If Green's theorem relates to Curl and the 2D Divergence theorem relates flux to Divergence, how can these two theorems mean exactly the same thing? thanks James

To my understanding, they don't mean the same thing (hence the reason why there are two different theorems and not just one). I think the article is trying to convey that the process and idea is the same, and computationally they don't really look any different. The last two concept checks are what show this. So conceptually they are different, and you probably should keep this in mind when solving a problem, but in reality you can "plug in" a flux integral into Green's theorem and out pops the 2D Divergence theorem. I hope this helps! And somebody please correct me if I am wrong.

In the last equation, why is the dA on the right side not there anymore?

This confused me, too. By checking the reasoning many times, I think it's just a writing mistake. Also, in one of the Concept Check question, the dA is there.

Main content

Course: Multivariable calculus > Unit 5

Lesson 8: Divergence theorem (articles)

2D divergence theorem

Google Classroom

This is the analog of Green's theorem, but for divergence instead of curl.

Background

Not strictly required, but helpful for a deeper understanding:

Formal definition of divergence

What we're building to

The 2D divergence theorem is to divergence what Green's theorem is to curl. It relates the divergence of a vector field within a region to the flux of that vector field through the boundary of the region.
Setup:
- $F (x, y)$ ‍ is a two-dimensional vector field.
  - $R$ ‍ is some region in the $x y$ ‍-plane.
  - $C$ ‍ is the boundary of $R$ ‍.
  - $\hat{n}$ ‍ is a function which gives outward-facing unit normal vectors to $C$ ‍.
The 2D divergence theorem says that the flux of $F$ ‍ through the boundary curve $C$ ‍ is the same as the double integral of $div F$ ‍ over the full region $R$ ‍.
$\underset{Flux integral}{\underset{⏟}{\int_{C} F \cdot \hat{n} d s}} = \iint_{R} div F d A$ ‍
The intuition here is that if $F$ ‍ represents a fluid flow, the total outward flow rate from $R$ ‍, as measured by the flux integral, equals the sum over all the little bits of outward flow at each point, as measured by divergence.
Often the component functions of $F (x, y)$ ‍ are given as $P (x, y)$ ‍ and $Q (x, y)$ ‍:
$F (x, y) = [\begin{matrix} P (x, y) \\ Q (x, y) \end{matrix}]$ ‍
In this case, once you write both integrals in terms of $P$ ‍ and $Q$ ‍, the 2D divergence theorem looks like this:
$\oint_{C} P d y - Q d x = \iint_{R} \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}$ ‍
Written in this form, it's easier to see that the 2D divergence theorem is secretly just saying the same thing as Green's theorem.

Intuition: Connecting two outward flow measures

The global view: Flux

Here, I am assuming you have already learned about two-dimensional flux, and what it represents. Namely, it gives the rate at which a flowing fluid passes through a curve, such as

C

. When that curve encloses some region, such as

R

, the flux is a measure of the rate at which fluid is exiting that region.

Given a vector field

F (x, y)

, representing the velocity vector field of the fluid, the flux of

F (x, y)

through

C

is measured with the following integral:

$\underset{Flux integral}{\underset{⏟}{\int_{C} F \cdot \hat{n} d s}}$ ‍

This integral walks over each point on the boundary

C

, and picks up the component of the vector from

F

which is in the direction of the outward-facing unit normal vector

\hat{n}

. The larger that value is, the faster fluid is flowing out of

R

at that point; the more negative it is, the more fluid is flowing in at that point.

The local view: Divergence

I am also assuming you have learned about a different measure of "outward flow" in fluid movements: Divergence. The divergence of

F (x, y)

is a function that tells you how much the fluid tends to diverge away from each point

(x, y)

The 2D divergence theorem connects these two ideas:

$\underset{Total outward flow from R}{\underset{⏟}{\overset{Flux integral}{\overset{⏞}{\int_{C} F \cdot \hat{n} d s}}}} = \underset{Sum of all little bits of outward flow}{\underset{⏟}{\iint_{R} div F d A}}$ ‍

Want a deeper understanding?

This intuition should feel very similar to the one behind Green's theorem, in which the total fluid rotation in a region equals the sum of all the little bits of rotation represented by

2d-curl F

$\underset{Total fluid rotation around R}{\underset{⏟}{\oint_{C} F \cdot d r}} = \underset{Sum over all little bits of rotation}{\underset{⏟}{\iint_{R} 2d-curl F d A}}$ ‍

However, for both Green's theorem and the 2D divergence theorem, talking about adding up little bits of rotation or outward flow is pretty vague. Although each provides a great intuition, they are not exactly rigorous math, are they?

In the article on Green's theorem, I stepped through a more precise line of reasoning for where the double integral of curl comes into play. This involved chopping up the region

R

, and seeing how certain line integrals canceled each other out along the slices through

R

An almost identical line of reasoning can be used to demonstrate the 2D divergence theorem. For anyone wishing to gain deeper insight, a good exercise would be to go back and walk through that same line of reasoning, but replace the line integral

\oint_{C} F \cdot d r

, which measures flow around

R

, with the flux integral

\oint_{C} F \cdot \hat{n} d s

, which measures flow out of

R

And if this deeper understanding is what you seek, I would also recommend going in armed with knowledge of the formal definition of divergence.

Proof: Flux integrals + Unit normal vector + Green's theorem

This exercise in deeper understanding is not necessary to prove the 2D divergence theorem. In fact, when you start spelling out how each integral is actually computed, you'll find that this theorem is really just saying the same thing as Green's theorem.

Start by writing out

F

in terms of the component functions

P (x, y)

and

Q (x, y)

$F (x, y) = [\begin{matrix} P (x, y) \\ Q (x, y) \end{matrix}]$ ‍

Applying the formula for a unit normal vector to the flux integral, here's another way to represent that flux integral.

$\int_{C} F \cdot \hat{n} d s = \int_{C} [\begin{matrix} P (x, y) \\ Q (x, y) \end{matrix}] \cdot \hat{n} d s$ ‍

Next, let's write out the unit normal vector explicitly.

Concept check: If we think of the vector

[\begin{matrix} d x \\ d y \end{matrix}]

as representing a tiny step in the counterclockwise direction around the curve

C

, with

d s = \sqrt{d x^{2} + d y^{2}}

as its magnitude, which of the following represents an outward facing unit normal vector?

Choose 1 answer:
(Choice A)
$\frac{1}{d s} [\begin{matrix} d x \\ d y \end{matrix}]$ ‍
(Choice B)
$\frac{1}{d s} [\begin{matrix} d y \\ - d x \end{matrix}]$ ‍
(Choice C)
$\frac{1}{d s} [\begin{matrix} - d y \\ d x \end{matrix}]$ ‍

The second answer choice is correct:
$\frac{1}{d s} [\begin{matrix} d y \\ - d x \end{matrix}]$ ‍
Both the second and the third answer choices give unit normal vectors, but only the second one is outward facing. If any of the following explanation seems unclear, consider reviewing the article on constructing a unit normal vector to a curve.
When constructing a unit normal vector, you think about taking the tangent vector $[\begin{matrix} d x \\ d y \end{matrix}]$ ‍ and rotating it $90^{\circ}$ ‍. In this case, since we are thinking of the tangent vector as being in the counterclockwise direction, and we want an outward-facing unit normal vector, we need to rotate this vector clockwise.
To rotate a vector by $90^{\circ}$ ‍, you swap the two components and make one of them negative. Personally, I can never remember which one to make negative for which direction of rotation, so I just try one and draw it out. For example, suppose you perform a transformation like this:
$[\begin{matrix} x \\ y \end{matrix}] \to [\begin{matrix} - y \\ x \end{matrix}]$ ‍
Drawing it out might look like this:
So evidently that gives a counterclockwise rotation. Since we need to go the other way for our outward facing normal vector, we negate the second coordinate:
$[\begin{matrix} d x \\ d y \end{matrix}] \to [\begin{matrix} d y \\ - d x \end{matrix}]$ ‍
Finally, to make this a unit vector, divide it by its magnitude:
$\frac{1}{d s} [\begin{matrix} d y \\ - d x \end{matrix}]$ ‍

Plugging this into our flux integral and simplifying, here's what we get:

\begin{aligned} \int_{C} [\begin{array}{c} P (x, y) \\ Q (x, y) \end{array}] \cdot \hat{n} d s & = \int_{C} [\begin{array}{c} P (x, y) \\ Q (x, y) \end{array}] \cdot (\frac{1}{d s} [\begin{array}{c} d y \\ - d x \end{array}]) d s \\ = \int_{C} P d y - Q d x \end{aligned}

Written in this form, we can directly apply Green's theorem.

Concept check: Which of the following is Green's theorem, where

C

represents a closed curve encompassing region

R

Choose 1 answer:
(Choice A)
$\oint_{C} P d x + Q d y = \iint_{R} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d A$ ‍
(Choice B)
$\oint_{C} P d x + Q d y = \iint_{R} (\frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x}) d A$ ‍

Concept check: What do you get when you apply Green's theorem to the flux integral

\int_{C} P d y - Q d x

Choose 1 answer:
(Choice A)
$\oint_{C} P d y - Q d x = \iint_{R} (\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}) d A$ ‍
(Choice B)
$\oint_{C} P d y - Q d x = \iint_{R} (\frac{\partial P}{\partial x} - \frac{\partial Q}{\partial y}) d A$ ‍

The first answer choice is correct:
$\oint_{C} P d y - Q d x = \iint_{R} (\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}) d A$ ‍
This might get confusing because both this expression and Green's theorem are using $P$ ‍'s and $Q$ ‍'s, but the roles of each are switched. Here's how the two look lined up:
$\begin{aligned} \underset{Green’s theorem}{\underset{⏟}{\oint_{C} P d x + Q d y = \iint_{R} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d A}} \\ \underset{Replace P with - Q, and Q with P}{\underset{⏟}{\oint_{C} - Q d x + P d y = \iint_{R} (\frac{\partial P}{\partial x} - \frac{\partial (- Q)}{\partial y}) d A}} \\ \underset{Rearranged form, and the answer to this question}{\underset{⏟}{\oint_{C} P d y - Q d x = \iint_{R} (\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}) d A}} \end{aligned}$ ‍

Notice, the expression inside the double integral of the answer to the last question is indeed the divergence of

F

$div F = div [\begin{matrix} P (x, y) \\ Q (x, y) \end{matrix}] = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}$ ‍

Using the 2D divergence theorem?

When it comes to translating between line integrals and double integrals, the 2D divergence theorem is saying basically the same thing as Green's theorem. So any of the actual computations in an example using this theorem would be indistinguishable from an example using Green's theorem (such as those in this article on Green's theorem examples).

However, the usefulness of learning the 2D divergence theorem is two-fold:

Conceptual benefit: It's a great way to deepen your understanding of flux, divergence, and Green's theorem.
Strategic benefit: Sometimes an example where Green's theorem is used lends itself more naturally to a divergence-based description. For example, if the line integral you want to compute begins its life as a flux integral, rather than expanding out this line integral to make it look like $\int P d x + Q d y$ ‍ and applying Green's theorem, you could recognize immediately that it's the same as doubly integrating divergence.

Summary

The 2D divergence theorem relates two-dimensional flux and the double integral of divergence through a region.
$\underset{Total outward flow from R}{\underset{⏟}{\int_{C} F \cdot \hat{n} d s}} = \underset{Sum of all little bits of outward flow}{\underset{⏟}{\iint_{R} div F d A}}$ ‍
Often the vector field $F (x, y)$ ‍ is expressed component-wise:
$F (x, y) = [\begin{matrix} P (x, y) \\ Q (x, y) \end{matrix}]$ ‍
In this case, here's how the 2D divergence theorem looks:
$\oint_{C} P d y - Q d x = \iint_{R} (\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}) d A$ ‍
In this form, it is easier to see that the 2D divergence theorem really just states the same thing as Green's theorem.

Want to join the conversation?

Sort by:

James Rae
Posted 7 years ago. Direct link to James Rae's post “If Green's theorem relate...”
If Green's theorem relates to Curl and the 2D Divergence theorem relates flux to Divergence, how can these two theorems mean exactly the same thing?
thanks James
Button navigates to signup pageComment on James Rae's post “If Green's theorem relate...”
(5 votes)
Answer
- Dexter Hughes
  Posted 7 years ago. Direct link to Dexter Hughes's post “To my understanding, they...”
  To my understanding, they don't mean the same thing (hence the reason why there are two different theorems and not just one). I think the article is trying to convey that the process and idea is the same, and computationally they don't really look any different. The last two concept checks are what show this.
  So conceptually they are different, and you probably should keep this in mind when solving a problem, but in reality you can "plug in" a flux integral into Green's theorem and out pops the 2D Divergence theorem.
  I hope this helps! And somebody please correct me if I am wrong.
  Button navigates to signup page
  (6 votes)
wcyi56
Posted 8 years ago. Direct link to wcyi56's post “In the last equation, why...”
In the last equation, why is the dA on the right side not there anymore?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- iamyoukou
  Posted 8 years ago. Direct link to iamyoukou's post “This confused me, too. By...”
  This confused me, too. By checking the reasoning many times, I think it's just a writing mistake. Also, in one of the Concept Check question, the dA is there.
  Button navigates to signup page
  (4 votes)
Matt
Posted a year ago. Direct link to Matt's post “I think there is a mistak...”
I think there is a mistake with the equation of the normal vector. I believe it should be n = 1/ds * [-dy; dx]. Yet the answer provided is n = 1/ds * [dy; -dx]
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Phan Đức Minh Đăng
  Posted 7 months ago. Direct link to Phan Đức Minh Đăng's post “Clicking on the "Show exp...”
  Clicking on the "Show explanation" option will help you on that
  Button navigates to signup page
  (1 vote)
festavarian2
Posted 2 years ago. Direct link to festavarian2's post “I'm not getting this. The...”
I'm not getting this. The flux integral measures flux as kg/sec.
Green's theorem measures work done (Joules). Where is the connection?
The article doesn't address this at all.
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer