# Divergence theorem proof (part 2) | Divergence theorem | Multivariable Calculus | Khan Academy

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--
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.