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# Divergence theorem proof (part 2)

Breaking up the surface integral. Created by Sal Khan.

## Want to join the conversation?

- At7:00, I understand why Sal crosses out the third integral for his situation (the cylinder). But isn't that taking out some generalization? What if the sides of the cylinder bulged out? Sort of like a barrel. The normal vectors on the third surface would not be orthogonal to the k vectors, right?(9 votes)
- Actually, if the cylinder bulged out like a barrel it would still be a type 1, 2 and 3 region. The (x,y) domain would be a slightly larger circle to represent the maximum diameter of the barrel. S1 would be the entire lower half of the barrel and S2 would be the entire upper half of the barrel. As S1 and S2 meet at the middle of the barrel, there would be no S3 to consider.(5 votes)

- Why is K dot N going to be zero for the third integral at7:00?(2 votes)
- So if "flux" would be like mass flow per unit area (kg/s*m^2), then the "flux across the boundary" is the change in momentum (kg*m/s)?(1 vote)
- If your aim is to calculate the rate at which mass flows through a surface, then the vector function along the surface is density times velocity. Its units are( kg/(s*m^2). Along each infinitesimal surface area, you multiply a component of the vector function in the direction of the normal vector by the area(with units m^2) to get kg/s. When you add all the contributions to the flux, you end up with the rate at which mass flows(kg/s), so the flux across the boundary is not a change in momentum.(2 votes)

## Video transcript

I think we're now ready to get
into the meat of the proof. In the last video, we
said if we can just prove that each of these
parts are equal to each other, we essentially have proven
that that is equal to that. Because here in
yellow is another way of writing the flux across
the surface and here in green is another way of writing
the triple integral over our region of
the divergence of f. And what I'm going
to do in this video and probably in
the next video is prove that these two are
equivalent to each other. And I'm going to prove
it using the fact that our original region
is a simple solid region, in particular is
a Type 1 region. And then that's
essentially going to be it because you can
use the exact same argument and the fact that it is a
Type 2 region to prove this, and the exact same
argument and the fact that it is a Type 3
region to prove that. So I'm going to assume
it's a Type 1, which I can. It's a Type 1, 2, and 3 region. So given the fact
that it's Type 1, I'm now going to prove this
relationship right over here. And then I'll leave it up to you
to do the exact same argument with a Type 2 and
a Type 3 region. So let's get going. So Type 1 region, just
to remind ourselves, a Type 1 region is
a region that is equal to the set of
all x, y's and z's such that the xy pairs are a member
of a domain in the xy plane, and z is bounded
by two functions. z's lower bound
is f1 of x and y. And that's going to be
less than or equal to z. And z's upper bound, we
can call it f2 of xy. And then let me close the
set notation right over here. And let me just draw a general
version of a Type 1 region. So let me draw my
x, y, and z-axes. So this is my z-axis. This is my x-axis. And there is my y-axis. And so we might have a
region D. So our region, I'll draw it as a little
circle right over here. This is our region D. And
for any xy in our region D, you can evaluate the function f. You can figure out
an f1, I should say. So this might define an f1,
which we can kind of imagine as a surface or a
little thing that's at the bottom of a
cylinder if you want. So every xy there,
when you evaluate or when you figure out
what the corresponding f1 of those points in
this domain would be, you might get a surface that
looks something like this. It doesn't have to be flat, but
hopefully this gives the idea. It doesn't have to
be completely flat. It can be curved
or whatever else. But this just shows you that
every xy, when you evaluate it right over here,
it gets associated with a point, this lower
bound surface right over here. And I'll draw a
dotted line to show that it's only for the
xy's in this domain. And then we have an upper bound
surface that might be up here. Give me any xy. When I evaluate f2, I
get this surface up here. And once again, they don't
have to look the same. This could be like a dome
or it could be slanted or who knows what it might be. But this will give
you the general idea. And then z fills up the region. Remember, the region isn't
just the surface of the figure, it's the entire
volume inside of it. So when z varies between that
surface and that surface, for any given xy in our domain,
we fill up the entire region. So we can fill up
this entire region. And this is the way I drew it. It looks like a
cylinder, but it doesn't have to be a cylinder like this. And these two surfaces might
touch each other, in which case there would be no
side of the cylinder. They could be lumpier than this. They might be
inclined in some way. But hopefully, this is
a good generalization of a Type 1 region. Now, a Type 1 region, you
can kind of think of it, it can be broken up
into three parts. It can be broken
up into surface-- or the surfaces of a Type
1 region, I should say, can be broken up
into three parts. Let's call that surface one. Let's call this right
over here surface two, the top of the cylinder
or whatever kind of lumpy top it might be. And let's call the side, if
these two surfaces don't touch each other, let's call
that surface three. There might not necessarily
even be a surface three if these two touch each other
as in the case of the sphere. But let's just assume that there
actually is a surface three. So if we're evaluating
the surface integral-- we'll think about this
triple integral in a second. But let's think about how
we can rewrite this surface integral right over here. This entire surface
is S1 plus S2 plus S3. So we can essentially
break this up into three separate
surface integrals. So let's do that. Just remember
we're just focusing on this part right over here. So the surface
integral of R times k dot n, the dot
product of k and n, ds can be rewritten as the-- let
me write it this way-- can be rewritten as the
surface integral over S2 of R times k dot n ds,
plus the surface integral-- I'm just breaking up
the surface here-- plus the surface integral
over S1 of R times k dot n ds plus the surface
integral over surface three of the same thing,
R times k dot n ds. Now, the way I've drawn it,
and this is actually the case, S3 is if those surfaces
don't touch each other. And for Type 1 situation
right over here, the normal vector
at any given point on the side of the cylinder
for this Type 1 region-- if there is this
in between region. There always isn't. In a sphere, there wouldn't
be this surface, in which case this would be 0. But if there is this surface in
a Type 1 region, the one that essentially connects the
boundaries of the top and the bottom, then
the normal vector will never have a k component. The normal vector will
always be pointing flat out. It will only have an
i and j component. So if you take this normal
vector right over here does not have a k component
and you're dotting it with a k vector, then the dot
product of two things that are orthogonal, the k
vector goes like that, you're going to get 0. So this thing is going
to be 0 because k dot n is going to be 0 in this
situation for this surface. k dot n is going
to be equal to 0. So this part right
over here simplifies to this right over here. Now in the next video what we
can do is essentially express these in terms of
surface integrals, but in terms of double
integrals over this domain right over here. We'll kind of evaluate
these surface integrals.