## 基础知识

### 第一类曲线积分

\begin{align} M &= \lim_{\lambda \to \infty} \sum_{i=1}^{+\infty} \mu (x_i,y_i) \Delta S_i\\ & = \int_{L}^{} \mathrm{d} m = \int_{L}^{} f(x,y,z) \mathrm{d}s \end{align}

#### 计算原理

$$L:\left\{\begin{array}{l} x=x(t) \\ y=y(t) \end{array}\right.$$

$$I = \int_{\alpha}^{\beta} f\left(x(t), y(t)\right) \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(x)\right)^{2}} \mathrm{d} t$$

### 向量场的积分

#### 原理

$$\left[\mathbf{F}\left(x_{i}^{*}, y_{i}^{*}, z_{i}^{*}\right) \cdot \mathbf{T}\left(x_{i}^{*}, y_{i}^{*}, z_{i}^{*}\right)\right] \Delta s_{i}$$

$$W=\int_{C} \mathbf{F}(x, y, z) \cdot \mathbf{T}(x, y, z) \mathrm{d} s=\int_{C} \mathbf{F} \cdot \mathbf{T} \mathrm{d} s$$

$$\int_{C} \mathbf{F} \cdot \mathrm{d} \mathbf{r}=\int_{a}^{b} \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}^{\prime}(t) \mathrm{d} t=\int_{C} \mathbf{F} \cdot \mathbf{T} \mathrm{d} s$$

#### 向量场的分解积分法

\begin{aligned} \int_{C} \mathbf{F} \cdot \mathrm{d} \mathbf{r} &=\int_{a}^{b} \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}^{\prime}(t) \mathrm{d} t \\ &=\int_{a}^{b}(P \mathbf{i}+Q \mathbf{j}+R \mathbf{k}) \cdot\left(x^{\prime}(t) \mathbf{i}+y^{\prime}(t) \mathbf{j}+z^{\prime}(t) \mathbf{k}\right) \mathrm{d} t \\ &=\int_{a}^{b}\left[P(x(t), y(t), z(t)) x^{\prime}(t)+Q(x(t), y(t), z(t)) y^{\prime}(t)+R(x(t), y(t), z(t)) z^{\prime}(t)\right] \mathrm{d} t \end{aligned}

$$\int_{C} \mathbf{F} \cdot \mathrm{d} \mathbf{r}=\int_{C} P \mathrm{d} x+Q \mathrm{d} y+R \mathrm{d} z \quad \text { where } \mathbf{F}=P \mathbf{i}+Q \mathbf{j}+R \mathbf{k}$$

#### 梯度定理

$$\int_{a}^{b} F^{\prime}(x) \mathrm{d} x=F(b)-F(a)$$

$$\int_{C} \nabla f \cdot \mathrm{d} \mathbf{r}=f(\mathbf{r}(b))-f(\mathbf{r}(a))$$

【例子】 现有重力场 $\mathbf {F}(\mathbf {x})=-\dfrac {GMm}{|\mathbf {r}|^{3}} \mathbf {r}$ ，质点质量为 $m$，从 $(3,4,12)$ 沿着某个光滑曲线移动到 $(2,2,0)$ ，求做的功。

\begin{aligned} W &=\int_{C} \mathbf{F} \cdot \mathrm{d} \mathbf{r}=\int_{C} \nabla f \cdot \mathrm{d} \mathbf{r} \\ &=f(2,2,0)-f(3,4,12) \\ &=\frac{GMm}{\sqrt{2^{2}+2^{2}}}-\frac{GMm}{\sqrt{3^{2}+4^{2}+12^{2}}}=GMm\left(\frac{1}{2 \sqrt{2}}-\frac{1}{13}\right) \end{aligned}

$$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$$

### 格林定理

$$\int_{C} P \mathrm{d} x+Q \mathrm{d} y=\iint_{D}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) \mathrm{d} A$$

$$\oint_{C} \mathbf{F} \cdot d \mathbf{r}=\iint_{D}(\operatorname{curl} \mathbf{F}) \cdot \mathbf{k} d A$$

【例子】 曲线 $C$ 和域 $D$ 如图。求 $\int_{C}^{} x^4 \mathrm {d} x + xy \mathrm {d} y$

$$\frac{\partial Q}{\partial x} = y\\ \frac{\partial P}{\partial y} = 0$$
\begin{aligned} \int_{C} x^{4} \mathrm{d} x+x y \mathrm{d} y &=\iint_{D}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) \mathrm{d} A=\int_{0}^{1} \int_{0}^{1-x}(y-0) \mathrm{d} y \mathrm{d} x \\ &=\int_{0}^{1}\left[\frac{1}{2} y^{2}\right]_{y=0}^{y=1-x} \mathrm{d} x=\frac{1}{2} \int_{0}^{1}(1-x)^{2} \mathrm{d} x \\ &\left.=-\frac{1}{6}(1-x)^{3}\right]_{0}^{1}=\frac{1}{6} \end{aligned}

### 旋度和散度

#### 旋度的定义

$$\operatorname{curl} \mathbf{F}=\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right) \mathbf{i}+\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right) \mathbf{j}+\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) \mathbf{k}$$

\begin{aligned} \nabla \times \mathbf{F} &=\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{array}\right| \\ &=\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right) \mathbf{i}+\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right) \mathbf{j}+\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) \mathbf{k} \\ &=\operatorname{curl} \mathbf{F} \end{aligned}

$$\operatorname{curl} \mathbf{F}=\nabla \times \mathbf{F}$$

#### 旋度定理

$$\iint_{S}\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)dydz+\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)dzdx+\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy=\oint_{\Gamma}Pdx+Qdy+Rdz$$

$$\int_{S} \nabla \times \mathbf{F} \cdot d\mathbf{S} = \oint_{\partial S} \mathbf{F} \cdot d \mathbf{r}$$

#### 散度定理

$$\iiint_{\Omega} \operatorname{div} \mathbf{A} d v=\oiint_{\Sigma} \mathbf{A} \cdot \mathbf{n} d S$$

## 题型

### 直接给出参数方程的

【例子】

$$\left\{\begin{array}{l} x=a \cos t \\ y=a \sin t \end{array}\right.$$

【分析与解答】

$$\begin{array}{l} \oint_L\left(x^{2}+y^{2}\right)^{n} \mathrm{d} s \\ =\int_{0}^{2 \pi}\left[(a \cos t)^{2}+(a \sin t)\right]^{n} \cdot \sqrt{a^{2} \sin ^{2} t+a^{2} \cos ^{2} t} \mathrm{d} t \\ =\int_{0}^{2 \pi} a^{2 n+1} \mathrm{d} t \\ =2 \pi a^{2 n+1} \end{array}$$

【例子】 Find the work done by the force field $\mathbf {F}(x, y)=x^{2} \mathbf {i}-x y \mathbf {j}$ in moving a particle along the quarter-circle $\mathbf {r}(t)=\cos t \mathbf {i}+\sin t \mathbf {j}, 0 \leqslant t \leqslant \pi / 2$

\begin{align} \int_{C}^{} \vec{F} \cdot \mathrm{d} \vec{r} &= \int_{0}^{\frac{\pi }{2}} \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) \mathrm{d}t\\ &= \int_{0}^{\frac{\pi }{2}} (\cos ^2t \vec{i} - \sin t \cos t \vec{j})(- \sin t \vec{i} + \cos t \vec{j}) \mathrm{d}t\\ &= \int_{0}^{\frac{\pi }{2}} (\cos ^2t, -\sin t \cos t) \cdot (-\sin t , \cos t) \mathrm{d} t \\ &= \int_{0}^{\frac{\pi }{2}} - 2\sin t \cos ^2 t \mathrm{d}t\\ &= 2 \left.\frac{\cos ^3 t}{3}\right|_0^\frac{\pi }{2}\\&= - \frac{2}{3} \end{align}

### 普通题

【例子】 计算 $\oint_{L}^{} \sqrt []{y} \mathrm {d} s$ ，其中 $L$ 是 $y = x^2$ 上的 $(0,0)$ 余 $(1,1)$ 之间的曲线。

【分析与解答】

$ds = \sqrt[]{1+(y'(X)^2)} \mathrm{d}x= \sqrt[]{1+4x^2} \mathrm{d}x$

\begin{align} \oint_{L}^{} \sqrt[]{y} \mathrm{d}s &= \int_{0}^{1} |x| \sqrt[]{1+4x^2} \mathrm{d}x\\ & \text {令} u = 1+4x^2 \text {则} du = 8x \mathrm {d} x\\ &= \int_{0}^{5} u ^{\frac{1}{2}}\frac{\mathrm{d}u}{8}\\ &= \frac{1}{8} \left.(\frac{u^{3/2}}{3/2} )\right|_0^5\\ &= \frac{1}{8} \frac{2}{3} 5^{\frac{3}{2}} - 1\\ &= \frac{5\sqrt{5}-1}{12} \end{align}