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## Divergence theorem proof

Current time:0:00Total duration:3:12

# Divergence theorem proof (part 5)

## Video transcript

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.