It seems to me like the formula presented here for the formal definition is a little too simple. |A|->0 does not necessarily mean that A is shrinking into a point. If A is a rectangle of width and height 1, |A|->0 when the width remains 1 and the height tends to 0.

That is correct. In another article the author notes it is important that all dimensions are shrinking and in that article they use the maximum dimension of the changing area. Just saying A shrinks is using a sort of short-hand. But, you are correct.

Shouldn't the limit be 𝛑r² → 0 since we're observing the limit of |Dr| → 0?

since pi is a constant it doesn't affect the limit. so we have r^2->0 and if r^2 goes to zero that means r must go to zero r->0 is the same thing

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Course: Multivariable calculus > Unit 5

Lesson 1: Formal definitions of div and curl (optional reading)

Formal definition of divergence in two dimensions

Google Classroom

Learn how line integrals are used to formalize the idea of divergence.

Background

If you haven't already, you may also want to read "Why care about the formal definitions of divergence and curl" for motivation.

What we're building to

In two dimensions, divergence is formally defined as follows:
$\begin{array}{r} div F (x, y) = lim_{| A_{(x, y)} | \to 0} \underset{Flux per unit area}{\underset{⏟}{\frac{1}{| A_{(x, y)} |} \overset{2d-flux through C}{\overset{⏞}{\oint_{C} F \cdot \hat{n} d s}}}} \end{array}$ ‍
$F (x, y)$ ‍ is a two-dimensional vector field.
$(x, y)$ ‍ is a specific point on the $x y$ ‍-plane.
$A_{(x, y)}$ ‍ is some region of the $x y$ ‍-plane which includes the point $(x, y)$ ‍. For example, you can think of this as being a circle centered at $(x, y)$ ‍.
$| A_{(x, y)} |$ ‍ is the area of $A_{(x, y)}$ ‍.
$lim_{| A_{(x, y)} | \to 0}$ ‍ indicates that we are considering the limit as the area $| A_{(x, y)} |$ ‍ goes to zero. In other words, as $A_{(x, y)}$ ‍ shrinks around the point $(x, y)$ ‍.
$C$ ‍ is the boundary of $A_{(x, y)}$ ‍.
$\hat{n} (x, y)$ ‍ is a function which gives an outward-facing unit normal vector at each point on $C$ ‍.
The line integral $\oint_{C} F \cdot \hat{n} d s$ ‍ represents the two-dimensional flux of $F$ ‍ through $C$ ‍.

There is a lot going on in this definition, but we will build up to it one piece at a time. The bulk of the intuition comes from the background understanding of flux.

"Outward flow" at a point doesn't really make sense

By this point you should have some idea of what divergence is trying to measure. When a vector field

F (x, y)

represents a fluid flow, divergence measures the tendency for the fluid to flow away from each point.

Khan Academy video wrapper

See video transcript

However, there's a disconnect between the idea of "outward flow" and divergence itself:

Divergence is a function which takes in individual points in space.
The idea of outward flow only makes sense with respect to a region in space. You can ask if a fluid flows out of a given region or into it, but it doesn't make sense to talk about fluid flowing out of a single point.

Formally defining divergence will involve using a flux integral, which measures the outward flow in a region, then taking the appropriate limit as this region shrinks around a specific point.

From a region to a point

In the article on two-dimensional flux, we had the following setup:

$F (x, y)$ ‍ is a vector-valued function representing the velocity vector field of some fluid.
$C$ ‍ is a closed loop in the $x y$ ‍-plane.
$\hat{n} (x, y)$ ‍ is a function that gives the outward unit normal vector at all points on the curve $C$ ‍.

I talked about how if you were tracking the mass of fluid in the region enclosed by the curve

C

, you could compute the rate at which mass is leaving the region using the following line integral:

$\begin{array}{r} \underset{Rate at which mass leaves region}{\underset{⏟}{- \frac{d (fluid mass in region)}{d t}}} = \underset{Flux integral}{\underset{⏟}{\oint_{C} F \cdot \hat{n} d s}} \end{array}$ ‍

This is called a "flux integral". If it is positive, fluid tends to be exiting the region, otherwise it tends to be entering the region. You can interpret this integral by imagining walking along the boundary

C

and measuring how much fluid tends to be exiting/entering the region at each point.

What if instead of measuring the change of mass, you wanted to know the change in density? Well, just divide this integral by the area of the region in question. Let's go ahead and give that region a name,

A

, and say that

| A |

is the area of the region.

$\begin{array}{r} \underset{Change in mass per unit area in A}{\underset{⏟}{- \frac{d (fluid density in region)}{d t}}} = \frac{1}{| A |} \oint_{C} F \cdot \hat{n} d s \end{array}$ ‍

To formally define divergence of

F

at a point

(x, y)

, we consider the limit of this change in density as the region shrinks around the point

(x, y)

There is no set-in-stone notation for this, but here's what I'll go with:

Rather than just writing $A$ ‍, write $A_{(x, y)}$ ‍ to emphasize that this region contains a specific point $(x, y)$ ‍.

This is important because as we start letting the region shrink, we don't want it to wander away from the point.

The expression "

| A_{(x, y)} | \to 0

" will indicate that we are considering the limit as the area of

A_{(x, y)}

goes to

0

, meaning

A_{(x, y)}

is shrinking around the point

(x, y)

With all this, here's how we write the formal definition of divergence:

$\begin{array}{r} div F (x, y) = \underset{\begin{array}{c} Region is shrinking \\ around (x, y) \end{array}}{\underset{⏟}{lim_{| A_{(x, y)} | \to 0}}} \overset{\begin{array}{c} Negative change of \\ fluid density in A_{(x, y)} \end{array}}{\overset{⏞}{\frac{1}{| A_{(x, y)} |} \underset{Flux through C}{\underset{⏟}{\oint_{C} F \cdot \hat{n} d s}}}} \end{array}$ ‍

"Simple" example: Constant divergence

Unlike other topics, the purpose of an example here is not to practice a skill that you will need. It is just to get a feel for what this relatively abstract definition actually looks like with a concrete function.

Let's use the quintessential "outward flowing" vector field in two dimensions:

$\begin{array}{r} F (x, y) = [\begin{array}{c} x \\ y \end{array}] \end{array}$ ‍

Concept check: Using the usual divergence formula, the one which arises from the notation

\nabla \cdot F

, what is the divergence of

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

\nabla \cdot F (x, y) =

Now let's see how the formal definition of divergence works in this case. Let's focus on the origin.

$(x, y) = (0, 0)$ ‍

And for our shrinking regions around this point, consider circles. Let

C_{r}

denote a circle of radius

r

centered at the origin, and

D_{r}

represent the region enclosed by that circle, where

D

stands for "Disk".

Notice, for all values of

r

, the disk

D_{r}

will contain the point

(0, 0)

, so this is indeed a good family of regions to use.

The formal definition of divergence at

(0, 0)

would then be written as follows:

$\begin{array}{r} div F (0, 0) = lim_{| D_{r} | \to 0} \frac{1}{| D_{r} |} \oint_{C_{r}} F \cdot \hat{n} d s \end{array}$ ‍

This is rather abstract, so let's start filling in the details of this integral.

Concept check: What is

| D_{r} |

$| D_{r} | =$ ‍

Concept check: Which of the following parameterizes

C_{r}

Choose 1 answer:
(Choice A)
$\begin{array}{r} [\begin{array}{c} \cos (t) \\ \sin (t) \end{array}] \end{array}$ ‍ as $t$ ‍ ranges from $0$ ‍ to $2 π$ ‍.
(Choice B)
$\begin{array}{r} [\begin{array}{c} r \cos (t) \\ r \sin (t) \end{array}] \end{array}$ ‍ as $t$ ‍ ranges from $0$ ‍ to $2 π$ ‍.

Concept check: Using this parameterization, what should we replace

d s

with in the integral

\int_{C_{r}} \dots d s

$d s =$ ‍
$d t$ ‍

If you are unfamiliar with this step, consider reviewing the articles on line integrals or arc length.
The value $d s$ ‍ represents a tiny bit of arc length on the curve $C_{r}$ ‍. Start by taking the derivative of the function parameterizing $C_{r}$ ‍:
$\begin{array}{r} \frac{d}{d t} [\begin{array}{c} r \cos (t) \\ r \sin (t) \end{array}] = [\begin{array}{c} - r \sin (t) \\ r \cos (t) \end{array}] \end{array}$ ‍
Then take its magnitude:
$\begin{aligned} \sqrt{(- r \sin (t))^{2} + (r \cos (t))^{2}} \\ = \sqrt{r^{2} (\sin^{2} (t) + \cos^{2} (t))} \\ = r \end{aligned}$ ‍
When you multiply this by $d t$ ‍, it gives you a little snippet of arc length:
$d s = r d t$ ‍

Concept check: Which of the following gives an outward facing unit normal vector

\hat{n}

C_{r}

Choose 1 answer:
(Choice A)
$\begin{array}{r} \hat{n} (x, y) = [\begin{array}{c} x / r \\ y / r \end{array}] \end{array}$ ‍
(Choice B)
$\begin{array}{r} \hat{n} (x, y) = [\begin{array}{c} x \\ y \end{array}] \end{array}$ ‍
(Choice C)
$\begin{array}{r} \hat{n} (x, y) = [\begin{array}{c} r x \\ r y \end{array}] \end{array}$ ‍

Applying all these answers to the expression we had before, here's what we get:

$\begin{aligned} div F (0, 0) \\ ⇓ \\ = lim_{| D_{r} | \to 0} \frac{1}{| D_{r} |} \oint_{C_{r}} F \cdot \hat{n} d s \\ = lim_{r \to 0} \frac{1}{π r^{2}} \int_{0}^{2 π} F (r \cos (t), r \sin (t)) \cdot \hat{n} (r \cos (t), r \sin (t)) r d t \\ = lim_{r \to 0} \frac{1}{π r^{2}} \int_{0}^{2 π} [\begin{array}{c} r \cos (t) \\ r \sin (t) \end{array}] \cdot [\begin{array}{c} (r \cos (t)) / r \\ (r \sin (t)) / r \end{array}] r d t \\ = lim_{r \to 0} \frac{1}{π r^{2}} \int_{0}^{2 π} \underset{r}{\underset{⏟}{(r \cos^{2} (t) + r \sin^{2} (t))}} r d t \\ = lim_{r \to 0} \frac{1}{π r^{2}} \int_{0}^{2 π} r^{2} d t \\ = lim_{r \to 0} \frac{1}{π r^{2}} 2 π r^{2} \\ = 2 \end{aligned}$ ‍

So the formal definition does actually match the formula

\nabla \cdot F

that we know and love. Well, at least for this specific example anyway.

I think you'll agree, though, that this is much more labor-intensive to compute. But the point of this formal definition is not to use it for actual computations. The point is that it does a much better job reflecting the idea of "outward fluid flow" in a mathematical formula. Having such a solid grasp of that idea will be helpful when you learn about Green's divergence theorem.

What about higher dimensions?

In the next article, I'll show how you can do essentially the same thing to define three-dimensional divergence using three-dimensional flux, which involves a surface integral.

Summary

Given a fluid flow, divergence tries to capture the idea of "outward flow" at a point. But this doesn't quite make sense, because you can only measure the change in fluid density of a region.
When talking about a region, the idea of "outward flow" is the same as flux through that region's boundary.
To adapt the idea of "outward flow in a region" to the idea of "outward flow at a point", start by considering the average outward flow per unit area in a region. This just means dividing the flux integral by the area of the region.
Next, consider the limit of this outward flow per unit area as the region shrinks around a specific point.
Putting this all into symbols, we get the following definition of divergence:
$\begin{array}{r} div F (x, y) = lim_{| A_{(x, y)} | \to 0} \underset{Flux per unit area}{\underset{⏟}{\frac{1}{| A_{(x, y)} |} \overset{2d-flux}{\overset{⏞}{\oint_{C} F \cdot \hat{n} d s}}}} \end{array}$ ‍

Want to join the conversation?

Sort by:

Jacob R
Posted 5 years ago. Direct link to Jacob R's post “It seems to me like the f...”
It seems to me like the formula presented here for the formal definition is a little too simple. |A|->0 does not necessarily mean that A is shrinking into a point. If A is a rectangle of width and height 1, |A|->0 when the width remains 1 and the height tends to 0.
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(8 votes)
Answer
- bussard.da
  Posted 5 years ago. Direct link to bussard.da's post “That is correct. In anot...”
  That is correct. In another article the author notes it is important that all dimensions are shrinking and in that article they use the maximum dimension of the changing area. Just saying A shrinks is using a sort of short-hand. But, you are correct.
  Button navigates to signup page
  (9 votes)
White
Posted 7 years ago. Direct link to White's post “I was about to ask about ...”
I was about to ask about the difference between Flux and Divergence on your article on Flux.

Brilliant article!
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(6 votes)
Answer
AndrewLi357
Posted 6 years ago. Direct link to AndrewLi357's post “Shouldn't the limit be 𝛑...”
Shouldn't the limit be 𝛑r² → 0 since we're observing the limit of |Dr| → 0?
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(2 votes)
Answer
- Fredde9311
  Posted 6 years ago. Direct link to Fredde9311's post “since pi is a constant it...”
  since pi is a constant it doesn't affect the limit. so we have r^2->0 and if r^2 goes to zero that means r must go to zero r->0 is the same thing
  Button navigates to signup page
  (8 votes)
Jose Molina
Posted 7 years ago. Direct link to Jose Molina's post “is there somewhere that I...”
is there somewhere that I can find a mathematical derivation or proof that the limit is equal to nabla dot F? a good place to see the same sort of proof for 2d and 3d curl equaling the 'cross product' nabla X F could be awesome too! great article btw.
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(3 votes)
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Benito Kestelman
Posted 8 years ago. Direct link to Benito Kestelman's post “In the simple positive di...”
In the simple positive divergence example video, it looks like all the vectors have equal magnitude, implying their partial derivatives are 0, so isn't the divergence also 0?
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(1 vote)
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- alek aleksander
  Posted 6 years ago. Direct link to alek aleksander's post “They have various directi...”
  They have various direction so their partial derivatives won't be zero except for the points lying on the x axis and their partial derivative over x (except for the origin) and analogous for the points on the y axis.
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  (1 vote)
ravishankar joshi
Posted 7 years ago. Direct link to ravishankar joshi's post “I cannot read the formal ...”
I cannot read the formal definition of divergence.
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(1 vote)
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Sot P
Posted 7 years ago. Direct link to Sot P's post “HELP NEEDED! In defining ...”
HELP NEEDED!
In defining the 2d flux integral we had to assume that the density was uniform and equal to 1. Now for the formal proof of divergence we are dividing our flux integral which represents the change in mass per unit time ( represents a mass quantity as the density was equal to 1 so area=mass) by area to get the change in density per unit time. Is this not a contradiction as the change in density would be 0 as we had to assume this to come up with the flux integral in the first place.

Could it be that we are assuming that the density of fluid when it flows out is constant but the density of the region itself is changing?
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(1 vote)
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- alek aleksander
  Posted 6 years ago. Direct link to alek aleksander's post “I don't really think you ...”
  I don't really think you have to assume this. Just keep in mind we're calculating divergence (etc.) "at the beginning of time". When all the density is equal to one. It does not matter that latter on the density of the escaping fluid will change. We're measuring rate of the flow in instantaneous time at the very start of the flow. You could also recognize rate of change of this rate of the flow (change of the density of the flowing fluid in time) but it won't affect the former. It's like speed and acceleration. If a train passing a semaphore is accelerating what is the rate of wagons passing the semaphore in a chosen moment? It's according to its current speed at this moment despite its acceleration. And when you have this rate you could translate it to a rate of disappearing wagons from the area before the semaphore, i.e. wagon density. ;)
  Hope this helped.
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  (1 vote)
Andrew
Posted 5 years ago. Direct link to Andrew's post “I just want to make sure ...”
I just want to make sure I understand correctly, if the divergence at a point is positive there is outward flow, but divergence at a point can also be negative for a net negative flow meaning an "inward" flow, right? It just seems repeatedly described as outward flow so I want to make sure that that's just over looked and that I'm not misinterpreting it.
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(1 vote)
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- Maheshwaran Narayanan
  Posted 3 years ago. Direct link to Maheshwaran Narayanan's post “Yes you are correct....Ne...”
  Yes you are correct....Negative divergence(negative outward) flow is just "inward flow"
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  (1 vote)
ricyuka
Posted 7 months ago. Direct link to ricyuka's post “In Summary section, at th...”
In Summary section, at the end of the 1st paragraph says "measure the change in fluid density of a region", what causes of the change in density? The change in the speed of fluid in different regions? Is this related to compressible fluids?
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(1 vote)
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