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
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
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.