Is R(x,y,z) supposed to indicate surface area or volume? I believe it should be surface area...

R is a 3D region, but it is neither area nor volume. Consider this: is Mike weight or height? He is a person, not a quantity. He has a weight and a height, but he himself is not equal to any number. R(x,y,z) just emphasizes that the region R must contain the point (x,y,z). To denote the volume, we can use |R| or |R(x,y,z)|, which you can sort of think of as the "magnitude" of the region. But strictly this is just a definition, just like "+" is defined as plus. We _can_ ask, however, whether R represents a 3D solid region or a 2D closed surface. I personally think R is a 3D region and S is the surface enclosing it. And I think |R| is the volume and Σ the surface area. dΣ is a tiny bit of the surface area (but still a quantity). Therefore, R and S are _things_ and |R| and Σ are _numbers_ (though you could argue that numbers are also things). Of course, it is important to note that different people use different notation, so don't take this too seriously. Always pay attention to context.

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Multivariable calculus

Course: Multivariable calculus > Unit 5

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

Formal definition of divergence in three dimensions

Google Classroom

Learn how surface integrals and 3D flux are used to formalize the idea of divergence in 3D.

Background

It is a short step between these two prerequisites, and understanding the formal definition of divergence in three dimensions. For that reason, I'm going to keep this article relatively short, assuming that you have the intuition behind both of those pieces of background knowledge.

What we're building to

The goal is to capture the intuition of outward fluid flow at a point in a mathematical formula.
In three-dimensions, divergence is defined using the following limit:

$\begin{array}{r} div F (x, y, z) = lim_{| R_{(x, y, z)} | \to 0} \overset{Average outward flow from R per unit volume}{\overset{⏞}{\frac{1}{| R_{(x, y, z)} |} \underset{Flux through the surface of R}{\underset{⏟}{\iint_{S} F \cdot \hat{n} d Σ}}}} \end{array}$ ‍

[Breakdown of terms]

There is quite a lot going on in this definition, but most of the complexity lies in that flux integral. If you understand that part, the rest comes from taking the limit with respect to a region shrinking around a point.

From a region to a point

Let's say you have a three-dimensional vector field.

$F (x, y, z) \leftarrow Three-dimensional vector field$ ‍

As always, think of this vector field as representing a fluid flow. The divergence

div F

tries to measure the "outward flow" of this fluid at each point. However, it doesn't quite make sense to talk about what it means for fluid to flow out of a point.

What does make sense is the idea of fluid flowing out of region. Specifically, picture some region

R

in the vector field.

Khan Academy video wrapper

See video transcript

Let's name the surface of this region "

S

". In the article on flux in three dimensions, I showed how you can measure the rate at which fluid is leaving this region by taking the flux of

F

over the surface

S

$\underset{Rate at which fluid exits R}{\underset{⏟}{- \frac{d (fluid mass in R)}{d t}}} = \underset{Flux surface integral}{\underset{⏟}{\iint_{S} F \cdot \hat{n} d Σ}}$ ‍

Here,

\hat{n} (x, y, z)

is a vector-valued function which returns the outward facing unit normal vector at each point on

S

Divergence itself is concerned with the change in fluid density around each point, as opposed mass. We can get the change in fluid density of

R

by dividing the flux integral by the volume of

R

. To denote the volume of

R

, put bars around it:

$| R | \leftarrow Volume of R$ ‍

So here's what rate at which fluid density changes inside

R

looks like:

$- \frac{d (fluid density in R)}{d t} = \frac{1}{| R |} \iint_{S} F \cdot \hat{n} d Σ$ ‍

The divergence of

F

at a point

(x, y, z)

is defined as the limit of this change-in-fluid-density expression as the region shrinks around the point

(x, y, z)

$div F (x, y, z) = \underset{R shrinks around (x, y, z)}{\underset{⏟}{lim_{R \to (x, y, z)}}} \frac{1}{| R |} \iint_{S} F \cdot \hat{n} d Σ$ ‍

In that equation, I wrote

R \to (x, y, z)

to communicate the idea of

R

shrinking around the point

(x, y, z)

. At the end of the day, all this notation is just a desperate attempt to communicate a heavily visual idea with symbols. You will see different authors use different notation. If you prefer, you could alternatively start by saying

R_{(x, y, z)}

is a region which contains the point

(x, y, z)

, then write the following:

$div F (x, y, z) = lim_{| R_{(x, y, z)} | \to 0} \frac{1}{| R_{(x, y, z)} |} \iint_{S} F \cdot \hat{n} d Σ$ ‍

I have a slight preference for this last notation, just because it makes it a bit easier to see the connection between

(x, y, z)

on the left hand side and the right hand side without relying so heavily on the context in which all the terms are defined.

Congratulations!

If you are at the point where you can understand this (rather complicated) definition, it is a good sign that you have a solid mental grasp of both divergence and surface integrals. It also means you are in a strong position to understand the divergence theorem, which connects this idea to that of triple integrals.

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Aneesh Vasudev
Posted 7 years ago. Direct link to Aneesh Vasudev's post “Is R(x,y,z) supposed to i...”
Is R(x,y,z) supposed to indicate surface area or volume? I believe it should be surface area...
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(3 votes)
Answer
- Alexander Wu
  Posted 7 years ago. Direct link to Alexander Wu's post “R is a 3D region, but it ...”
  R is a 3D region, but it is neither area nor volume. Consider this: is Mike weight or height? He is a person, not a quantity. He has a weight and a height, but he himself is not equal to any number.
  
  R(x,y,z) just emphasizes that the region R must contain the point (x,y,z). To denote the volume, we can use |R| or |R(x,y,z)|, which you can sort of think of as the "magnitude" of the region. But strictly this is just a definition, just like "+" is defined as plus.
  
  We can ask, however, whether R represents a 3D solid region or a 2D closed surface. I personally think R is a 3D region and S is the surface enclosing it. And I think |R| is the volume and Σ the surface area. dΣ is a tiny bit of the surface area (but still a quantity).
  
  Therefore, R and S are things and |R| and Σ are numbers (though you could argue that numbers are also things).
  
  Of course, it is important to note that different people use different notation, so don't take this too seriously. Always pay attention to context.
  Button navigates to signup page
  (16 votes)
Raffaele Formica
Posted a year ago. Direct link to Raffaele Formica's post “Since S is a "closed" sur...”
Since S is a "closed" surface, why is no circle-sign in the middle of the double sigma? Is the circle-sign to be included only with single integrals?
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(3 votes)
Answer
- 1564538
  Posted 4 months ago. Direct link to 1564538's post “You can put a circle-sign...”
  You can put a circle-sign around a double integral to indicate a closed surface, but you don't have to. The circle sign just makes it explicit that the surface is closed, which is helpful when you want to express integral equations conceptually without having to get too into the math. A good example of this are Maxwell's equations. People rarely use the full equations for computations, but instead use them to concisely describe electromagnetism.
  
  My guess as to why there's no circle sign here is that this article is concerned with a formal definition, not a conceptual explanation, and so the circle-sign is a little too hand-wavy.
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  (1 vote)
apndolby297
Posted 4 years ago. Direct link to apndolby297's post “What are the dimensions o...”
What are the dimensions of the physical quantity F(x,y,z) represents?
they should be mass/unit area.... according to me.
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(1 vote)
Answer
apndolby297
Posted 4 years ago. Direct link to apndolby297's post “What does d sigma in the ...”
What does d sigma in the expression imply?
is it the same as dxdy?
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(1 vote)
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mattiacasc
Posted 2 years ago. Direct link to mattiacasc's post “Is there a way to go back...”
Is there a way to go back and forth between a surface and its volume? For example, assuming I was given a parametrization for that red blob surface, how would i set up a triple integral to find the contained volume?
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(1 vote)
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