曲线与曲面积分
XantC
曲线积分
对弧长的曲线积分
定义
\[ \int_Lf(x,y)\mathrm{d}s=\lim_{\lambda \to 0}\sum_{i=1}^nf(\xi_i,\eta_i)\Delta s_i \]
计算
\[ \begin{cases} x=\varphi(t)\\ y=\psi(t) \end{cases},\;t_1\leq t\leq t_2\colon \]
\[ \int_Lf(x,y)\mathrm{d}s=\int_{t_1}^{t_2}f(\varphi(t), \psi(t))\sqrt{(\varphi(t)^{'})^2+(\psi(t)^{'})^2}\mathrm{d}t \]
对坐标的曲线积分
定义
\[ \int_L\boldsymbol{F}\mathrm{d}\boldsymbol{r}=\int_LP(x,y)\mathrm{d}x+Q(x,y)\mathrm{d}y \]
计算
\[ \begin{cases} x=\varphi(t)\\ y=\psi(t) \end{cases},\;t_1\leq t\leq t_2\colon \]
\[ \int_LP(x,y)\mathrm{d}x+Q(x,y)\mathrm{d}y=\int_{t_1}^{t_2}(P(\varphi(t), \psi(t))\varphi^{'}(t)+Q(\varphi(t), \psi(t))\psi^{'}(t))\mathrm{d}t \]
两类曲线积分关系
\[ \boldsymbol{\tau}=(\cos\alpha, \cos\beta) \]
\[ \int_L\boldsymbol{A}\mathrm{d}\boldsymbol{r}=\int_L\boldsymbol{A}\cdot\boldsymbol{\tau}\mathrm{d}s \]
证明
\[ \boldsymbol{r}=(\varphi(t), \psi(t)) \]
\[ 令\cos\alpha=\frac{\varphi^{'}(t)}{\sqrt{(\varphi^{'}(t))^2+(\psi^{'}(t))^2}},\cos\beta=\frac{\psi^{'}(t)}{\sqrt{(\varphi^{'}(t))^2+(\psi^{'}(t))^2}} \]
\[ \begin{align*} \int_{L}(P\cos\alpha+Q\cos\beta)\mathrm{d}s&=\int_{L}(P\cos\alpha+Q\cos \beta)\sqrt{(\varphi^{'}(t))^2+(\psi^{'}(t))^2}\mathrm{d}t\\ &=\int_L (P\varphi^{'}(t)+Q\psi^{'}(t))\mathrm{d}t\\ &=\int_L P\mathrm{d}x+Q\mathrm{d}y \end{align*} \]
格林公式
\[ \iint_{D}(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})\mathrm{d}x\mathrm{d}y=\oint P\mathrm{d}x+Q\mathrm{d}y \]
意义
\[ 把关于区域D内一切点的二重积分，转化成了只与区域边界有关的曲线积分 \]
证明
把区域D切成很多个小正方形，再对其中每一个的边界求曲线积分，会发现互相重合的积分抵消了，只剩下最外围的一圈。所以对任意区域D边界的曲线积分，就可以把它算成所有正方形的曲线积分元素之和。
\[ 对于单个正方形M_1M_2M_3M_4\colon M_1(a_0,b_0),M_2(a_1,b_0),M_3(a_1,b_1),M_4(a_0,b_1) \]
\[ \begin{align*} \int_{L_0}P\mathrm{d}x+Q\mathrm{d}y&=\int_{M_1M_2}P\mathrm{d}x+\int_{M_2M_3}Q\mathrm{d}y+\int_{M_3M_4}P\mathrm{d}x+\int_{M_4M_1}Q\mathrm{d}y\\ &=\int^{a_1}_{a_0}P(x,b_0)\mathrm{d}x+\int^{b_1}_{b_0}Q(a_1,y)\mathrm{d}y+\int^{a_0}_{a_1}P(x,b_1)\mathrm{d}x+\int^{b_0}_{b_1}Q(a_0,y)\mathrm{d}y\\ &=\int^{b_1}_{b_0}[Q(a_1,y)-Q(a_0,y)]\mathrm{d}y-\int^{a_1}_{a_0}[P(x,b_1)-P(x,b_0)]\mathrm{d}x \\ &=\int^{b_1}_{b_0}[\int^{a_1}_{a_0}\frac{\partial Q}{\partial x}\mathrm{d}x]\mathrm{d}y-\int^{a_1}_{a_0}[\int^{b_1}_{b_0}\frac{\partial P}{\partial y}\mathrm{d}y]\mathrm{d}x\\ &=\iint_{cube}(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})\mathrm{d}x\mathrm{d}y \end{align*} \]
\[ \therefore \oint_LP\mathrm{d}x+Q\mathrm{d}y= \iint_{D}(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})\mathrm{d}x\mathrm{d}y \]
应用
曲线积分与路径无关条件
\[ \frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x} \]
证明
\[ \int_{L_1} P\mathrm{d}x+Q\mathrm{d}y=\int_{L_2} P\mathrm{d}x+Q\mathrm{d}y \]
\[ \Rightarrow \int_{L_1+\overline{L_2}}P\mathrm{d}x+Q\mathrm{d}y=0 \]
\[ \Rightarrow \int_{L_1+\overline{L_2}}(\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x})\mathrm{d}x\mathrm{d}y=0 \]
\[ \therefore \frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x} \]
全微分求积分
\[ u(x,y)=P(x,y)\mathrm{d}x+Q(x,y)\mathrm{d}y \]
\[ 条件\colon \frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x} \]
积分路线法
\[ \begin{align*} u(x,y)&=\int^{(x,y)}_{(x_0,y_0)}P(x,y)\mathrm{d}x+Q(x,y)\mathrm{d}y\\ &=\int^{(x,y_0)}_{(x_0,y_0)}P(x,y_0)\mathrm{d}x+\int^{(x,y)}_{(x,y_0)}Q(x,y)\mathrm{d}y \end{align*} \]
偏积分法
\[ \begin{cases} u=\displaystyle\int P(x,y)\mathrm{d}x+\varphi(y)\\ u=\displaystyle\int Q(x,y)\mathrm{d}y+\psi(x) \end{cases} \]
曲线积分基本定理
\[ \int_L \nabla f\cdot\mathrm{d}\boldsymbol{r}=f(\boldsymbol{r}(b))-f(\boldsymbol{r}(a)) \]
证明
\[ 令\boldsymbol{r}=(\varphi(t), \psi(t)),\; a\leq t\leq b \]
\[ \begin{align*} \nabla f\cdot\frac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t}&=(f_x^{'},f_y^{'})\cdot(\varphi^{'}(t), \psi^{'}(t))\\ &=f_x^{'}\varphi^{'}(t)+f_y^{'}\psi^{'}(t)\\ &=\frac{\mathrm{d}f}{\mathrm{d}t} \end{align*} \]
\[ \Rightarrow \nabla f\cdot \mathrm{d}\boldsymbol{r}=\mathrm{d}f \]
\[ \Rightarrow \int_L \nabla f\cdot\mathrm{d}\boldsymbol{r}=\int^{\boldsymbol{r}(b)}_{\boldsymbol{r}(a)}\mathrm{d}f=f(\boldsymbol{r}(b))-f(\boldsymbol{r}(a)) \]
曲面积分
对面积曲面积分
定义
\[ \iint_{\Sigma}f(x,y,z)\mathrm{d}S=\lim_{\lambda \to 0}\sum^n_{i=1}f(\xi_i,\eta_i,\zeta_i)\Delta S \]
计算
\[ \iint_{\Sigma}f(x,y,z)\mathrm{d}S=\iint_{D_{xy}}f(x,y,z(x,y))\sqrt{1+z_x^{'2}+z_y^{'2}}\mathrm{d}x\mathrm{d}y \]
对坐标曲面积分
定义
\[ \iint_{\Sigma}\boldsymbol{A}\cdot\mathrm{d}\boldsymbol{S}=\iint_{\Sigma}P(x,y,z)\mathrm{d}y\mathrm{d}z+Q(x,y,z)\mathrm{d}z\mathrm{d}x+R(x,y,z)\mathrm{d}x\mathrm{d}y \]
计算
\[ \iint_{\Sigma}P(x,y,z)\mathrm{d}y\mathrm{d}z=\pm\iint_{D_{yz}}P(x(y,z),y,z)\mathrm{d}y\mathrm{d}z \]
\[ \iint_{\Sigma}Q(x,y,z)\mathrm{d}z\mathrm{d}x=\pm\iint_{D_{xz}}P(x,y(x,z),z)\mathrm{d}z\mathrm{d}x \]
\[ \iint_{\Sigma}R(x,y,z)\mathrm{d}x\mathrm{d}y=\pm\iint_{D_{xy}}P(x,y,z(x,y))\mathrm{d}x\mathrm{d}y \]
两类曲面积分关系
\[ \iint_{\Sigma}\boldsymbol{A}\cdot\boldsymbol{n}\mathrm{d}S=\iint_{\Sigma}\boldsymbol{A}\cdot\mathrm{d}\boldsymbol{S} \]
高斯公式
\[ {\int\kern{-7pt}\int \kern{-24mu} \bigcirc}_{\Sigma}P\mathrm{d}y\mathrm{d}z+Q\mathrm{d}z\mathrm{d}x+R\mathrm{d}x\mathrm{d}y=\iiint_{\Omega}(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z})\mathrm{d}V \]
\[ {\int\kern{-7pt}\int \kern{-24mu} \bigcirc}_{\Sigma}\boldsymbol{A}\cdot\mathrm{d}\boldsymbol{S}=\iiint_{\Omega}\text{div}\; \boldsymbol{A}\mathrm{d}V \]
通量
\[ \Phi=\iint_{\Sigma}\boldsymbol{A}\cdot\boldsymbol{n}\mathrm{d}S=\iint_{\Sigma}\boldsymbol{A}\cdot\mathrm{d}\boldsymbol{S}=\iint_{\Sigma}P\mathrm{d}y\mathrm{d}z+Q\mathrm{d}z\mathrm{d}x+R\mathrm{d}x\mathrm{d}y \]
散度
\[ \text{div}\; \boldsymbol{A}=\nabla\cdot\boldsymbol{A}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z} \]
意义
- 指出立体图形边界和内部的关系
- 指出通量等于散度
- 计算曲面积分时，把三个二重积分转化成了一个三重积分
证明阿基米德原理
\[ \begin{align*} \boldsymbol{F}&=({\int\kern{-7pt}\int \kern{-24mu} \bigcirc}\rho gz\cos\alpha\mathrm{d}S,{\int\kern{-7pt}\int \kern{-24mu} \bigcirc}\rho gz\cos\beta\mathrm{d}S,{\int\kern{-7pt}\int \kern{-24mu} \bigcirc}\rho gz\cos\gamma\mathrm{d}S)\\ &=(0,0,\rho g\iiint_{\Omega}\mathrm{d}V)\\ &=(0,0,\rho gV) \end{align*} \]
斯托克斯公式
\[ \begin{align*} \oint_{L}P\mathrm{d}x+Q\mathrm{d}y+R\mathrm{d}z&=\iint_{\Sigma}\begin{vmatrix} \mathrm{d}y\mathrm{d}z&\mathrm{d}z\mathrm{d}x&\mathrm{d}x\mathrm{d}y\\ \displaystyle\frac{\partial}{\partial x}&\displaystyle\frac{\partial}{\partial y}&\displaystyle\frac{\partial}{\partial z}\\P&Q&R\\ \end{vmatrix}\\ &=\iint_{\Sigma}\begin{vmatrix} \cos\alpha&\cos\beta&\cos\gamma\\ \displaystyle\frac{\partial}{\partial x}&\displaystyle\frac{\partial}{\partial y}&\displaystyle\frac{\partial}{\partial z}\\P&Q&R\\ \end{vmatrix}\mathrm{d}S \end{align*} \]
\[ \oint_L \boldsymbol{A}\cdot\boldsymbol{\tau}\mathrm{d}s=\iint_{\Sigma}\mathbf{rot}\;\boldsymbol{A}\cdot\boldsymbol{n}\mathrm{d}S \]
证明
\[ \iint_{\Sigma}(\frac{\partial P}{\partial z}\mathrm{d}z\mathrm{d}x-\frac{\partial P}{\partial y}\mathrm{d}x\mathrm{d}y)=\iint_{\Sigma}(\frac{\partial P}{\partial z}\cos\beta-\frac{\partial P}{\partial y}\cos\gamma)\mathrm{d}S \]
\[ \because \cos\beta=\frac{-f^{'}_y}{\sqrt{1+f^{'2}_x+f^{'2}_y}},\;\cos\gamma=\frac{1}{\sqrt{1+f^{'2}_x+f^{'2}_y}} \]
\[ \therefore \cos\beta=-f^{'}_y\cos\gamma \]
\[ \begin{align*} \Rightarrow \iint_{\Sigma}(\frac{\partial P}{\partial z}\mathrm{d}z\mathrm{d}x-\frac{\partial P}{\partial y}\mathrm{d}x\mathrm{d}y)&=-\iint_{\Sigma}(\frac{\partial P}{\partial z}f^{'}_y\cos\gamma+\frac{\partial P}{\partial y}\cos\gamma)\mathrm{d}S\\ &=-\iint_{D_{xy}}(\frac{\partial P}{\partial z}f^{'}_y+\frac{\partial P}{\partial y})\mathrm{d}x\mathrm{d}y\\ &=-\iint_{D_{xy}}(\frac{\partial}{\partial y}P[x,y,f(x,y)])\mathrm{d}x\mathrm{d}y\\ &=\oint_L P\mathrm{d}x \end{align*} \]
\[ \therefore \oint_L P\mathrm{d}x+Q\mathrm{d}y+R\mathrm{d}z=\iint_{\Sigma}(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z})\mathrm{d}y\mathrm{d}z+(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x})\mathrm{d}z\mathrm{d}x+(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})\mathrm{d}x\mathrm{d}y \]
环流量
\[ \oint_L\boldsymbol{A}\cdot\boldsymbol{\tau}\mathrm{d}s=\oint_L\boldsymbol{A}\mathrm{d}\boldsymbol{r}=\oint_L P\mathrm{d}x+Q\mathrm{d}y+R\mathrm{d}z \]
旋度
\[ \mathbf{rot}\;\boldsymbol{A}=\nabla\times\boldsymbol{A}=\begin{vmatrix} \boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k}\\ \displaystyle\frac{\partial}{\partial x}&\displaystyle\frac{\partial}{\partial y}&\displaystyle\frac{\partial}{\partial z}\\ P&Q&R \end{vmatrix} \]