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# Divergence theorem proof (part 4)

## Video transcript

Sometimes when you're doing
a large multipart proof like this, it's easy
to lose your bearings. We're trying to prove
the divergence theorem. Well, we started off
just rewriting the flux across the surface and
rewriting the triple integral of the divergence. And we said, well,
if we can prove that each of these components
are equal to each other, we will have our proof done. And what I said is that
we're going to use the fact this is a Type 1 region
to prove this part, the fact that it's a Type 2
region to prove this part, and the fact that it's a Type
3 region to prove this part. In particular, I'm
going to show the fact that, if it is a Type 1
region, we can prove this. And you can use the
exact same argument to prove the other
two-- in which case, the divergence theorem
would be correct. So we're focused on
this part right here. And in particular, we're focused
on evaluating this surface integral. And to evaluate that
surface integral, we broke up the entire
surface into three surfaces-- that upper bound on z,
the lower bound on z, and kind of the side
of-- if you imagine some type of a funky cylinder. These don't have to be
flat tops and bottoms. And for a Type 1
region, you don't even have to have this side surface. This is only for the case where
the two surfaces don't touch each other within this
domain right over here. So we broke it up
into three surfaces. But we said, look,
the normal vector along the side right
over here is never going to have a k component. And so when you take the dot
product with the k unit vector, this is just going to cross out. So our surface
integral simplified to the surface integral over s2
and the surface integral of s1. And in the last video, we
evaluated the surface integral of s2-- or at least, we turned
it into a double integral over the domain. Now, I'm going to do the
same exact thing with s1. So let's just remind ourselves
what we are concerned with. We want to re-express s1. This surface
integral-- I should say we want to re-express the
surface integral over s1. It was r of x, y, and z. Up here we just wrote r, but
this is making it explicit that r is a function
of x, y, and z. Times k dot n ds. And the way we evaluate
any surface integral is we want to have
a parameterization for our actual surface. So let's introduce
a parameterization for our surface. Let's say that
s1-- and I'll use-- the letter O is kind
of a weird one to use. It looks like a 0,
and we're already using it for our normal vector. Let's use e. So the parameterization of
our surface could be x times i plus y times j plus-- and
since it's a Type 2 region, our surface is a function
of x and y-- plus f1-- which is a function
of xy-- times k. We see f1 right over here. That's our lower
bound on our region. And then, this is for
all of the xy pairs that are a member of our
domain in question. So we have our parameterization. Now, we can think
about how we can write this right
over here, the nds. And so n times ds-- so
we've done this multiple, multiple, multiple
times-- which is actually the same thing as ds. We've done this multiple times. This is equal to the cross
product of the parameterization in one direction with
respect to one parameter. And then, that cross with
respect to the other parameter. And then, that times da. But we want to make sure
we get the order right. So I'm going to
claim that this is going to be the partial of our
parameterization with respect to y crossed with the partial
of our parameterization with respect to x. And then, we have times da. But we need to make sure this
has the right orientation. Because, for this bottom
surface, remember, we need to be pointed straight
down, outward from the region. So if we're going
in the y direction, the partial with respect
to y is like that. The partial with respect
to x is like that. And if you use the
right-hand rule, your thumb will point
downward, like that. I could draw my right hand. My index finger
could go like that. My middle finger
would bend like this. I don't really care what
my other two fingers do. And then, my thumb
would go downwards. So this is the right order,
which is a different order, or it's the opposite
ordering, as we did last time. You'll see, we'll just
get a negative value, but let's just work
it through just to be a little bit
more convincing. So this business
right over here is going to be equal to-- let's
write our i, j, and k unit vectors. It's going to be all
of that times da. So the partial of
e with respect to y is 0, 1, partial of
f1 with respect to y. Partial with respect
to x is going to be 0, 1, partial of
f1 with respect to x. And then, when you
evaluate this whole thing-- and so let me draw a
little dotted line here-- this is going to be equal
to some business times i minus some other
business times j. And on our k
component, you're going to have 0 times 0
minus 1 times k. So minus k, and then,
all of that times da. And now, the reason why I didn't
take even worry about what these things are
going to be is I'm going to have to take the
dropped product with k. So this whole thing--
all of this business right over here-- I can now
express in the xy domain. I can now write is going to be
equal to the double integral. And let me do it in
that same purple color because that was the original
color for that surface integral. It's going to be the double
integral over our parameter's domain, in the xy
plane of r of x-- let me write this a little
bit cleaner-- r of xy. Instead of z, I'm going to
write z over that surface is f1 of x and y, so that
we have everything in terms of our parameters. Times all of this
business k dotted this. Well, what is k dotted this? Well, the dot product of k
and negative k is negative 1. So we're just going to
be left with negative da. So we'll put the
da out front here. And we have a negative
da right over here. So we have now expressed
this surface integral as the sum of two
double integrals. Let me find it. The sum of this, the sum of
that and that right over there. And actually, let
me just rewrite it just so that we can
make everything clear. So this surface integral. So let me write it
right over here. The surface integral over our
entire surface of r times k dot n ds is equal to
the double integral-- and I'll do this
in a new color-- is equal to the double
integral in the domain d of this thing
minus this thing da. So I'll write r of
xy and f2 of xy, and that's that minus this--
minus r of xy and f1-- be careful here-- f1 of xy. That's this thing
right over here. All of that times da. Now, we just showed
this is equal to this. All we have to show now
is that this is also equal to that same expression. And we will have proven
this for the Type 1 case. And we can use the exact
same argument for the Type 2 and Type 3 cases and feel good
that the divergence theorem is about to be proven for
a simple solid region.