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## Multivariable calculus

### Course: Multivariable calculus > Unit 5

Lesson 7: 3D divergence theorem# Explanation of example 1

Intuition as to why we got no net flux in the last worked example. Created by Sal Khan.

## Want to join the conversation?

- At1:18why didn't he draw the arrows the other way?. In that case the flux (out) would be equal to the flux (in) and there would be 0 flux, right?(8 votes)
- I agree. Since positive divergence is defined in a direction out of a surface, a negative divergence would in this case mean that the arrows on the negative x-axis would point in the opposite direction.(1 vote)

- I completely understand Sal's intuitive argument explaining why the eventual result of this problem is 0. However, when Sal offers up the possibility of changing the bounds of "x" (and thus our region R) to show that the flux would not equal 0 for this new region, does it not contradict the idea that Gauss' theorem (the divergence theorem) can only be used for symmetric regions? In my calculus text, R is defined as a "symmetric elementary region." Is this new region actually "symmetric" & "elementary" and I'm actually lost in translation here?(3 votes)
- A Gaussian surface need not be symmetric, only closed. I guess that one often chooses a symmetrical surface because it's easier to do calculations with.(3 votes)

- If all of the arrows are pointing outward in the original problem, why isn't the total outward flux equal to 2*(5/6)?(3 votes)
- The arrows he drew were just examples of the normal direction. The direction that flux would have to travel in to be considered "outward flux".

Sal did not draw the direction that mass was traveling in. To do that he would need to draw the vector field F (the velocity of the mass), that would be difficult to do by hand for this example.(1 vote)

- Would the solution change if F is not a divergent?(1 vote)
- at1:18those arrows should be pointing inwards correct? That would denote a negative divergence as particles are coming into the surface.(1 vote)
- He was just drawing the direction that mass/stuff has to travel in to be considered "outward flux" (basically the normal vector, he drew the normal vector direction when z=0).

If you wanted to see the direction that mass traveled in when it crossed the surface you would need to draw the vector field F, which would be difficult to do by hand in this example.

Also remember divergence is a scalar field, not a vector field (it has no direction).(1 vote)

- Would anyone know if there are videos on transport theorems ? I'm a vector calculus student trying to get some help on Reynolds transport and flux transport theorem(1 vote)
- Is this condsidered as Calculus 3(1 vote)
- Would it be valid to recognize right away that, because the integral with respect to x is being evaluated over an odd function (2x) and the bounds of this integration are a = -b and b=b, the integral is zero knowing that this is an inherent property of integrals of this type? In other words, once the bounds and flux are known, can all this calculation be skipped and the drawing be left un-drawn by calling this property? I feel like it would save a lot of calculus time if this is generalized.(1 vote)
- You can, but you need to make sure that you are actually integrating over an odd function. If for instance, the bounds of integration of the z-integral were not the even function 1 - x^2, then the final integral might not go over an odd function.(1 vote)

- Is it only because de base is symetric(one half in negative x et one half in positive x) that it's equals zero or because the entire volume is symetric? if the entire volume wasn't symétrique it wouldn't have been zero right?(1 vote)
- Given a divergence of 2x, if the volume of our region is not symmetric about the yz plane, then the flux of F across the surface will be none-zero since the positive divergence on one side of the yz plane cannot completely cancel the negative divergence on the other side owing to a lack of symmetry.(1 vote)

- Will you always get a flux or a divergence equal to zero when the shape is symmetric about an axis?(0 votes)
- the eventual result of this problem is 0 because the divergence (derived from the given vector field) happened to equal 2x, which creates a coincidence that the end result is zero. If you were given a different vector field from which you would derive a different divergence you could find an answer not equal to 0.(1 vote)

## Video transcript

In the last video we used
the divergence theorem to show that the
flux across this surface right now, which is
equal to the divergence of f along or summed up
throughout the entire region, it's equal to 0. And what I want to do in this
video is think a little bit. When you get an
answer like 0, you want to think about why is it? This is saying that there's
no net flux across the surface right over here. Or if you sum up all of the
divergences in this volume, you are getting 0. So why is that? Well, the simple way
to think about it is, when we took the divergence
of f, this vector field f is hard to visualize,
but the divergence of f is fairly easy to visualize. The divergence is
equal to 2 times x. So over here you're
going to get, as you go further and
further in this direction, as x becomes larger,
your divergence becomes more and more positive. So you have kind of a
divergence of 2 right over here. You have a divergence
of 1 along that line. And you have a divergence
of 0 right there. And that's also true,
obviously, as you go higher. Because you're just
changing the z. You're not changing the x. So all over here you
have positive divergence. Over there you have
positive divergence, and not just along that plane. But if you go in the
x direction as well, this whole region of space,
you have positive divergence, I guess you could say, in the
positive x side of our octant. But then as you go on that
side, on the other side, you have negative divergence. And this diagram is symmetric
with respect to the zy plane. And so those divergences
cancel them out. You would have had
a positive flux across the surface or a
positive value right over here. Instead of calculating
it for this region, we had calculated
for region that was just between x is 0 and 1. So let's just think
about that region. So that region that would have
been cut off right over here. That would have been cut
off right over there. And so the back--
I guess you could say the back wall of that,
the back wall of this, would have been the zy plane. Now if we care
about this volume-- so we're essentially
eliminating the rest of it. So let me try to eliminate
it as best as I can, change the colors. So if I eliminate that part
of it over there and all maybe even what we
see, all of that-- I should have deleted that first. So if we eliminate
the back part of it and we're just dealing
with it when x is positive, then our entire solution
that we did in the last video would have been the
exact same, except now x is going to vary between--
instead of negative 1 and 1, it'll vary between 0 and 1. And so our bounds
of integration, x is going to go between 0 and 1. And then in that situation,
our final answer-- this part, this would be between 0 and 1. That would all be 0. And we would be left
with 3/2 minus 1/2. 3/2 minus 1/2 is 1 minus 1/6,
which is just going to be 5/6. And so when you just think
about this part of it-- this side of it, this one
that I've just drawn-- you had a positive flux of 5/6. On the other side you had
a negative flux of 5/6. And then they canceled out. One way to interpret
that is, if we thought about the entire surface, like
what we saw in the last video, then you had an aggregate
inward flux in the last video. That's why it was negative. And then it's completely
offset by an outward flux right over here, the positive. You had a negative divergence
in that other region, and you have a positive
divergence that completely offsets it in this
region right over here.