第一型曲线和曲面积分

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

第一型曲线和曲面积分

曲线的弧长

定义 1.
$L$是$\mathbb{R}^3$中的一条曲线，在$L$上依次从起点$A$到终点$B$取分点

\[A=P_0, P_1,\cdots, P_n=B \]

称为对曲线$L$的一个分割，记为$T$。 用线段连接$T$中的每相邻两点得到$L$的$n$条弦$P_{i-1}P_i$，这$n$条弦组成$L$的一条内接折线
% 曲线弧长 \begin{tikzpicture}[x=2cm, y=2cm, global scale=0.7, 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] {$P_2$} -- (1.0, 1.65) node[above] {$P_{i-1}$} -- (1.45, 1.55) node[above] {$P_i$} -- (1.9,1.35) node[below] {$P_n=B$}; \end{tikzpicture}

记

\[\|T\|=\max_{1\leq i\leq n}|P_{i-1}P_i| , s(T)=\sum_{i=1}^n|P_{i-1}P_i| \]

分别表示最长弦的长度与折线的总长度。

若

\[\lim_{\|T\|\to0}s(T)=s \]

则称$L$是可求长的，并将极限$s$称为$L$的弧长。即对$\forall \epsilon>0$, $\exists \delta>0$, 当$\|T\|<\delta$时，满足

\[|s(T)-s|<\epsilon, \]

定理 1.
$\mathbb{R}^3$中的曲线$L$有参数方程

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

或$\vec r=\vec r(t), t\in[\alpha,\beta]$。若$L$为光滑曲线（即$x(t)$, $y(t)$, $z(t)$均有连续的导数，且不全为$0$），则$L$是可求长的，且弧长为

\[s=\int_{\alpha}^{\beta} \sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2} dt \]

1

证明.

$P_0=(x(\alpha),y(\alpha),z(\alpha))$到动点$P(t)=(x(t),y(t),z(t))$的弧长为

\[s(t)=\int_\alpha^t \sqrt{(x'(\tau))^2+(y'(\tau))^2+(z'(\tau))^2}d\tau \]

因为$L$是光滑的，则$s(t)$可导，且

\[\frac{ds}{dt}=\sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2} \]

则有

\[\begin{aligned} ds=&\sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}dt \\ =&\sqrt{dx^2+dy^2+dz^2} \end{aligned} \]

为弧长微元

  • 弧长微元$ds$就是函数$\vec r(t)$的微分$d\vec r$的模，即
    \[ds=|d\vec r|=|\vec r'(t)|dt \]
    与前面通过微元分析得到的弧长微元是一致的。
  • 直接利用微元分析，也可以得到弧长为
    \[s=\int_{\alpha}^{\beta} |d\vec r| =\int_{\alpha}^{\beta} \sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2} dt \]
    • 相比于定理，这样的处理简单得多。
    • 用类似的方法，来讨论曲面面积
  • 当$L$光滑时，则
    \[\frac{ds}{dt}=\sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}>0 \]
    所以，$s$是$t$的增函数。
  • 这样，存在反函数$t=t(s)$，且有
    \[\dfrac{dt}{ds}=\frac1{\sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}} \]
  • 将$t=t(s)$代入曲线方程
    \[\vec r=\vec r(t)=\begin{pmatrix} x(t) \\ y(t) \\ z(t) \end{pmatrix} =\begin{pmatrix} x(t(s)) \\ y(t(s)) \\ z(t(s)) \end{pmatrix} =\vec r(s) \]
  • 方程$\vec r=\vec r(s), 0\leq s\leq s_0$（$s_0$为$L$的弧长）以弧长$s$为参数，称为曲线$L$的自然方程
  • 在自然方程下，
    \[\frac{|d\vec r|}{ds}=\frac{\sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}dt}{\sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}dt}=1 \]
  • 在自然方程下，$\vec r(s)$对$s$的微商是单位向量， 是曲线$L$在点$r(s)$处的单位切向量，并指向弧长增加的方向。
  • 若这个切向量的方程余弦为$(\cos\alpha,\cos\beta,\cos\gamma)$，则有
    \[\frac{dx}{ds}=\cos\alpha, \frac{dy}{ds}=\cos\beta, \frac{dz}{ds}=\cos\gamma \]

例 1. 求螺旋线

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

的弧长。
\begin{tikzpicture}[x=(215:1cm/sqrt 2), y=(0:1cm), z=(90:1cm),scale=1] %coordinates \draw[-stealth] (-1,0,0)--(2,0,0) node[above,font={\footnotesize}] {$x$}; \draw[-stealth] (0,-1.5,0)--(0,2,0) node[right,font={\footnotesize}] {$y$}; \draw[-stealth] (0,0,-0.9)--(0,0,3) node[below left,font={\footnotesize}] {$z$}; \draw[thick,blue] plot[variable=\t,domain=0:pi*2,smooth] ({cos(\t r)},{sin(\t r)},{0.2*\t}); \draw[thin,gray] plot[variable=\t,domain=pi*2:pi*4.5,smooth] ({cos(\t r)},{sin(\t r)},{0.2*\t}); \node at(0,1.5,0.5) {$L$}; \end{tikzpicture}

解. 例1.

\[\begin{aligned} ds =& \sqrt{dx^2+dy^2+dz^2} \\ =& \sqrt{ R^2(-\sin t)^2+R^2\cos^2t+k^2} dt \\ =& \sqrt{R^2+k^2}dt \end{aligned} \]

因此，弧长为

\[s=\int_0^{2\pi}dx=2\pi\sqrt{R^2+k^2} \]

第一型曲线积分

$L$为一段非均匀线材，其线密度为$f(x,y,z)$

求总质量。

% 第一型曲线积分 \begin{tikzpicture}[x=2cm, y=2cm, 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); \fill[red] (0,0) circle(1pt) node[left] {$P_0=A$} (-0.1,0.5) circle(1pt) node[left] {$P_1$} (0.23,1.1) circle(1pt) node[left] {$P_2$} (1.0, 1.65) circle(1pt) node[above] {$P_{i-1}$} node[below right] {$\Delta s_i$} (1.45, 1.55) circle(1pt) node[above] {$P_i$} (1.9,1.35) circle(1pt) node[below] {$P_n=B$}; \end{tikzpicture}

从$A$到$B$依次插入$A=P_0$,$P_1$,$\cdots$,$P_n=B$，将$L$分为$n$小段。 记第$i$段弧长为$\Delta s_i$，在第$i$上任取一点$M_i$，则第$i$段的质量近似为$f(M_i)\Delta s_i$，所以总质量近似为

\[\sum_{i=1}^n f(M_i)\Delta s_i \]

这个Riemann和的极限就是$L$的总质量了。

定义 2.
$L$为$\mathbb{R}^3$上可求长曲线，$f(x,y,z)$定义在$L$上。 用$P_0$,$P_1$,$\cdots$,$P_n$把$L$分为$n$段，第$i$段弧长为$\Delta s_i$， $M_i$为第$i$段上任一点，Riemann和

\[S(f,M)=\sum_{i=1}^nf(M_i)\Delta s_i \]

记$\lambda=\max \Delta s_i$，若极限 $\displaystyle\lim_{\lambda\to 0}S(f,M)$ 存在，且与$M_i$的选取无关。 则称此极限为$f(x,y,z)$在曲线$L$上的第一型积分，记为

\[\int_Lf(x,y,z)ds=\lim_{\lambda\to0}\sum_{i=1}^nf(M_i)\Delta s_i \]

从定义上看，第一型曲线积分只是把区间上的定积分中的区间换成了曲线。 因此，第一型曲线积分与定积分有类似的性质

  • 被积函数的线性性、保序性，对积分曲线的可加性等
    \[\int_{L_1}f(x,y,z)ds + \int_{L_2}f(x,y,z)ds = \int_{L_1+L_2} f(x,y,z)ds \]
\[\begin{aligned} \int_{L}(\alpha f(x,y,z)+&\beta g(x,y,z) )ds \\ = & \alpha \int_{L}f(x,y,z)ds + \beta \int_{L} g(x,y,z)ds \end{aligned} \]

定理 2.
光滑曲线$L$的参数方程

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

$f(x,y,z)$为定义在$L$上的连续函数， 则$f(x,y,z)$在曲线$L$上的第一型曲线积分存在，且

\[\begin{aligned} &\int_Lf(x,y,z)ds \\ &=\int_{\alpha}^{\beta}f(x(t),y(t),z(t))\sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}dt \end{aligned} \]

证明. 曲线$L$的弧长为$S$，则以弧长为参数，有

\[\int_L f(x,y,z)dx = \int_0^S f(x(t(s)), y(t(s)), z(t(s))) ds \]

在平面上，弧长微元为$ds=\sqrt{dx^2+dy^2}$

对于平面曲线$y=y(x)$, $x\in[a,b]$，有

\[\int_Lf(x,y)ds=\int_a^b f(x,y(x))\sqrt{1+(y'(x))^2}dx \]

极坐标下的平面曲线$r=r(\theta)$, $\theta\in[\alpha,\beta]$，有

\[\begin{cases} x=r(\theta)\cos\theta \\ y=r(\theta)\sin(\theta) \end{cases} \]
\[\int_Lf(x,y)ds=\int_\alpha^\beta f(r,\theta)\sqrt{r(\theta)^2+(r'(\theta))^2}d\theta \]

在极坐标下，

\[\begin{aligned} dx=& (r'(\theta)\cos\theta + r(\theta)(-\sin\theta)) d\theta \\ dy=& (r'(\theta)\sin\theta + r(\theta)(\cos\theta)) d\theta \end{aligned} \]

因此

\[\begin{aligned} ds=&\sqrt{dx^2+dy^2} \\ =&\sqrt{ (r'(\theta))^2+ (r(\theta))^2}d\theta \end{aligned} \]

例 2. 计算第一型曲线积分

\[\int_L\sqrt{x^2+y^2}ds \]

其中$L$为圆周$x^2+y^2=ax$, $a>0$

\begin{tikzpicture}[x=2cm, y=2cm, global scale=0.5] \draw[->] (0,-1.1)--(0,1.1); \draw[->] (-0.1,0)--(2.3,0); \draw (1,0) circle (1); \draw[->, red] (0,0) -- node[sloped, below, black] {$a \cos\theta$} (-30:{2*cos(30)}); \draw[dashed,blue] (-30:{2*cos(30)})--(2,0); \node[below] at (-30:{2*cos(30)}) {$(r,\theta)$}; \fill (2,0) circle (1pt) node[font=\small, below right] {$a$}; \draw[red] (-30:0.3) arc [start angle=-30, end angle=0, radius=0.3] node [right, pos=0.5] {$\theta$}; \end{tikzpicture}

采用极坐标表示$L$，

\[r(\theta) = a \cos\theta, \theta\in[-\frac{\pi}2,\frac{\pi}2] \]

例 3. 计算

\[\int_L |y|ds \]

其中，$L$为双纽线 $(x^2+y^2)^2=a^2(x^2-y^2)$

% 画 lemniscate 伯努利双纽线 \begin{tikzpicture}[x=2cm, y=2cm, global scale=0.9, baseline=0] %\draw[very thin,color=gray] (-1.5,-1.2) grid (1.5,1.22); \draw[->, red] (-1.1,0) -- (1.1,0) node[right] {$x$}; \draw[->, red] (0,-0.5) -- (0,0.5) node[above] {$y$}; \draw[domain=-pi:pi, samples=101, color=blue, thick] plot ({cos(\x r)/(sin(\x r)^2+1)},{sin(\x r)*cos(\x r)/(sin(\x r)^2+1)}); \fill[orange] (1,0) circle(1pt) node[below right] {$a$}; \draw[orange, dashed] (0,0) -- (45:0.6); \draw[orange, dashed] (0,0) -- (-45:0.6); \end{tikzpicture}

$L$的极坐标表示为

\[r^4=a^2r^2\cos(2\theta) \]

因而，左半边方程为

\[r^2=2\cos(2\theta), \theta\in[-\frac{\pi}4,\frac{\pi}4] \]

例 4. 计算

\[\int_L (x^{\frac43}+y^{\frac43})ds \]

其中，$L$为内摆线 $x^{\frac23}+y^{\frac23}=a^{\frac23}$

%内摆线 $x^{\frac23}+y^{\frac23}=a^{\frac23}$ \begin{tikzpicture}[scale = 1.2] %\clip (-5,-5) rectangle (6,6);%只在这个区域内画图 \draw[->] (-0.2,0) -- (1.2,0);% node[right] {$x$}; \draw[->] (0,-0.2) -- (0,1.2) node[left] {$y$}; % 利用内置函数作图 \draw[domain=0:360, samples=50, color=blue, thick] plot ({cos(\x)^3},{sin(\x)^3}); %\draw[domain=0:2.6, color=red, dashed] plot (\x,\x); \end{tikzpicture}

$L$的参数表示

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

2.

例 5. 计算

\[\int_L x^2ds \]

其中，$L$为圆周 $x^2+y^2+z^2=a^2$, $x+y+z=0$

% 球面和平面的交线 \begin{tikzpicture}[x={(215:3cm/sqrt 3)},y={(3cm,0cm)},z={(0cm,3cm)}, 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) --({-0.5*sqrt(2)},{0.5*sqrt(2)},0) node[red, right] {$(\frac{\sqrt2}2,-\frac{\sqrt2}2,0)$}; \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}

5.

曲面面积

$S$为$\mathbb{R}^3$空间中的曲面，其参数方程为

\[\vec r=\vec r(u,v) =\begin{pmatrix} x(u,v) \\ y(u,v) \\ z(u,v) \end{pmatrix} , (u,v)\in D \]

$S$光滑（$x(u,v),y(u,v),z(u,v)\in C^1$且$|\vec r'_u\times \vec r'_v|\neq0$）。

前面已经得到面积微元为

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

则$S$的面积为

\[S=\iint\limits_D |\vec r'_u\times \vec r'_v|dudv \]

由

\[\begin{aligned} |\vec r'_u\times \vec r'_v|^2 =(\vec r'_u\times\vec r'_v) \cdot (\vec r'_u\times\vec r'_v) \\ =(\vec r'_u)^2(\vec r'_v)^2-(\vec r'_u\cdot\vec r'_v)^2 \end{aligned} \]

记

\[\begin{aligned} E=(\vec r'_u)^2=(x'_u)^2+(y'_u)^2+(z'_u)^2 \\ G=(\vec r'_v)^2=(x'_v)^2+(y'_v)^2+(z'_v)^2 \\ F=\vec r'_u \cdot\vec r'_v =x'_u x'_v+ y'_u y'_v+ z'_u z'_v \end{aligned} \]
\[dS=|\vec r'_u\times\vec r'_v|dudv=\sqrt{EG-F^2}dudv \]
\[S=\iint\limits_D \sqrt{EG-F^2}dudv \]

例 6. 求半径为$R$的球的表面积

解. 取球面的参数表示

\[\vec r(\theta,\phi)= \begin{cases} x= R\sin\theta\cos\phi \\ y= R\sin\theta\sin\phi \\ z= R\cos\theta \end{cases}, (\theta,\phi)\in[0,\pi]\times[0,2\pi] \]

得到

\[\begin{aligned} \vec r'_\theta =& (R\cos\theta \cos\phi, R\cos\theta\sin\phi, -R\sin\theta) \\ \vec r'_\phi =& (-R\sin\theta\sin\phi, R\sin\theta\cos\phi, 0) \end{aligned} \]
\[\begin{aligned} E = &\vec r'_\theta \cdot \vec r'_\theta = R^2 \\ G = & \vec r'_\phi \cdot\vec r'_\phi = R^2\sin^2\theta \\ F = &\vec r'_\theta \cdot\vec r'_\phi = 0 \end{aligned} \]

6. 面积微元为

\[ dS=\sqrt{EG-F^2}d\theta d\phi =\sqrt{R^2\cdot R^2\sin^2\theta-0}d\theta d\phi \]

因而，球面的面积为

\[ \sigma(S)=\int_0^{\pi} d\theta \int_0^{2\pi} R^2|\sin\theta| d\phi =4\pi R^2 \]
  • 若$S$为显式曲面$ z=z(x,y)$，则参数化为
    \[\vec r=\vec r(x,y)=\begin{pmatrix} x \\ y \\ z(x,y) \end{pmatrix} , (x,y)\in D \]
    得到
    \[E=1^2+(z'_x)^2 , G=1^2+(z'_y)^2, F=z'_x z'_y \]
    这样，
    \[dS=\sqrt{1+(z'_x)^2+(z'_y)^2}dxdy \]
    \[S=\iint\limits_D \sqrt{1+(z'_x)^2+(z'_y)^2}dxdy \]

对于上半球面$z=\sqrt{R^2-x^2-y^2}, x^2+y^2\leq R^2$，有

\[\begin{aligned} dS=&\sqrt{1+(z'_x)^2+(z'_y)^2}dxdy \\ =&\sqrt{1+\frac{x^2}{R^2-x^2-y^2}+\frac{y^2}{R^2-x^2-y^2}}dxdy \\ =&\frac{R}{\sqrt{R^2-x^2-y^2}}dxdy \end{aligned} \]

则球面积为

\[\begin{aligned} S=&\iint_{x^2+y^2\leq R^2}\frac{R}{\sqrt{R^2-x^2-y^2}}dxdy \\ =&\int_0^R \int_0^{2\pi} \frac{R}{\sqrt{R^2-r^2}}r d\theta dr =-2\pi R \sqrt{R^2-r^2}\Big|_{r=0}^{R} \end{aligned} \]
  • 若$S$为隐式曲面$F(x,y,z)=0$，且$F'_z\neq0$，则
    \[z'_x=-\frac{F'_x}{F'_z} , z'_y=-\frac{F'_y}{F'_z} \]
    则
    \[s=\iint\limits_D\frac{\sqrt{(F'_x)^2+(F'_y)^2+(F'_z)^2}}{|F'_z|}dxdy \]

例 7. 求球面$x^2+y^2+z^2=R^2$ 被柱面$x^2+y^2=Rx$所截的曲面$S$的面积。

% 球体$x^{2}+y^{2}+z^{2}\leq 4a^2$被圆柱面$x^2+y^2=2ax$（$a>0$）所截得 \begin{tikzpicture}[x=(215:2em/sqrt 2), y=(0:2em), z=(90:2em),thick, global scale=0.8] %coordinates \draw [-stealth, black!75, thin] (0,0,0) -- (3,0,0) node [above] {$x$}; \draw [-stealth, black!75, thin] (0,0,0) -- (0,3,0) node [above] {$y$}; \draw [-stealth, black!75, thin] (0,0,0) -- (0,0,3.5) node [left] {$z$}; \def\R{2} \pgfmathsetmacro{\r}{\R/2} \def\a{125} %angle %sphere \draw plot[variable=\t,domain=0:pi*0.5,smooth]({\R*cos(\t r)},{\R*sin(\t r)},0) -- plot[variable=\t,domain=0:pi*0.5,smooth](0,{\R*cos(\t r)},{\R*sin(\t r)}) -- plot[variable=\t,domain=0:pi*0.5,smooth]({\R*sin(\t r)},0,{\R*cos(\t r)}); %\node[below right=-2pt and -3pt] at(\R,0,0) {$2a$}; \node[below right=-2pt and -3pt] at(0,\R,0) {$R$}; %\node[below right] at({\R*cos(30)},{\R*sin(30)},0) {$x^2+y^2+z^2=4a^2$}; %cylinder \draw plot[variable=\t,domain=0:pi,smooth]({\r+\r*cos(\t r)},{\r*sin(\t r)},0); \draw plot[variable=\t,domain=0:pi,smooth]({\r+\r*cos(\t r)},{\r*sin(\t r)},{\R*1.5}); \draw (\R,0,0)--(\R,0,{\R*1.5})--(0,0,{\R*1.5}); \draw ({\r+\r*cos(\a)},{\r*sin(\a)},0)--({\r+\r*cos(\a)},{\r*sin(\a)},{\R*1.5}); %\node[above right=-2pt and 5pt] at(0,0,{\R*1.5}) {$(x-a)^2+y^2=a^2$}; %intersection %\visible<3->{ \draw[thick] plot[variable=\t,domain=0:pi,smooth]({\r+\r*cos(\t r)},{\r*sin(\t r)},{2*\r*sin(\t/2 r)})-- plot[variable=\t,domain=0:pi*0.5,smooth]({\R*sin(\t r)},0,{\R*cos(\t r)}); \shade[ball color=gray!50,opacity=0.6] plot[variable=\t,domain=0:pi,smooth]({\r+\r*cos(\t r)},{\r*sin(\t r)},{2*\r*sin(\t/2 r)})-- plot[variable=\t,domain=0:pi*0.5,smooth]({\R*sin(\t r)},0,{\R*cos(\t r)})--cycle; \draw[thick] plot[variable=\t,domain=0:\a,smooth]({\r+\r*cos(\t)},{\r*sin(\t)},{2*\r*sin(\t/2)})-- ({\r+\r*cos(\a)},{\r*sin(\a)},0)-- plot[variable=\t,domain=0:\a,smooth]({\r+\r*cos(\a-\t)},{\r*sin(\a-\t)},0); \shade[ball color=gray!60,opacity=0.7] plot[variable=\t,domain=0:\a,smooth]({\r+\r*cos(\t)},{\r*sin(\t)},{2*\r*sin(\t/2)})-- ({\r+\r*cos(\a)},{\r*sin(\a)},0)-- plot[variable=\t,domain=0:\a,smooth]({\r+\r*cos(\a-\t)},{\r*sin(\a-\t)},0); %} %\visible<4->{ \draw[<-] ({\r*0.5},0,{\R*0.5}) to[out=80,in=190] (0,{\r*1.5},{\R*1.25}) node[right] {$z=\sqrt{R^2-x^2-y^2}$}; %} %\visible<5->{ \node[above] at({\r+\r*cos(50)},{\r*sin(50)},0) {$D$}; %} \end{tikzpicture} % 球面和圆柱面的交线 \tdplotsetmaincoords{70}{120} \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] {$z$}; % 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)}); \end{tikzpicture}

例 8. 求曲面$\displaystyle (x^2+y^2+z^2)^2=2a^2xy$的面积

解. 取球坐标$\begin{cases}x=r\sin\theta\cos\phi \\ y=r\sin\theta\sin\phi \\ z=r\cos\theta\end{cases}$，则空间曲线可以表达为

\[r=f(\theta, \phi), (\theta,\phi)\in D \]

可以得到面积微元为

\[\sqrt{EG-F^2}d\theta d\phi =\sqrt{(f'_\phi f)^2+(f^2\sin\theta)^2+(f'_\theta f\sin\theta)^2}d\theta d\phi \]

第一型曲面积分

定义 3.
$S\in\mathbb{R}^3$为一可求面积的曲面，$f(x,y,z)$定义在$S$上。将$S$任意分为$n$块小曲面$S_1,S_2,\cdots,S_n$，每块面积为$\Delta S_i$。$\forall M_i\in S_i$， Riemann和

\[R(M,f)=\sum_{i=1}^n f(M_i)\Delta S_i \]

记所有小曲面的最大半径为$\lambda$，若$\displaystyle\lim_{\lambda\to0}R(M,f)$存在，且与$M_i$的选取无关，则称$f(x,y,z)$在曲面$S$上的第一型曲面积分存在，记为

\[\iint\limits_S f(x,y,z)ds=\lim_{\lambda\to0}R(M,f) \]

定理 3.
$\mathbb{R}^3$中曲面$S$有参数方程

\[r(u,v)=\begin{cases} x=x(u,v) \\ y=y(u,v) \\ z=z(u,v) \end{cases} , (u,v)\in D \]

$D$为$Ouv$平面的有界闭区域。若$f(x,y,z)$在$S$上连续， 则$f(x,y,z)$在$S$上的第一型曲面积分存在，且

\[\begin{aligned} \iint\limits_S f(x,y,z)ds =&\iint\limits_D f(x,y,z) |r'_u\times r'_v|dudv \\ =&\iint\limits_D f(x,y,z) \sqrt{EG-F^2}dudv \\ \end{aligned} \]
  • 若$S$为显式表达的曲面$z=f(x,y)$， $(x,y)\in D$，则
    \[\begin{aligned} &\iint\limits_S f(x,y,z)ds \\ &=\iint\limits_D f(x,y,z(x,y))\sqrt{1+(z'_x)^2+(z'_y)^2}dxdy \end{aligned} \]

例 9. 积分

\[\begin{aligned} \iint\limits_S (x^2+y^2+z^2)ds , \iint\limits_P (x^2+y^2+z^2)ds \\ \end{aligned} \]

其中$S: x^2+y^2+z^2=a^2$，为球面 ；

$P: x+y+z=a$在第一象限部分

% 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] {$A(a,0,0)$} circle (0.6pt); \fill[fill=red] (B) node[color=blue,below right=2pt and -3pt] at (B) {$B(0,a,0)$} circle (0.6pt); \fill[fill=red] (C) node[color=blue,right=5pt] at (C) {$C(0,0,a)$} 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}

9

例 10. 求积分

\[\iint\limits_S zds \]

$S$为曲面$x^2+z^2=2az$, $a>0$被曲面$z=\sqrt{x^2+y^2}$所割下的部分

% 球面和圆柱面的交线 \tdplotsetmaincoords{70}{100} \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] {$z$}; % 柱面 \draw[color=cyan] plot[domain=0:2*pi] ({(1/2)*cos(\x r)}, -1, {(1/2)*sin(\x r)+(1/2)}); \draw[ color=cyan] plot[domain=0:2*pi] ({(1/2)*cos(\x r)}, 1, {(1/2)*sin(\x r)+(1/2)}); \draw[ color=cyan] (0,-1,0) -- (0,1,0) (0,-1,1) -- (0,1,1); % 锥面 \draw[thick,densely dashed] (0,0,0) -- (0,1,1) (0,0,0) -- (0,-1,1); \draw[densely dashed] plot[domain=0:2*pi] ({sin(\x r)},{cos(\x r)},1); % 两个曲面的交线 \draw[densely dashed,thick,color=red] plot[domain=0.5*pi:1.5*pi, samples=50] ({1/(1+cos(\x r)^2)*cos(\x r)}, {1/(1+cos(\x r)^2)*sin(\x r)}, {1/(1+cos(\x r)^2)}); \draw[thick,color=red] plot[domain=1.5*pi:2.5*pi, samples=50] ({1/(1+cos(\x r)^2)*cos(\x r)}, {1/(1+cos(\x r)^2)*sin(\x r)}, {1/(1+cos(\x r)^2)}); % 被割下的部分 \foreach \x in {0,1,...,9}{ \pgfmathparse{\x * 9 + 10} \fill[cyan!\pgfmathresult!white, opacity=0.5] ({1/(1+cos(\x*pi/20 r)^2)*cos(\x*pi/20 r)}, {1/(1+cos(\x*pi/20 r)^2)*sin(\x*pi/20 r)}, {1/(1+cos(\x*pi/20 r)^2)}) -- ({1/(1+cos(\x*pi/20 r)^2)*cos(\x*pi/20 r)}, {-1/(1+cos(\x*pi/20 r)^2)*sin(\x*pi/20 r)}, {1/(1+cos(\x*pi/20 r)^2)}) -- ({1/(1+cos((\x+1)*pi/20 r)^2)*cos((\x+1)*pi/20 r)}, {-1/(1+cos((\x+1)*pi/20 r)^2)*sin((\x+1)*pi/20 r)}, {1/(1+cos((\x+1)*pi/20 r)^2)}) -- ({1/(1+cos((\x+1)*pi/20 r)^2)*cos((\x+1)*pi/20 r)}, {1/(1+cos((\x+1)*pi/20 r)^2)*sin((\x+1)*pi/20 r)}, {1/(1+cos((\x+1)*pi/20 r)^2)}) -- cycle; } \end{tikzpicture}

例10. 取柱面的参数表示$\begin{cases}x=a\sin\theta \\ y=y \\ z=a+a\cos\theta\end{cases}$

需要确定题目中$(\theta, y)$的范围。

柱面被锥面所截，将柱面的参数方程代入锥面的方程，

\[(a+a\cos\theta)^2 - a^2\sin^2\theta = y^2 \]

得到

\[\begin{aligned} &2\cos^2\theta+2\cos\theta \geq 0 \\ &y=\pm\sqrt2 a \sqrt{\cos^2\theta+\cos\theta} \end{aligned} \]

10.

chap7-3-ex-8

例 11. 求积分

\[\iint\limits_S (xy+yz+zx)ds \]

$S$为曲面$z=\sqrt{x^2+y^2}$的锥面被$x^2+y^2=2ax$所割下的部分

% 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] {$z$}; % 柱面 \draw[color=cyan] plot[domain=0:2*pi] ({(1/2)*cos(\x r)+(1/2)}, {(1/2)*sin(\x r)}, 0); \draw[ color=cyan] plot[domain=0:2*pi] ({(1/2)*cos(\x r)+(1/2)}, {(1/2)*sin(\x r)}, 1); \draw[densely dashed, color=cyan] (0,0,0) -- (0,0,1) (0.5,-0.5,0) -- (0.5,-0.5,1) (0.5,0.5,0) -- (0.5,0.5,1) (1,0,0) -- (1,0,1); % 锥面 %\draw[thick,densely dashed] % (0,0,0) -- (0,1,1) % (0,0,0) -- (0,-1,1); \draw[densely dashed] plot[domain=0:2*pi] ({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,...,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; } % 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}

例10： 取曲面的显示表达式$z=\sqrt{x^2+y^2}, (x,y)\in D$，则

\[\begin{aligned} &\iint_S (xy+yz+zx) dS \\ =& \iint_D (xy+(x+y)\sqrt{x^2+y^2}) \sqrt{1+(z'_x)^2+(z'_y)^2} dxdy \\ =& \iint_D (xy+(x+y)\sqrt{x^2+y^2}) \sqrt{2} dxdy \\ \end{aligned} \]

由题意， $D$由曲线$x^2+y^2=2ax$所围成。

观察被积表达式，可以使用极坐标变换来计算二重积分。

11.

chap7-3-ex-9

若$S$为平面，则

\[r=r(u,v)=\begin{pmatrix} x(u,v) \\ y(u,v) \\ 0 \end{pmatrix} , (u,v)\in D \]

则

\[|r'_u\times r'_v|=\left|{\frac{\partial (x,y)}{\partial (u,v)}}\right| \]
\[ds=\left|{\frac{\partial (x,y)}{\partial (u,v)}}\right|dudv \]

重心

设物质分布在空间$V$上，它的密度分布函数为$\rho(x,y,z)$。 则质心坐标$(x_G, y_G, z_G)$满足

\[\begin{cases} x_G=&\dfrac{\iiint_V x \rho(x,y,z) dxdydz}{\iiint_V \rho(x,y,z) dxdydz} \\ y_G=&\dfrac{\iiint_V y \rho(x,y,z) dxdydz}{\iiint_V \rho(x,y,z) dxdydz} \\ z_G=&\dfrac{\iiint_V z \rho(x,y,z) dxdydz}{\iiint_V \rho(x,y,z) dxdydz} \\ \end{cases} \]

  • 设物质分布在光滑曲面$S$上，它的面密度分布函数为$\rho(x,y,z)$

    则质心坐标$(x_G, y_G, z_G)$满足
    \[\begin{cases} x_G=&\dfrac{\iint_S x \rho(x,y,z) dS}{\iint_S \rho(x,y,z) dS} \\ y_G=&\dfrac{\iint_S y \rho(x,y,z) dS}{\iint_S \rho(x,y,z) dS} \\ z_G=&\dfrac{\iint_S z \rho(x,y,z) dS}{\iint_S \rho(x,y,z) dS} \\ \end{cases} \]

  • 设物质分布在光滑曲线$L$上，它的线密度分布函数为$\rho(x,y,z)$

    则质心坐标$(x_G, y_G, z_G)$满足
    \[\begin{cases} x_G=&\dfrac{\int_L x \rho(x,y,z) ds}{\int_L \rho(x,y,z) ds} \\ y_G=&\dfrac{\int_L y \rho(x,y,z) ds}{\int_L \rho(x,y,z) ds} \\ z_G=&\dfrac{\int_L z \rho(x,y,z) ds}{\int_L \rho(x,y,z) ds} \\ \end{cases} \]

例 12. 求摆线

\[\begin{cases} & x=a(t-\sin t) \\ & y=a(1-\cos t) \end{cases} , t\in[0,\pi] \]

的重心

\begin{tikzpicture} [scale=0.6] \begin{axis}[axis lines=middle,width=6cm, height={}, scale only axis=true,axis equal=true, trig format plots=rad,variable=t] \addplot[red,thick,domain=0:pi,samples=100] ({t-sin(t)},{1-cos(t)}); % 画图 \addplot[dashed,blue,domain=pi:2*pi,samples=100] ({t-sin(t)},{1-cos(t)}); % 画图 \end{axis} \end{tikzpicture}

目录

本节读完

例 13. $S$为柱面$x^2+y^2=1$在$z=0$, $z=2$之间的部分，求

\[\iint_S\frac{y+z}{x^2+y^2+z^2}dS \]

例 14. $L$为连接点$A(-2,1)$与原点$O$的线段和圆周$x^2+y^2=-2y$第四象限的部分，求

\[\int_L \sqrt{x^2+y^2}ds \]

13.