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# Divergence theorem proof (part 1)

Setting up the proof for the divergence theorem. Created by Sal Khan.

Video transcript

Let's now prove the divergence
theorem, which tells us that the flux across the
surface of a vector field-- and our vector field
we're going to think about is F. So the flux
across that surface, and I could call
that F dot n, where n is a normal vector
of the surface-- and I can multiply
that times ds-- so this is equal to the
trip integral summing up throughout the volume
of that region, summing up that volume
of the divergence of F. And we've done several videos
explaining the intuition here, but now we are actually
going to prove it. And of course, we're times
each little differential cube of volume. And we're going to make
an assumption here. We're going to assume
that we're dealing with a simple solid region. And what this means,
more formally, is that the region
we're thinking about can be a type I, type
II, and type III region. I should say it is
all of the three. So it is types I, II, and III. And there are videos that go
into what each of these regions are, but a lot of
the basic shapes fall into this
simple solid region. Like a sphere or a
cylinder of some kind, they can be types I,
type II, and type III. And for a lot of situations that
aren't simple solid regions, you can break them up
into simple solid regions. But let's just prove it for
this case right over here. So let's just assume
that our vector field F can be written as P, which is
a function of x, y, and z times i, plus Q, which is
a function of x, y, and z times j, plus R,
which is a function of x, y, and z times k. So let's think about what each
of these sides of the equation would come out to be. Well, first of all, what
is going to be F dot n? So let's think about
that a little bit. F dot n is going to be
equal to this component right over here times
n's i-component, plus this component right over
here times n's j-component, plus this component here
times n's k-component. So we could write
it as P times-- or I'll just write P
open parentheses, the dot product of i and
n, and let me make sure I write i as a unit vector. Now, I want to be clear. What's going to happen
right over here? If you take the dot
product of i and n, you're just going to get
the i-component, the scaling factor of the i-component
of the n normal vector, and we're just going to
multiply that times P. So that's essentially the
product of the x-components, or I guess you could say the
magnitude of the x-components. And then to that, we are going
to add Q times j dotted with n. And once again, when
you dot j with n, you get the magnitude
of the j-component of the normal vector
right over there, and then times-- or
plus, I should say, plus R times k dotted with n. This isn't how we
normally see it, but I think it's reasonable to
say that this is actually true. This right over
here is going to be equal to P times the magnitude
of the i-component of n's normal vector, which is exactly
what we want in a dot product. This is the same thing
for the j-component. This is the same thing
for the k-component. And you can try it out Define
n as equal to m times i plus n times j plus o times k,
or something like that, and, you'll see that this
actually does work out fine. So how could we simplify this
expression right up here? Well, we could rewrite
the left-hand side as-- so the surface
integral of F-- now let me write
it multiple ways. F dot ds, which is equal to
the surface integral of F dot n times the scalar ds is
equal to the double integral of the surface of all of this
business right over here, is equal to the double integral
over the surface of-- let me just copy and paste that--
of all of that business. And I just noticed that I forgot
to put the little unit vector symbol, a little caret
right over there. Put some parentheses, and
then we are left with our ds. And then this, all
of this, can be rewritten as the
surface integral of P times this business. And I'll just do it in
the same color-- of P times the dot product
of i and n ds, plus the surface integral of
Q times the dot product of j and n ds, plus the surface
integral of R times the dot product of k and n-- I
Forgot a caret-- k and n ds. So I just broke it up. We were taking the
integral of this sum, and so I just rewrote it as
the sum of the integrals. So that's the left-hand
side right over here. Now let's think about
the right-hand side. What is the divergence of F? And actually I'm going to
take some space up here. What is the divergence of F? Well, the divergence of F
based on this expression of F is just going to
be-- let me just write it over here real small. The divergence of
F is going to be the partial of P
with respect to-- let me do this in a new
color, because I'm using that yellow too much. The divergence of
F is going to be the partial of P with respect
to x, plus the partial of Q with respect to y, plus the
partial of R with respect to z. So this triple integral
right over here could be written as the triple
integral of the partial of P with respect to x, plus the
partial of Q with respect to y, plus the partial of
R with respect to z. Well, this thing, once
again, instead of writing it as the triple
integral of this sum, we could write it as a
sum of triple integrals. So this thing
right over here can be rewritten as
the triple integral over our
three-dimensional region. Actually, let me copy
and paste that so I don't have to keep rewriting it. So it's going to be equal to the
triple integral of the partial of P with respect to x dv
plus the triple integral of the partial of Q with respect
to y dv plus, once again, triple integral of the partial
of R with respect to z dv. So we've essentially restated
our divergence theorem. This is our surface integral,
and the divergence theorem says that this needs to
be equal to this business right over here. We've just written
it a different way. And so what I'm going to
do, in order to prove it, is just show that each of
these corresponding terms are equal to each other, that
these are equal to each other, that these are
equal to each other, and that these are
equal to each other. And in particular, we're going
to focus the proof on this. And we're going to use the fact
that our region is a type I region. It's a type I, type
II, and type II. But we're going to use the
fact that it's a type I region to prove that these
two things are equivalent. And then you can use
the fact that it's also a type II and type III region to
make the exact same argument as to why this is equal to this
and why this is equal to that.