Mathematics - Multivariable Calculus - Lecture 22

Uploaded by UCBerkeley on 24.11.2009

OK, how are you guys doing?

Are you ready for Thanksgiving break?
All right.
Yeah, me too.
We still have one hour and a half to go.
One hour, twenty minutes to be precise.
The initial motivation for the material, which we are
discussing now was our desire to generalize Green's theorem,
so as to apply it to more general regions in the
three-dimensional space.
And last time we made the first steps towards that goal.
We introduced the curl of a vector field.
So now we are almost ready to state Stokes' theorem and it's
going to take the following shape.
On the right-hand side we're going to have something we've
learned before and understand fairly well by now, I hope.
And that's a line integral of a vector field over the
boundary of a surface in a three-dimensional space.
So we're talking about things like spheres or maybe a better
example, which we discussed before was a hemisphere, an
upper hemisphere like this.
So a sphere, as we discussed already many times is a
two-dimensional object.
I'm not talking about the interior of the sphere I'm
talking about just the surface of the sphere.
It is a surface, hence a two-dimensional object.
But it's a kind of two-dimensional object
which really lives in the three-dimensional space
and not on the plane.
Not on the two-dimensional space because it's curved.

So this surface has a boundary, so this surface m-- m will
stand for membrane, you'll see why.
We'll talk about it later.
No cellphones.
I'm actually impressed how rarely we hear these
sounds so keep it up.
This surface has a boundary.
And that's what I call here b of f.
So in this case, if you really think in terms of the upper
hemisphere it is a circle.
In any case it's a curve and over that curve we can take
this line integral where f is a vector field.
So that's the left-hand side.
And the left-hand side is kind of easy in the sense that we
already know what it means.
And the left-hand side is going to involve a double
integral over m itself.
And now we can say the integral of what?
It's going to be the integral of we learned
last time, the curl of f.
If you have a vector field f you can define a new vector
field, which we call curl of f.
So this has already been defined.
Now we have to learn how to integrate vector
fields over surfaces.
So that notation here represents a surface integral.
An integral of a vector field over the surface, also known as
a flux and today we will be define integrals of this type.
So that finally we'll be able to say what we mean by the
left-hand side and at least we'll be able to make the
statement of Stokes' theorem, which we will then
discuss next week.
So that's the plan.
We would like to learn surface integrals.

Now this is by the way about the final exam, which again,
will be on December 19 at Hearst Gym.
No more jokes about it.
And now I put the materials for the final on my homepage so you
can see the reveiw problems and various other things here, OK?
I will put the solutions of the review problems probably
sometime during next week.
So let me erase it now.
So what we'd like to talk about is surface integrals.

And as always when we talk about things like that
it's always good to think in terms of analogies.
And an analogy to a surface integral would be an integral
over a curve, over a one-dimensional object.
Something which we now understand fairly well.
So let's revisit first, line integrals.
Revisit integrals over curves.
One of the aspects of this story, something that I know
you have found puzzling at first is the fact that actually
the way we thought about it, the way we talked about it
there are two types of integrals over curves.
One type is the integral of a function and it's an
arc length type integral.
In this case, we have a curve c and we integrate a function.
And the expression for this is f ds.
So f here is a function on this curve and this ds represents
the length of a very small piece of that curve.
So what is the meaning of this integral?
The meaning of this integral is that if f is equal to 1 we
get arc length of the curve.
And if f is not equal to 1 in general, but it takes
nonnegative values we can interpret this function
as a density function.
For example, mass density function and we can think of a
curve as being a metal wire like this and that function
would then represent the density of mass.
Some parts of it could be heavier, some parts lighter.
And this integral would correspond to the total mass.
That's the interpretation of such an integral.
So that's the first type.
And the second type, which actually we have studied more
and which turn out to be more important for our purposes, in
particular for formulating things like Stokes' theorem or
Green's theorem, or fundamental theorem of line integrals, for
that matter, are integrals of vector fields.
Line integrals of vector fields.

So let me recall that.
So in this case we have a vector field f and we are
computing its line intergral, which we denote like this.
And that also had a nice physical interpretation.
It's not a mass, but rather it is work done by force.
If f represents a force vector field, then this represents
the work done by this force.
Examples of such vector fields are the force of gravity,
which we talked about before.
Also the electric field and the magnetic field.
And we talked about Maxwell's equations last time, you had
those two vector fields, e and b, right?
So these would be examples of such vector fields and such a
line integral for such a vector field would actually have a
very concrete physical interpretation as the
work done by this force.
For example, electric field, the work done by it in moving
an electron from one place to another along this curve c.
So those were the two types.
The reason I am reminding you of all this is because surface
integrals, which we are going to study today can also
be broken into these two different 2-D categories.
One is the analog of the first type here and the second one is
analog of the second type here.
In other words the first one, the first type of surface
integrals will apply to functions.
Will integrate functions and the second time will
apply to vector fields.
We'll be integrating vector fields.
And in both cases they will have nice interpretation very
similar to the interpretation we've learned for
line integrals.
Having said that, having recalled the integrals over
curves I want to now talk about surface integrals.

For surface integrals we also have two types.
Now I have to say from the beginning that this looks a
little bit disconcerting, the way we talk about this.
That there are two types and they kind of seem unrelated.
When we talked about these guys, the integrals over curves
I actually tried to explain that in a way these are two
special cases of kind of most general integrals over curves.
So from the point of view of the big picture these are
actually not so unrelated, they are actually quite
close to each other.
But for the purposes of this course, for the purposes of the
kind of results we want to obtain and the kind of
applications we have in mind, it's just easier to introduce
them separately and kind of think about them as
two different types.
One applies to functions, one applies to vector fields.
And the same actually will be true here for
surface integrals.
We'll kind of divide surface integrals into these two big
groups, but I just want to tell you that actually from the
point of view of the big picture this is unnatural
and there is actually a more natural theory.
There's a more natural point of view from which both of these
integrals appear as just special cases of some
general construction.
Nonetheless, we'll do just fine by kind of defining
them into two separate ways.
So for surface integrals we also have two types.
Or the way we'll introduce them here, two types.

So the first type will be an integral of a function, surface
integral of a function.
So we will have a surface m again, like in Stokes' theorem
on that board, but we will be integrating not a vector field,
but a function and it will be represented in this way.
So this is not yet the kind of integral which will show
up in Stokes' theorem.
This is sort of a preliminary integral which is like type one
integral in the case of curves.
And this is something which we will use to define surface
integrals for vector fields.
The reason why we define them first is because they are
easier to define than the other guys in some sense and also, we
will actually use this theory to define the integrals
of the second type.
So what does this integral represent?
Well just like in the case of curves it represents
something which is sort of inherent to this curve.
The quantity which is to the surface in this case, so here
there was a curve and inherent quantity for a curve
is its arc length.
Or if you have a general density function,
the total mass.
And likewise here, this will represent for f equal 1,
this will represented the surface area.
Surface area of m.
And for f not equal to 1, but nonnegative, or maybe positive
is better, this will correspond to total mass if we interpret
this function as a density function.

So let's talk about the special case when f is
actually equal to 1.
So in this case we will denote this integral just like
this and this will simply correspond to the area of m.
So let's discuss how to find the area of m.
This is not entirely new.
Is not an entirely new subject for us, because at the very
beginning of this course we actually talked about
revolution surfaces, and we talked about the areas
of revolution surfaces.
So we actually know one example, at least one example
of this kind of integral.
But now we will talk about areas of more general surfaces.
Not just the revolution surfaces.
Revolution surface, remember, was obtained when you rotated
the graph of a function in one variable around the x-axis,
or maybe around the y-axis.
But this is only a very special kind of surface.
It's like a vase, if you know what I mean.
Like a vase is a revolution , surface if it's a symmetrical
vase, like a round vase.
Is the revolution surface of a curve around this, well,
the z-axis, in this case.
If you think about vertical axis as the z-axis.
But not every surface is like this.
There are even vases, which are not exactly symmetrical.
For example, it could have squares.
So we would like to have a formula which would enable
us to calculate the area of any surface.
So what are the necessary steps for that?
Orr how should we approach this?
It's the same drill as before.
We have a piece of surface like this and we would
like to compute its area.
And as always, what we need to do is we need to break it into
small pieces and evaluate the area of each of the small
pieces and then take the sum.
And then the sum, when the partition becomes finer will
actually gives us an integral.
But how would we actually break it into small pieces?
Our priority is not clear.
So in the case of curves if you had a curve like this you also
want to break it into smaller pieces and take the small
lengths of each piece and take the sum.
And the way we learned to do this was through
This was in fact the most efficient way to talk about
curves in the first place because a general curve was
best represented algebraically if you expressed it
in a parametric form.
In other words, we would introduce an auxiliary
parameter t and identify an interval from-- so this is
parameter t-- from a to b.
We would identify this curve with this interval, right?
And algebraically we'd just write that x is a function of
t, y is a function of t and well in two dimensions that
would be enough, but in three dimensions we would
also have z equal 0.
And then we'll say t is between a and b.
So when we write these formulas what we're doing is we are
identifying points on this interval, on this straight line
interval, which is very simple with this complicated curve.

Once we do that there is a natural way to break into
pieces because we can just break this interval into
pieces of equal length.
Say 1,000 pieces, 100 pieces or in the beginning maybe
10 or 100, or 1,000.
For example, this one will correspond to this part here
and then we'll try to find the lengths of this one and then
the next one and so on.
And so the result of this calculation was an integral
of the first type.
Now instead of a curve we have a surface.
So we should apply the same approach.
We should try to parameterize this surface.
This is something which we have actually managed
to avoid up till now.
Because you know, this is something we talked about
before, but we never actually got around to try to write
surfaces in the same way or present surfaces in the same
way we've been representing curves and the reason
is very simple.
Because surface is two-dimensional, so this m,
let's call it m, its dimension is 2 and so it requires two
independent parameters.
Two independent parameters.

But only one equation.
Only one equation.
So whenever we could define a surface by equations, we said
it's better to define by equations than try to
find parameterization.
Because one is better than two.
One is easier to handle than two.
One equation is easier to handle than two parameters.
Notice for curves it's the opposite situation because the
curve is one-dimensional so it requires one auxiliary
parameter or two equations.
So that's why curves usually show up through parametric
On rare occasions we will talk about representing curves
by two equations or as intersection of two surfaces.
But now it's finally time to use the first approach and to
parameterize a surface by two parameters now.
So how is this going to work?
Well in exactly the same way as in the case of curves.
We will choose two auxiliary parameters, which oftentimes
we'll call the, u and v, but again, it's just
a choice of notation.
You can choose whatever letters you like. u and v.
And we will try to identify this piece of a curved surface
in the three-dimensional space like this.
With a flat domain in a two-dimensional space of
these two parameters.
What I mean is that somewhere in the background we will
have a two-dimensional coordinate system with
coordinates u and v.
Note that this is the analog of that line with the coordinate
t, which I drew on the top blackboard.
And the analog of that interval from a to b now will be a
domain, a two-dimensional domain on this auxiliary plane.
So it is going to look something like this.
Now of course, there's more choice for such a domain .
There are certainly more pictures we can draw like this
than pictures of intervals.
Interval is an interval.
I mean, the only things which we can play with is just the
endpoints, but they all look the same.
Here the domains look different.
They have different shapes.
Now you can of course ask, why would we bother to
identify this and this?
The reason is that this object is much simpler to
handle than this object.
You have to realize what this object represents.
This object represents something which actually does
not fit on the blackboard.
It can only live in three-dimensional space.
In other words, without bending and squashing it I wouldn't
be able to fit it in two-dimensional blackboard.
Here I have drawn it on the two-dimensional plane, but
that's just a projection.
So in fact, usually to emphasize the curviness
of this object I would kind of draw a grid.
This kind of curvy grid to kind of indicate
that it has this shape.
And this is in stark contrast to the structure
of this object.
So this guy is actually flat.
This guy actually fits in the two-dimensional plane and
that's why it's to our advantage to try to
identify this curvy object with this flat one.
So that's the first idea.
And once we do that we can then try to calculate
the surface area.
Now since I drew it like this it kind of looks strange that I
establish 1 to 1 correspondence between this one and this one.
So of course it would be more natural if in fact the flat
object, which was on the plane that corresponds to it would
be something more like a rectangle.
Then the picture would make a lot more sense.
Here I drew just a most general region that you
could have a priori.
But just to simplify our pictures let's assume that in
this particular instance we will have a rectangular domain.
This is not the most general one, but this for the sake of
the argument, just to do an example.
So if we do that for this guy there is a natural grid because
just like for the interval there is a natural grid because
we can just break it into equal pieces like this.
Once we do that, if we identify this and this we'll be able to
also give a grid to this picture.
You see what I mean?
Just as before.
So when I say identify it means that I would try to
map this flat rectangle to this curvy domain.
The curviest surface in the three-dimensional space.
So that for instance, this interval would go here and
this one would go here.
This one would go here and this one would go here.
But not only that, also each of these horizontal lines,
horizontal intervals would go to these curves and all the
vertical ones would go to those curves.
So if I have already a parameterization like this
then I would be able to break it into smaller pieces.
And then that would be the first step in getting
the integral.

So let's talk about examples of this kind of parameterization.

So this simplest one appears when our surface is a
graph of a function.

If the surface, m, is part of a graph of a function,
z equals f of x,y.
In this case we can just choose x and y as the
auxiliary parameters.
Or if you will, we can just say x is u and y
is v and z is f of u,v.
So here's is a parameterization.
I have represented each of the three variables as a function
of my new auxiliary variables, u and v.

So this is just like parameterizing curves, the
difference being that now the functions on the right-hand
side depend on two variables, u and v instead of just one
variabel, namely t in the previous case.
That's the first example.
The second example is example of a plane.
Because after all, plane is also a surface.
Plane is two-dimensional and in fact plane is the
simplest possible surface.
So in the case of planes we're used to representing planes.
So this is 2 and this is 1.

We are used to representing planes by equations.
Exactly what I just said now, which is that surfaces
oftentimes are easier to represent by a single equation.
But now we're talking about parameterization, which is
important for integrals.
So let's talk about parameterizing a surface, which
is very easy because think about-- in fact, let me use
my favorite surface.
So in the case of a surface, in the case of a curve if I had--
sorry, if I had a line, to know a line I need to know a point
and I need to know a direction vector.
For a plane, one vector is not enough.
Because one vector would just give me this line.
If I want to get the entire plane, I should also
pick a second vector.
Which is kind of transversal to this one.
So if I do that, if I pick two vectors-- let's call it r1 and
r2 -- then any other vector on this plane can be obtained as
a super position of some multiple of this one and
some multiple of this one.
Or as some combination of this one and this one.
I can use that to find a parametric representation
for this plane.
So let me draw it on this picture.
So this is my coordinate system.
This would be some vector r0 and then I would have two
vectors, r1 and r2 and so a general point on this plane
would correspond to some vector r of uv and it can be found by
using this one plus a super position of r1 and r2.
So it would be r0 plus ur1 plus vr2.
Then if I write this in coordinates because as
always r is just x,y,z.
Has three coordinates.
So likewise for r0, r1, and r2.
So say r1 is k1, l1, and m1 and same for r2 then
I can actually write each of them as a function of u and v.
So x would be x0, the first component of r0
plus uk1 plus vk2.
y would be y0 plus ul1 plus vl2 and z would be z0
plus um1 plus vm2.
Here these numbers, k1, l1, m1 and k2, l2, m2 are given and
the variables are u and v.
So now I have been able to express each point on this
plane as the function of two variables.
That's exactly what I'm talking about in general.
Except in general these functions will be
more complicated.
Here they're linear functions in u and v.
In other words, the degree is [UNINTELLIGIBLE]
of degree 1, or less than or equal to 1.
Something slightly more interesting corresponds
to a sphere.

For a sphere we could use spherical coordinates.

For a sphere we could actually apply spherical coordinates.
So we use as auxiliary variable, auxiliary variable
will be theta and phi and we would write is r sine phi
cosine theta. y is r sine phi sine theta and z
is r cosine phi.
Where r is a number, r is the radius of a sphere is fixed.
So you see the difference between that spherical
coordinates and this parameterization is whereas
for spherical coordinates we had rho here.
So there were three variables, rho, phi, and theta and those
variables gave us all points on the three-dimensional space.
And now we are talking about points just on the sphere.
So that actually leaves only two independent variables, phi
and theta and rho is fixed because rho is equal to a fixed
number for all points on the sphere.
You have a question.
The question is, what does this plane represent?
This picture means like that plane.
No that's a plane in a three-dimensional space,
but now I am relating it to a uv plane.
It's actually easier maybe to think about it in this way, so
here's a plane and this plane is in the three-dimensional
space, which os our classroom and now I would like to
parameterize all points on this plane by points of
another plane, which say will be this table.
And this table will have coordinates u and v and if I
have a point u,v here they will correspond to a
point here, namely.
That will be the point for which this vector-- that would
be this point-- for which this vector is ur1 plus vr2.
Meaning that this is ur1-- actually didn't draw it very
well-- I have to do in a parallel way like this.
It's moving that's why I'm not making a good picture.
So this is vr2.
Does that make sense?
So once I fix u and v on an auxiliary plane I
can get a point here.
Namely, I just draw the vector u times r1.
That's this one.
Then v times r2.
I think their sum, that's the point.
And this way I can get to all points of this plane,
of this tilted plane.
This pair of numbers, u and v provide me a unique address
system for all points on this green plane and that's what
we're doing in general.
It's just that iin general our surface is not
flat, it's curved.
But we are finding an addresses system for all points
of that surface.
Just the way I do here for a plane.
So for instance, in the case of a sphere, you know,
here's a sphere.
Let me again draw an upper part of the sphere because
it's easier to see, to analyze like this.
So the sphere of radius r, some number, let's
say r is equal to 5.
So spere of radius 5.
So this number is fixed, but now there are many points here
and I would like to address each point on this upper
hemisphere by some unique pair of numbers.
I call those numbers phi and theta and I denote them just by
using the same rule as I use for spherical coordinates.
So for example, if I want the entire sphere as would
correspond to the limits theta from 0 to 2 pi and phi from 0--
sorry phi from 0 to pi.
This is a whole sphere.

If on the other hand I want just the upper hemisphere this
would mean that theta is between 0 and 2 pi and phi
is between 0 and pi over 2.
I put the limit pi over 2 because I don't want to go
more than upper hemisphere.
So going beyond pi over 2 would correspond to
the lower hemisphere.
And then even if you think about this cone, if you just
want the part of the cone for example, this top part-- think
about an ice cream cone.
So that would be the ice cream, the part of the ice cream.
So that would correspond to theta between 0 and 2 pi and
phi between 0 and pi over 4.
So by choosing limits in the appropriate way I can describe
different regions on the sphere by using spherical coordinates.
And this is the third example.
Now once I have a parameterization I can get to
work and try to calculate the surface area.
And now I just have to look at this picture and what I need to
do is I need to find a way to express the area of this one
little curvy rectangle on the surface, which will correspond
to a particular rectangle in the parameterizing domain.
So here let's say would be delta u, the lengths would
be delta u and this would be delta v.
And I would like to find the area of this, which is delta s.
And this actually is something which we have already done.
We have done this when we talked about double integrals
in general coordinate systems.
When we talked about that we did exactly
the same calculation.
The only difference was that actually our surfaces at that
time were still part of the plane, but the calculation
is exactly the same.
If you remember we took the tangent vectors here, r sub u
and r sub v and we took the cross product, which
represented the area of the parallelogram.
So the answer is the same as before.
As in the old calculation for double integrals.

So you can review that if you like.
Just to save time I'll give you the answer and the answer is
that this is the absolute value of ru cross rv detla u delta v.
The area of this little piece.
So that's the formula.
What do I mean by ru and rv?
So r of uv is just a position vector of the point x,y,z.
So that's just x of uv i plus y of uv j plus z of uv k.
And if I want to take r sub u that just means taking the
derivative of x with repsect to u times i plus derivative of y
with resoect to u times j plus derivative of z with
respect to u times k.
So what I mean by x sub u is just dx/du but it's just
easier to write it.
It's faster to write it like this.
And likewise, for r sub v is x sub v i.
So you see now to find the area and of course, I would like to
emphasis this is not the exact answer.
This is an approximate answer just as before.
This is an approximate answer for this little surface area.
And what we need to do after this we have to take the sum
of all of this over all pieces of this partition.
And then take the limit when the partition becomes finer.
So then we get the surface area.
So this way we find that the area of this membrane, of
this two-dimensional surface is a double integral.
But now, it's a double integral over d you see.
This is d. d is a domain which is a flat domain.
The whole purpose of this discussion was to replace
a curvy surface by a flat surface.
And to replace a curvy integral, which we are trying
to define by a flat integral.
A flat integral we already know.
We studied months ago.
It has been awhile.
So it's something which we understand fairly well, so the
whole point of discussion is to define the area as an integral,
but over a flat domain, which we understand.
The question is, integral of what?
And now we know the integral of what, because we know
the elementary area.
And it's the same game as before.
Once you know the elementary object, you know the formula
for the elementary object, which in this case is the
area of this little piece.
And it's given in this way in terms of delta u and delta v.
Then you know that when you take the limit this will
get replaced by du dv, and this will be the function.
equal to this?
I would love to explain it, but I'm afraid that if I do that
then I won't have time to explain other things.
So I have to make a choice.
The reason why I'm not explaining this is because it's
verbatim the explanation for general coordinate systems.
Which actually I didn't do.
I didn't explain in great detail either,
I have to confess.
But I explained enough of it to convince you that
that's the right answer.
But think about the fact that, remember the fact that the
cross product of two vectors always represents the area of
the rectangle which is formed by them.
So what I don't here is that the calculation is done
in the following way.
I approximate this curvy rectangle by a flat rectangle
formed by the tangent vectors.
And then once I have the two vectors the area of the
rectangle spanned by them is just a cross product.
So the only question is, what are these two vectors?
And you know, when you do tangent vectors
you take derivatives.
So if you actually look at this question you will see right
away and actually you have all the information you need to be
able to find this because we talked about tangent vectors
two lines and these are actually lines.
Oh, sorry line, curves is a tangent vectors to curves and
you will see that these two tangent vectors are r sub u
delta u cross-- and the second one is r sub v delta v.
So actually the answer comes out to this.
Comes out to be this.
But then I just take the delta u and delta v, which are
numbers out of the cross product and that's how
I get this formula.
Maybe if you look at this formula it will
be more convincing.
So having said all of this is the formula I get. r
sub u cross r sub v dudv.
Which you can also write, is more appropriate
to write as dA.
You know, the little piece of area on this flat region.
So that's the formula for the area.
And now we can actually look back at all of those examples
that we have studied here and try to figure out what the
answer is going to be.
And the point is that-- well let's do it in this case.
In this case, r sub u, you have to take the derivative of
these three functions with respect to u.
So this derivative is 1, so it's i.
This derivative is 0 because derivative of v with
resepct to u is 0.
So I'll just skip it.
And then derivative of this one is f sub u times k and r sub v
is actually j plus f sub v k.
Now I can actually find this expression by simply taking
the cross product of these two vectors.
So what do I get?

So r sub u cross r sub v, just use the usual rule so you'll
have i, j, k and then you'll have 1, 0, f
sub u, 0, 1, f sub v.

And that's what?

That's negative f sub u i negative f sub v j plus k.
So it's norm is the square root of f sub u squared plus
f sub v squared plus 1.
That's the function which in this particular case I would
have to put here and integrate it over my flat domain,
parameterizing my surface.
And that would give me the surface area.
So it's very straightforward.
Likewise, you could calculate the answer for sphere.
For sphere you just have to do the same kind of calculation.
You have to write, find out what r sub phi is and what r
sub theta is, put this together, calculate
the cross product.
This has already been done, so I'll just give you the answer.
And in fact, the answer you can guess, because we already
know the Jacobian for the spherical coordinates.
And this is going to be exactly the same formula as the formula
for the Jacobian of spherical coordinates, except that now we
fixed the radius of the sphere.
The rho is fixed.
So the answer is rho squared times sine phi.
The answer was rho squared times sine phi, but now it's
R squared times sine phi.
Because it's fixed.
So again, if you're working with a sphere, this is a
function you'll have to insert in this integral to obtain
the spheres surface area.
So that's surface area.
This completes the case when your function is 1.
So this is by definition, and the definition-- usually we
write definition from left to right but now I write
definition from right to left.
So this is the definition for the integral over m.

This was our first task.
To calculate the integral, calculate the surface area as
an integral over a flat region, so this is the answer.
The next step is very easy.
The next step is we want to now calculate not just an integral
of ds, but we want to integrate an arbitrary density function.
For example, think about the mass.
If it's aluminum foil, which has you know, some parts of
heavier, some parts lighter so the mass is distributed in an
uneven way described by this function f, so then this
would be the mass.
How to calculate this integral?
Well, very easy.
You just insert this f in this integral and that's it.
And this completes the definition of the first
type of integral.
And you know, the reason why I'm kind of going so fast over
this is because I would like you to appreciate the fact that
this is really parallel to our derivation for example, of the
arc length of a general curve.
Or the formula for the mass of a general curve with a
given density function.
It is exactly the same derivation, what has changed is
that now the surfaces parameter is by two parameters,
by two variables.
Whereas, before there was only one.
But the idea of the calculation is exactly the same.
Any questions about this?
So that completes the double integrals of the first kind.
And now finally, we're getting to something more interesting,
which is the double integrals of the second kind.
Where we will not be integrating functions over
surfaces, we'll be integrating vector fields, which is what is
relevant to the Stokes' theorem in particular.

And now, before I give you a technical definition or I give
you the formal definition I would like to motivate it by
considering a real-life problem where this kind of
integral shows up.
And again, I want you to appreciate the parallel
with line integrals of vector fields.
For line integrals of vector fields we talked about
work done by force.
And then we tried to see what this means and how to
represent it by an integral.
That's how we came up with the notion of line integral
of a vector field.
So here we're going to do the same thing except now of course
it's not going to be work done by force, but what it's going
to be is it's going to be the flux of a vector field.
Or the rate of flow of a vector field.
OK, this is slightly trickier.
I have to say, just slightly trickier than the other one,
but very transparent nonetheless.
So here's the real-life situation we can imagine.
Imagine that there is a pipe and you have the flow of
water through the pipe or any other liquid.
I'm not going to make any suggestions.
And you would like to find the rate of the flow
through a given.
Think about a membrane.
Think of this invisible membrane and would like to know
how much water will flow through this membrane
per second.
That's what I mean by rate of flow.
So for that we have to know the density, which would amount
of water per second.
Or per hour, whatever, per unit of time.
And I'm assuming here that my membrane is strictly
perpendicular to the flow.
So it is perpendicular to this pipe.
I'm looking really at the flow through this section of the
pipe which is just obtained by cutting it a perpendicular way.
So what do I need to know?
I need to know the density function.
The density of water.
I need to know the velocity, This is mass density.
So this is mass per cube of a measure of length.
Let's say meters.
Then I have a velocity vector whose length is the lengths
of the velocity vector.

So this is the velocity vector.
And the lengths of the velocity vector is measured in what?
In distance, so I say meters per second.
So what do I need to do?
And I also have the delta s, the area of this section,
of this membrane.
This is the area, so this is meters squared.
So what I need to do is I need to just multiply this.
So the rate of flow is equal to rho times v, right?
Times delta s.
That's pretty clear.
Because over a given period of time it will move, say
the water that will move through this membrane
will correspond to this part of the pipe, which
we find by multiplying.
By taking the lengths of this velocity vector and multiplying
by time and multiplying by the area of this membrane.
So I think this is pretty clear.
So now I would like to make it more complicated.
First we have to make things more complicated then
they become easier.
That's always the idea, right?
And here's what I want to do.
I would like to now understand the flow, not through this one,
which is kind of perpendicular, but through another
one which it tilted.
So another membrane which is tilted like this.
So what has changed now?
Well the point is what has changed is that what will
contribute now to the flow through this membrane is not v.
It's the component of v, which is perpendicular
to this membrane.
So this little membrane will have a normal vector n.
Let's call it n.

And what will matter now is not v, but the
component of v along n.
Because you can decompose the velocity vector, so let me draw
this picture one more time.
This is my membrane, this is n and this is v.
So v has two components now.
One is the normal component, which is let's say this
one, and the other was a tangential component.
So we can think that the flow of the water is combined.
The flow of the water through this membrane is combined
of two different effects.
One is that some water flows in a perpendicular way and water
kind of flows in a super positional flow in the
perpendicular direction and you know, parallel to the membrane.
And of course, what is parallel to the membrane is irrelevant.
Nothing flows if the flow is just like this there will be
no flow through the membrane.
So the only thing that you have that is going to contribute is
just a normal component of this vector.
And what is the component?
So this is v, so let's remember again, those products.
So the normal component, the lengths of normal component of
v, is just v times the cosine the angle, which is formed
by these two vectors.
That's this normal component.
And that's just v dot n.
So my more general formal is going to be like this.
So the flow rate is going to be rho times
v dot n times delta s.
I hope I've convinced you that in this idealized situation,
where you have the uniform flow of this liquid, the velocity
vector is the same everywhere.
And the normal vector is the same everywhere that the rate
of flow is given by this formula.
And now I generalize.
But before I generalize there's a question.
Right, but n is a unit normal vector.
I'm sorry I should have said that.
The question is do I need to put-- here I assume that
the length of n is 1.
So it's a unit normal vector.
When I take the dot product, it's going to be -- oh,
you're also saying this.
Sorry, you're right.
Thank you for correcting me.
First of all I should have put the length of v.
The magnitude of v and second, I would normally have to also
put the magnitude of this vector.
But I assume it to be equal to 1.
It's the unit normal vector if you want.
So now we're ready to generalize it.
We generalize it by saying that it could be an arbitrary
membrane and also the velocity could be different in
different places.
The velocity will just be given by some vector field.
So what answer are we going to get now?
It's the same idea as in the case of our calculation
of the surface area.
We break a general membrane into small pieces and for
each small piece we apply the idealized scenario.
Where we assume that approximately the velocity
is the same everywhere.
And the normal vector is approximately the same.
So then we use this formula.
But after this we have to take the sum over all of those
pieces and when we take the sum we take the limit when the
partition becomes finer.
That's what gives us the integral of the second type.
And you can't read off the expression for this integral
from this formula.
As always, just look at the formula for the elementary
calculation, which we get get from the idealized
situation like this.
And just replace delta s by ds where now we already know what
ds is because we are able to calculate ds by using
By using this ru cross rv.
So a general membrane, the flow rate just looking at this
formula is-- now let me put rho outside because rho
is just a number.

Actually, it might as well depend, so let
me do something else.
Let me say that f is rho times v.
The mass density function could of course in general be
nonconstant, that's fine.
And v, the velocity of the vector field is not constant,
but let me just isolate this, the product of the two as a
particular vector field.
The formula will be nicer this way.
Well if I do that this can be written as f dot n
know times delta s.
So what I end up with is an integral over m of f dot n ds.
But now, what is n?
Now I would like to make sense of this integral by using
a parameterization.

This is already the definition because once I take the dot
product I start with a vector field, but my surface has its
own vector field attached to it.
So it's a kind of inherent vector field attached
to the surface.
Namely, the unit normal vector at each point.
And I take the dot product.
And I get a function.
And functions I have already learned how to integrate.
Since I've learned how to [UNINTELLIGIBLE]
functions this actually already makes sense.
But let me spell it out in more detail by using
So suppose now that my surface is actually parameterized
in this way.
Then I can actually find a normal vector and the normal
vector is going to be precisely this ru cross rv, which
we have used here.
As a vector it is normal to the surface at this point.
It is normal because remember as I said, in this calculation
you have to take a cross product of two tangent vectors.
When you take the cross product of two tangent vectors you get
something which is perpendicular to them and
therefore perpendicular to the surface.
This is a normal vector.
This guy is a normal vector.
But in a previous calculation we just took its norm, and we
saw that the norm is this.
So this gives me a way to actually obtain a very
explicit formula because this is a normal vector.
But unfortunately, it's not normalized, which
is a horrible pun.
So it's normal in the sense that it is perpendicular.
But it's not normalized in the sense that its norm
is not equal to 1.
But now I will normalize it, so we'll get a normalized
normal vector.
I think it's because our language, our everyday language
is too poor to express some mathematical concepts.
That's why it sounds really bad, so let me just try the
formula instead without saying anything more.
The formula is that n is ru cross rv divided
by its magnitude.
If you have a vector and you divide it by its norm or
magnitude you obtain a vector, which was norm
1 as always the case.
So this is what I should put here instead of n.
And also instead of ds I have to put my old formula, it has
already been erased, but I want to remind you maybe here that
the integral over m of f ds was equal to the integral over d
where d is this region of r sub u cross r sub v dA.

In other words, and I forgot the function f.
In other words, ds becomes dA times this guy.
So what's going to happen now and we'll have a very nice
consolation, we will have double integral over d
of f dot ru cross rv.
Then we'l have to dvide by this.
ru cross rv and then we'll have to multiply by ru cross rv dA.
You see?
So it's kind of lucky because the same quantity appears in
the numerator and denominator.
And it's nonzero.
This is actually something which we have to take care of,
because it should be nonzero because otherwise it would
not be a parameterization.
Think about this as like the Jacobian.
It is a Jacobian, if you think about it.
And in fact not surprisingly the formulas we get for
it, say for spherical coordinates is a Jacobian.
And the Jacobian should be nonzero.
For otherwise, this is not going to be a good
So that's why actually it's nonzero, and we have the
right to cancel them out.
When we cancel them out we finally get a
very nice formula.
Namely we get this.
Finally, we have come up with the definition of the integral
of a vector field across a membrane, across a
two-dimensional surface.
If this surface is parameterized by a
domain, D, this integral is simply given by this formula.
By this double integral over D.
What it is, on the other hand, what it is in a more canonical
way, is an integral over m with respect to the measure of the
surface area of this function f dot n.
But since we don't want to write it each time we will
actually write it-- we'll use a new notation for it.
We will call it f dot ds where over ds we put an arrow.
In other words, we say that ds, as of vector by
definition is n times ds.
If we do that then we can combine these two guys into
single symbol ds with a vector over it.
And what we're doing is we're taking the dot
product of f with ds.
Let me summarize.
We're trying to understand what to make of an integral of an
vector field across a membrane.
And a good analogy is a line integral of a vector field.
In the case of a line integral of a vector field we had a
different interpretation because in the case of a line
integral of a vector field it was all about pushing
something along the curve.
So what mattered was a tangential component of the
vector field tangential to the curve.
What matters now is the perpendicular component because
we are talking about the flow.
The flow through, not along the surface.
There's no such thing as a flow along the surface.
It could be something along the curve.
You cannot have something along the surface.
It has to be across the surface.
If it's across the surface what matters is the
perpendicular component.
The normal component, which is what we get by taking the dot
product of our vector field with n.
So we're trying to find this and this is called the surface
integral of a vector field or a flux of a vector field through
or across the surface.
And this is very important.
It's not along, it's across.
This is an important thing.
An important distinction between curves and surfaces.
Of course, you field along the curve.
For surfacees, this is the membrane, you go
across the membrane.
That's why it's a dot product with a normal.
So this integral is given by this formula. f is a vector
field and the meaning of this integral is given
by this formula.
Once you parameterize your membrane m, by domain d, like
here you can actually evaluate this integral in a very
straightforward way by simply taking this cross product
and taking the dot with f.
For example, let's suppose that we are in this situation.
We are in this situation where our ru cross rv is
given by this formula.
This I recall is the situation where m is part of a
graph of a function f.
And your vector field f, say is some pi plus qj plus rk.
In this case we can use this formula.
So what we have to do is we simply have to take the dot
product of this vector and this vector and so what do we get?
That's going to be integral over d of minus p f sub u minus
q f sub v plus r times dA.

Very explicit formula.
If you have to do a problem like this you're given
everything you need.
You're given the vector field, p,q,r.
You're given your parameterization in
this particular case.
Although, I have to warn you that this a very special case
when the parameterization is like this.
This is the case when your surface is part of a graph.
So you're given function f.
All you need to do is to find the partial derivatives of f
with respect to the two variables.
And after this, you assemble all of this data into this
formula and you get a double integral and double integrals
you know how to do.
You can use any method you like.
In fact, I wanted to do an example, but I'm-- OK.
Let's do a very quick example just to see how this works.
And then I actually have something to show you.
Something fun.
Not that this has not been fun.
I have to say, the way I present this material is not
the most economical and I feel a little awkward because this
is not the best possible way to present it.
It looks a little bit ad hoc again, because first you
introduced the surface area.
That was fine.
Surface area was fine and the mass is fine, but defining this
flux in terms of the surface area integral of the first
type that's not the most optimal way.
There is a much nicer way to explain this and I will explain
it if not next week then during the review lecture for sure.
I will give you a much nicer perspective on this type of
integral and then you will see right away where
these formulas come from.
It will not look anymore so ad hoc.
But for now this will suffice.
This will give you a reasonable definition
which you can work with.
And you can calculate fluxes and these surface
integrals very easily.
Like in this case.
Here's one example.
Find the flux of the vector field f which is minus xi
minus yj plus z squared k.
And s is part of the cone z equals square root of x
squared plus y squared.
Between the planes z equal 1 and z equal 2.

So let's do this calculation.
First of all, in this formula above let's just say
u is x and v is y.
I mean, after all in this particular case we don't have
to pretend that this is some new variables, some
auxiliary variables.
These are our x and y.
So in this case, the function f is this.
This is f of x,y.
So I'm just substituting in that formula.
So what I get is that f dot ds over this membrane--and by the
way, it's very easy to draw it.
This is a cone.
We are talking about the cone like this and you use the
parts which is confined between two planes.
So it's just one.

I mean, just the surface of the cone, but between
this plan and this plane.
It's z equal 1 and z equal 2.

Think about ice cream cones.
So this is going to be the integral over-- you will have
to the parameterization and the natural parameterization would
be by its projection onto the xy-plane.
And under this projection this will go to an annulus,
somebody said annulus, right?
Very good.
But you don't have to say it quite so loudly.
Anyway, so you get this annulus and then you have to
integrate over this annulus.
And annulus, of course, has a very nice parameterization.
Which is that you have r and theta, the polar coordinates,
r between 1 and 2 and theta is between 0 and 2 pi.

So you can always convert this into iterated
intergral very easily.
But now integral of what?
We have to take the dot product of their so what is the
p f sub x, so what is p?
P is negative x.

Some reason-- right.
So when' you do that what you're going to end up
with is negative x.
So you first of all have, overall it's a negative sign.
Then you have negative x.
That's this negative x.
Then you have to take the derivative of this.
And derivative of this you know as well as I do it's x
divided by square root of x squared plus y squared.
Then the next one is negative y, well, first of all it's
minus, overall minus because there's an overall minus here.
Then you have to take this negative y and the derivative,
which is y divided by square root of x squared
plus y squared.
And finally, you have to take this times just
take this d squared.

Let me make it a little bit easier, so you have the last
one is d squared divided by square root of x squared
plus y squared.
Actually, no, it's OK.
Let's just do it this way.
So you get plus d squared dA.
And after this you have to substitute the
square of this in here.
So that would be just x squared plus y squared.
So here you've got a very simple formula, which you can
now a very simple integral that you can of course take the
sum and then you cancel out.
You get square root of x squared plus y squared
in the numerator.
Here you get just x squared plus y squared, so you just
calculate it by using the following coordinates
very easily.
So this is how you do it.
Now there is one issue which is left unanswered here and that
is the issue of the choice of this vector n and let me
illustrate it in this example.
In this example we are talking about this surface.
This part, but now the point is which n am I talking about?
Am I talking about this n, which goes downward or am I
talking about this n, which goes upward.
So there is this one issue which we have
not yet talked about.
And to make everything precise we have to-- in other words, to
make the definition work we have to say which normal
vector we choose.
And the proper terminology for this is orientation.
This unit normal vector to our surface, or unit normal
vector field to our surface, is called orientation.
It tells us sort of which side is the front and
which side is the back.
And as always, there are two different choices.
It's kind of parallel to the choice orientation of a curve.
But again, for a curve orientaton goes
along the curve.
And for surface it goes across, it's perpendicular
to the surface.
So there are two different orientations in this case.
And we have to say which one.
So always, on the problems which you will solve on this
homework, it will have to say which orientation
you have to choose.
When it says downward it means that the z component of
the vector is negative.
When it says upward it means that the z component of
the vector is positive.
In this particular case we use the normal vector for which
the z component is 1.
So that means that this formula refers to the
upward orientation.
And then of course the really fun part, which I unfortunately
don't have time to talk about is about surfaces which
have no orientation.
And this is something really cool, which we'll have
to talk about next week.
But for now, happy Thanksgiving and see you next week.