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## Multivariable calculus

### Course: Multivariable calculus > Unit 5

Lesson 9: Proof of Stokes' theorem# Stokes' theorem proof part 2

Figuring out a parameterization of our surface and representing dS. Created by Sal Khan.

## Want to join the conversation?

- At2:32is dxdy = dS, where dS is not a vector but represents a small patch on the surface S?

Never mind, I understand now that |rx X ry| dxdy = dS, which makes perfect sense since that will be a point in dxdy times the area of the corresponding "patch" on S, so dS represents changes in the area of "patches" on S with respect to dxdy.(3 votes) - Is there a better or (at least) another way to see if the vector ds is pointing in the right direction other than the right-hand rule Pr. Khan begins to mention at3:30?(2 votes)
- A vector that points toward increasing x crossed with a vector that points toward increasing y will always result in a vector that points toward increasing z.(2 votes)

## Video transcript

Let's now parameterize
our surface. And then, we can figure out what
ds would actually look like. And so I will define my
position vector function for our surface as r. And I'm going to make it a
function of two parameters because we're going to have
to define a surface right over here. And I can actually use
x and y as my parameters because the surface
can be defined as a function of x and y. So I'll say that
my parameterization is going to be a
function of x and y. And in my i direction,
it's going to be x times i. In my j direction, it's
going to be y times j. And then, in the k
direction, well, that's just going to be z. And z is a function of x and y. And whenever you do a
parameterization of a surface, you have to think
about, well, what are the constraints on the
domain for your parameters? And so the constraints of
the domain for my parameters, I'm going to say that
every pair of xy-- every xy coordinate-- it has
to be a member-- this is the symbol for
member-- it has to be a member of this little
region right over here. We could call this the
domain of the parameters. It has to be a member of r. Actually, we assume
that up here. And actually, I
should have written this as the coordinates
of xy-- the xy pairs that are a member of r-- that's going
to help define our surface. An xy that is not a
member of r, then we're not going to consider
that the z of that xy as part of the surface. Only the z of xy's where the
xy's are part of this region. So now that we have
a parameterization for our surface,
we're ready to start thinking about what ds might be. And we need to think about
this a little bit carefully. So first, I'll just
write something. And then, we can confirm whether
this really will be the case. ds-- and we've seen
this before about why this is the case--
is going to be the cross product of the partial
derivative of this with respect to each of the parameters
times the little chunk of area in that domain. You could view it as the partial
of r of this parameterization with respect to x crossed with
the partial of r with respect to y. And then, that whole
thing-- and we actually want this to be a
vector, not just the absolute value or the
magnitude of this vector. We actually want
this to be a vector. That thing times-- we could
put in some order dx dy or we could write dy dx. And if we want to
be general just to say that it's a little
differential of our region right over here,
we can just write-- instead of writing dx dy or
dy dx-- we will write da. And the reason why I said
we need to be careful is we need to make sure, based
on how we've parameterized this position vector function,
based on how we parameterized it, whether this cross
product really points in the right direction,
the direction that we need to be oriented in. Because, remember,
if we're traversing a boundary like
this, we want to make sure that the surface is
oriented the right way. And the way we
think about-- if we were to twist a cap like
this, the cap would move up. Or if you were to walk around
this boundary in this direction with the surface to your left,
your head would point up. And so we need to make sure
that this vector, which really defines the orientation
of the surface, is definitely going to be
pointed up or above the surface as opposed to going
below the surface. And so let's think
about that a little bit. The partial derivative
with respect to x-- well, as x gets bigger,
it's going to go in that direction
along the surface. And the partial with respect
to y, as y gets bigger, it's going to go in that
direction along the surface. If I take r cross y-- and we
could use the right-hand rule here. We put our index
finger in the direction of the first thing we're
taking the cross product of, our middle finger in the
direction of the second thing. So just like that. We bend our middle finger. We don't care what the
other fingers are doing. So I'll just draw
them right there. Then, the thumb will go in the
direction of the cross product. So in this case, the thumb
is going to point up. And that's exactly what
we wanted to happen. So this actually is
the right ordering. The partial with respect to y
cross the partial with respect to x actually would
not have been right. That would have given us
the other orientation. We would have done that
if this boundary actually went the other way around. But this is the right
orientation given the way that we are going to
traverse the boundary. Now, with that out
of the way, let's actually calculate
this cross product. So let me just rewrite it. So this is going to
be equal to-- well, I'll just focus on the
cross product right now. The partial of r with respect to
x crossed with the partial of r with respect to y is
going to be equal to-- and we've done this
many times already. We'll just do it in
more general terms now. It's going to be equal
to the determinant of this matrix-- i, j, and k. Let me do those in
different colors. I think that's a helpful
way of thinking about it. So i, j, and k. Actually, I'll use
this magenta color. i, j, and k. Let me write this
line right over here. We want to write the different
components of the partial with respect to x. And so the partial of the i
component with respect to x is just going to be 1. The partial of the j
component with respect to x is going to be 0. And the partial of
this with respect to x, well, we can just write
that as a partial derivative of the function z
with respect to x. And so r sub x-- or the
partial with respect to x-- is the vector 1, i
plus 0, j plus z sub x, k. And we'll do the same thing
for this piece right over here. The partial with respect
to y-- it's i component, well, this is going to be 0. It's j component
is going to be 1. The partial of this
with respect to y is 1. And the partial of
this with respect to y is the partial of z
with respect to y. And actually, I forgot to
write k right over here in our parameterization. So now with all
of this set up, we are ready to figure out
what the cross product is. The cross product is going to
be equal to-- so our i component is going to be 0 times
the partial with respect to y minus 1 times the partial
of z with respect to x. So we get negative partial
of z with respect to x. And then, checkerboard pattern. We'll have minus j times--
so ignore that column, that row-- 1 times the partial
with respect to y. So that's z sub y--
the partial of z with respect to y--
minus 0 times this. So we're just left with
that right over there. And then, finally, we're
going to have plus k. And here we have 1
times 1 minus 0 times 0. So it's going to be k times 1. So we can just
write k over there. And so we can write,
the cross product really is just equal to negative z sub
x times i minus z sub y times j plus k.