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## Multivariable calculus

# Divergence theorem proof (part 3)

Evaluating the surface integral. Created by Sal Khan.

## Want to join the conversation?

- why did we take the S3 such that it's normal vector has no k^ component where as there can be the case when it's not so, for example section of a cone above xy-plane. Also is it necessary that we should be taking surface such that it's domain for f1 and f2 are same, as in this video we have taken for both f1 ans f2 as D.(3 votes)
- if it was a cone, the pointy part would be S2, which would touch S1, so there would not be an S3 (and thus no k^ component). As for having the domains be the same, yes, to be a type I, II, or III, the two functions must both be applied to the same region.(3 votes)

- What video(s) can I watch to understand the intuition for the equation at1:40?(2 votes)
- Why for the t(x,y) function the K component only incorporates the f2 surface function and not the f1?(2 votes)
- At1:30Sal takes two partial derivatives: t with respect to x and t with respect to y, crosses them and multiples with dA. I understand why we would multiply with dA, just not where the partials of t come from and why we are even crossing them(1 vote)
- Why isn't the normal vector, n, a unit vector? Shouldn't we divide tx X ty over its magnitude?(1 vote)

## Video transcript

Let's now evaluate this green
surface integral right over here. And to do that, we need
to parameterize surface 2. So let's say surface
2 can be represented by the vector position function. I'll just call it t-- t
for two-- so t-- which is a vector-- it's going to
be a function of x and y. Those are going are going
to be our parameters. And it's going to
be equal to-- and we can do this, once again,
because our surface is a function of x and y. So it's going to be equal
to x times i plus y times j plus f2 of xy times
k for all the xy's that are a member of our domain. Now, with that
out of the way, we can re-express
what k dot n ds is. So let me write this over here. k dot n ds is equal to-- and
we could put parentheses here that we're going to
take this dot product. That's, at least, how I
like to think about it. This is the exact
same thing as k dotted with the cross
product of the partial of t-- let me make this clear. I'm going to do
the magenta color. The partial of t with respect
to x crossed with the partial of t with respect to y times
a little chunk of our area-- times da. A little chunk of our
area in the xy domain. We've done this
multiple times as we evaluated surface integrals. And we got the intuition
for why this works. And so then, we're
essentially just evaluating the surface integral. And the one thing
we want to make sure is make sure this has
the right orientation. Because, remember, in order for
the divergence theorem to be true, the way we've
defined it is, all the normal vectors
have to be outward-facing. So for this top surface,
the normal vector has to be pointing straight up. Not necessarily straight up. At least, upwards. If this is a curve, it wouldn't
be necessarily straight up. But it needs to be up kind
of outward-facing, like that. On the sides, it would be
outward-facing like that. And down here, it
would be outward-facing going in the general
downwards direction. So let's just make sure
that this is upwards-facing. If we're changing
with respect to x, we're going in this direction. Changing with respect to y,
we're going in that direction. Take the right-hand rule with
the cross product-- index finger there, middle finger
there, your right thumb, I should say, will
go straight up. So it goes in the
right direction. So this would be an
upward-pointing vector. So we got the right
orientation for our surface. Now, let's think
about what this is. And it's important to
realize we could calculate all of the components
of this, but then we're just going to take the dot
product of that with k. So really, we care about
the k component only, but I'll work it out. This is equal to k
times a matrix i, j, k of the partial of
t with respect to x. Well, the partial
of t with respect to x-- I'll do this
in blue-- is going to be 1, 0, and the partial
of f2 with respect to x. And then the partial
of t with respect to y is just going to be
0, 1, and the partial of f2 with respect to y. And then, of course, we
have to multiply times da. And this is all going to be
equal to unit vector k dotted with-- and I don't even
have to work it all out. It's going to be
something times the i unit vector minus
something-- checkerboard pattern-- minus something times
something else necessarily-- times our j unit vector. And now, we can think
about the k unit vector. So the k unit vector is going
to be 1 times 1 minus 0. So it's just going to be
plus the k unit vector. We know it's 1 times
the k unit vector. And of course, we have
our da out right here. But when you take
this dot product, you only are left
with the k components. And it's essentially
just 1 times 1. You end up with a
scalar quantity of 1. All of this business
just simplified to da. So now, we can rewrite
our surface integral. And we're going to rewrite
it in the xy domain now-- in our parameters domain. So our surface integral
right up here-- so this will be
good for this video. And then, we'll do the same
thing with this surface here. Just making sure that we
get the orientation right. So this surface
integral, s2-- and I'll even rewrite a little bit--
s2, which is a function. r is a function of x, y,
and z times k dot n ds-- I just rewrote all of
this right over here-- is equivalent to
the double integral over our parameters domain--
which is just D-- of r of x, y, z times all of this business. All of this business
just simplified to da. And since I want to write it
in terms of my parameters, I'll write it as R of xy. And while we're on that
surface, z is equal to f2. So it's xy and f2 of xy. And then, all of this business
we just saw simplified to da. So you might be
saying, hey, Sal. It didn't look like you
simplified it a lot. But at least now put it in terms
of a double integral instead of a surface integral. So at least, in my mind,
that is a simplification. In the next video, we're
going to the exact same thing with this-- just making sure
that our vectors are oriented properly. And we could just
introduce a negative sign to make sure that they are. And then, we're going to think
about the triple integrals and try to simplify those.