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# Divergence theorem proof (part 5) | Divergence theorem | Multivariable Calculus | Khan Academy

Now we can work on the triple integral part of our problem
or our proof, this right over here.
And so I can rewrite that.
So this is the triple integral over our region, which
we're assuming is a type I region, of the partial of R
with respect to Z, which we can write like this.
It doesn't matter.
Partial of r with respect to do z.
And I'll dV.
And we can rewrite this as-- we can
assume we're going to integrate with respect to Z first.
So I'm going to integrate with respect to Z.
Let me do that in another color.
I'm going integrate with respect to Z first.
The lower bound on Z in our type I region, the lower bound
is f1.
The upper bound is f2.
So we're going to integrate from f1 of x, y to f2 of x, y.
And I'm going to integrate the partial of R with respect to Z.
So let me do that in that same yellow color--
partial of R with respect to Z.
And then I have dZ, and then I'll
have to integrate with respect to y and x or with respect
to x and y.
So it's dx dy or dy dz.
I can just write that as dA.
So what you could think of it-- we
can evaluate the yellow part.
And then we're just going to take the double integral
over the x, y domain.
So this is just going to be over the x, y domain.
Let me put some brackets here just
to make it clear what we're going to do.
So all we're doing is we're integrating with respect to Z
first, and we have the bounds there.
Well, this is pretty straightforward.
This is all going to be equal to-- I'll
write the outside first-- the double integral
over the domain.
And I have the dA right over here.
Actually, let me give myself some real estate-- dA.
Well, what's the antiderivative of this?
This is just R, and this is just R, or R of x, y,
z evaluated when Z is f1-- or when Z is f2.
And from that, we evaluate when Z is f1.
So this is just going to be R of x, y and z,
and we evaluate when Z is equal to that.
And from that, we subtract when Z is equal to that.
So that's going to be equal to-- so R of x, y
z evaluated when Z is equal to that is R of x, y, f2 of x, y.
And from that, we need to subtract R
when Z is this-- minus R of x, y f1 of x, y,
and then make sure that we got our parentheses.
Now, this is exactly what we saw in the last video.
It is exactly that, which shows that this is exactly this.
So when we assumed it was a type I region,
we got that this is exactly equal to this.
You do the exact same argument with the type II region
to show that this is equal to this, type III region
to show this is equal to that, and you
have your divergence theorem proved.
And we can consider ourselves done.