Shouldn't you put parentheses around "Pdx+Qdy"? Like ∮(Pdx+Qdy)=∬(∂Q/∂x−∂P/∂y)dA

No, unfortunately writing ∮ Pdx + Qdy is common notation that you should expect to see. I don't understand why people use it––maybe someone else could comment on that––but I think it's terrible notation. It is used fairly often, though, so you just have to learn to recognize it.

I don't understand the P and Q notation. Why Pdx + Qdy? Are the dx and dy the differential vector components of dr?

Yes exactly, that is what it is. dr = idx+jdy, where i j are the cartesian basis vectors.

First, it is hard to read the technical definitions, specifically the end. I hope this gets looked into and resolved soon. Second, I don't fully realize why the area R(x,y) is significantly greater than the error factor in the 2-d curl.

The formal definition of 2-D curl uses a line integral to calculate the "net rotation" of some fluid in a region. As the region shrinks and becomes closer to some point (x, y) (and its area approaches 0), the approximation of the line integral gets closer and closer to the curl at (x, y). The error factor keeps getting smaller as the R(x,y) shrinks. You can read about the formal definition here: https://www.khanacademy.org/math/multivariable-calculus/greens-theorem-and-stokes-theorem/formal-definitions-of-divergence-and-curl/a/defining-curl

In the equation: ∬ 2d-curl F dA R What is dA? Do we have to replace dA by the derivative of the area of region R to evaluate the integral?

dA is the infinitesimal area element. In rectangular coordinates, dA is dxdy (or dydx). In polar coordinates, dA is r dr d theta (or r d theta dr). Note, in particular, that the double integral over a region R of a constant function is just that constant times the area of R. This can be used to simplify some calculations.

Main content

Multivariable calculus

Course: Multivariable calculus > Unit 5

Lesson 3: Green's theorem (articles)

Green's theorem

Google Classroom

Green's theorem relates the double integral curl to a certain line integral. It's actually really beautiful.

Background

Not strictly required, but helpful for a fuller understanding:

Formal definition of curl in two dimensions

Other resources

You can find examples of how Green's theorem is used to solve problems in the next article. Here, I will walk through what I find to be a beautiful line of reasoning for why it is true. You can find a different perspective in Sal's video on the topic.

One lesson, four times the payoff

Green's theorem is one of four major theorems at the culmination of multivariable calculus:

Green's theorem
2D divergence theorem
Stokes' theorem
3D Divergence theorem

Here's the good news: All four of these have very similar intuitions. So if you really get to the point where you feel Green's theorem in your bones, you're already most of the way there to understanding these other three!

What we're building to

Setup:
- start color #0c7f99, start bold text, F, end bold text, end color #0c7f99 is a two-dimensional vector field.
  - start color #bc2612, R, end color #bc2612 is some region in the x, y-plane.
  - start color #bc2612, C, end color #bc2612 is the boundary of that region, oriented counterclockwise.
Green's theorem states that the line integral of start color #0c7f99, start bold text, F, end bold text, end color #0c7f99 around the boundary of start color #bc2612, R, end color #bc2612 is the same as the double integral of the curl of start color #0c7f99, start bold text, F, end bold text, end color #0c7f99 within start color #bc2612, R, end color #bc2612:
\iint, start subscript, start color #bc2612, R, end color #bc2612, end subscript, start text, 2, d, negative, c, u, r, l, end text, start bold text, F, end bold text, d, A, equals, \oint, start subscript, start color #bc2612, C, end color #bc2612, end subscript, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, dot, d, start bold text, r, end bold text
You think of the left-hand side as adding up all the little bits of rotation at every point within a region start color #bc2612, R, end color #bc2612, and the right-hand side as measuring the total fluid rotation around the boundary start color #bc2612, C, end color #bc2612 of start color #bc2612, R, end color #bc2612.
Often times start color #0c7f99, start bold text, F, end bold text, end color #0c7f99 is written component-wise as follows:
start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, left parenthesis, x, comma, y, right parenthesis, equals, P, left parenthesis, x, comma, y, right parenthesis, start bold text, i, end bold text, with, hat, on top, plus, Q, left parenthesis, x, comma, y, right parenthesis, start bold text, j, end bold text, with, hat, on top
In terms of P and Q, here's what Green's theorem looks like:
\oint, start subscript, start color #bc2612, C, end color #bc2612, end subscript, P, d, x, plus, Q, d, y, equals, \iint, start subscript, start color #bc2612, R, end color #bc2612, end subscript, left parenthesis, start fraction, \partial, Q, divided by, \partial, x, end fraction, minus, start fraction, \partial, P, divided by, \partial, y, end fraction, right parenthesis, d, A

Fluid rotation around a boundary

As you read on, the picture to have in your head is a blob in a vector field.

start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, left parenthesis, x, comma, y, right parenthesis is the function for the vector field. And, as you're probably getting used to if you've read other articles like this involving vector fields, imagine that start color #0c7f99, start bold text, F, end bold text, end color #0c7f99 represents a fluid flow.
start color #bc2612, R, end color #bc2612 is some region in the x, y-plane. In practice, and in problems, it will be some well-defined shape like a circle or the boundary between two graphs, but while thinking abstractly I like to just draw it as a blob.
start color #bc2612, C, end color #bc2612 is the boundary of start color #bc2612, R, end color #bc2612, oriented counterclockwise. Remember that orientation, because it actually matters when you solve problems. Counterclockwise. Are you remembering it? Counterclockwise.

Concept check: How can you interpret the following line integral in terms of a fluid flow?

\oint, start subscript, start color #bc2612, C, end color #bc2612, end subscript, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, dot, d, start bold text, r, end bold text

(Remember, in a line integral through a vector field, the term d, start bold text, r, end bold text represents a tiny step along the curve, as a vector, which in this case will always point in the counterclockwise direction.)

Choose 1 answer:
(Choice A)
It will be positive if the fluid has an overall counterclockwise rotation around the boundary of start color #bc2612, R, end color #bc2612, and negative if that overall rotation is clockwise.
(Choice B)
It will be positive if the fluid tends to flow out of the region start color #bc2612, R, end color #bc2612, and positive if the fluid tends to flow into start color #bc2612, R, end color #bc2612.

Here's one way to think about the line integral \oint, start subscript, start color #bc2612, C, end color #bc2612, end subscript, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, dot, d, start bold text, r, end bold text : Image rowing a boat around the line start color #bc2612, C, end color #bc2612, counterclockwise.

At each point of your journey, the vector d, start bold text, r, end bold text gives the direction of your motion. The dot product start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, dot, d, start bold text, r, end bold text will be positive at points where the fluid flow is with you, and negative at points where it's against you.

On the whole, the line integral \oint, start subscript, start color #bc2612, C, end color #bc2612, end subscript, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, dot, d, start bold text, r, end bold text adds up all these dot products to tell you if the flow was generally helpful or burdensome.

So this line integral is positive when the fluid flow has a general counterclockwise tendency around the boundary start color #bc2612, C, end color #bc2612 (meaning it was generally helpful), and it will be negative if it has a clockwise tendency (generally burdensome).

Bringing the boundary to the interior

Green's theorem is all about taking this idea of fluid rotation around the boundary of start color #bc2612, R, end color #bc2612, and relating it to what goes on inside start color #bc2612, R, end color #bc2612. Conceptually, this will involve chopping up start color #bc2612, R, end color #bc2612 into many small pieces. In formulas, the end result will be taking the double integral of start text, 2, d, negative, c, u, r, l, end text, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99.

Cut the region

Imagine chopping up the region start color #bc2612, R, end color #bc2612 with a line straight down the middle, giving two subregions start color #bc2612, R, start subscript, 1, end subscript, end color #bc2612 and start color #bc2612, R, start subscript, 2, end subscript, end color #bc2612:

Name the boundaries of these two regions start color #0d923f, C, start subscript, 1, end subscript, end color #0d923f and start color #a75a05, C, start subscript, 2, end subscript, end color #a75a05. What happens if we take the line integral of start color #0c7f99, start bold text, F, end bold text, end color #0c7f99 around these two boundaries, and add them up?

\oint, start subscript, start color #0d923f, C, start subscript, 1, end subscript, end color #0d923f, end subscript, start color #0c7f99, F, end color #0c7f99, dot, d, start bold text, r, end bold text, plus, \oint, start subscript, start color #a75a05, C, start subscript, 2, end subscript, end color #a75a05, end subscript, start color #0c7f99, F, end color #0c7f99, dot, d, start bold text, r, end bold text

Notice, these line integrals will cancel out along the vertical line cut that you made. Namely, the integral around start color #0d923f, C, start subscript, 1, end subscript, end color #0d923f goes "up" this line, while the integral around start color #a75a05, C, start subscript, 2, end subscript, end color #a75a05 integrates "down" this line. (Remember, when performing a line integral in a vector field, changing the direction along a curve multiplies your result by minus, 1).

This means the sum of our two integrals is the same as just going around the full boundary start color #bc2612, C, end color #bc2612.

\oint, start subscript, start color #0d923f, C, start subscript, 1, end subscript, end color #0d923f, end subscript, start color #0c7f99, F, end color #0c7f99, dot, d, start bold text, r, end bold text, plus, \oint, start subscript, start color #a75a05, C, start subscript, 2, end subscript, end color #a75a05, end subscript, start color #0c7f99, F, end color #0c7f99, dot, d, start bold text, r, end bold text, equals, \oint, start subscript, start color #bc2612, C, end color #bc2612, end subscript, start color #0c7f99, F, end color #0c7f99, dot, d, start bold text, r, end bold text

Cut it again

You could do this one more time, maybe with a horizontal cut this time:

If you integrate around the boundaries of the resulting four subregions, the integrals will all cancel out along the cuts you made in the interior of start color #bc2612, R, end color #bc2612:

In a formula, this means the sum of the line integrals around all four subregions end up just equalling the line integral around the full region:

$\displaystyle \oint_\redE{C_1} \blueE{F} \cdot d\textbf{r} + \oint_\redE{C_2} \blueE{F} \cdot d\textbf{r} + \oint_\redE{C_3} \blueE{F} \cdot d\textbf{r} + \oint_\redE{C_4} \blueE{F} \cdot d\textbf{r} = \oint_\redE{C} \blueE{F} \cdot d\textbf{r}$ \oint, start subscript, start color #bc2612, C, start subscript, 1, end subscript, end color #bc2612, end subscript, start color #0c7f99, F, end color #0c7f99, dot, d, start bold text, r, end bold text, plus, \oint, start subscript, start color #bc2612, C, start subscript, 2, end subscript, end color #bc2612, end subscript, start color #0c7f99, F, end color #0c7f99, dot, d, start bold text, r, end bold text, plus, \oint, start subscript, start color #bc2612, C, start subscript, 3, end subscript, end color #bc2612, end subscript, start color #0c7f99, F, end color #0c7f99, dot, d, start bold text, r, end bold text, plus, \oint, start subscript, start color #bc2612, C, start subscript, 4, end subscript, end color #bc2612, end subscript, start color #0c7f99, F, end color #0c7f99, dot, d, start bold text, r, end bold text, equals, \oint, start subscript, start color #bc2612, C, end color #bc2612, end subscript, start color #0c7f99, F, end color #0c7f99, dot, d, start bold text, r, end bold text

I should emphasize that this only works if we make sure that all the boundaries start color #bc2612, C, start subscript, 1, end subscript, end color #bc2612, comma, dots, comma, start color #bc2612, C, start subscript, 4, end subscript, end color #bc2612 are oriented the same way. Otherwise, they might not cancel each other out along the cuts. It's common to think of counterclockwise as being the positive direction, so think about everything as being oriented counterclockwise.

Cut it many, many times

You might be able to see where I'm going with this. Imagine chopping of the region start color #bc2612, R, end color #bc2612 into many many tiny pieces, start color #bc2612, R, start subscript, 1, end subscript, end color #bc2612, comma, dots, comma, start color #bc2612, R, start subscript, n, end subscript, end color #bc2612. Orient all of their boundaries start color #bc2612, C, start subscript, 1, end subscript, end color #bc2612, comma, dots, comma, start color #bc2612, C, start subscript, n, end subscript, end color #bc2612 counterclockwise, and integrate the function start color #0c7f99, start bold text, F, end bold text, end color #0c7f99 over each one.

The integrals will cancel out along all the cuts inside start color #bc2612, R, end color #bc2612 itself. This is because for any cut, one of the integrals will go along it in one direction, while another goes along it in the other direction. In the end, the only parts where these integrals don't cancel are the pieces of the boundary start color #bc2612, C, end color #bc2612.

This means adding up the line integrals along the tiny boundaries of the pieces will give the same result as just integrating across the full region:

sum, start subscript, k, equals, 1, end subscript, start superscript, n, end superscript, left parenthesis, \oint, start subscript, start color #bc2612, C, start subscript, k, end subscript, end color #bc2612, end subscript, start color #0c7f99, F, end color #0c7f99, dot, d, start bold text, r, end bold text, right parenthesis, equals, \oint, start subscript, start color #bc2612, C, end color #bc2612, end subscript, start color #0c7f99, F, end color #0c7f99, dot, d, start bold text, r, end bold text

Integrating curl

So... why am I doing this? It's because there is another way to interpret each of these line integrals around a tiny piece using two-dimensional curl. Pick one of those pieces and zoom in on it.

Let start color #bc2612, R, start subscript, k, end subscript, end color #bc2612 be the piece you chose, with boundary start color #bc2612, C, start subscript, k, end subscript, end color #bc2612.
Let vertical bar, start color #bc2612, R, start subscript, k, end subscript, end color #bc2612, vertical bar represent the area of start color #bc2612, R, start subscript, k, end subscript, end color #bc2612, which we are thinking of as being some very small number.
Let start color #a75a05, left parenthesis, x, start subscript, k, end subscript, comma, y, start subscript, k, end subscript, right parenthesis, end color #a75a05 be some point sitting inside this piece, any point really.

The fluid rotation around this piece due to start color #0c7f99, start bold text, F, end bold text, end color #0c7f99 can be measured with the line integral \oint, start subscript, start color #bc2612, C, start subscript, k, end subscript, end color #bc2612, end subscript, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, dot, d, start bold text, r, end bold text. Think tiny row boat. But since this is a really small piece, there is another multivariable calculus concept that measures fluid rotation: Curl.

This line integral can be approximated by taking the start text, 2, d, negative, c, u, r, l, end text of start color #0c7f99, start bold text, F, end bold text, end color #0c7f99 at any point within start color #bc2612, R, start subscript, k, end subscript, end color #bc2612, and multiplying it by the (tiny) area vertical bar, start color #bc2612, R, start subscript, k, end subscript, end color #bc2612, vertical bar:

$\displaystyle \underbrace{ \oint_\redE{C_k} \blueE{\textbf{F}} \cdot d\textbf{r} }_{\substack{ \text{Integral around a} \\ \text{tiny piece $\redE{R_k}$} }} \approx \left( \text{2d-curl}\,\blueE{\textbf{F}} \underbrace{ \goldE{(x_k, y_k)} }_{\text{Point in $\redE{R_k}$}} \right) \underbrace{|\redE{R_k}|}_{\text{Area of $\redE{R_k}$}}$

Also, and this is important, the smaller start color #bc2612, R, start subscript, k, end subscript, end color #bc2612 is, the better this approximation is.

[More technical explanation using the formal definition of curl]

Adding up these approximations over all of the tiny pieces start color #bc2612, R, start subscript, k, end subscript, end color #bc2612, here's what you get:

$\displaystyle \sum_{k = 1}^n \left( \oint_\redE{C_k} \blueE{\textbf{F}} \cdot d\textbf{r} \right) \approx \sum_{k = 1}^n \left( \text{2d-curl}\,\blueE{\textbf{F}} \underbrace{ \goldE{(x_k, y_k)} }_{\text{Point in $\redE{R_k}$}} \,|\redE{R_k}| \right)$ sum, start subscript, k, equals, 1, end subscript, start superscript, n, end superscript, left parenthesis, \oint, start subscript, start color #bc2612, C, start subscript, k, end subscript, end color #bc2612, end subscript, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, dot, d, start bold text, r, end bold text, right parenthesis, approximately equals, sum, start subscript, k, equals, 1, end subscript, start superscript, n, end superscript, left parenthesis, start text, 2, d, negative, c, u, r, l, end text, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, start underbrace, start color #a75a05, left parenthesis, x, start subscript, k, end subscript, comma, y, start subscript, k, end subscript, right parenthesis, end color #a75a05, end underbrace, start subscript, start text, P, o, i, n, t, space, i, n, space, start color #bc2612, R, start subscript, k, end subscript, end color #bc2612, end text, end subscript, vertical bar, start color #bc2612, R, start subscript, k, end subscript, end color #bc2612, vertical bar, right parenthesis

Taking the conclusion from the previous section, the left-hand side above is the same as a single line integral around the full boundary of start color #bc2612, R, end color #bc2612, so we can rewrite this approximation as follows:

$\displaystyle \oint_\redE{C} \blueE{\textbf{F}} \cdot d\textbf{r} \approx \sum_{k = 1}^n \left( \text{2d-curl}\,\blueE{\textbf{F}} \underbrace{ \goldE{(x_k, y_k)} }_{\text{Point in $\redE{R_k}$}} \,|\redE{R_k}| \right)$ \oint, start subscript, start color #bc2612, C, end color #bc2612, end subscript, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, dot, d, start bold text, r, end bold text, approximately equals, sum, start subscript, k, equals, 1, end subscript, start superscript, n, end superscript, left parenthesis, start text, 2, d, negative, c, u, r, l, end text, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, start underbrace, start color #a75a05, left parenthesis, x, start subscript, k, end subscript, comma, y, start subscript, k, end subscript, right parenthesis, end color #a75a05, end underbrace, start subscript, start text, P, o, i, n, t, space, i, n, space, start color #bc2612, R, start subscript, k, end subscript, end color #bc2612, end text, end subscript, vertical bar, start color #bc2612, R, start subscript, k, end subscript, end color #bc2612, vertical bar, right parenthesis

Now take a close look at the sum on the right-hand side.

It includes a scalar-valued function, start text, 2, d, negative, c, u, r, l, end text, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99
The sum is taken over many tiny pieces start color #bc2612, R, start subscript, k, end subscript, end color #bc2612 of a two-dimensional region, start color #bc2612, R, end color #bc2612.
For each piece within the sum, the function is evaluated on a point inside that piece, then multiplied by its area.

Sound familiar? This is all the recipe for a double integral! (If this does not sound familiar, consider taking a look at this article on double integrals).

In particular, if you imagine chopping up the region start color #bc2612, R, end color #bc2612 more and more finely, you can replace the sum above with a double integral of start text, 2, d, negative, c, u, r, l, end text, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99 over start color #bc2612, R, end color #bc2612:

$\displaystyle \sum_{k = 1}^n \left( \text{2d-curl}\,\blueE{\textbf{F}} \underbrace{ \goldE{(x_k, y_k)} }_{\text{Point in $\redE{R_k}$}} \,|\redE{R_k}| \right) \rightarrow \iint_\redE{R} \text{2d-curl}\,\blueE{\textbf{F}} \,\redE{dA}$ sum, start subscript, k, equals, 1, end subscript, start superscript, n, end superscript, left parenthesis, start text, 2, d, negative, c, u, r, l, end text, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, start underbrace, start color #a75a05, left parenthesis, x, start subscript, k, end subscript, comma, y, start subscript, k, end subscript, right parenthesis, end color #a75a05, end underbrace, start subscript, start text, P, o, i, n, t, space, i, n, space, start color #bc2612, R, start subscript, k, end subscript, end color #bc2612, end text, end subscript, vertical bar, start color #bc2612, R, start subscript, k, end subscript, end color #bc2612, vertical bar, right parenthesis, right arrow, \iint, start subscript, start color #bc2612, R, end color #bc2612, end subscript, start text, 2, d, negative, c, u, r, l, end text, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, start color #bc2612, d, A, end color #bc2612

Putting everything together, here's what we have:

$\begin{aligned} \oint_\redE{C} \blueE{\textbf{F}} \cdot d\textbf{r} &= \sum_{k = 1}^n \left( \oint_\redE{C_k} \blueE{\textbf{F}} \cdot d\textbf{r} \right) \\\\ &\approx \sum_{k = 1}^n \left( \text{2d-curl}\,\blueE{\textbf{F}} \underbrace{ \goldE{(x_k, y_k)} }_{\text{Point in $\redE{R_k}$}} \,|\redE{R_k}| \right) \\\\ &\to \iint_\redE{R} \text{2d-curl}\,\blueE{\textbf{F}} \,\redE{dA} \end{aligned}$

This is actually more than a mere approximation, the line integral around the boundary equals the double integral of two-dimensional curl:

\oint, start subscript, start color #bc2612, C, end color #bc2612, end subscript, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, dot, d, start bold text, r, end bold text, equals, \iint, start subscript, start color #bc2612, R, end color #bc2612, end subscript, start text, 2, d, negative, c, u, r, l, end text, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, start color #bc2612, d, A, end color #bc2612

[Wait, how did that approximation turn into an equality?]

This marvelous fact is called Green's theorem. When you look at it, you can read it as saying that the rotation of a fluid around the full boundary of a region (the left-hand side) is the same as looking at all the little "bits of rotation" inside the region and adding them up (the right-hand side).

Alternative notation

It is very common to see Green's theorem written like this:

\oint, start subscript, start color #bc2612, C, end color #bc2612, end subscript, P, d, x, plus, Q, d, y, equals, \iint, start subscript, start color #bc2612, R, end color #bc2612, end subscript, left parenthesis, start fraction, \partial, Q, divided by, \partial, x, end fraction, minus, start fraction, \partial, P, divided by, \partial, y, end fraction, right parenthesis, d, A

This is just spelling out the dot product in the left-hand side line integral, as well as the curl in the right-hand side double integral. For whatever reason, it is common to use the letters P and Q to denote the components of the vector-valued function start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, left parenthesis, x, comma, y, right parenthesis:

$\displaystyle \blueE{\textbf{F}}(x, y) = P(x, y)\hat{\textbf{i}} + Q(x, y)\hat{\textbf{j}} = \left[ \begin{array}{c} P(x, y) \\ Q(x, y) \end{array} \right]$

In the next article, you will find examples of how this formula can be used to make either line integrals or double integrals simpler.

Summary

You can think about the line integral \oint, start subscript, start color #bc2612, C, end color #bc2612, end subscript, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, dot, d, start bold text, r, end bold text as measuring the rotation of the fluid flow represented by the vector field start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, left parenthesis, x, comma, y, right parenthesis around the curve start color #bc2612, C, end color #bc2612. It is conventional to think of counterclockwise rotation as being positive, in which case start color #bc2612, C, end color #bc2612 should be oriented counterclockwise.

Imagine chopping up the two-dimensional region start color #bc2612, R, end color #bc2612 enclosed by start color #bc2612, C, end color #bc2612 into many tiny pieces. Name the boundaries of these pieces start color #bc2612, C, start subscript, 1, end subscript, end color #bc2612, comma, dots, comma, start color #bc2612, C, start subscript, n, end subscript, end color #bc2612, and orient them all counterclockwise. Then adding up the line integrals of start color #0c7f99, start bold text, F, end bold text, end color #0c7f99 around each piece-boundary start color #bc2612, C, start subscript, k, end subscript, end color #bc2612 boils down to the same thing as the line integral around the full boundary start color #bc2612, C, end color #bc2612.
sum, start subscript, k, equals, 1, end subscript, start superscript, n, end superscript, left parenthesis, \oint, start subscript, start color #bc2612, C, start subscript, k, end subscript, end color #bc2612, end subscript, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, dot, d, start bold text, r, end bold text, right parenthesis, equals, \oint, start subscript, start color #bc2612, C, end color #bc2612, end subscript, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, dot, d, start bold text, r, end bold text
That is to say, the little line integrals cancel out along all the cuts within start color #bc2612, R, end color #bc2612
As you consider smaller and smaller pieces, the line integral around each tiny piece can be approximated using two-dimensional curl:

$\displaystyle \underbrace{ \oint_\redE{C_k} \blueE{\textbf{F}} \cdot d\textbf{r} }_{\substack{ \text{Integral around a} \\ \text{tiny piece $\redE{R_k}$} }} \approx \left( \text{2d-curl}\,\blueE{\textbf{F}} \underbrace{ \goldE{(x_k, y_k)} }_{\text{Point in $\redE{R_k}$}} \right) \underbrace{|\redE{R_k}|}_{\text{Area of $\redE{R_k}$}}$

Adding up these little "bits of curl" using a double integral over start color #bc2612, R, end color #bc2612, and applying the fact that the sum of the line integrals cancels out along interior cuts, you get Green's theorem:

\oint, start subscript, start color #bc2612, C, end color #bc2612, end subscript, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, dot, d, start bold text, r, end bold text, equals, \iint, start subscript, start color #bc2612, R, end color #bc2612, end subscript, start text, 2, d, negative, c, u, r, l, end text, start color #0c7f99, start bold text, F, end bold text, end color #0c7f99, start color #bc2612, d, A, end color #bc2612

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Alexander Wu
Posted 7 years ago. Direct link to Alexander Wu's post “Shouldn't you put parenth...”
Shouldn't you put parentheses around "Pdx+Qdy"? Like ∮(Pdx+Qdy)=∬(∂Q/∂x−∂P/∂y)dA
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- William Cabell
  Posted 6 years ago. Direct link to William Cabell's post “No, unfortunately writing...”
  No, unfortunately writing ∮ Pdx + Qdy is common notation that you should expect to see. I don't understand why people use it––maybe someone else could comment on that––but I think it's terrible notation. It is used fairly often, though, so you just have to learn to recognize it.
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Matt A360
Posted 4 years ago. Direct link to Matt A360's post “I don't understand the P ...”
I don't understand the P and Q notation. Why Pdx + Qdy? Are the dx and dy the differential vector components of dr?
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(3 votes)
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- Viktor Schwarz
  Posted 4 years ago. Direct link to Viktor Schwarz's post “Yes exactly, that is what...”
  Yes exactly, that is what it is. dr = idx+jdy, where i j are the cartesian basis vectors.
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Aakash Amish
Posted 7 years ago. Direct link to Aakash Amish's post “First, it is hard to read...”
First, it is hard to read the technical definitions, specifically the end. I hope this gets looked into and resolved soon.
Second, I don't fully realize why the area R(x,y) is significantly greater than the error factor in the 2-d curl.
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- DhereDhere
  Posted 7 years ago. Direct link to DhereDhere's post “The formal definition of ...”
  The formal definition of 2-D curl uses a line integral to calculate the "net rotation" of some fluid in a region. As the region shrinks and becomes closer to some point (x, y) (and its area approaches 0), the approximation of the line integral gets closer and closer to the curl at (x, y). The error factor keeps getting smaller as the R(x,y) shrinks.
  
  You can read about the formal definition here: https://www.khanacademy.org/math/multivariable-calculus/greens-theorem-and-stokes-theorem/formal-definitions-of-divergence-and-curl/a/defining-curl
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Alexander Wu
Posted 7 years ago. Direct link to Alexander Wu's post “I found somewhere that sa...”
I found somewhere that said "vector integral identities," and it included some identities not mentioned here. There was ∬n × ∇f dΣ = ∮f d r, ∭∇ × F dV = ∯n × F dΣ, ∭∇f dV = ∯n F dΣ, and something called Green's first and second identities. What are these?
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(3 votes)
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lgqnty
Posted 7 months ago. Direct link to lgqnty's post “How do we know that the e...”
How do we know that the error of the approximation is significantly smaller than r?
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SSchoeler99
Posted 3 years ago. Direct link to SSchoeler99's post “In the equation: ∬ 2d-cur...”
In the equation:
∬ 2d-curl F dA
R
What is dA? Do we have to replace dA by the derivative of the area of region R to evaluate the integral?
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- Ian Pulizzotto
  Posted 3 years ago. Direct link to Ian Pulizzotto's post “dA is the infinitesimal a...”
  dA is the infinitesimal area element.
  In rectangular coordinates, dA is dxdy (or dydx).
  In polar coordinates, dA is r dr d theta (or r d theta dr).
  
  Note, in particular, that the double integral over a region R of a constant function is just that constant times the area of R. This can be used to simplify some calculations.
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Chuck
Posted 4 months ago. Direct link to Chuck's post “Can you do an article in ...”
Can you do an article in the future about Green's first and second identities?
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(1 vote)
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festavarian2
Posted a year ago. Direct link to festavarian2's post “Why does he say--> 2D cur...”
Why does he say--> 2D curl F at a point, times the tiny area, is a "scalar valued function". Isn't curl by definition a vector? Why isn't this a vector valued function? Isn't Curl a cross product resulting in a vector?
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