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Current time:0:00Total duration:22:14

In the last video, we finished
off with these two results. We started off just thinking
about what it means to take the partial derivative of
vector-valued function, and I got to these kind of, you might
call them, bizarre results. You know, what was the whole
point in getting here, Sal? And the whole point is so that
I can give you the tools you need to understand what
a surface integral is. So let's just think about,
let's draw the st plane, and then see how it gets
transformed into this surface r. So let's do that. Let's say that is the t-axis,
and let's say that that is the s-axis, and let's say that our
vector-valued function, our positioned vector-valued
function, is defined from s's between a and b, I'm just
picking arbitrary boundaries, and between t being
equal to c and d. So the area under question, if
you take any t and any s in this rectangle right here, it
will be mapped to part of that surface. And if you map each of these
points, you will eventually get the surface r. So let me draw r
in 3 dimensions. A surface in 3D. So that is our x-axis, that
is our y-axis, and then that is the z-axis. And just as a bit of a
reminder, it might look something like this. If we were to, this point right
here, where s is equal to a, and t is equal to c, remember,
we're going to draw the surface indicated by the position
vector function s, r of s and t. So this point right here, when
s is a and t is c, maybe it maps to, I'm just, you
know, that point! Right there. When you take a and c, and you
put it into this thing over here, you're just going to get
the vector that points at that. So you could say, it'll
give you a position vector that'll point right at that
position, right there. And then, let's say that this
line, right here, if we were to hold s constant at a, and if we
were to just vary t from c to d, maybe that looks
something like this. I'm just drawing some
arbitrary contour there. Maybe if we hold t constant at
c, and vary s from a to b, maybe that'll look at
something like that. I don't know. I'm just trying to
show you an example. So this point right here would
correspond to that point right there, when you put it into the
vector-valued function r, you would get a vector that points
to that point, just like that. And this point right here in
purple, when you evaluate r of s and t, it'll give you a
vector that points right there, to that point over there, and
we could do a couple of other points, just to get an idea of
what the surface looks like, although I'm trying to keep
things as general as possible. So maybe, let me do it
in this bluish color. This, if we hold p at d and
vary s from a to b, we're going to start here. This is when t is d and s is a. And when you vary it, maybe
you get something like that. I don't know. So this point right here would
correspond to a vector that points to that point, right
there, and then finally this line, or this, if we hold s at
b and vary t between c and d, we're going to go from that
point, to that point. So it's going to look something
like this-- oh, sorry, we're going to go from this
point to that point. We're holding s at b, varying
t from c to d, maybe it looks something like that. So our surface, we went from
this nice rectangular area in the ts plane, and it gets
transformed into this wacky-looking surface. We could even draw some
other things right here. Let's say we get some
arbitrary value. Let me pick a new color,
I'll do it in white. Or a new noncolor. And let's say if we hold s at
that constant value and we vary t, maybe that will look
for something like this. Maybe that we'll look something
like, well, I don't know. Maybe it'll look
something like that. So you get an idea of what
the surface might look like. Now, given this, I want
to think about what these quantities are. And then when we visualize what
these quantities are, we'll be able to kind of use these
results of the last videos to do something that I
think will be useful. So let's say that we
pick arbitrary s and t. So this is the point, let me
just we pick it right here. That is the point s, t. s comma t. If you were to put those values
in here maybe it maps to, and I want to make sure I'm
consistent with everything I've drawn, maybe it maps to
this point right here. Maybe it maps to that
point right there. So this point right here,
that is r of s and t. For a particular s and t. I mean, I could put little
subscripts here, but I want to be general. I could call this a, well,
I already used a and b. I could call this x and y,
this would be r of x and y. It would map to that
point, right there. So that right there,
or that right there. Now let's see what happens
if we take, if we move just in the s direction. So this is we could
do that as s. Now let us move forward
by a differential, by a super small amount of s. So this right here, let's
call that a s plus a super small differential in s. That's right there. So that point. Let me do that in a better
color, in this yellow. So that point right there
is the point s plus my differential of s. I could write delta s, but
I wanted a super small change in s, comma t. And what is that going
to get mapped to? Well, if we apply these two
point in r, we're going to get something that maybe
is right over there. And I want to be very clear. This right here, that is
r of s plus ds comma t. That's what that is. That's the point when we just
shift s by a super small differential, this distance
here, you can view as ds. It's a super small change in s. And then when we map it or
transform it, or put that point into our vector-valued
function, let me copy and paste the original vector-valued
function, just so we have a good image of what we're
talking about this whole time. Let me put it right down there. So just to be clear what's
going on, when we took this little blue point right here,
this s and t, and we put the s and t values here, we get a
vector that points to that point on the surface,
right there. When you add a ds to your
s-values, you get a vector that points at that
yellow point right there. So going back to the results we
got in the last presentation, or the last video,
what is this? r of s plus delta s, or r of s
plus ds, the differential of s, t, well, that is
that, right there. That is the vector that
points to that position. This right here is a vector
that points to this blue position. So what is the difference
of those two vectors? And this is a bit of
basic vector math, you might remember. The difference of these two
vectors, head to tails. The difference of these
two vectors is going to be this vector. If you subtract this vector
that vector, you're going to get that vector, right there. A vector that looks
just like that. So that's what this is
equal to, that vector. And it makes sense. This blue vector plus the
orange vector, this vector, right here, plus the orange
vector, is equal to this vector. It makes complete sense. Heads to tails. So that's what that represents. Now let's do the same
thing in the t direction. I'm running out of color. I'll go back to the pink,
or maybe the magenta. So we had that s and t. Now if we go up a little bit,
in that direction, let's say that that is t, so this is the
point s, t plus a super small change in t, that's that
point right there. This distance right there is
dt, you can view it that way. If you put s and t plus dt into
our vector value function, what are you to get? You're going to get a vector
that maybe points to this point, right here. Maybe I'll draw it right here. Maybe it points to this
point, right here. A vector that points
right there. So that will be mapped to a
vector that points to that position, right over there. Now by the same argument that
we did on the s-side, this point, or the vector that
points to that, that is r of st plus dt. That is the exact same thing
as that right there, and of course, this, we already saw. This is the same thing
as that over there. So what is that vector
minus this blue vector? The magenta vector
minus the blue vector? Well, once again, this
is hopefully a bit of a review of adding vectors. It's going to be a vector
that looks like this. I'll do it in white. It's going to be a vector
that looks like that. And you can imagine, if you
take the blue vector plus the white vector, the blue vector
plus this white vector is going to equal this purple vector. So it makes sense, if the
purple vector minus the blue vector is going to be equal
to this white vector. So something interesting
is going on here. I have these two, this is a
vector that is kind of going along this parameterized
surface, as we changed our s by a super small amount. And then this is a vector that
is going along our surface if we change our t by a
super small amount. Now, you may or may not
remember this, and I've done several videos where
I showed this to you. But the magnitude, if I take 2
vectors, and I take their cross product, so if I take the cross
product of a and b, and I take the magnitude of the resulting
vector-- remember, when you take the cross product, you
get a third vector that is perpendicular to both of these. But if you were just to take
the magnitude of that vector, that is equal to the area
of a parallelogram, defined by a and b. What do I mean by that? Well, if that is vector a and
that is vector b, that's a and that is b, if you were to just
take the cross product of those two, you're going to
get a third vector. You're going to get a third
vector that's perpendicular to both of them, it kind
of popped out of the page. That would be a cross b. But the magnitude of this, so
if you just take a cross product, you're going
to get a vector. But then if you take the
magnitude of that vector, you're just saying, how
big is that vector, how long is that vector? That's going to be the
area of the parallelogram defined by a and b. And I've proved that in the
linear algebra videos, maybe I'll prove it again in this. I mean it's because-- well, I
won't go into that in detail. I've done it before, don't want
to make this video too long. So the parallelogram defined by
a and b, you just imagine a, and then take another kind of
parallel version of a, is right over there, and another
parallel version of b is right over there. So this is the parallelogram
defined by a and b. So, going back to our surface
example, if we were to take the cross product of this orange
vector and this white vector, I'm going to get the surface
area, I'm going to get the area of the parallelogram, defined
by these two vectors. So if I take the parallel
to that one, it will look something like this, and then a
parallel to the orange one it will look something like that. So if I take the cross product
of that and that, I'm going to get the area of that
parallelogram. Now you might say, this is a
surface, you're taking a straight-up parallelogram,
but remember, these are super small changes. So you can imagine, a surface
can be broken up into super small changes in
parallelograms, or infinitely many parallelograms. And the more parallelograms you
have, the better approximation of the surface you're
going to have. And this is no different than
when we first took integrals. We approximated the area
under the curve with a bunch of rectangles. The more rectangles
we had, the better. So let's call this little
change in our surface d sigma, for a little change, for a
little amount of our surface. And we could even say that, you
know, the surface area of the surface will be the infinite
sum of all of these infinitely small d sigmas. And there's actually a
little notation for that. So surface area is equal to, we
could integrate over the surface, and the notation
usually is a capital sigma for a surface as opposed to a
region or-- so you're integrating over the surface,
and you do a double integral, because you're going in
two directions, right? A surface is a kind of a
folded, two-dimensional structure. And you're going to take
the infinite sum of all of the d sigmas. So this would be
the surface area. So that's what a d sigma is. Now we just figured out, we
just said, well, that d sigma can be represented, that value,
that area, of that little part of the surface, of that
parallelogram, can be represented as a cross product
of those two vectors. So let me write here. And this is not
rigorous mathematics. The whole point here is to give
you the intuition of what a surface integral is all about. So we can write that d sigma is
equal to the cross product of the orange vector and
the white vector. The orange vector is
this, but we could also write it like this. This was the result
from the last video. I'll write it in orange. So the partial of r with
respect to, I'm running out of space, with respect to s, ds,
and it's going-- well, d sigma is going to be the magnitude
of the cross product, not just the cross product. The cross product by itself
will just give you a vector, and that's going to be useful
when we start doing vector-valued surface
integrals, but just think about it this way. So this orange vector is
the same thing is that. And we're going to take
the cross product of that with this white vector. This white vector is the same
thing as that, which we saw, which was the same
thing as this. The partial of r with
respect to t, dt. And we saw, if we take the
magnitude of that, that's going to be equal to our little small
change in area, or the area of this little
parallelogram over here Now, you may or may not
remember that if you take these, so let's just be clear. This and this, these
are vectors, right? When you take the partial
derivative of a vector-valued function, you're still
getting a vector. This ds, this is a number. That's a number and
that's a number. And you might remember, when
we, in the linear algebra or whenever you first saw taking
cross products, taking the cross product of some
scalar multiple. You can take the scalars out. So if we take this number and
that number, we essentially factor them out of
the cross product. This is going to be equal to
the magnitude of the cross product of the partial of r,
with respect to s, crossed with the partial of r with respect
to t, and then all of that times these two
guys, over here. Times ds and dt. So I wrote this here, hey,
maybe our surface area, if we were to take the sum of all of
these little d sigmas, but there's no obvious way
to evaluate that. But we know that all of the d
sigmas, they're the same thing as if you take all of the
ds's and all of the dt's. So you take all of the
ds's, all of the dt's. So this is a ds
times a dt, right? A ds times a dt, ds times
a dt is right there. If we multiply this times the
cross product of the partial derivative, this times this is
going to give us this area. So if we summed up all of this
times this, or this times this, if we summed them up over this
entire region, we will get all of the parallelograms
in this region. We will get the surface area. So we can write-- I know this
is all a little bit convoluted. And you need to kind of
ponder a little bit. Surface intervals, at least in
my head, are one of the hardest things to really visualize, but
it all hopefully makes sense. So we can say that this thing
right over here, the sum of all of the little parallelograms on
our surface, or the surface area, is going to be equal to,
instead of taking the sum over the surface, let's take the sum
of all the ds times dt's over this region right here. And of course, we're also
going to have to take this cross product here. And we know how to do that. That's a double integral. for going to take the double
integral over this, we could call it this region, or
this area, right here. That area is the same thing
as that whole area, right over there, of this thing. I'll just write it in yellow. Of the cross product of the
partial of r with respect to s, and the partial of
r with respect to t. ds and dt. And so you literally just take,
and it seems very convoluted in how you're going to actually
evaluate it, but we were able to express this thing called a
surface-- well, this is a very simple surface integral-- in
something that we can actually calculate. And in the next few videos, I'm
going to show you examples of actually calculating it. Now, this right here will only
give you the surface area. But what if, at every point
here-- so over here what we've done in both of these
expressions, is we're just figuring out the surface area
of each of these parallelograms and then adding them all up. That's what we're doing. But what if, associated with
each of those little parallelograms, we had some
value, where that value is defined by some third
function f of x, y, z? So every parallelogram, it's
super small, it's around a point, you can say it's maybe
the center of it, doesn't have to be the center. But maybe the center of it
is at some point in three-dimensional space, and if
you use some other function, f of x, y, and z, you'll get the
value of that point. And what we want to do is
figure out what happens if for every one of those
parallelograms, we were to multiply it times the value of
the function at that point? So we could write it this way. So this is where, you can
imagine, the function is just one. We're just multiplying each of
the parallelograms by one. But we could imagine we're
multiplying each of the little parallelograms by f of x, y,
and z, d sigma, and it's going to be the exact same thing,
where this is each of the little parallelograms, we're
just going to multiply it by f of x, y, and z there. So we're going to integrate
it over the area, over that region, of f of x, y, and z,
and then times the magnitude of the partial of r with respect
to f, crossed with the partial of r with respect to t, ds, dt. And of course, we're
integrating with respect to s and t. Hopefully we can express this
function in terms of s and t, and we should be able
to, because we have a parameterization there. Wherever we see an x
there, it's really x is a function of s and t. y is a function of s and t. z is a function of s and t. And this might look super
convoluted and hard. And the visualizations for
this, of why you'd want to do this, it has
applications in physics. It's a little hard
to visualize. It's easier just to visualize
the straight-up surface area. But we're going to see the next
few videos that it's a little hairy to calculate these
problems, but they're not too hard to do. That you just kind of
have to stick with them.