[examination skills] memory method and simple derivation of Green formula

Video Explanation ： Memory method and simple derivation of Green formula
When you learn Green's formula, you will find that the form given in books is not easy to remember .
You may have the following questions
Forget which is positive, counterclockwise or clockwise ？
Forget the P,Q Who should be partial derivative ？
Forget who subtracts who after the partial derivative ？

This article is divided into two parts , The first part is to transform Green's formula into a form that is easier to remember .
The second part is a simple derivation of Green's formula , If you really can't remember in the examination room , It can also be done through 2-3 The calculation of minutes comes from the Green formula .

First we need to know , Green's formula is a bridge between closed curve integral and double integral .
$$\oint \stackrel{\text{Green's formula}}{\leftrightarrow }\iint$$

Write it out completely
$${\oint }_{L}Pdx+Qdy={\iint }_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy$$

Memory method

How to remember counterclockwise , Take out your right hand , Give yourself a thumbs up , We postgraduate candidates are all awesome , Then four fingers bend in the direction （ Anti-clockwise ） Is the positive direction , The students who have the courage to take the postgraduate entrance examination are very good ！

Then, how to remember the form in the integrand function of the double integral ？
You can write it in the form of a determinant
$$\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=\left|\begin{array}{cc}\frac{\partial }{\partial x}& \frac{\partial }{\partial y}\\ P& Q\end{array}\right|$$
The form of determinant is very regular , Above are two partial derivatives , The following is also in the order of integration P,Q
If you are familiar with Hamiltonian operator （Nabla operator ） Students can also remember this form
$$\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=\left|\begin{array}{cc}\frac{\partial }{\partial x}& \frac{\partial }{\partial y}\\ P& Q\end{array}\right|=\left|\nabla \times \left(P,Q\right)\right|$$

Simple deduction

If you can't remember in the examination room , The integrand function in the double integral can also be derived in a simple way .
Use the simplest curve , Counterclockwise rectangle , Let the coordinates of the lower left corner be $(x_0,y_0)$, The coordinates in the upper right corner are $(x_1,y_1)$

Do a line integral over it , It can be disassembled into 4 Segment calculation
$${\oint }_C={\int }_{C1}+{\int }_{C2}+{\int }_{C3}+{\int }_{C4}$$

about $C1$ paragraph ,$y=y_0,dy=0$
$${\int }_{C1}Pdx+Qdy={\int }_{x_0}^{x_1}P\left(x,y_0\right)dx$$
about $C2$ paragraph ,$x=x_1,dx=0$
$${\int }_{C2}Pdx+Qdy={\int }_{y_0}^{y_1}Q\left(x_1,y\right)dy$$
The same can be
$${\int }_{C3}Pdx+Qdy={\int }_{x_1}^{x_0}P\left(x,y_1\right)dx=-{\int }_{x_0}^{x_1}P\left(x,y_1\right)dx$$
$${\int }_{C4}Pdx+Qdy={\int }_{y_1}^{y_0}Q\left(x_0,y\right)dy=-{\int }_{y_0}^{y_1}Q\left(x_0,y\right)dy$$

Then the integral of the whole line is
$${\oint }_{L}Pdx+Qdy={\int }_{x_0}^{x_1}P\left(x,y_0\right)dx+{\int }_{y_0}^{y_1}Q\left(x_1,y\right)dy-{\int }_{x_0}^{x_1}P\left(x,y_1\right)dx-{\int }_{y_0}^{y_1}Q\left(x_0,y\right)dy$$
Merge those with the same integral limit
$${\oint }_{L}Pdx+Qdy={\int }_{x_0}^{x_1}\left[P\left(x,y_0\right)-P\left(x,y_1\right)\right]dx+{\int }_{y_0}^{y_1}\left[Q\left(x_1,y\right)-Q\left(x_0,y\right)\right]dy$$
The subtraction in the integrand can be written as a definite integral
$$P\left(x,y_0\right)-P\left(x,y_1\right)={\int }_{y_1}^{y_0}{P}_y\left(x,y\right)dy=-{\int }_{y_0}^{y_1}{P}_y\left(x,y\right)dy$$
$$Q\left(x_1,y\right)-Q\left(x_0,y\right)={\int }_{x_0}^{x_1}{Q}_x\left(x,y\right)dx$$

So it can be written in the form of a double integral
$${\oint }_{L}Pdx+Qdy=-{\int }_{x_0}^{x_1}dx{\int }_{y_0}^{y_1}{P}_y\left(x,y\right)dy+{\int }_{y_0}^{y_1}dy{\int }_{x_0}^{x_1}{Q}_x\left(x,y\right)dx$$

Because our area is rectangular , So it's easy to swap the order of integrals
$$-{\int }_{x_0}^{x_1}dx{\int }_{y_0}^{y_1}{P}_y\left(x,y\right)dy+{\int }_{y_0}^{y_1}dy{\int }_{x_0}^{x_1}{Q}_x\left(x,y\right)dx=\underset{D}{\iint }\left[Q_x\left(x,y\right)-P_y\left(x,y\right)\right]dxdy$$
So we can get the form of Green's formula
If the title is given clockwise , Then do it clockwise , The end result will be
$$\underset{D}{\iint }\left[P_y\left(x,y\right)-Q_x\left(x,y\right)\right]dxdy$$

In fact, we can also divide the region into small rectangles （ The Tongji book proves that the region is cut horizontally and vertically ）, In this way, the Green's formula of any curve can be derived , Interested students can refer to this article
kaysen School leader ： The most popular and thorough explanation in the history of Green's formula
Cut the area into small rectangles , Take out each rectangle and convert it from line integral to double integral

Because the adjacent rectangular line integrals will cancel each other , So adding up the line integrals of the small rectangles is the line integral of the peripheral curves , The double integral of the small rectangle adds up to the double integral of the whole region