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Studying for a test? Prepare with these 11 lessons on Green's, Stokes', and the divergence theorems.

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# Explanation of example 1

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.