Concise Notes on Integrals - Types of Curve Integrals of the Second Kind

Types of curve integrals of the second kind

一、Classification of curve integrals of the second kind

01 Plane curve integrals of the second kind

02 Curve integrals of the second kind in space

二、A representation of the curve integral of the second kind

00 Preparation for analysis

$ds\cdot \cos \alpha =|PT|\cos \alpha =|PQ|=dx$,$ds=\sqrt{{\tan }^2\alpha +1}\cdot dx$

$\vec{a}=\{x_1,y_1,z_1\}$,$x_1=\vec{a}\cdot \vec{i}=|\vec{a}|\cos \alpha$,

$y_1=\vec{a}\cdot \vec{j}=|\vec{a}|\cos \beta$,$z_1=\vec{a}\cdot \vec{k}=|\vec{a}|\cos \gamma$

01 Plane curve integrals of the second kind

一般形式：${\int }_{{\Gamma }_{AB}}(\vec{A}(P)\cdot \overrightarrow{T^0}(P))ds={\int }_{{\Gamma }_{AB}}(\vec{A}\cdot \overrightarrow{T^0})ds$

若 ${\Gamma }_{AB}\subset {\mathrm{R}}^2$,$P(x,y)\in {\Gamma }_{AB}$,$\vec{T}$ Curves in the plane $\Gamma$ 上点 $(x,y)$ 处,The unit vector of the tangent $\overrightarrow{T^0}$ 与指定的方向一致.

$\overrightarrow{T^0}=\{\cos \alpha ,\cos \beta \}$ ,$\overrightarrow{T_0}\cdot ds\stackrel{△}{=}d\vec{s}=\{\cos \alpha ,\cos \beta \}=\{dx,dy\}$ ,$\vec{A}(x,y)=\{P(x,y),Q(x,y)\}$

$$\begin{array}{rl}①& {\int }_{{\Gamma }_{AB}}(\vec{A}\cdot \overrightarrow{T^0})ds\\ ②& ={\int }_{{\Gamma }_{AB}}\vec{A}\cdot d\vec{s}\\ ③& ={\int }_{{\Gamma }_{AB}}(P(x,y)\cos \alpha +Q(x,y)\cos \beta )ds\\ & ={\int }_{{\Gamma }_{AB}}(P\cos \alpha +Q\cos \beta )ds\\ ④& ={\int }_{{\Gamma }_{AB}}P(x,y)dx+Q(x,y)dy\\ & ={\int }_{{\Gamma }_{AB}}Pdx+Qdy(用得较多)\\ ⑤& ={\int }_{{\Gamma }_{AB}}P(x,y)dx+{\int }_{{\Gamma }_{AB}}Q(x,y)dy\end{array}$$

02 Curve integrals of the second kind in space

一般形式：${\int }_{{\Gamma }_{AB}}(\vec{A}(P)\cdot \overrightarrow{T^0}(P))ds={\int }_{{\Gamma }_{AB}}(\vec{A}\cdot \overrightarrow{T^0})ds$

若 ${\Gamma }_{AB}\subset {\mathrm{R}}^3$,$P(x,y,z)\in {\Gamma }_{AB}$,$\vec{T}$ Curves in space $\Gamma$ 上点 $(x,y,z)$ 处,

The unit vector of the tangent $\overrightarrow{T^0}=\{\cos \alpha ,\cos \beta ,\cos \gamma \}$ 与指定的方向一致.

$\overrightarrow{T^0}=\{\cos \alpha ,\cos \beta ,\cos \gamma \}$ ,$\overrightarrow{T_0}\cdot ds\stackrel{△}{=}d\vec{s}=\{\cos \alpha ,\cos \beta ,\cos \gamma \}=\{dx,dy,dz\}$

$\vec{A}(x,y)=\{P(x,y,z),Q(x,y,z),R(x,y,z)\}$
$$\begin{array}{rl}①& {\int }_{{\Gamma }_{AB}}(\vec{A}\cdot \overrightarrow{T^0})ds\\ ②& ={\int }_{{\Gamma }_{AB}}\vec{A}\cdot d\vec{s}\\ ③& ={\int }_{{\Gamma }_{AB}}(P(x,y,z)\cos \alpha +Q(x,y,z)\cos \beta +R(x,y,z)\cos \gamma )ds\\ & ={\int }_{{\Gamma }_{AB}}(P\cos \alpha +Q\cos \beta +R\cos \gamma )ds\\ ④& ={\int }_{{\Gamma }_{AB}}P(x,y,z)dx+Q(x,y,z)dy+R(x,y,z)dz\\ & ={\int }_{{\Gamma }_{AB}}Pdx+Qdy+Rdz(用得较多)\\ ⑤& ={\int }_{{\Gamma }_{AB}}P(x,y,z)dx+{\int }_{{\Gamma }_{AB}}Q(x,y,z)dy+{\int }_{{\Gamma }_{AB}}R(x,y,z)dz\end{array}$$

三、Calculation of curve integrals of the second kind

01 Plane curve integrals of the second kind

If you directly calculate the two-curve integral,Which one to turn into？

If the second kind of curve integral is calculated directly,No matter which form is given, it must be turned into the first④种,把曲线 ${\Gamma }_{AB}$ represented as a parametric equation：

${\Gamma }_{AB}:\{\begin{array}{l}x=x(t)\\ y=y(t)\end{array}$ 找出起点 $A$ 对应的参数 $t_A$,找出终点 $B$ 对应的参数 $t_B$ .
$$\begin{array}{rl}& {\int }_{{\Gamma }_{AB}}Pdx+Qdy={\int }_{t_A}^{t_B}[P(x(t),y(t))x^{\prime }(t)+Q(x(t),y(t))y^{\prime }(t)]dt\end{array}$$
对于平面曲线 ${\Gamma }_{AB}$,in five“Plane curve integrals of the second kind“的基础上,The equation of the curve has the following situations.

(1) Unary function type

形式：${\Gamma }_{AB}:y=\phi (x),x\in [a,b]$,( $x=x$ ) Special parametric equations

(2) Constant function type

形式：${\Gamma }_{AB}:y=a(常数)$, ( $x=x$ ) Special parametric equations

(3) Unary inverse function type

形式：${\Gamma }_{AB}:x=\psi (y),y\in [c,d],\psi ’(x)$ 连续 ( $y=y$ ) Special parametric equations

(4) Inverse function constant type

形式：${\Gamma }_{AB}:x=a(常数),y\in [c,d],\psi ’(x)$ 连续 ( $y=y$ ) Special parametric equations

(5) Polar coordinate type

形式：${\Gamma }_{AB}:r=r(\theta ),\theta \in [\alpha ,\beta ],r^{\prime }(\theta )$ 连续

​ $\Rightarrow \{\begin{array}{l}x=r(\theta )\cos \theta \\ y=r(\theta )\sin \theta \end{array}\theta \in [\alpha ,\beta ]$ ,$x^{\prime 2}(\theta )+y^{\prime 2}(\theta )=r^2(\theta )+r^{\prime 2}(\theta )$

(6) Inverse polar type

形式：${\Gamma }_{AB}:\theta =\theta (r),r\in [a,b],{\theta }^{\prime }(r)$ 连续

​ $\Rightarrow \{\begin{array}{l}x=r\cos \theta (r)\\ y=r\sin \theta (r)\end{array}r\in [a,b]$

02 Curve integrals of the second kind in space

(1) Closed curve type

形式：${\oint }_{L}Pdx+Qdy+Rdz$

① 直接计算

Calculate directly if you can,$L:\{\begin{array}{l}x=x(t)\\ y=y(t)\\ z=z(t)\end{array}$ ,Find the parameter corresponding to the starting point $t_0$,Find the parameter corresponding to the end point $t_1$ .into a definite integral of the parameters.

② 斯托克斯公式

(2) Unclosed curve type

形式：${\int }_{{\Gamma }_{AB}}Pdx+Qdy+Rdz$

① 直接计算

Calculate directly if you can,${\Gamma }_{AB}:\{\begin{array}{l}x=x(t)\\ y=y(t)\\ z=z(t)\end{array}$ 找出起点 $A$ 对应的参数 $t_A$,找出终点 $B$ 对应的参数 $t_B$ .into a definite integral of the parameters.

② 牛顿-莱布尼兹公式

若找到 $u(x,y,z)$​,使 $du=Pdx+Qdy+Rdz$,

${\int }_{{\Gamma }_{AB}}Pdx+Qdy+Rdz={\int }_{A(x_0,y_0,z_0)}^{B(x_1,y_1,z_1)}du=u(x,y,z){|}_{A(x_0,y_0,z_0)}^{B(x_1,y_1,z_1)}$

${\int }_{A(x_0,y_0,z_0)}^{B(x_1,y_1,z_1)}Pdx+Qdy+Rdz={\int }_{x_0}^{x_1}P(x,y_0,z_0)dx+{\int }_{y_0}^{y_1}Q(x_1,y,z_0)dy+{\int }_{z_0}^{z_1}R(x_1,y_1,z)dz$

③ 路径无关

Find the line simply connected $\Omega$ 使 ${\Gamma }_{AB}\subset \Omega$,有 $P,Q,R$ 偏导数连续,且 $\mathrm{rot}\vec{A}\equiv 0$,Knowing has nothing to do with paths.

${\int }_{A(x_0,y_0,z_0)}^{B(x_1,y_1,z_1)}P(x,y,z)dx+Q(x,y,z)dy+R(x,y,z)dz$

$={\int }_{x_0}^{x_1}P(x,y_0,z_0)dx+{\int }_{y_0}^{y_1}Q(x_1,y,z_0)dy+{\int }_{z_0}^{z_1}R(x_1,y_1,z)dz$

将 $x_1$ 换成 $x$,同理有：

$u(x,y,z)={\int }_{A(x_0,y_0,z_0)}^{B(x_1,y_1,z_1)}Pdx+Qdy+Rdz+C$ （与路径无关）

$={\int }_{x_0}^{x}P(x,y_0,z_0)dx+{\int }_{y_0}^{y}Q(x,y,z_0)dy+{\int }_{z_0}^{z}R(x,y,z)dz+C$

四、Eight types of problem solving for curve integral of the second kind of plane

Four equivalent conditions for path-independent integration of plane curves using Green's formula,Discuss the following types.

01 Closed curve type

形式：${\oint }_{\Gamma }P(x,y)dx+Q(x,y)dy$, $\Gamma$ is a closed curve,along the positive direction.

首先需要求出 $\frac{\partial Q}{\partial x},\frac{\partial P}{\partial y}$ ,Then there are the following discussions of three different situations：

① 若 $P$,$Q$ 在 $\Gamma$ 包围的区域 $D$ is continuous and has continuous first-order partial derivatives,Use Green's formula：

${\oint }_{\Gamma }P(x,y)dx+Q(x,y)dy=\underset{D}{\iint }\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy$ ( The double integral is required to be easily calculated )

② 在 $\Gamma$ 包围的区域 $D$ 内部有”洞“（所谓”洞“,主要指的是 $P$,$Q$ Not defined in this area）

在”洞“的外部, $P$,$Q$ The partial derivatives of are continuous and have $\frac{\partial Q}{\partial x}\equiv \frac{\partial P}{\partial y}$ ,

则 ${\oint }_{{\Gamma }^{+}}Pdx+Qdy={\oint }_{{\Gamma }_1^{+}}Pdx+Qdy,\Gamma ,{\Gamma }_1$ surround the same”洞“,同方向.

Justify the conclusion：
$$\begin{array}{rl}& 证明:{\oint }_{{\Gamma }^{+}+{\Gamma }_1^{-}}Pdx+Qdy=\underset{D_1}{\iint }(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dxdy=\underset{D_1}{\iint }0dxdy=0\\ & {\oint }_{{\Gamma }^{+}}Pdx+Qdy+{\oint }_{{\Gamma }_1^{-}}Pdx+Qdy=0\\ & {\oint }_{{\Gamma }^{+}}Pdx+Qdy-{\oint }_{{\Gamma }_1^{+}}Pdx+Qdy=0\\ & {\oint }_{{\Gamma }^{+}}Pdx+Qdy={\oint }_{{\Gamma }_1^{+}}Pdx+Qdy\end{array}$$
常用场景：
$$\begin{array}{rl}& P,Q分母为a{x}^2+b{y}^2(a>0,b>0,常数)\\ & 如\Gamma Surrounds\; the\; originO(0,0),此时0为洞.\\ & 若不在O点时,\\ & P,QPartial\; derivatives\; are\; continuous\; and\frac{\partial Q}{\partial x}\equiv \frac{\partial P}{\partial y}\\ & 选取{\Gamma }_1：a{x}^2+b{y}^2=R^2(R>0,常数)\\ & 经常R=1,{\Gamma }_1与\Gamma 同方向\\ & 化成{\Gamma }_{1}Curve\; integrals\; of\; the\; second\; kind\; on\; .\\ & The\; denominator\; at\; this\; timea{x}^2+b{y}^2=R^2,化简后的P,QExpressions\; made\; simple(科学瘦身)\\ & 此时,偏导数连续,Green\text{'}s\; formula\; can\; be\; used.\end{array}$$
③ 如果 $\Gamma$ The definite integral expressed as a parametric equation simplifies into a one-variable function is easy to integrate,也可以直接计算.

02 Unclosed curve type

形式：${\int }_{{\Gamma }_{AB}}P(x,y)dx+Q(x,y)dy$, ${\Gamma }_{AB}$ is a non-closed curve in the plane.

首先需要求出 $\frac{\partial Q}{\partial x},\frac{\partial P}{\partial y}$ ,Then there are the following discussions of three different situations：

① 如果 $(x,y)\in {\Gamma }_{AB}$,$\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}$ 连续,in the area shown in the figure $D$ 上,$\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}$ 连续,

$\overrightarrow{AC}:y=y_0(x=x)Special\; parametric\; equations\{\begin{array}{l}A:x_0\\ C:x_1\end{array}$ ,

$\overrightarrow{CB}:x=x_1(y=y)Special\; parametric\; equations\{\begin{array}{l}C:y_0\\ B:y_1\end{array}$ .
$$\begin{array}{rl}& {\int }_{{\Gamma }_{AB}}P(x,y)dx+Q(x,y)dy={\int }_{A(x_0,y_0)}^{B(x_1,y_1)}Pdx+Qdy\\ & ={\int }_{\overrightarrow{AC}}P(x,y)dx+Q(x,y)dy+{\int }_{\overrightarrow{CB}}P(x,y)dx+Q(x,y)dy\\ & ={\int }_{x_0}^{x_1}P(x,y_0)dx+{\int }_{y_0}^{y_1}Q(x_1,y)dy\end{array}$$
② 若 $(x,y)\in {\Gamma }_{AB}$,$\frac{\partial Q}{\partial x}\not\equiv \frac{\partial P}{\partial y}$ ,

${\int }_{{\Gamma }_{AB}}Pdx+Qdy={\int }_{{\Gamma }_{AB}+l}Pdx+Qdy-{\int }_{l}Pdx+Qdy$

其中 $l$ A simple curve is required,Often a directed straight line segment,The former uses Green's formula,The latter is calculated directly.

③ 如果 ${\Gamma }_{AB}$ can be expressed as a parametric equation,Definite integrals of unary functions that are transformed into parameters,容易计算,也可以直接计算.

03 求原函数

形式：求 $P(x,y)dx+Q(x,y)dy$ 的原函数

Recall the unary primitive function：若 $F^{\prime }(x)=f(x)$,称 $F(x)$ 是 $f(x)$ 在区间 $I$ 上的一个原函数.

Speaking of primitive functions of two variables：Must be the original function of a pair of functions,满足 $u=u(x,y),\frac{\partial u}{\partial x}=P(x,y),\frac{\partial u}{\partial y}=Q(x,y)$ ,

​ 即满足 $dF(x)=f(x)dx\Rightarrow du=Pdx+Qdy$ .

什么条件下 $P,Q$ 有原函数？

Find a simply connected region $D$,$P,Q$ 在 $D$ The upper partial derivative is continuous,且 $\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y},(x,y)\in D$ .

By path independent line,know③Article established,存在 $D$ a binary function on ,使得：
$$\begin{array}{rl}& u(x,y)={\int }_{A(x_0,y_0)}^{B(x_1,y_1)}P(x,y)dx+Q(x,y)dy+C\\ & ={\int }_{x_0}^{x}P(x,y_0)dx+{\int }_{y_0}^{y}Q(x,y)dy+C\\ & 使得,du=Pdx+Qdy,即\frac{\partial u}{\partial x}=P,\frac{\partial u}{\partial y}=Q\end{array}$$
称 $u(x,y)$ 是 $P(x,y)dx+Q(x,y)dy$ 的 (全体) 原函数.

04 Solve full differential equations

形如 $P(x,y)dx+Q(x,y)dy=0$,If there is a simply connected region $D$,使 $\frac{\partial Q}{\partial x}\equiv \frac{\partial P}{\partial y}$ 连续,In this case this equation is called a total differential equation.

由 (3) 知,$u(x,y)={\int }_{x_0}^{x}P(x,y_0)dx+{\int }_{y_0}^{y}Q(x,y)dy$,

由 $Pdx+Qdy=0$,$\iff du(x,y)=0$$\iff u(x,y)=C(C常,C\in R)$

$\iff {\int }_{x_0}^{x}P(x,y_0)dx+{\int }_{y_0}^{y}Q(x,y)dy=C$ 是方程的通解.

Total differential quadrature（全微分方程）

设函数 $P(x,y),Q(x,y)$ on a simply connected region $D$ 有连续导数,且
$$\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}$$
则 $Pdx+Qdy$ 是某个函数 $u$ 的全微分：
$$u(x,y)={\int }_{\left(x_0,y_0\right)}^{(x,y)}Pdx+Qdy\leftarrow (u的求法)$$
若 $P(x,y)dx+Q(x,y)dy$ is the total differential of a function of two variables,称方程 $Pdx+Qdy=0$ 为全微分方程.

求出原函数 $u$,则解为 $u=C$ .
若 $P(x,y)\mathrm{dx}+Q(x,y)\mathrm{dy}$ is not the total differential of a binary function,方程 $Pdx+Qdy=0$ 的解法：

Find the integral factor $\mu$,so that the equation becomes $\mu P(x,y)dx+\mu Q(x,y)dy=0$ becomes a total differential equation.
$$\begin{array}{rl}& Common\; integral\; factors:\frac{1}{x^2},\frac{1}{y^2},\frac{1}{xy},\frac{1}{x^2{y}^2},\frac{1}{x^2+y^2}\\ & Combination\; patchwork:\\ & 例如,求解方程①\left(x-\sqrt{x^2+y^2}\right)dx=-ydy(Grouping\; and\; combining\; make\; up\; the\; full\; differential,Observe\; what\; factors\; need\; to\; be\; multiplied\; during\; the\; process)\\ & (xdx+ydy)-\sqrt{x^2+y^2}dx=0\Rightarrow d\left(x^2+y^2\right)-\sqrt{x^2+y^2}dx=0\\ & Multiply\; by\; what\; makes\; the\; latter\; term\; fully\; differentiated,The\; former\; term\; is\; still\; total\; differential?若将x^2+y^2看作u,should\; be\; multiplied\; by\phi (u)\\ & In\; fact,\; as\; long\; as\; each\; group\; makes\; up\; the\; full\; differential,The\; equation\; is\; solved\\ & 求解方程②y(1+xy)dx+x(1-xy)dy=0\end{array}$$

05 Points include reference questions

形式：$P(x,y)dx+Q(x,y)dy$ 中,求 $P,Q$ Alphabetical constants in .

$P(x,y)dx+Q(x,y)dy$ 中,$P,Q$ The expression contains the letter constant to be evaluated,Find this letter constant.

Find a simply connected region $D$,$P,Q$ 在 $D$ The upper partial derivative is continuous,根据题目条件,验证 ①$\sim$③One of the articles is established,

从而第④Article established,即有 $\frac{\partial Q}{\partial x}\equiv \frac{\partial P}{\partial y}$ ,Find the value of the letter constant from it.

06 牛顿-莱布尼兹公式

若 $P,Q$ 在单连通区域 $D$ 上连续,${\Gamma }_{AB}\subset D$,且存在 $D$ 上的二元函数 $u(x,y)$,使 $du=Pdx+Qdy$,
$${\int }_{{\Gamma }_{AB}}Pdx+Qdy={\int }_{AB}du=u(x,y){|}_{A(x_0,y_0)}^{B(x_1,y_1)}=u(x_1,y_1)-u(x_0,y_0)$$

07 求面积

There is a closed area $D$ 的边界为 $\Gamma$ along the positive direction,$D$ 的面积为 $S$,则 $S={\int }_{D}1d\sigma =\frac{1}{2}{\oint }_{{\Gamma }^{+}}-ydx+xdy$ Calculate directly on the right.

08 物理应用

Find a point $M$ 在变力 $\vec{F}$ Acts along the curve ${\Gamma }_{AB}$ 由 $A$ 移动到 $B$ 所作的功 $W$,$W={\int }_{{\Gamma }_{AB}}(\vec{F}\cdot \overrightarrow{T^0})ds$ .