曲线积分

Curve Integral

一、第一类曲线积分（对弧长的曲线积分）

Line Integral with Respect to Arc Length 对弧长的曲线积分，标量场

曲线形构件的质量，利用弧微分转换为定积分的计算

基本定义

$${\int }_{L}f(x,y)\mathrm{d}s=\underset{\lambda \to 0}{\lim }\sum _{i=1}^{n}f({\xi }_i,{\eta }_i)\Delta s_i$$

$L$ 为积分弧段

实际计算

\begin{cases} x=\varphi(t) \\ & (\alpha\leq t\leq \beta)\\ y=\psi(t) \end{cases} \quad \int _{L}f(x,y)\, \mathrm{d}s = \int _{\alpha}^{\beta} f[\varphi(t),\psi(t)]\sqrt{ \varphi'^{2}(t)+ \psi'^{2}(t) } \, \mathrm{d} t \end{align}$$ $$\begin{align} y & =\psi(x) \;\; (x_{0}\leq x\leq X) \quad \int _{L}f(x,y)\, \mathrm{d}s = \int _{x_{0}}^{X} f[x,\psi(x)]\sqrt{ 1+\psi'^{2}(x) }\, dx \\ x & =\varphi(y) \;\; (y_{0}\leq y\leq Y) \quad \int _{L}f(x,y)\, \mathrm{d}s = \int _{y_{0}}^{Y} f[\varphi(y),y]\sqrt{ 1+\varphi'^{2}(y) }\, dy \end{align}$$ 1. 确定积分曲线的表达式 2. 确定积分变量，以及积分变量的取值范围 3. 被积函数换元，并确定弧微分 $\mathrm{d}s$ ### 二、第二类曲线积分（对坐标的曲线积分） **Line Integral with Respect to the Coordinate** 第二类曲线积分 **向量场** > 变力沿曲线所作的[[功\|功]] #### 基本定义 $$\begin{align} \int _{L}P(x,y)\, dx =\lim\limits_{ \lambda \to 0 } \sum\limits_{i=1}^{n}P(\xi_{i},\eta_{i})\Delta x_{i} \\ \int _{L}Q(x,y)\, dy =\lim\limits_{ \lambda \to 0 } \sum\limits_{i=1}^{n}Q(\xi_{i},\eta_{i})\Delta y_{i} \end{align}$$ #### 实际计算 转化为[[定积分\|定积分]]进行计算 1. 换元 2. 确定积分限 3. 换积分变量：利用弧微分 二维情形： $x=\varphi (t),y=\psi (t)$ $$\begin{align} \int _{L} \mathbf{F}(x,y)\, \cdot\mathrm{d} \mathbf{r} & =\int _{L} P(x,y)\, \mathrm{d}x +Q(x,y) \,\mathrm{d}y=\int _{\alpha}^{\beta} \left\{P[\varphi(t),\psi(t)]\varphi'(t)+Q[\varphi(t),\psi(t)]\psi'(t) \right\} \, \mathrm{d}t \end{align}$$ 三维情形：$x=\varphi (t),y=\psi (t),z=\omega(t)$ $$\begin{align} & \int _{\Gamma} \mathbf{A}(x,y,z)\cdot \mathrm{d} \mathbf{r}=\int _{\Gamma} P(x,y,z)\mathrm{d}x+Q(x,y,z)\mathrm{d}y+R(x,y,z)\mathrm{d}z \\ & =\int _{\alpha}^{\beta} \left\{P[\varphi(t),\psi(t),\omega (t)]\varphi'(t)+Q[\varphi(t),\psi(t),\omega (t)]\psi'(t)+R[\varphi(t),\psi(t),\omega (t)]\omega'(t) \right\} \, \mathrm{d}t \end{align}$$ ### 三、两类曲线积分之间的联系 $$\begin{align} \int _{L} \mathbf{A}\cdot \mathrm{d}\mathbf{r}= \int _{L} \mathbf{A}\cdot \boldsymbol{\tau}\mathrm{d}s \end{align}$$ $$\begin{align} \int _{L} P\mathrm{d}x+Q\mathrm{d}y= \int _{L} (P\cos \alpha+Q\cos \beta)\, ds \end{align}$$ $$\begin{align} \int _{L} P\mathrm{d}x+Q\mathrm{d}y+R\mathrm{d}z= \int _{L} (P\cos \alpha+Q\cos \beta+R\cos\gamma)\, ds \end{align}$$