Mathematics - Multivariable Calculus - Lecture 25


Uploaded by UCBerkeley on 08.12.2009

Transcript:
Hey guys, how are you doing?

Are you ready for your exams?
No?
Yeah, it was much nicer at the end of August, you
know, the semester was
just beginning, and the weather was so nice.
Now, it's winter time in Berkeley, which means
like 60 degrees, 55.

So, today is the review lecture.
OK.
In the last few lectures, I gave you an overview from a
kind of theoretical point of view of what this
course is about.
And, in particular, this last chapter that we have studied.
And today, I would like to focus more on the
practical aspects of it.
And, basically on the question of what do you need to know to
do well on the final exam, OK.
So, I'll go over that most important points of the
material which we'll focus on in the final exam.
Which, again, will be on December 19.
And, I'm not going to go over all the information again,
everything is available online, here, and also on B space.
I will also have last minute office hours on the day
before, on the 18th.
In case, you know, Sunday don't see the difference between curl
and divergence, and you panic the night before, you can come
to see me on the 18th and we'll sort it out, I'll
sort it out for you.
OK.
Also, I want to tell you one more time, that the exam
will be at Hearst Gym in three different rooms.
OK.
And, the room is determined by your section.
It's determined by who your GSI is.
And, your GSIs have already announced the rooms for you,
but you can find all this information again
here and here.
OK?
But you should be aware of where your exam room is.
It's important that you are at the right place.
You don't end up in -- you know, it's Hearst Gym, so it's
big, and it's easy to get lost.
So be sure you know which room you should show up at.
OK.
Any questions about the organizational
aspect of the final?
Okay, good.
So, now let's talk about the material.
I already said that this exam will focus on the material
after the second midterm, what we call the third season.
And, but it doesn't mean that you should forget what happened
in the first two seasons, meaning that certainly
problems, solving all the problems, will require skills
which you learned before.
You should not forget all this stuff, right.
Especially multiple integrals are very important part of
this, of this material.
So, I think it's a good idea to go over that old material.
Perhaps not devote as much time as the material of the last
chapter, which studied since the last exam.
But, you should probably review it nonetheless.
Any questions from this part of the audience?
So, I would like to break it into the material.
So, I will only talk about the material since the last exam.
But again, don't take it as meaning that you should
not know anything about what happened before.
You should, but the focus will be on this material.
OK.
That's why I'm going over this material now.
Alright, so the first thing I would like to talk
about is integrals.
In this part of the course, we have talked about different
kinds of integrals.
And, I would like to summarize again what those integrals are,
and what kind of information you need to be able
to compute them.
So, the first thing I want to focus is just the integrals.
What kind of integrals.
Well, we integrate over curves, and we integrate over surfaces.
So, there are two types of integrals.
First of all, depending on the dimension of the domain.
Let me talk first about integrals over curves.
So here, we have two type of integrals also.
Type I integral is an integral of a function, so this an
integral that looks like this fds, where f is a
function over curve.
So, a curve here could be over -- could be a curve on the
plane that is R2 or in space.
On the plane or in space.
But, the way we handle this integral is very similar.
Now, how do we compute such an integral.
To compute such an integral, the first step is to
perameterize your curve.
Perameterize the curve.
So, which by the way, was the very first subject that we
learned in this course.
So, this is a good example of what I mean when I say, you
know, revisit, review the material from the
previous chapters.
Because, certainly to be able to do such an integral, you
have to know how to parameterize curves.
And, that's something we learned at the very beginning.
OK.
So, parameterizing the curve means introducing
an auxiliary parameter, which we usually call
t, but you may call it something else if you like.
In writing each of the coordinates as functions of
this auxiliary parameter, so, you would have, x is a function
of this paremeter t, y is a function of this parameter
t, and z is a function of this parameter t.
That would be in the case when the curve is in space.
If it's on the plane, you would only have x and y of course.
Right.
And once you do that, this is integral, and then you have to
say, that t, what is the range for this auxiliary parameter t,
and the range will be between some numbers a and b.
So, then this integral will be equal to the integral of f,
where instead of x, y and z, you insert those three
functions: x of t, y of t, and z of t.
Right?
And, you integrate from a to b.
But, in addition, you also have to insert a factor
which has to do with the arc length of the curve.
And that would be the following: x prime of t
squared, plus y prime of t squared, plus z prime of t
squared, and finally dt.
And again, if you are on the plane, you just don't have the
z variable, you just erase this, and you get the
formula for the plane.
Now, what does this represent.
This represents a mass of say, a thin wire in the shape of
this curve, if is a density function, is mass
density function.
Also, remember that if f is equal to 1.
If f is equals to 1 in other words, you just
have this square root.
This integral represents simply the arc length.

So, this integral measures the total mass or total length of
this object of this curve.
That's what this integral is about.
It's very easy to distinguish this integral from the next
type of integral, because you see that in this integral,
you integrate functions.
Whereas, the second type of integral, you integrate
vector fields.
So, the second type you have an integral with vector field,
rather than a function.
F is a vector field.
So, as such it has components: i, in front of i you have p, in
front of j you have q, then you have a third component r.

If this is a vector field in space.
If it's on the plane, again, you only have two components.
OK.
So, vector field is more than a function.
It has three components in space, two components
on the plane.
And, this integral can also be written as Pdx
plus Qdy plus Rdz.
This is the same, because the R here stands for dz equals
dxi plus dyj plus dzk.

And, so if you take the dot product of this two, you
end up with this form.
So these are two different ways to write a line
integral of a vector field.
Either this way or that way.
It's the same.
Okay, in the case of a line integral of a vector field
there is an additional complication, which we don't
have for integrals of this type.
Integral of this type, it's enough to just write this.
You have to specify the curve, you have to specify the
function, and that's it.
But, in the case of a line integral of a vector field,
you have to specify additional piece of data.
So, which I kind of, I wanted to emphazie.
Here you need orientation.

Of course, when I say you need to specify, what I really mean
is that I need to specify.
In other words, when I, if I give you problem, compute this
line integral, I have to tell you with respect to
which orientation.
OK.
If I don't tell you, it means that the problem
is not well posed.
Unless, you know, there's something said, which
should enable you to find the orientation.
In other words, this is not well defined, if you just have
the vector field and the curve.
You have to also have orientation on this curve.
What do I mean by orientation.
A curve is going to be, to look like this.
So it goes from some point a to point b.
And, orientation means the direction.
Which way do with traverse this curse.
So, there are two possibly orientations.
Here is one orientation, and here's another orientation.
You can go from a to b, you
can go from b to a.
And, of course, the answer, if you choose the first
orientation, is going to be just the negative of the answer
if you use, choose the second orientation.
But, it's a different answer, unless the integral is just 0,
you will get a different answer depending on which
orientation you choose.
So, orientation has to be specified, and you have to
follow that orientation.
You have to be careful when you set up such an integral, to
make sure that you are computing it in the
right direction.
You are doing the integral, oriented in the right way.
Of course, here I have drawn a curve, which has two endpoints.
There are also curves which do not have endpoints,
closed curves.
Here's an example of a closed curve.
In the case of a closed curve -- it also has two orientation,
which we usually call clockwise and counterclockwise, right.
So, counterclockwise is this one.
There is also a somewhat misleading terminology in
the book, which is called positive orientation.
Which I will prefer not to use because it's a matter
of convention, right.
So, let's just call it counterclockwise
or a clockwise.
In the book, counterclockwise is also called positive.
But, I think there's potential for misunderstanding if
you use that terminology.
So, let's just stick to clockwise, counterclockwise.
I think it's fairly, it's very self-explanatory.

OK.
Now, let's suppose you have this.
You have chosen such an orientation.
What will be, what will, how to compute this integral.
So, you use again parameterization like this, but
now it's important to make sure that your orientation, that
you've chosen, agrees with the orientation on the
auxiliary of parameter t.
This t is actually, just belongs to the one-dimensional
space to the line.
And, the line is oriented, because points on the line
correspond to numbers, and numbers are oriented, we can
say which numbers is bigger between any two numbers.
So, when we write it like this, it means that we identified
the curve with the interval from a to b.
And, the interval from a to b is always oriented from
a to b, like this, right.
So, let's suppose that under this parameterization, you
choose orientation like this.
So, it goes from this point to this point.
And, this point is t equals a, and this point is t equals b.
OK.
Let's suppose that.
In this case, this integral is computed as an integral from
a to b of P x prime, t dt.
Well, let's just write like this.
Plus Q y prime plus R z prime dt.

Here is a possible trick question.

Let's suppose I will ask you on the exam to compute such a line
integral, where the curve is oriented in a different way.
So, if you don't pay attention, you would say, okay.
You wound find your parameterization, where
again, let's say t equals a corresponds to this and t
equals b corresponds to this.
Let's suppose.
And you, look if you don't pay attention to this stuff, to the
orientation, you might want to write it again like this.
That would be wrong, you see.
Because, you have to make sure that with respect to that
orientation, which you are told to use, the parameterization is
such that, you know, the lowest one is the initial point, the
lowest value a is initial point and the largest value b
corresponds to the endpoint.
In this picture, that's the case.
And, that's why we end up with this integral from a to b.
In other words, the integral that I got with respect to
t has this orientation.
Which comes from the orientation of the curve.
But, if the orientation is like this, you have to go from t
equal b to t equal a, right.
Because, you have to go from this point to this point.
This point corresponds to t equals b, this point
corresponds to t equals a.
That's why, so in this case, then you should write it as
from b to a, and then the same thing.
Then of course what is integral from b to a, integral from b to
a, is the same as negative integral from a to b.
So, you will actually end up with minus this integral.
OK.
Is that clear?
Ask me if it's not clear.
This is a saddle point, which you have to remember.
Now, what does this integral represent?
Interpretation of this integral is that it represents work done
by force, this is work done by force F along c from the
point a to the point b.
You see, from this, if you think about this
interpretation, it becomes more clear why there's a sign,
because when you talk about the work done by force, the force
is moving an object or it may be resisting movement
of the object.
And, so it's important whether you go from a
to b or from b to a.
If the force, let's say the force comes from the wind, and
the wind is blowing this way, and so the wind is carrying
me this way, so the wind does perform a certain
amount of work, right.
And this work is positive.
But what if the wind is blowing this way, but I'm still,
I still insist on going there, okay.
Because I'm stubborn, so I go against the wind.
So, in that case, I do work not the wind, and so the wind is
resisting me, therefore the work of the wind is
negative, you see.
So, it's important to know not just the trajectory of the
object, but it's also important, you know, whether
the relative position of the force and the trajectory.
In other words, the orientation of the trajectory relative
to the direction of the vector field, you see.
So, it becomes more clear that if I go this way, if I go this
way against the wind, it's a negative work.
If I go this way with the wind, it would be positive, OK.
You have a question.
[INAUDIBLE].
That's right, that's right.
When I write it like this, so the question is about this PQR.
In this formula I spelled it out, I wrote f is a
function of x, y, z.
But now, x, y, z have become functions of t.
So, I substituted these three functions into f.
Likewise in this formula, I didn't write just this time,
but what I mean by this is, and the same, similar for Q and R.
OK.

So, the orientation is important, change of
orientation results in an appearance of a sign.
So, that's, that's pretty much all about the setup
of integrals for curves.
So, next we move on to integrals over surfaces.
Integrals over surfaces.

So, what do we need to know here.
Here again there are two types of integrals that
we are interested in.
And, there is the a lot of similarity between the first
type for surfaces and the first type for curves, and the second
type for surfaces and the second type for curves.
So the first, integrals of the first type over surfaces are
integrals are integrals of functions.
So, here again f is a function.
In this case, you have a surface m and it
is embedded in R3.
You could also have a surface embedded in R2, that would
be just a region on the plane, like this, right.
But that's, so that's integrating over such regions
with the subject of the earlier chapter of course, chapter 15.
They said that's the usual doubling integrals.
So, here the novelty is when you look at surfaces in R3,
which we dno not see it in R2, which cannot fit in a plane.
But only fit in a three-dimensional space, like
a sphere or part of a sphere.
So, what I'm talking about is something this, like
a cap, a cap like this.
That will be m.
So, what does, how do I compute this.

Again, to compute, I have to parameterize my surface.
Parameterize.
Because it is now two-dimensional, I have to use
two auxiliary parameters, which we usually call u and v.
And, so we write, x is some function of u and v, y is some
function of u and v and z is some function of u and v; u and
v run over points on some domain already on a plane.
So, what I'm doing is I'm trying to identify a curve
surface like this, like this m with a flat surface like
this, which I call d.
So each point here will correspond to a point here.
And, it's a one-to-one correspondence.
That's what this parameterization means.
Once you have this parameterization, you have a
formula for this integral.
To write this formula, I will use the following notation.
I will call R of u,v the vector field, which is obtained from
this parameterization by interpreting each of the three
functions x, y, and z as components of a three vector.
So, you'll have x of u,v i plus y of u,v j plus z of u,v k.

This is vector R.
And so I can differentiate this vector.
For example, R sub u will simply mean x sub u times
i plus y sub u times j plus z sub u times k.
So, again, this stands for partial derivative of
x with respect to u.
This is a partial derivative, partial derivative
with respect to u.
So, this is another example of something we use in
this part of the course.
We're building on some material that we've learned before.
Because, of course we learned partial derivatives
earlier, right.
Before the second exam.
So again, you have to know, for example, how to compute
partial derivatives.
Without knowing how to compute partial derivatives, you will
not be able to compute such intervals.
Okay, so once we have this, we also have R of v,
define in a similar way.
You just put derivatives with respect to v instead of u.
Then the formula is the following.
I actually made, I only drew one sign of integration, I
should have drawn two signs of integration.
For double integrals, we draw two signs.
Well, it's not a big deal.
I will not deduct points for this, since I've
made the same mistake.
I'll go easy on you if you put one integration sign instead
of two for double integrals.
But, for triple integrals, you should put more than one.
For double integrals, I will be lenient on double integrals
in terms of how many integral signs you put.
OK.
Let me put both two signs here though.
So, integral over this m, you'll be able to write as
a double integral over d.
Where here, you'll put f where again you put x of
uv, y of uv, and z or uv.
Then you will put R sub u cross R sub v.
This is a cross product, again something we learned before.
And, then you take the magnitude of this vector, dA.
dA is the usual measure of integration on uv plane.
So this term, ru cross rv is just like this term, which
you have four integrals of functions over curves.
The first type of it.
It's an analog, it's an analog of that.
That ru cross rv absolute value is an analog of this.
OK.
So, now the question really becomes how do
you parameterize it.
Once you parameterize it, once you parameterize it, it's very
easy, because you just need to compute derivatives, and you
take the cross-product, and you end up with just an
ordinary double integral.
But the question is how do you compute this.
So, well there are at least two special cases,
which you should know.
The first special case is when your surface
is part of a graph.
It's part of a graph of a function.
In this case, that formula simplifies it.
And in fact, it's a good idea to remember, to
remember this formula.
Or, you will actually have a cheat sheet, so you can just
write it on the cheat sheet.
You don't have to remember.
Okay, so special cases.
The first special case is when the surface, M, is
part of the graph of some function g of x, y.
So, in this case, what we can do is say that x is u, y
is v and z is g of u,v.
Or we can stop pretending that u and v are two additional
auxiliary variables, and just call them what they are.
I mean, let's just call this auxiliary variables x and y, so
then parameterization will be in terms of x and y, and it
will have x is x, y is y and z is g of x,y.

So, in this case actually, you can compute this cross-product.
This has been done.
I did it in class, I wrote this formula in class,
the computation explained in the book.
You must have done it many times from homework.
So, you might as well just remember this formula, and the
formula is like this it's minus dg over dx minus dg over dy, so
this i and this is k if I remember it correctly.
Please correct me if I'm wrong.
I think it's correct.
This formula you just get by applying, by applying
that general formula in this particular case.
It's just that when you compute the cross-product so you have
to compute this, you know, the determinant of a 3 by 3 matrix.
It's just that the matrix becomes very simple, simplifies
it so it's easy to compute.
I'm sorry?
[INAUDIBLE]
I'm sorry.
That's right.
I did not close the parenthesis, for
which I apologize.
Ok.
Please tell me because we want our world-wide audience to see
only the correct formulas.

And now, if you have this, you can compute it's
magnitude very easily.
And, so it's just going to be the sum, square root of the sum
of squares of this components.
So, it's like this.
OK.
So, we actually have not just the magnitude, but we actually
have a formula for the cross-product, which I
will use in a minute.
OK.
That is the first special case.
There is a second special case, when your surface
is part of a sphere.
The second special case is m is part of the sphere, of
a sphere of some radius.
Let's call this radius R.
So, this radius R is just a number.
So, you got a sphere, which is of this radius, and your m
could be just the entire sphere .
It can be the entire sphere.
Or, it could be a hemisphere, or it could be part of a
sphere, which is within, confined in a cone, like an
ice cream cone, right, but that's just the surface
of the ice cream.
So, in each of these cases, a good way to parameterize
such a surface is to use spherical coordinates.
Spherical coordinates we learned before.
Spherical coordinates give us an alternative coordinate
system for the entire three-dimensional space.
And, you got three coordinates, we got rho, phi, and theta.
But now, we are on a sphere.
We are on sphere of radius R.
So, from the point of view of spherical coordinates, this
coordinate, rho, is fixed.
Rho, the coordinate of spherical [UNINTELLIGIBLE],
is equal to R.
But, we still have two remaining spherical
coordinates, which we can use to parameterize our surface.

Which is good, because surface is two-dimensional, so it
requires two independent coordinates.
And phi and theta give us these two coordinates.
So, in other words, u and v here are phi and theta.
For example, if it's the entire sphere, you will have theta
going from 0 to 2 pi, and phi going from 0 to pi, which is
the full range of fi and theta.
If it's the upper hemisphere, again theta will go from 0 to
2 pi, and phi will go from 0 to pi over 2.
And here, theta is from 0 to 2 pi, and phi, well it depends
how much ice cream you get.
So, it's in your interest to have it as large
as possible, I guess.
Although, then it becomes more difficult to hold it.
So, let's just settle on the middle ground, pi over 4.
OK.
So,that's how you, this is an example of parameterization.
And now, in this case, there is a formula for, let me erase
this, for R, u cross Rv.

But, now we don't have u and v, we have phi and theta.
And, so R phi cross R theta, in this case, and I'll just write
the formula for the magnitude.
It's actually what you would expect it to be.
It's exactly the formula for the Jacobian, except that you
have to set rho equals R.
R squared times sin phi
[INAUDIBLE]
Right, so the question is about the cone.
How do I measure phi.
Phi is measured from the z axis, right.
So, 0 to pi over 4 would be, this kind of range.
Yep, it works.
Although, I have to say in this picture it looks a little
bit less than pi over 4.
I don't want to short change you with ice cream.
But, this is just an approximate picture.
Not the best day to eat ice cream at Berkeley though,
perhaps earlier in the semester will be more appropriate.
So, that's a formula for this.
And, let's go back to this double integral.
So, that's all you need.
You need this guy, you need your function, where you stick
those expressions, coming from your parameterization, and
then you end up with a simple double interval.
So you compute it.
What is the interpretation of this.
What is the meaning of this.
The meaning, just like in a case of Type I intervals for
curves is that it represents something that has to do
with the area or mass.
Something when you average density over your surface.
So the typical example is that this represents mass of say
aluminum foil in the shape of M, where f, this function
f, which appears in that integral, f is the mass
density function.
Now, it doesn't have to be mass, it could be any quantity
which you can measure, which can have a density.
For instance, charge.
This foil could be charged.
But, in a complicated way, so that some parts are
more charged than others.
So, you have to integrate, you have to average them out.
And, so you can have a, f may be the charge density function,
not necessarily mass density function.
So, tehn you will be calculating total charge.
By the way, if it is mass, you could also talk about the
momenta of the mass, which means that you would be
multiplying under this integral f by x or y or z, right.
So, you can also find the center of mass for this
membrane, for this surface.
So, if somebody, if a question is phrased in the following
way, find the charge of an aluminum foil in the shape of
such-and-such surface, if charged density function is
such-and-such function.
Then, you immediately know that you have to do this integral.
So that's the integral of the first kind or Type I
integral over surfaces.
And next we go to Type II integrals.
Type II integrals are integrals for vector fields.

In this case, we are given a vector field F.
And, we are computing this, which is oftentimes called,
sometimes called surface integral of a vector field, but
oftentimes it's also called flux, flux of F across M.
So, if you see the word flux it refers to the integral
of the second type.
And, I explained what it means, it means the rate of flow,
for example, of a fluid through this membrane.
If F is a velocity vector field.
So, this integral actually was defined originally as follows.
We said that we have to take the dot product of F with the
unit normal vector field along M.
And, this will be a function, and then we take its integral
as a Type I integral.
So, this definition already makes clear that we need
orientation, we need a choice of n.
So just like integrals of second type the curves are not
really well defined if you just specify the vector
field and the surface.
Integrals of the second type are also not well defined if
you just specify vector field and surface.
You have to specify orientation on your surface.
So, it's exactly the same.
We need orientation.
And, orientation should be given to you if you have
to solve a problem.
If you are given a problem where you have to compute flux
of a vector field across a given surface, the problem has
to stay what is orientation, or there should be a way to
find which orientation it's supposed to be.
Now, again there are two different types of orientation,
there are two different orientations, in case the
surface is orientable, but of course we will be dealing
with orientable surfaces.
I explained to you a Mobius strip.
Remember a Mobius strip is an example of a surface
which is not orientable.
But, we will be working with orientable surfaces.
So, in orientable surfaces we will have two different types,
two different orientations, which, they will
differ by a sign.
So, how do you find out which one, which orientation to use.
So, let me first of all you rewrite this to
make it computable as a double integral.

So, again I'm assuming that my surface is parameterize
in this way, by this functions of u and v.
OK.
And in this case, this double integral will be written as
follows: the double integral over this domains D, which you
have there, and here you have F dot ru cross rv.
I mean, in principle you could try to compute this integral by
using this n, and there are different ways, there are
different ways to try to find this normal vector, OK,
a unit normal vector.
So, you could actually apply this formula, then compute this
integral as a Type I integral.
You could compute as a Type I integral, but I think, I
suggest, I strongly suggest, that you use it as, that you
use this formula for it, OK.
This, because, the point is that there are many ways to
find a normal vector after.
For instance, you know, we talked about normal
vectors for graphs and functions for example.
But what we need to do after this, you have to first
of all normalize it.
You have to get a normal vector of unit plane.
You have to normalize it.
And, then in addition, you will have to compute this
ds, which itself is a rather complicated tasks, right.
You have to compute this ds, which we here compute as the
magnitude of our u crossed rv.
The nice thing about this formula is that this formula is
obtained from this formula, where instead of n you take
this ru crossed rv, which we normalized by dividing
by its magnitude.
So, what happens is that there is some consolation.
This norm, this magnitude get canceled by the same magnitude
in the denominator.
So, the formula actually simplifies.
That's why, I think that this formula is nicer, so it is easy
to get confused if you use a different type of formula.
I mean you can do it at your own risk if you like, if you
know what I mean, but I just think this is the
best way to do it.
In the framework, for the kind of problems that we
are dealing with here.
So, here is the formula, but how do you see if this is the
right answer, and not its negative.
You see, so this is a good question.
How to see that you get the right orientation
in this formula.
In other words, you could have a problem, say where you are
asked to compute, to use orientation, to compute with
respect to the orientation, upward orientation.

So let's say you are asked to use upward orientation.
How do you know that this formula gives you the integral
with the upward orientation and not the downward orienation.
For example, right.
So, first of all what do we mean by upward orientation.
It means that you have to choose n of the form something
times i plus something times j, where this doesn't matter, plus
something, which is positive, times k.
This is upward.
And if I put less than 0 here, it will be downward.
So, that's the terminology.
Upward vector is a vector whose k component or z
component is positive.
And, downward vector is a vector which has a
negative z component.
OK.
So, let's say you're ask to compute the integral over a
surface, assuming the upward orientation, and you have this
formula, and the natural question is does this formula
give me the upper orientation or it gives me a
downward orientation.
We have to check that, because if it does give me the
upward orientation, I will get the right answer.
If it gives me downward orientation, then I would get
the minus sign, I will get negative of the right answer.
So, it means I will get the wrong answer.
If I get the wrong answer, it means that I don't get the full
score on this problem, right.
So, how do you see that.
Well, here's an example.
Let's look at some special cases where this actually can
be computed more explicitly.
And, the first case is when m is part of the graph.
So, for example, let's look at the case when, for example,
M is part of a graph of the function g of x,y.
In this case, we have already computed our x cross our y.

It is written on this formula.
This is our x cross our y.
So, let's me just copy this formula here.
So, this formula, in this formula the coefficient in
front of i and the coefficient in front of j depends on this
function that you have, of which your, you know, which
gives you this graph, right.
So this coefficient depends.
But this coefficient is always equal to 1.
So this coefficient is 1.
So, it's positive.
So, that means if you are using this formula, for rx cross ry,
right, if you are using this formula for rx cross ry, then
you are getting orientation, or you are computing, this formula
is computing the integral with respect to the upward
orientation, you see.
So, this is an important point that upwardly it could be, this
gives you a normal vector, not normalized, not [? unit ?], but
it gives you a normal vector.
And, by looking at this vector, you can find out with
respect to which orientation you are now computing.

And then you have to match that with what you are asked to do.
In this case, rx cross ry.
So, you have to look more closely at this vector, and
see, what does it look like [UNINTELLIGIBLE]?
In this particular case, the simplest information we can
infer by this vector is the fact that coefficient
in front of k is 1.
So, it is upward, it's always upward.
So, now you match, you say OK so what am I asked to do.
I'm asked to do double integral, surface integral
with respect to the upward orientation.
Oh, is my orientation upward.
Yes it is.
So, this is the right formula: F dot rx cross ry is
the right formula.
It is the right formula.
For upward orientation.
But, what if I want to trick you and I ask you on the test,
I ask you to compute a double integral, compute a surface
integral with respect to the downward orientation.
If you don't pay attention, you just, you have copied this
formula on your cheat sheet right, and you copied this
formula on your cheat sheet, and, so you think you
are in good shape.
You just applied this formula and you get an answer, but you
get the wrong answer, because you did not pay attention
to what was asked.
You were asked to compute with respect to downward
orientation, but you computed with respect
to upward orientation.
Just simply, because the formula you are using
always gives you upward.
But, I have the right to ask you to compute with respect
to downward orientation.
You know, both integrals makes sense.
It's my right and my prerogative, in fact, to ask
you to define any, to ask you to compute with respect to any
orientation that I like.
So, in that case, you have to be clever, and you
have to put a minus sign.

So, this is the right formula.
This is for downward orientation.

Because putting this minus sign, you can interpret it
as putting the minus sign in front of rx cross ry.
If I put it in front of rx cross ry, I will get plus here,
plus here and most important, I get minus 1 here.
If I get minus 1 here, I got orientation vector
which is downward.
So, that matches what I am asked to do, right.
To do the downward orientation.
So, that's why this is the correct formula.
Is this clear?
Ask me.
Go ahead.
[INAUDIBLE]
Right, I'll give another example in a minute for when
its not upward or downward.
[INAUDIBLE]
Very good.
What if the coefficient is 0.
So, OK, in this case is it never happens, right.
Because in this case, it's just always like this.
So, here is the next level of trick question.
Which is that, suppose that some ru cross rv, OK, but
for some other u and v, and suppose it just happens
that the component of the k factor is 0, right.
In this case, I should not be asking you to compute in
respect to upward or downward orientation, because if it's k
component is 0, it means it's parallel to the x, y plane.
So, it's neither upward nor downward.
And, saying that compute with respect to upward orientation
would be misleading.
In other words, ambiguous.
It will not give you any information.
Because, neither of the two vectors is upward
or downward, right.
So, in this case, you actually have to use another way
to describe it, OK.
So what are other ways to say, we have to describe in words,
you know, there are two possibilities.
You have to be able to distinguish between them.
If the k component is non-0, you can distinguish between
the sign of k, right.
Positive sign or negative sign.
But, if it is 0, for example, you can talk about the sign of
the coefficient in front j.
Even in this case, here is another trick
question I could ask.
Instead of saying compute this integral, even in this case
when it's part of a graph, instead of saying, it's the
upward or downward, I could say compute it with respect to the
orientation which is to the left.
So, what does it mean.
What should you do if you see that.
Don't panic.
If the orientation is described in terms of left, right, it
means that we're talking about the y coordinate, right.
Instead of the z coordinate, which would be k and I've got
the y coordinate, which would be j.
So, to the left would mean that the coefficient is negative,
and to the right would mean that the coefficient in front
of j is positive, right.
Now, of course in this case it's not clear, it depends on
the function g, sometimes, just the fact that there is a minus
sign here, doesn't mean anything because a function
could be negative.
So, the total sign would be positive, and so, right.
But, in the particular example, which you are computing, you
should be able to see if it's positive or negative
over that range.
So, what I'm saying is, what you are interested in is just
the piece of, of a graph.
So, you are working with a particular domain D in uv, and
you have to see whether over this domain certain coefficient
is positive or negative, right.
That's all it is.
And, for example, suppose you find that ru cross rv, let me
now switch to the more general setting where you actually have
coordinates u and v and its not necessarily graph of a
function, but more general surface.
So, you could have ru cross rv is equal to some, you know,
some -- A of u,v i plus B of u,v j plus C of u,v k.
This is what you will compute by taking this partial and
taking the cross-product.
So, you'll get three functions, and then you
just have to look closely.
What do the it look like.
Well, let's say this is u squared plus v squared, OK.
So, it's not 1 but it's going to be positive.
Well, the only point where it will not be
positive will be 0,0.
So, let's suppose that your domain does not
include point 0,0.
So, then it's always positive.
If you look, if you see that, you know, that the coefficient
in front of j is positive, which means that the
orientation vector is pointing to the right.
Because, because this is x, well, of course when we say, we
say that, it's all relative to how we organize our
coordinate system.
What do we mean by left and right.
But there is a way which we're kind of used to, which would be
like x goes this way and y goes this way, and this
is a z,right.
So, so in this case to the right means that it's a
positive number in front of j.
To the left means it's a negative number.
So, in this case it's going to be a vector, not necessarily
like this, but maybe like this.
So, this is to the right, one and for the left it's this way.
Is that clear?
There is another way to describe orientation by using
the words outward and inward.
So, usually, this is often used spheres or for closed domains.
OK.
So, that's another way.
That should also be pretty self-explanatory.
So, at the end of the day, you have to compute the formula
by using ru cross rv.
But, the bottom line is what you get is either the correct
answer, or either the answer which, if you put in, if you
stick it here, you get the correct answer, or
it's negative.
So, it becomes fine-tuning issue.
Is it correct, or should I put a minus sign.
So, then you match the description of that
orientation, which is given in wards -- upward, downward,
left, right, to what you got.
And, usually it should be fairly easy to see.
And, then that's how you see whether you got
the right answer.
So, you use that formula without any modification, or
you put a negative sign.
That's all it is.
Any questions?
Yes.
[INAUDIBLE]
Right.
So what if the sign changes.
If the sign changes, of course then the description would not
be correct, I mean well define.
The description should be such that if it says to the right,
it means that that coefficient should stay positive or
negative, whatever, right.
Okay, alright.
So, that's about the set up of the integrals.
And now, I want to say a few words about the connection
between different integrals.
These are the formulas which we've learned.
So, this is 2, formulas.
So, we have essentially learned 3 different formulas.
We have the Fundamental Theorem for line integrals.

We have Stokes Formula, and then we have Divergence Theory.
OK.
So now, I have explained -- oh yeah.
Also, Green's Theorem is a special case of this, when m
actually belongs to the plane is a special case this.
So, that's why I didn't single it out.
So, I already explained the kind of general theory behind
this, and how you could think of all of this formulas, a
special cases of one and only formula.

what I would like to focus now is the practical
aspects of this.
How can you apply these formulas to computing
integrals of the type that we have just discussed.
So here, I would like to emphasize, first of all, the
point aspect of these formulas.
In these formulas, always, there is a trade-off between
the sort of a boundary, the sort of a geometric boundary
of a domain, and a sort of derivative of the office
that you are integrating.
It's a trade-off.
On the left, it's a general domain of integration, but
the integrand is special.
In the first formula, you're computing only line integrals
of vector fields of this form.
The gradients, the conservative vector fields.
In this formula, integrals only of curl, of something which is
a curl of another vector field.
Here, you compute a function, which is divergence
of some vector field.
So, that kind of derivative of other objects.
But, the domains are most general, most general curve,
most general surface, most general three-dimensional
solid.

On the other hand, on the right-hand side you have the
most general object, here a general function here a general
vector field, here also a general vector field.
But, your integrating over domains, which are kind of
derivatives of the domains on the left-hand side.
Where in the geometric world, derivative means boundary.
So, this is a very important point to remember when you
decide which formulsa is appropriate for this or that
exercise, or for this or that parameterization.
So on the right-hand side, you've got general integrand,
but special domain, mainly boundary.
Domain.
Here it's derivative, here it's boundary.
OK.
So, the most -- so don't try to use the formula, where for
example, where just on this grounds you can rule it out.
In other words, or don't use it in a straightforward way.
Use it in a more creative way.
So let me give you just two examples of how this
formulas could be used.
So, example 1.
So, let's suppose you have, you're asked to compute a line
integral of a vector field, but the curve over which you have
to compute is very complicated.

So, that's the first indication that you should replace
the domain of integration.
OK.
So, in this case, let's say you were asked to compute this.
So, the clever thing to do is to replace this domain by
this curve, which has the same endpoints, right.
And, then you can argue that this line integral is equal to,
let's call this c1 and let's call this c2, this line
integral is equal to integral over c2 over F dr plus the
double integral of curl F ds, where you integrate over the in
interior of this domain.
You see, you can use this formula.
Let's call it D or M, let's call it M.
OK.
So, the reason is the following that if you take this to the
left-hand side, you will get negative c2, right.
I'm writing it like this, because, I prefer to
say this is equal to this plus something.
But, I could put this integral to the left with minus sign,
which means I'm integrating over c1 minus c2, and then you
have to see that this is a boundary of M; c1 minus c2
would be going with this orientation, that would
be boundary of M.
Now, if I do it like this, this brings up another issue, which
is the issue of orientation in this formula.
So, I'm using this formula, but in both cases, I have integrals
which depend on orientation.
And, we have this rule for orientations.
OK.
Which we discussed already many times.
So, orientation on the left and on the right have to match in
this way where if you walk on the boundary on the
surface in your left.
So, the way I did it right now would mean that the orientation
on this surface would have to go that way.
OK.
So, here it would be the surface where orientation is
going through the black board.
The usual orientation, which we, you normally consider,
would be this orientation towards us, right.
And, that orientation, which we would get
with counterclockwise.
But, here I have chosen orientation which goes
clockwise, well because I also put minus c2
so it reverse this.
And so this will be the orientation, let's call it
away from us, not towards us, but way from us.
Because, if somebody would be walking, if their feet are on
the blackboard, but their head points that way, so they would
be walking in that classroom.
They'll be like Spider Man, and they would be walking also on
this bizarre world, which is behind this blackboard, which
we don't even know what's going on over there, right.
But, that person would be walking like this.
The surface will be on their left.
So, normally if I were to walk like this, I would have to walk
counterclockwise for the surface to be on my left.
So, if I wanted to have orientation here, towards us, I
would have to, for this formula to be true, I would have to
choose up as the orientation.
But, so I'm putting sort of several things into example,
because I'm out of time essentially, but I'm trying
to sort of pack a lot of information into
this one example.
So, you should really look at it more carefully, and
different aspects of this, OK.
The orientations have to be compatible.
And there is one more thing which I'm doing here, which is
that I'm replacing a complicate path by an easy path, a simple
path, and the trade-off is that this guy is equal to this one.
But, there's a price to pay, and the price is the double
integral over this membrane, over this sort of film.
Think of it sort of like a film that you get from soap.
It's like a frame and there is a kind of soap.
So, you have to integrate over that, the curl.
The best possible scenario, where this works,
is when curl is 0.
If curl is 0 there is no price to pay, and, in fact, this
complicated integral over this complicated curve is equal, is
just straight equal to the integral here.
But, sometimes it's not 0, sometimes it's not 0, but
it could be very simple.
So, it's not such a big price to pay sometimes.
You see what I mean.
So, you have to, you have to see different options, the
different options that you have.
So, this is one example, where I am utilizing
formula number 2.
And, let me give you one more example, which is
kind of similar, but for formal number three.
So, maybe I should emphasize here, that in this formulas,
orientations have to be compatible.

This is important.
I'm not going to explain one more time, because we spent a
lot of time talking about this in the previous few lectures.
So, example number 2.
Example number 2, is kind of a similar trade-off, but where
you have to compute a double integral.
So let's say you have this upper hemisphere, and then you
also have this disc, this base of this upper hemisphere.
And, together they form the boundary of the
upper half of this bowl.
So, let's call this B, upper half of a bowl.
So, the last formula will tell you that the integral of
divergence of some vector field E over this B is equal to the
flux through the surface, through the boundary.
And, the boundary now consists of tow things.
Let's call this M1 and let's call this M2.
So, the formula then will be that it's double integral
over M1 plus double integral the over M2.
But, I have to say with respect to which orientation, and the
rule in the divergence theorem is that the orientation is
outward, which here would mean like this, and here
would mean like this.
Here it's outward, actually is upward on top part, and on
the bottom outward is actually downward.
But, both are outward.
So, that's what I mean here and here.
So, how can we utilize this formula.
For example, what you can do is you can express this as the
difference between this and this.
So, this way you can express a complicated integral over
something very curved, this sphere, or upper hemisphere.
In terms of two flat integrals, one of them is over a
three-dimensional solid, which actually may be quite simple --
depending on what -- I made a mistake like this. dV.
It could actually be 0, this divergence.
It could be 0.
So, then this actually drops out.
And, then you can express this guy as a negative
of this guy, right.
Or, even if it's non-0, it could be something very
simple, like 1, function 1 or something.
So, then it's easier to compute.
And, this one is certainly easy to compute easier to compute
than this one, because it's an integral over a flap region.
So, you have this trade-off again.
Similar to the trade-off discussed in the previous
example -- between an integral over one part of the boundary,
and the integral over the other part of the boundary, with a
surplus or price to pay coming from the integral
of the divergence.
So, these are very typical examples of the kind of things
that you need to know to be able to apply these formulas.
So, here's what we're going to do now.
I'm going to stop, and we're going to do a
course evaluations.
It will just take a few minutes.
Okay.
Come on guys.
Come on over, so we'll start distributing them.
And, we'll just take a few minutes, and after this,
after we're done with course the evaluations, I'll be
here for office hours.
And, so you can ask me questions about anything,
well related to the course.
Okay.