第12章曲线积分和曲面积分12.1 场的概念与例子12.2 曲线积分12.2.1 第一类曲线积分：标量场中的曲线积分12.2.2 第二类曲线积分：二维向量场中的曲线积分12.2.2.1 封闭曲线上的第二类曲线积分12.2.2.2 两类曲线积分之间的关系12.2.2.3 特殊情况：梯度场12.2.2.4 格林公式12.3 曲面积分12.3.1 通量与曲面积分12.3.2 高斯公式12.3.3 Stokes 公式12.4 总结与联系

第12章曲线积分和曲面积分

Tip

本章我们将综合运用之前所学的多元函数微分和积分工具来解决来源于物理中的核心问题, 如做功和通量等物理量的数学定义和计算, 并揭示曲线和曲面积分之间的重要而深刻的联系.

12.1 场的概念与例子

Important

场 $\mathbb{R}^2$ $\mathbb{R}^3$ ) 中的每个点都赋予一个量 (可以是标量, 向量等), 那么我们就得到了一个场. 根据赋予的量分类, 我们有

标量场(scalar field): 空间每个点对应一个标量.
向量场(vector field): 空间每个点对应一个向量.

Note

判断下列场是标量场还是向量场

引力场：向量场
密度场：标量场
温度场：标量场
电磁场：向量场
流场：向量场

Important

多元函数 $f(\mathbf{x}), \mathbf{x} \in \mathbb{R}^n$ 表示. 如

$f(x, y)$ ;
$f(x, y, z)$ .

多元向量值函数 $\mathbf{F}(\mathbf{x}), \mathbf{x} \in \mathbb{R}^n$ $\mathbf{F}$ 写成如下的分量形式.

$\mathbf{F}(x, y) = P(x, y)\hat{\mathbf{i}} + Q(x, y)\hat{\mathbf{j}}$ ;
$\mathbf{F}(x, y, z) = P(x, y, z)\hat{\mathbf{i}} + Q(x, y, z)\hat{\mathbf{j}} + R(x, y, z)\hat{\mathbf{k}}$ $P, Q, R$ $x, y, z$ -轴上的分量.

Note

几个二维向量场的例子

例1

$\mathbf{F}(x, y) = 2\hat{\mathbf{i}} + 3\hat{\mathbf{j}}$

例2

$\mathbf{F}(x, y) = x\hat{\mathbf{i}}$ [补充图]

例3

$\mathbf{F}(x, y) = x\hat{\mathbf{i}} + y\hat{\mathbf{j}}$

例4

$\mathbf{F}(x, y) = y\hat{\mathbf{i}} + x\hat{\mathbf{j}}$ [补充图]

Warning

三维向量场 $\mathbb{R}^3$ 空间中每一点都有一个三维向量, 图像从略.

12.2 曲线积分

12.2.1 第一类曲线积分：标量场中的曲线积分

Tip
图

\int_{L} f (x, y) d x

$L$ $f(x,y)$ $L$ 上的标量场，则积分可表示为极限形式：

\int_{L} f (x, y) d x = lim_{N \to \infty} \sum_{i = 1}^{N} f (x_{i}, y_{i}) d s_{i}

$L$ $x = x(t)$ $y = y(t)$ $x(t),y(t)$ $[t_0,t_1]$ $x'^2(t)+y'^2(t)\neq 0$ ,则弧长

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

$t$ 的定积分：

\int_{L} f (x, y) d s = \int_{t_{0}}^{t_{1}} f (x (t), y (t)) \cdot \sqrt{[x^{'} (t)]^{2} + [y^{'} (t)]^{2}} d t

例1 计算曲线积分

\int_{L} \sqrt{y} d s

$L$ $y = x^2$ $O(0,0)$ $B(1,1)$ 之间的一段弧（如图所示）解 $L$
$y = x^2 \quad (0 \leq x \leq 1)$
给出，因此

\begin{aligned} \int_{L} \sqrt{y} d s & = \int_{0}^{1} \sqrt{x^{2}} \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x \\ = \int_{0}^{1} x \sqrt{1 + 4 x^{2}} d x \\ = {[\frac{1}{12} (1 + 4 x^{2})^{3 / 2}]}_{0}^{1} \\ = \frac{1}{12} (5 \sqrt{5} - 1) \end{aligned}

例2 $\int_{\Gamma} (x^2 + y^2 + z^2) \, ds$ $\Gamma$ $x = a \cos t、y = a \sin t、z = kt$ $t$ $0$ $2\pi$ 的一段弧。解

\begin{aligned} \int_{Γ} (x^{2} + y^{2} + z^{2}) d s & = \int_{0}^{2 π} [(a \cos t)^{2} + (a \sin t)^{2} + (k t)^{2}] \sqrt{(- a^{2} \sin t)^{2} + (a \cos t)^{2} + k^{2}} d t \\ = \int_{0}^{2 π} (a^{2} + k^{2} t^{2}) \sqrt{a^{2} + k^{2}} d t \\ = \sqrt{a^{2} + k^{2}} {[a^{2} t + \frac{k^{2}}{3} t^{3}]}_{0}^{2 π} \\ = \frac{2}{3} π \sqrt{a^{2} + k^{2}} (3 a^{2} + 4 π^{2} k^{2}) \end{aligned}

12.2.2 第二类曲线积分：二维向量场中的曲线积分

Tip
图设向量场为

\vec{F} (x, y) = P (x, y) \hat{i} + Q (x, y) \hat{j},

$L$ $W$ 定义为：

W = \int_{L} \vec{F} \cdot d \vec{r} = lim_{N \to \infty} \sum_{i = 1}^{N} {\vec{F}}_{i} \cdot Δ {\vec{r}}_{i},

将点积展开并分离坐标分量：

\begin{aligned} W & = \int_{L} \vec{F} \cdot d \vec{r} = \sum_{i = 1} {\vec{F}}_{i} \cdot Δ {\vec{r}}_{i} \end{aligned}

$\mathrm{d}\vec{r}=(\mathrm{d}x,\mathrm{d}y)^T,\Delta\vec{r}=(\Delta x,\Delta y)^T$ ,从而

\begin{aligned} W & = \sum_{i} (P Δ x_{i} + Q Δ y_{i}) \\ = \int_{L} P (x, y) d x + Q (x, y) d y . \end{aligned}

$L$ $x = x(t)$ $y = y(t)$ $x(t),y(t)$ $[t_1,t_2]$ $x'^2(t)+y'^2(t)\neq 0$ ，则

Δ {\vec{r}}_{i} = [x^{'} (t) \hat{i} + y^{'} (t) \hat{j}] d t .

且

\begin{aligned} W & = \int_{t_{1}}^{t_{2}} [P (x, y) x^{'} (t) + Q (x, y) y^{'} (t)] d t \\ = \int_{t_{1}}^{t_{2}} [P (x (t), y (t)) x^{'} (t) + Q (x (t), y (t)) y^{'} (t)] d t . \end{aligned}

例1 $W = \int_{L} \vec{F} \cdot \mathrm{d}\vec{r}$ $F=-y\hat{i}+x\hat{j}$ $x=t,y=t^2$

解

\begin{aligned} W & = \int_{L} \vec{F} \cdot d \vec{r} \\ = \int_{L} - y d x + x d y \\ = \int_{0}^{1} - y d t + x \cdot 2 t d t \\ = \int_{0}^{1} (- t^{2} + 2 t^{2}) d t \\ = \frac{1}{3} \end{aligned}

Note: $L$ 若改为

{\begin{cases} x = s i n θ \\ y = s i n^{2} θ \end{cases}, 0 \leq θ \leq \frac{π}{2}

积分结果不变 $L$ 的方向

例2 $\int_{L} xy\mathrm{d}x$ $L$ $y^2 = x$ $A(1, -1)$ $B(1, 1)$ 的一段弧图 解法一
$x$ 的定积分来计算：

\begin{aligned} \int_{L} x y d x & = \int_{A O} x y d x + \int_{O B} x y d x \\ = \int_{1}^{0} x (- \sqrt{x}) d x + \int_{0}^{1} x \sqrt{x} d x \\ = 2 \int_{0}^{1} x^{\frac{3}{2}} d x \\ = \frac{4}{5} \end{aligned}

解法二
$y$ 的定积分来计算

\begin{aligned} \int_{L} x y d x & = \int_{- 1}^{1} y^{2} y (y^{2})^{'} d y \\ = 2 \int_{- 1}^{1} y^{4} d y \\ = 2 {[\frac{y^{5}}{5}]}_{- 1}^{1} \\ = \frac{4}{5} \end{aligned}

例3 $M(x,y)$ $F$ $F$ $M$ $O$ $F$ $A(a,0)$ $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $B(0,b)$ $F$ $W$ 图由题意，力的表达式为：

\vec{F} = - k (x i + y j),

$k > 0$ 为比例常数。功的表达式为：

W = \int_{A B} \vec{F} \cdot d r = - k \int_{A B} (x d x + y d y) .

利用椭圆的参数方程：

{\begin{cases} x = a \cos θ, \\ y = b \sin θ, \end{cases} θ \in [0, \frac{π}{2}],

计算微分：

d x = - a \sin θ d θ, d y = b \cos θ d θ .

代入积分式：

\begin{aligned} W & = - k \int_{0}^{\frac{π}{2}} [a \cos θ (- a \sin θ) + b \sin θ (b \cos θ)] d θ \\ = k (a^{2} - b^{2}) \int_{0}^{\frac{π}{2}} \sin θ \cos θ d θ \\ = \frac{1}{2} k (a^{2} - b^{2}) \int_{0}^{\frac{π}{2}} \sin 2 θ d θ \\ = \frac{1}{2} k (a^{2} - b^{2}) {[- \frac{\cos 2 θ}{2}]}_{0}^{\frac{π}{2}} \\ = \frac{k}{2} (a^{2} - b^{2}) . \end{aligned}

12.2.2.1 封闭曲线上的第二类曲线积分

Tip
图

\vec{F} (x, y) = y \hat{i} + x \hat{j}

\oint_{C} \vec{F} \cdot d \vec{r} = \int_{C_{1}} \vec{F} \cdot d \vec{r} + \int_{C_{2}} \vec{F} \cdot d \vec{r} + \int_{C_{3}} \vec{F} \cdot d \vec{r}

$\int_{C_1}\vec{F} \cdot \mathrm{d}\vec{r}=0$ $C_2$ ,其参数方程为

{\begin{cases} x = a c o s θ \\ y = a s i n θ \end{cases}, 0 \leq θ \leq \frac{π}{4}

从而

{\begin{cases} d x = - a s i n θ d θ \\ d y = a c o s θ d θ \end{cases}

则

\begin{aligned} \int_{C_{2}} \vec{F} \cdot d \vec{r} & = \int_{0}^{\frac{π}{4}} (- y a s i n θ + x a c o s θ) d θ \\ = \int_{0}^{\frac{π}{4}} (- a^{2} s i n^{2} θ + a^{2} c o s^{2} θ) d θ \\ = a^{2} \int_{0}^{\frac{π}{4}} \frac{1 + c o s 2 θ - 1 + c o s 2 θ}{2} d θ \\ = \frac{a^{2}}{2} \int_{0}^{\frac{π}{4}} c o s 2 θ d θ \\ = \frac{a^{2}}{2} {[s i n 2 θ]}_{0}^{\frac{π}{4}} \\ = \frac{a^{2}}{2} \end{aligned}

$C_3$ ,其参数方程为

{\begin{cases} x = \frac{\sqrt{2}}{2} a - \frac{\sqrt{2}}{2} t \\ y = \frac{\sqrt{2}}{2} a - \frac{\sqrt{2}}{2} t \end{cases}, 0 \leq t \leq a

从而

{\begin{cases} d x = - \frac{\sqrt{2}}{2} d t \\ d y = - \frac{\sqrt{2}}{2} d t \end{cases}

则

\begin{aligned} \int_{C_{3}} y d x + x d y & = - \frac{\sqrt{2}}{2} \int_{C_{3}} (\frac{\sqrt{2}}{2} a - \frac{\sqrt{2}}{2} t) d t + (\frac{\sqrt{2}}{2} a - \frac{\sqrt{2}}{2} t) d t \\ = \int_{0}^{a} (t - a) d t \\ = {[\frac{1}{2} t^{2} - a t]}_{0}^{a} \\ = - \frac{a^{2}}{2} \end{aligned}

$I=0+\frac{a^2}{2}-\frac{a^2}{2}=0$

12.2.2.2 两类曲线积分之间的关系

Tip
图

W = \int_{L} \vec{F} \cdot d \vec{r} = \int_{L} \vec{F} \cdot \vec{τ} d s

例1 (1)保守场 $\vec{F}=x\hat{i}+y\hat{j}$ $W=\int_L\vec{F}\cdot \mathrm{d}\vec{r}$ $L$ 如下图所示：图解：

W = \int_{L} \vec{F} \cdot d \vec{r} = 0

(2)非保守场 $\vec{F}=-y\hat{i}+x\hat{j}$ $W=\int_L\vec{F} \cdot \vec{\tau}\mathrm{d}s$ $L$ 如下图所示：图 解法一：

\begin{aligned} W & = \int_{L} \vec{F} \cdot \vec{τ} d s \\ = \int_{L} | F | d s \\ = R \cdot 2 π R \\ = 2 π R^{2} \end{aligned}

解法二：曲线的参数方程为

{\begin{cases} x = R c o s θ \\ y = R s i n θ \end{cases}

则

{\begin{cases} d x = - R s i n θ d θ \\ d y = R c o s θ d θ \end{cases}

\begin{aligned} W & = \int_{0}^{2 π} (- y \cdot - R s i n θ + x \cdot R c o s θ) d θ \\ = \int_{0}^{2 π} (R^{2} s i n^{2} θ + R^{2} c o s^{2} θ) d θ \\ = \int_{0}^{2 π} R^{2} d θ \\ = 2 π R^{2} \end{aligned}

例2 $\int_L y^2 \, dx$ $L$
$a$ $A(a,0)$ $x$ $B(-a,0)$ 的直线段。解 (1)参数方程为：

{\begin{cases} x = a \cos θ, \\ y = a \sin θ \end{cases}, 0 \leq θ \leq π

\begin{aligned} \int_{L} y^{2} d x & = \int_{L} y^{2} d x + 0 d y \\ = \int_{0}^{π} - a^{2} \sin^{2} θ \cdot a \sin θ d θ \\ = a^{3} \int_{0}^{π} (1 - c o s^{2} θ) d c o s θ \\ = a^{3} [{[c o s θ]}_{0}^{π} - \frac{1}{3} {[c o s^{3} θ]}_{0}^{π}] \\ = - 2 a^{3} + \frac{2}{3} a^{3} \\ = - \frac{4}{3} a^{3} . \end{aligned}

(2)直线段方程为：

{\begin{cases} x = x \\ y = 0 \end{cases}, - a \leq x \leq a

\int_{L} y^{2} d x = \int_{L} 0 d x = 0.

Note:路径相关

例3 $\int_{L} 2xy \, dx + x^2 \, dy$ $L$
$y = x^2$ $O(0,0)$ $B(1,1)$
$x = y^2$ $O(0,0)$ $B(1,1)$
$OAB$ $O(0,0)$ $O,A,B$ $(0,0),(1,0),(1,1)$ 解 (1)这段弧可以表示为：

{\begin{cases} x = x \\ y = \sqrt{x} \end{cases}, 0 \leq x \leq 1

从而

{\begin{cases} d x = d x \\ d y = \frac{1}{2} \frac{1}{\sqrt{x}} d x \end{cases}

\begin{aligned} \int_{L} 2 x y d x + x^{2} d y & = \int_{0}^{1} (2 x \cdot \sqrt{x} + x^{2} \cdot \frac{1}{2} \frac{1}{\sqrt{x}}) d x \\ = \int_{0}^{1} (2 x^{\frac{3}{2}} + \frac{1}{2} x^{\frac{3}{2}}) d x \\ = {[x^{\frac{5}{2}}]}_{0}^{1} \\ = 1. \end{aligned}

(2)

\int_{L} 2 x y d x + x^{2} d y = 1

$L_1:OA$ $y = 0，x \in [0, 1]，\mathrm{d}y = 0$ ：

\int_{L_{1}} 2 x y d x + x^{2} d y = \int_{0}^{1} 0 d x = 0.

$L_2:AB$
$x = 1，y \in [0, 1]，\mathrm{d}x = 0$ ：

\int_{L_{2}} 2 x y d x + x^{2} d y = \int_{0}^{1} 1^{2} d y = {[y]}_{0}^{1} = 1.

最终结果：

\int_{O A B} 2 x y d x + x^{2} d y = \int_{L_{1}} 2 x y d x + x^{2} d y + \int_{L_{2}} 2 x y d x + x^{2} d y = 1.

Note:路径无关

12.2.2.3 特殊情况：梯度场

$\exist f(x,y),s.t.$

\frac{\partial f}{\partial x} = P (x, y) ， \frac{\partial f}{\partial y} = Q (x, y)

即

d f = P d x + Q d y

则

\begin{aligned} \int_{L} P (x, y) d x + Q (x, y) d y & = \int_{L} d f \\ = f (x (t_{2}), y (t_{2})) - f (x (t_{1}), y (t_{1})) \\ = f (终) - f (起) \end{aligned}

在物理上，

W = f (终) - f (起)

$\vec{F}=(P,Q)=\nabla f=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y})$ $\vec{F}$ $f$ 的梯度场 $\vec{F}$ 都是某函数的梯度场

Fundamental Thm of Line Integral:
$\int_{L} P (x, y) d x + Q (x, y) d y = f (x (t_{2}), y (t_{2})) - f (x (t_{1}), y (t_{1}))$
类似于牛顿-莱布尼茨公式

Tip
$\vec{F}$ $f$ 的梯度场,会有什么结论？以单连通区域为前提

积分与路径无关
$\vec{F}$ $\oint_L\vec{F} \cdot \mathrm{d}\vec{r}=0$ $L$ $\int_{C_1}\vec{F} \cdot \mathrm{d}\vec{r}+\int_{C_2}\vec{F} \cdot \mathrm{d}\vec{r}=\int_{C_1}\vec{F} \cdot \mathrm{d}\vec{r}-\int_{-C_2}\vec{F} \cdot \mathrm{d}\vec{r}=0$

Tip
$\vec{F} = (P(x,y), Q(x,y))$ 是否为梯度场 $f(x,y)$ $\vec{F}$ 可以表示为该函数的梯度，即：

\nabla f = \frac{\partial f}{\partial x} d x + \frac{\partial f}{\partial y} d y,

$\vec{F}$ $\vec{F}$ 的分量满足：

P = \frac{\partial f}{\partial x}, Q = \frac{\partial f}{\partial y} .

$f(x,y)$ 的二阶混合偏导数必须满足：

\frac{\partial^{2} f}{\partial x \partial y} = \frac{\partial^{2} f}{\partial y \partial x} .

$\vec{F}$ 的分量代入，可得：

\frac{\partial}{\partial y} (\frac{\partial f}{\partial x}) = \frac{\partial}{\partial x} (\frac{\partial f}{\partial y}) ⟹ \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} .

因此， $\vec{F}$ 是梯度场的条件为：

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

例1 $<y^2,0>$ 是否为梯度场解

P_{y} = 2 y \neq Q_{x} = 0

$<y^2,0>$ 不是梯度场

例2 $<2xy,x^2>$ 是否为梯度场解

P_{y} = 2 x = Q_{x} = 2 x

$<2xy,x^2>$ 是梯度场

例3 $<y,x>$ 是否为梯度场解

P_{y} = 1 = Q_{x} = 1

$<y,x>$ 是梯度场

[!TIP]有问题 $f$ $\vec{F}=(y,x),f(0,0)=c$

\begin{aligned} f (x, y) & = \int_{C_{1}} P d x + Q d y + \int_{C_{2}} P d x + Q d y + f (0, 0) \\ = \int_{0}^{x} 0 d x + \int_{0}^{y} x d y + c \\ = x y + c \end{aligned}

若

\vec{F} (x, y) = [(x^{2} + a x y + 3), (3 y^{2} - 2 x^{2})]

P_{y} = a x = Q_{x} = - 4 x

$a=-4$

\vec{F} (x, y) = [(x^{2} - 4 x y + 3), (3 y^{2} - 2 x^{2})]

Tip
$curl$ 定义：

c u r l \vec{F} = Q_{x} - P_{y} = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}

$curl\vec{F}=0\Leftrightarrow 保守场（梯度场）$ $\vec{F}=a\hat{i}+b\hat{j},curl\vec{F}=0$ 图

$\vec{F}=x\hat{i}+y\hat{j}$ 此时

\frac{\partial x}{\partial y} = 0, \frac{\partial y}{\partial x} = 0

因此，

c u r l \vec{F} = \frac{\partial y}{\partial x} - \frac{\partial x}{\partial y} = 0

图

$\vec{F}=-y\hat{i}+x\hat{j}$ 此时

c u r l \vec{F} = \frac{\partial x}{\partial x} - \frac{\partial (- y)}{\partial y} = 1 + 1 = 2

Tip
$curl{\vec{F}}$ $curl{\vec{F}}$ $\vec{F}$ $\vec{F}$ $curl{\vec{F}}$ $curl{\vec{F}}$ 的分布与性质，更高效地求解复杂曲线积分问题，揭示向量场在空间中的动态特征。图

Tip
格林公式(Green's theorem)

\oint_{C} \vec{F} d \vec{r} = \iint_{R} c u r l \vec{F} d A

\oint_{C} P d x + Q d y = \iint_{R} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d A

其中

$C:$ 闭曲线，逆时针
$\vec{F}:$ 连续可微向量场 $\vec{F}$ 连续

12.2.2.4 格林公式

Tip
定理 $D$ $L$ $P(x,y)$ $Q(x,y)$ $D$ $\oint_C P\mathrm{d}x+ Q\mathrm{d}y=\iint_R(Q_x-P_y)\mathrm{d}x\mathrm{d}y$ $L$ $D$ 的取正向的边界曲线. 上述公式称为格林公式. 证明 $Q=0$

\int_{C} P d x = \iint_{R} - P_{y} d A

$P=0$ ，则有

\int_{C} Q d y = \iint_{R} - Q_{x} d A

(2)简化区域图如图可得：

\begin{matrix} \int_{C_{1}} P d x = \iint_{R_{1}} - P_{y} d A \\ \int_{C_{2}} P d x = \iint_{R_{2}} - P_{y} d A \end{matrix}

由于

\begin{matrix} \int_{C} P d x = \int_{C_{1}} P d x + \int_{C_{2}} P d x \\ \iint_{R} - P_{y} d A = \iint_{R_{1}} - P_{y} d A + \iint_{R_{2}} - P_{y} d A \end{matrix}

因此，

\int_{C} P d x = \iint_{R} - P_{y} d A

(3)主要部分要证：

\oint_{C} P d x = \iint_{R} - P_{y} d A

$C:simple$ $R:vert\ simple$ 图由图可得：

\oint_{C} P d x = \int_{C_{1}} P d x + \int_{C_{2}} P d x + \int_{C_{3}} P d x + \int_{C_{4}} P d x

其中，

\begin{matrix} \int_{C_{1}} P d x = 0, \int_{C_{3}} P d x = 0 \\ \int_{C_{2}} P d x = \int_{a}^{b} P (x, f_{1} (x)) d x \\ \int_{C_{4}} P d x = \int_{b}^{a} P (x, f_{2} (x)) d x = - \int_{a}^{b} P (x, f_{2} (x)) d x \end{matrix}

因此，

\oint_{C} P d x = - \int_{a}^{b} [P (x, f_{2} (x)) - P (x, f_{1} (x))] d x

又因为，

\begin{aligned} \iint_{R} - P_{y} d x d y & = - \int_{a}^{b} [\int_{f_{1} (x)}^{f_{2} (x)} \frac{\partial P}{\partial y} d y] d x \\ = - \int_{a}^{b} [P (x, f_{2} (x)) - P (x, f_{1} (x))] d x \end{aligned}

得证.

例1

\vec{F} = - y \hat{i} + x \hat{j}

解

c u r l \vec{F} = 1 + 1 = 2

因此，

\iint_{D} c u r l \vec{F} d A = \iint_{D} 2 d A = 2 A r e a = \oint_{C} - y d x + x d y

例2 计算

\oint_{L} x^{2} y d x - x y^{2} d y

$L$ $x^2 + y^2 = a^2$ 解 $P = x^2 y$ $Q = -xy^2$ ，则

\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} == - y^{2} - x^{2}

由格林公式得：

\begin{aligned} \oint_{L} x^{2} y d x - x y^{2} d y & = - \iint_{D} (x^{2} + y^{2}) d x d y \\ = - \int_{0}^{2 π} d θ \int_{0}^{a} ρ^{3} d ρ \\ = - \frac{π}{2} a^{4} \end{aligned}

例3 计算

\oint_{L} \frac{x d y - y d x}{x^{2} + y^{2}}

$L$ $L$ 解 $P = \frac{-y}{x^2 + y^2},Q = \frac{x}{x^2 + y^2}$ $x^2 + y^2 \neq 0$ 时，有

\frac{\partial Q}{\partial x} = \frac{y^{2} - x^{2}}{(x^{2} + y^{2})^{2}}, \frac{\partial P}{\partial y} = \frac{y^{2} - x^{2}}{(x^{2} + y^{2})^{2}}

$L$ $D$ $(0,0)\notin D$ 由格林公式，

\oint_{L} \frac{x d y - y d x}{x^{2} + y^{2}} = 0

$(0,0)\in D$ $r > 0$ $l: x^2 + y^2 = r^2$ $D$ $L$ $l$ $D_1$ $D_1$ 应用格林公式，方向均为逆时针：

\oint_{L} \frac{x d y - y d x}{x^{2} + y^{2}} - \oint_{l} \frac{x d y - y d x}{x^{2} + y^{2}} = 0

于是

\begin{aligned} \oint_{L} \frac{x d y - y d x}{x^{2} + y^{2}} & = \oint_{l} \frac{x d y - y d x}{x^{2} + y^{2}} \\ = \int_{0}^{2 π} \frac{r^{2} (\cos^{2} θ + \sin^{2} θ)}{r^{2}} d θ \\ = 2 π \end{aligned}

例4

\oint_{C} x d y = \iint_{R} 1 d A = A r e a

12.3 曲面积分

12.3.1 通量与曲面积分

Tip
图

\begin{aligned} \int_{C} \vec{F} \cdot \vec{n} d s & = lim_{Δ s \to 0} \sum_{i} F (x_{i}, y_{i}) \cdot \vec{n_{i}} Δ s_{i} \\ = lim_{Δ s \to 0} \sum_{i} [- Q_{i} Δ x_{i} + P_{i} Δ y_{i}] \end{aligned}

F l u x = \int_{C} - Q (x, y) d x + P (x, y) d y

$\vec{F}=(P,Q)^T$ $\vec{F}^{\perp}=(-Q,P)$ $(P,Q)$ 垂直从而，

F l u x = \int_{C} {\vec{F}}^{⊥} \cdot d \vec{r}

格林公式的flux表示：对

F l u x = \oint_{C} - Q (x, y) d x + P (x, y) d y

应用格林公式，可得

\oint_{C} - Q (x, y) d x + P (x, y) d y = \iint_{D} (\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}) d x d y

由散度定义：

d i v \vec{F} = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}) \cdot (P, Q) = \nabla \cdot \vec{F}

最终得到通量形式：

F l u x = \iint_{D} \nabla \cdot \vec{F} d A

例1 图

\vec{F} = x \hat{i} + y \hat{j}

解由于

\begin{aligned} d i v \vec{F} & = \nabla \cdot \vec{F} \\ = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \\ = 1 + 1 \\ = 2 \end{aligned}

因此，

\begin{aligned} \int_{C} \vec{F} \cdot \vec{n} d s & = \iint_{D} d i v \vec{F} d A \\ = 2 \iint_{D} d A \\ = 2 A r e a \end{aligned}

例2 $div\vec{F}=0$ $\vec{F}$ $\nabla \cdot \vec{u}=0$ Navier-Stokes方程(不可压形式)：

\frac{\partial \vec{u}}{\partial t} + \vec{u} \cdot \nabla \vec{u} = - \frac{1}{ρ} \nabla p + ν \nabla^{2} \vec{u} + \vec{g}

$v=\frac{\mu}{\rho}$ $P$ $g$ ：外力（重力）

3D情形 $line \to surface$ 物理学家视角：图

F l u x = \iint_{S} \vec{F} \cdot \vec{n} d s = lim_{Δ s \to 0} \sum_{i} \vec{F} (x_{i}, y_{i}, z_{i}) \cdot \vec{n_{i}} Δ s_{i}

$\mathrm{d}s \neq \mathrm{d}A=\mathrm{d}x\mathrm{d}y$ $\vec{F}(x_i,y_i,z_i)\cdot \vec{n_i}$ $\Delta s_i$ 是底面积

数学家视角: 先看一个简单的例子：图

\begin{matrix} \vec{F} = x \hat{i} + y \hat{j} + z \hat{k} \\ \vec{n} = \frac{1}{a} (x, y, z)^{T} \end{matrix}

可得

\vec{F} \cdot \vec{n} = | F | = a

因此，

\iint_{S} \vec{F} \cdot \vec{n} d S = a \iint_{S} d S = 4 π a^{3}

（特殊情形）再来看一般情形：图

\vec{F} = P (x, y, z) \hat{i} + Q (x, y, z) \hat{j} + R (x, y, z) \hat{k}

\iint_{S} \vec{F} \cdot \vec{n} d S = \iint_{S} R (x, y, z) d x d y + Q (x, y, z) d y d z + P (x, y, z) d z d x

其中，

\begin{matrix} \iint_{S} R (x, y, z) d x d y = \iint_{D_{x} y} R (x, y, z (x, y)) d x d y \\ \iint_{S} Q (x, y, z) d y d z = \iint_{D_{x} z} Q (x, y (x, z), z) d x d z \\ \iint_{S} P (x, y, z) d z d x = \iint_{D_{y} z} P (x (y, z), y, z) d y d z \end{matrix}

\begin{aligned} \iint_{S} \vec{F} \cdot \vec{n} d S & = \iint_{S} (\vec{F_{1}} + \vec{F_{2}} + \vec{F_{3}}) \cdot \vec{n} d S \\ = \iint_{S} \vec{F_{1}} \cdot \vec{n} d S + \iint_{S} \vec{F_{2}} \cdot \vec{n} d S + \iint_{S} \vec{F_{3}} \cdot \vec{n} d S \end{aligned}

$\vec{F_1}=(0,0,R(x,y,z))^T$ $\vec{v_1}=(\Delta x,0,\frac{\partial z}{\partial x}\Delta x),\vec{v_2}=(0,\Delta y,\frac{\partial z}{\partial y}\Delta y)$ 则

\begin{aligned} \vec{n} \cdot d S & = \vec{v_{1}} \times \vec{v_{2}} \\ = | \begin{array}{c} \hat{i} & \hat{j} & \hat{k} \\ Δ x & 0 & \frac{\partial z}{\partial x} Δ x \\ 0 & Δ y & \frac{\partial z}{\partial y} Δ y \end{array} | \\ = (- \frac{\partial z}{\partial x}, - \frac{\partial z}{\partial y}, 1) Δ x Δ y \end{aligned}

从而

\begin{aligned} \iint_{S} (\vec{F_{1}}) \cdot \vec{n} d S & = \iint_{S} R (x, y, z) d x d y \\ = \iint_{D_{x y}} R (x, y, z) d x d y \end{aligned}

$\vec{F_2}=(0,Q(x,y,z),0)^T$ $S$ $(x,y),(y,z),(x,z)$ 参数化

12.3.2 高斯公式

Tip

\begin{aligned} Flux & = ∯_{Σ} P d y d z + Q d z d x + R d x d y & = ∭_{D} (\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}) d V \\ = ∯_{Σ} \vec{F} \cdot \vec{n} d S & = ∭_{D} d i v \vec{F} d V = ∭_{D} \nabla \cdot \vec{F} d V \end{aligned}

例1

\vec{F} = x \hat{i} + y \hat{j} + z \hat{k}

解

\begin{matrix} d i v \vec{F} = 1 + 1 + 1 = 3 \\ ∯_{Σ} \vec{F} \cdot \vec{n} d S = ∭_{D} 3 d V = 3 \cdot \frac{4}{3} π a^{3} = 4 π a^{3} \end{matrix}

例2 图利用高斯公式计算曲面积分

\iint_{Σ} (x - y) d x d y + (y - z) x d y d z

$\Sigma$ $x^2 + y^2 = 1$ $z = 0, z = 3$ $\Omega$ 的整个边界曲面的外侧解因为

P = (y - z) x, Q = 0, R = x - y,

\frac{\partial P}{\partial x} = y - z, \frac{\partial Q}{\partial y} = 0, \frac{\partial R}{\partial z} = 0,

利用高斯公式把所给曲面积分化为三重积分，再利用柱面坐标计算三重积分，得

\begin{aligned} ∯_{Σ} (x - y) d x d y + (y - z) x d y d z & = ∭_{Ω} (y - z) d x d y d z \\ = ∭_{Ω} (ρ s i n θ - z) ρ d ρ d θ d z \\ = \int_{0}^{2 π} d θ \int_{0}^{1} ρ d ρ \int_{0}^{3} (ρ s i n θ - z) d z \\ = - \frac{9 π}{2} \end{aligned}

例3 利用高斯公式计算曲面积分

\iint_{Σ} (x^{2} \cos α + y^{2} \cos β + z^{2} \cos γ) d S,

$\Sigma$ $x^2 + y^2 = z$ $z = 0, z = h \ (h > 0)$ $\cos \alpha, \cos \beta, \cos \gamma$ $\Sigma$ $(x,y,z)$ 解 $\Sigma$ $\Sigma_1$ $z = h(x^2 + y^2 \leq h^2)$ $\Sigma$ $\Omega$ ，利用高斯公式，得

\begin{aligned} ∯_{Σ + Σ_{1}} (x^{2} \cos α + y^{2} \cos β + z^{2} \cos γ) d S & = 2 ∭_{Ω} (x + y + z) d v \\ = 2 \iint_{D_{x y}} d x d y \int_{\sqrt{x^{2} + y^{2}}}^{h} (x + y + z) d z \end{aligned}

$D_{xy} = \{ (x, y) \mid x^2 + y^2 \leq h^2 \}$ 。注意到

\iint_{D_{x y}} d x d y \int_{0}^{\sqrt{h^{2} - x^{2} - y^{2}}} (x + y) d z = 0,

得

\iint_{Σ + Σ_{1}} (x^{2} \cos α + y^{2} \cos β + z^{2} \cos γ) d S = \iint_{D_{x y}} (h^{2} - x^{2} - y^{2}) d x d y = \frac{1}{2} π h^{4} .

而

\iint_{Σ_{1}} (x^{2} \cos α + y^{2} \cos β + z^{2} \cos γ) d S = \iint_{Σ_{1}} z^{2} d S = \iint_{D_{x y}} h^{2} d x d y = π h^{4} .

因此，

\iint_{Σ} (x^{2} \cos α + y^{2} \cos β + z^{2} \cos γ) d S = \frac{1}{2} π h^{4} - π h^{4} = - \frac{1}{2} π h^{4} .

例4 $u(x,y,z)$ $v(x,y,z)$ $\Omega$ 上具有一阶及二阶连续偏导数，证明

∭_{Ω} u Δ v d x d y d z = ∯_{Σ} u \frac{\partial v}{\partial n} d S - ∭_{Ω} (\frac{\partial u}{\partial x} \frac{\partial v}{\partial x} + \frac{\partial u}{\partial y} \frac{\partial v}{\partial y} + \frac{\partial u}{\partial z} \frac{\partial v}{\partial z}) d x d y d z,

$\Sigma$ $\Omega$ $\frac{\partial v}{\partial n}$ $v(x,y,z)$ $\Sigma$ $\Delta = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$ 称为拉普拉斯 (Laplace) 算子。这个公式叫做格林第二公式。证明因为方向导数

\frac{\partial v}{\partial n} = \frac{\partial v}{\partial x} \cos α + \frac{\partial v}{\partial y} \cos β + \frac{\partial v}{\partial z} \cos γ,

$\cos \alpha, \cos \beta$ $\cos \gamma$ $\Sigma$ $(x,y,z)$ 处的外法线向量的方向余弦。于是曲面积分

\begin{aligned} ∯_{Σ} u \frac{\partial v}{\partial n} d S & = ∯_{Σ} u (\frac{\partial v}{\partial x} \cos α + \frac{\partial v}{\partial y} \cos β + \frac{\partial v}{\partial z} \cos γ) d S \\ = ∯_{Σ} [(u \frac{\partial v}{\partial x}) \cos α + (u \frac{\partial v}{\partial y}) \cos β + (u \frac{\partial v}{\partial z}) \cos γ] d S \end{aligned}

利用高斯公式，即得

\begin{aligned} ∯_{Σ} u \frac{\partial v}{\partial n} d S & = ∭_{Ω} [\frac{\partial}{\partial x} (u \frac{\partial v}{\partial x}) + \frac{\partial}{\partial y} (u \frac{\partial v}{\partial y}) + \frac{\partial}{\partial z} (u \frac{\partial v}{\partial z})] d x d y d z \\ = ∭_{Ω} u Δ v d x d y d z + ∭_{Ω} (\frac{\partial u}{\partial x} \frac{\partial v}{\partial x} + \frac{\partial u}{\partial y} \frac{\partial v}{\partial y} + \frac{\partial u}{\partial z} \frac{\partial v}{\partial z}) d x d y d z . \end{aligned}

将上式右端第二个积分移至左端便得所要证明的等式.

12.3.3 Stokes 公式

Tip

三维空间中的曲线积分 图

\begin{matrix} \vec{F} = P (x, y, z) \hat{i} + Q (x, y, z) \hat{j} + R (x, y, z) \hat{k} \\ Δ \vec{r} = Δ x \hat{i} + Δ y \hat{j} + Δ z \hat{k} \end{matrix}

从而

\begin{matrix} \vec{F} \cdot Δ \vec{r} = P Δ x + Q Δ y + R Δ z \\ \int_{C} \vec{F} \cdot d \vec{r} = \int_{C} P d x + Q d y + R d z \end{matrix}

例1 计算

\int_{C} \vec{F} \cdot d \vec{r}

其中，

\begin{matrix} \vec{F} = (y z, x z, x y) \\ C : x = t^{3}, y = t^{2}, z = t, 0 \leq t \leq 1 \end{matrix}

解由题可知

d x = 3 t^{2} d t, d y = 2 t d t, d z = d t

从而

\begin{aligned} \int_{C} \vec{F} \cdot d \vec{r} & = \int_{C} y z d x + x z d y + x y d z \\ = \int_{0}^{1} (t^{3} \cdot 3 t^{3} + t^{4} \cdot 2 t + t^{5}) d t \\ = \int_{0}^{1} 6 t^{5} d t \\ = t^{6} |_{0}^{1} \\ = 1 \end{aligned}

例2(路径无关) 图计算

\int_{C} \vec{F} \cdot d \vec{r}

解

\int_{C} \vec{F} \cdot d \vec{r} = \int_{C_{1}} \vec{F} \cdot d \vec{r} + \int_{C_{2}} \vec{F} \cdot d \vec{r} + \int_{C_{3}} \vec{F} \cdot d \vec{r}

其中，

\int_{C_{1}} \vec{F} \cdot d \vec{r} = 0, \int_{C_{2}} \vec{F} \cdot d \vec{r} = 0, \int_{C_{3}} \vec{F} \cdot d \vec{r} = \int_{0}^{1} 1 d z = 1

曲线积分基本定理 $f(x,y,z)$ 满足

d f = P d x + Q d y + R d z = \frac{\partial f}{\partial x} d x + \frac{\partial f}{\partial y} d y + \frac{\partial f}{\partial z} d z

则

\int_{C} P d x + Q d y + R d z = \int_{C} d f = f (B) - f (A)

保守场的条件为：

\begin{aligned} f (x, y, z) \Rightarrow \frac{\partial^{2} f}{\partial x \partial y} & = \frac{\partial^{2} f}{\partial y \partial x} \Rightarrow \frac{\partial}{\partial x} Q & = \frac{\partial}{\partial y} P \Rightarrow Q_{x} & = P_{y} \\ \frac{\partial^{2} f}{\partial y \partial z} & = \frac{\partial^{2} f}{\partial z \partial y} \Rightarrow \frac{\partial}{\partial y} R & = \frac{\partial}{\partial z} Q \Rightarrow R_{y} & = Q_{z} \\ \frac{\partial^{2} f}{\partial x \partial z} & = \frac{\partial^{2} f}{\partial z \partial x} \Rightarrow \frac{\partial}{\partial x} R & = \frac{\partial}{\partial z} P \Rightarrow R_{x} & = P_{z} \end{aligned} (*)

$f(x,y,z)$ $\vec{F}=\nabla f$

$curl\vec{F}=(R_y-Q_z)\hat{i}+(P_z-R_x)\hat{j}+(Q_x-P_y)\hat{k}$

\begin{matrix} \nabla \times \vec{F} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{matrix} | = (R_{y} - Q_{z}) \hat{i} + (P_{z} - R_{x}) \hat{j} + (Q_{x} - P_{y}) \hat{k} \end{matrix}

以下条件等价

$curl\vec{F}=0$
梯度场
条件(*)
路径无关
环路为零

例1

P = y z, Q = x z, R = x y

解

\begin{matrix} Q_{x} = z, P_{y} = z \\ R_{y} = x, Q_{z} = x \\ R_{x} = y, P_{z} = y \end{matrix}

$curl\vec{F}=0$

$curl \vec{F}$ 的意义 图

\vec{F} = (- y, x, 0)

\begin{matrix} c u r l \vec{F} = \nabla \times \vec{F} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ - y & x & 0 \end{matrix} | = 0 \cdot \hat{i} + 0 \cdot \hat{j} + 2 \cdot \hat{k} \end{matrix}

12.4 总结与联系

$\nabla : (del,nabla)=(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z})^T$

	3D	2D
grad	$\nabla f=(f_x,f_y,f_z)^T$	$\nabla f=(f_x,f_y,)^T$
div	$\nabla \cdot \vec{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}$	$\nabla \cdot \vec{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}$
curl	$\nabla \times \vec{F}= \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\P & Q & R \end{matrix}$	$\nabla \times \vec{F}= \begin{matrix} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} \\ P & Q \end{matrix}$
$\iiint_Df(x,y,z)\mathrm{d}v$	$\iint_S\vec{F}\cdot \vec{n}\mathrm{d}S$	$\int_C\vec{F}\cdot \mathrm{d}\vec{r}$
体积	flux	work
多重积分	$\iint_SP\mathrm{d}x\mathrm{d}y+Q\mathrm{d}y\mathrm{d}z+R\mathrm{d}y\mathrm{d}x$	$\int_CP\mathrm{d}x+Q\mathrm{d}y+R\mathrm{d}z$

3D:Gauss公式

∭_{D} d i v \vec{F} d v = ∯_{S} \vec{F} c d o t \vec{n} d S

3D:Green公式

\iint_{D} c u r l \vec{F} d A = \oint_{C} \vec{F} \cdot d \vec{r}

3D:Stokes公式

w o r k = \oint_{C} P d x + Q d y + R d z = \oint_{C} \vec{F} \cdot d \vec{r} = \iint_{S} (c u r l \vec{F}) \vec{n} d S = \iint_{S} \nabla \times \vec{F} \cdot \vec{n} d S

例1 $\oint_{\Gamma} z\mathrm{d}x + x\mathrm{d}y + y\mathrm{d}z$ $\Gamma$ $x+y+z=1$ $\Sigma$ 上侧的法向量符合右手规则. 解根据斯托克斯公式，有

\oint_{Γ} z d x + x d y + y d z = \iint_{Σ} d y d z + d z d x + d x d y .

分别计算各投影积分：

\iint_{Σ} d y d z = \iint_{D_{y z}} d σ = \frac{1}{2},

\iint_{Σ} d z d x = \iint_{D_{z x}} d σ = \frac{1}{2}

\iint_{Σ} d x d y = \iint_{D_{x y}} d σ = \frac{1}{2},

$D_{yz}$ $D_{zx}$ $D_{xy}$ $\Sigma$ $yOz$ $zOx$ $xOy$ 坐标面上的投影区域。
因此，曲线积分的值为：

\oint_{Γ} z d x + x d y + y d z = \frac{3}{2} .

例2 利用斯托克斯公式计算曲线积分

I = \oint_{Γ} (y^{2} - z^{2}) d x + (z^{2} - x^{2}) d y + (x^{2} - y^{2}) d z,

$\Gamma$ $x + y + z = \frac{3}{2}$ $\{(x,y,z) \mid 0 \leq x, y, z \leq 1\}$ $Ox$ 轴正向看去的逆时针方向。解 $\Sigma$ $x + y + z = \frac{3}{2}$ $\Gamma$ 围成的部分，其单位法向量为

n = \frac{1}{\sqrt{3}} (1, 1, 1),

$\cos \alpha = \cos \beta = \cos \gamma = \frac{1}{\sqrt{3}}$ .根据斯托克斯公式，有

\begin{matrix} I = \iint_{Σ} | \begin{matrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y^{2} - z^{2} & z^{2} - x^{2} & x^{2} - y^{2} \end{matrix} | d S = - \frac{4}{\sqrt{3}} \iint_{Σ} (x + y + z) d S \end{matrix}

$\Sigma$ $x + y + z = \frac{3}{2}$ ，因此：

I = - \frac{4}{\sqrt{3}} \cdot \frac{3}{2} \iint_{Σ} d S = - 2 \sqrt{3} \iint_{D_{x} y} \sqrt{3} d x d y = - 6 σ_{x y} .

$\Sigma$ $xOy$ $D_{xy}$ 的面积为：

σ_{x y} = 1 - 2 \times \frac{1}{8} = \frac{3}{4},

故

I = - \frac{9}{2} .

第12章 曲线积分和曲面积分

12.1 场的概念与例子

12.2 曲线积分

12.2.1 第一类曲线积分：标量场中的曲线积分

12.2.2 第二类曲线积分：二维向量场中的曲线积分

12.2.2.1 封闭曲线上的第二类曲线积分

12.2.2.2 两类曲线积分之间的关系

12.2.2.3 特殊情况：梯度场

12.2.2.4 格林公式

12.3 曲面积分

12.3.1 通量与曲面积分

12.3.2 高斯公式

12.3.3 Stokes 公式

12.4 总结与联系

第12章曲线积分和曲面积分