7, declare function={% rhoone(\a,\t)=sqrt(\a^2*cos(2*\t));}, ] % 上面，用 x=2cm, y=2cm 来设置x,y方向的单位长度，缺省是1cm \draw[domain=0:360, color=blue, thick, smooth]%, samples=101] plot ({2*cos(\x)^3},{sin(\x)^3}); \draw[->] (-2.1,0) -- (2.1,0) node[right] {$x$}; \draw[->] (0,-1.1) -- (0,1.1) node[right] {$y$}; \end{tikzpicture}

(3) 双纽线 $(x^2+y^2)^2=a^2(x^2-y^2)$

%双纽线 $(x^2+y^2)^2=a^2(x^2-y^2)$ \begin{tikzpicture}[x=3cm, y=3cm] \draw[-latex] (-1.1,0) -- (1.2,0) node[below] {$x$}; \draw[-latex] (0,-0.6) -- (0,0.6) node[below left] {$y$}; \draw[red, thick, variable=\x] plot[domain=-45:45] (\x:{sqrt(cos(2*\x ))}); \draw[red, thick, variable=\x] plot[domain=135:225] (\x:{sqrt(cos(2*\x ))}); \fill (1,0) circle(1pt) node[below right] {$a$}; \end{tikzpicture}

例 10. 求外摆线（$m<1$为外圆与内圆的半径比）的一拱与对应的圆弧所围成的图形的面积

\[\begin{cases} x=a[(1+m)\cos(mt)-m\cos(1+m)t] \\ y=a[(1+m)\sin(mt)-m\sin(1+m)t] \end{cases} \]
%外摆线 \begin{tikzpicture}[x=4cm, y=4cm, global scale=0.7] \draw[-latex] (0,0) -- (1.3,0) node[below] {$x$}; \draw[-latex] (0,0) -- (0,1.3) node[right] {$y$}; \draw[black, thick] plot[domain=0:2*pi, variable=\t] ({1.2*cos(deg(0.2*\t))-0.2*cos(1.2*deg(\t))},{1.2*sin(0.2*deg(\t))-0.2*sin(1.2*deg(\t))}); \node[above, black] at (36:1.2) {$L$}; \draw[red] (1,0) arc [start angle=0, end angle=72, radius=1] node[pos=0.5, below left] {$C$}; % 画一个外接圆 \draw[green] ({1.2*cos(24)}, {1.2*sin(24)}) circle[radius=0.2]; \draw[green,dashed] (0,0)--({1.2*cos(24)}, {1.2*sin(24)}); \draw[green,dashed] ({1.2*cos(24)}, {1.2*sin(24)})-- +({0.2*cos(24*6+180)}, {0.2*sin(24*6+180)}); \draw[blue] (0,0) -- node[above] {$A$} (1,0); \draw[blue] (0,0) -- node[above left] {$B$} (72:1); \end{tikzpicture}

ex7-5-waibai

例 11. 求笛卡尔叶形线围成的面积

\[x^3+y^3=3axy \]
\begin{tikzpicture}[x=3cm, y=3cm, global scale=0.7] \draw[-latex] (-0.5,0) -- (1.5,0) node[below] {$x$}; \draw[-latex] (0,-0.5) -- (0,1.5) node[right] {$y$}; \draw[red, thick] plot[domain=-10:100, variable=\t, samples=200] (\t:{3*sin(\t )*cos(\t )/((cos(\t ))^3+(sin(\t ))^3)}); \end{tikzpicture}

参数化，令$y=tx$，有

\[x=\frac{3at}{1+t^3}, y=\frac{3at^2}{1+t^3}, t\in[0,+\infty) \]
%笛卡尔叶形线 descartes \begin{tikzpicture}[x=3cm, y=3cm, global scale=0.7] \draw[-latex] (0,0) -- (1.3,0) node[below] {$x$}; \draw[-latex] (0,0) -- (0,1.8) node[right] {$y$}; \draw[red, thick] plot[domain=-0.1:15, variable=\t, samples=200] ({3*\t/(1+\t*\t*\t)},{3*\t*\t/(1+\t*\t*\t)}); \draw[blue, thick] plot[domain=-10:-4.5, variable=\t, samples=200] ({3*\t/(1+\t*\t*\t)},{3*\t*\t/(1+\t*\t*\t)}); \end{tikzpicture}

取极坐标，

\[r(\cos^3\theta+\sin^3\theta) = 3a\cos\theta\sin\theta, \theta\in[0,\frac{\pi}2] \]

则$r(\theta)=\frac{3a\cos\theta\sin\theta}{\cos^3\theta+\sin^3\theta}$，面积为

\[\frac12\int_0^{\frac{\pi}2} r^2(\theta)d\theta =\frac{9a^2}2\int_0^{\frac{\pi}2} \left(\frac{\cos\theta\sin\theta}{\cos^3\theta+\sin^3\theta}\right)^2 d\theta \]

目录

例

例 12. 本节读完

12.

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