## TTU Math2450 Calculus3 Secs 13.6 - 13.7

• 0:00 - 0:03
PROFESSOR: Do you have
any kind of questions?
• 0:03 - 0:05
There were a few questions
• 0:05 - 0:11
Casey, you have that problem
we need to the minus D?
• 0:11 - 0:12
STUDENT: Yeah.
• 0:12 - 0:13
PROFESSOR: The minus D?
• 0:13 - 0:16
Let's do that in class, because
there were several people who
• 0:16 - 0:18
faced that problem.
• 0:18 - 0:22
You said you faced it, and
you got it and can I cheat?
• 0:22 - 0:26
Can I take your work so I
can present it at the board?
• 0:26 - 0:27
• 0:27 - 0:28
STUDENT: OK, um.
• 0:28 - 0:31
PROFESSOR: So I know
we've done this together.
• 0:31 - 0:34
I don't even
remember the problem.
• 0:34 - 0:35
How was it?
• 0:35 - 0:37
Homework problem.
• 0:37 - 0:40
STUDENT: She knows it.
• 0:40 - 0:40
[INAUDIBLE]
• 0:40 - 0:44
• 0:44 - 0:45
STUDENT: But she knows it.
• 0:45 - 0:48
PROFESSOR: They'll work with you
to be minus M. Can you tell me,
• 0:48 - 0:48
Casey?
• 0:48 - 0:50
Can I tell the statement?
• 0:50 - 0:53
STUDENT: Well, a black
guy's [INAUDIBLE].
• 0:53 - 0:56
PROFESSOR: If you find it, give
it to me and I'll give you \$2.
• 0:56 - 0:57
• 0:57 - 0:58
[INAUDIBLE]
• 0:58 - 1:00
PROFESSOR: How is it's this one.
• 1:00 - 1:01
STUDENT: No.
• 1:01 - 1:02
PROFESSOR: Oh, no.
• 1:02 - 1:03
It's not this one.
• 1:03 - 1:04
STUDENT: Can I have it from you?
• 1:04 - 1:06
I won't give you anything.
• 1:06 - 1:07
STUDENT: Um, it's doable.
• 1:07 - 1:10
[INAUDIBLE]
• 1:10 - 1:13
PROFESSOR: So can somebody with
me now, that's my handwriting.
• 1:13 - 1:14
STUDENT: Yeah, I know.
• 1:14 - 1:15
It is weird.
• 1:15 - 1:15
PROFESSOR: OK.
• 1:15 - 1:16
All right.
• 1:16 - 1:18
So the problem says--
• 1:18 - 1:18
STUDENT: Do you have
a problem with me?
• 1:18 - 1:19
PROFESSOR: Any--
• 1:19 - 1:21
STUDENT: And then we're all
here for her. [INAUDIBLE].
• 1:21 - 1:23
Doesn't it feel like
[INAUDIBLE] kind of a bit?
• 1:23 - 1:25
PROFESSOR: It's the one
that has X of D equals
• 1:25 - 1:29
into the minus the cosign D.
• 1:29 - 1:30
STUDENT: Oh, [INAUDIBLE].
• 1:30 - 1:32
PROFESSOR: Y of T, and you
• 1:32 - 1:37
I understand that you
love this problem.
• 1:37 - 1:44
And so you've had this type
of pathing to grow compute.
• 1:44 - 1:49
The pathing to grow with respect
to the [INAUDIBLE] fellow man
• 1:49 - 1:51
well meant that
in life is slowly
• 1:51 - 1:54
because nobody
[INAUDIBLE] with you.
• 1:54 - 1:58
• 1:58 - 2:04
And to go over C
of the integer will
• 2:04 - 2:08
be a very nice friend of
yours, [INAUDIBLE] explain it,
• 2:08 - 2:11
but of course, they are both
functions of T in general,
• 2:11 - 2:14
and you will have
the DS element,
• 2:14 - 2:15
and what does this mean?
• 2:15 - 2:17
S is [INAUDIBLE].
• 2:17 - 2:22
It means that you are
archic element should
• 2:22 - 2:26
be expressed in terms of what?
• 2:26 - 2:28
Who in the world is
the archeling infinite
• 2:28 - 2:30
decimal element.
• 2:30 - 2:32
It's the speed times the t.
• 2:32 - 2:33
STUDENT: Say it again?
• 2:33 - 2:35
PROFESSOR: It's the speed.
• 2:35 - 2:36
STUDENT: And what was the speed?
• 2:36 - 2:42
R in front of T. [INAUDIBLE].
• 2:42 - 2:43
Right?
• 2:43 - 2:46
So you will have to
transform this path integral
• 2:46 - 2:50
into an integral, respected
T, where T takes values
• 2:50 - 2:53
from a T0 to a T1.
• 2:53 - 2:59
And I don't want to give
• 2:59 - 3:00
STUDENT: It's OK.
• 3:00 - 3:02
PROFESSOR: OK, now I'll do
the same thing all over again,
• 3:02 - 3:07
and you control me and then
if I do something wrong,
• 3:07 - 3:08
you've done me.
• 3:08 - 3:13
And what were the--
what was the path?
• 3:13 - 3:15
Specified as what?
• 3:15 - 3:18
STUDENT: XYZ? [INAUDIBLE].
• 3:18 - 3:20
PROFESSOR: Yeah, the path was--
• 3:20 - 3:22
STUDENT: [INAUDIBLE].
• 3:22 - 3:25
PROFESSOR: T equals
from zero to pi over 2.
• 3:25 - 3:27
I have to write it down.
• 3:27 - 3:30
• 3:30 - 3:35
So let us write the--
[INAUDIBLE] are from the T.
• 3:35 - 3:44
The speed square root of is
from the T squared plus Y
• 3:44 - 3:50
prime of T squared, because
the sampling occurred.
• 3:50 - 3:54
Before we do that, we have
• 3:54 - 3:55
X prime and Y prime.
• 3:55 - 3:59
• 3:59 - 4:01
And of course
that's product rule,
• 4:01 - 4:03
and I need a better marker.
• 4:03 - 4:04
STUDENT: [INAUDIBLE].
• 4:04 - 4:05
PROFESSOR: Yes, sir?
• 4:05 - 4:07
STUDENT: Do you think
the arc too is really
• 4:07 - 4:09
taken as an arc [INAUDIBLE]?
• 4:09 - 4:09
This
• 4:09 - 4:10
PROFESSOR: This is the--
• 4:10 - 4:13
STUDENT: Because we take-- I
would consider it as a path
• 4:13 - 4:16
function that looks like
an arc, or, like, thinking
• 4:16 - 4:19
that it's missing one
• 4:19 - 4:20
That's fine.
• 4:20 - 4:22
PROFESSOR: No, no, no
no, no, no, no, no.
• 4:22 - 4:22
no.
• 4:22 - 4:23
OK, let me explain.
• 4:23 - 4:28
So suppose you are
[INAUDIBLE] arc in plane
• 4:28 - 4:30
and this is your r of t.
• 4:30 - 4:31
STUDENT: Oh, OK.
• 4:31 - 4:33
PROFESSOR: And that's
called the position vector,
• 4:33 - 4:36
and that's x of t, y of t.
• 4:36 - 4:36
OK.
• 4:36 - 4:38
• 4:38 - 4:42
Velocity vector would be
in tangent to the curve.
• 4:42 - 4:45
Suppose you go in this
direction, counterclockwise,
• 4:45 - 4:48
and then our prime of
t will be this guy.
• 4:48 - 4:51
And it's gonna be
x prime, y prime.
• 4:51 - 4:53
And we have to
find its magnitude.
• 4:53 - 4:56
And its magnitude
will be this animal.
• 4:56 - 4:59
So the only thing here
is tricky because you
• 4:59 - 5:03
will have to do this
carefully, and there
• 5:03 - 5:06
will be a simplification
coming from the plus
• 5:06 - 5:08
and minus of the binomial.
• 5:08 - 5:11
So a few people missed it
because of that reason.
• 5:11 - 5:14
So let's see what
we have-- minus
• 5:14 - 5:17
e to the minus t, first
prime, times second one
• 5:17 - 5:25
prime plus the first one
prime times the second prime.
• 5:25 - 5:26
Good.
• 5:26 - 5:28
We are done with this first guy.
• 5:28 - 5:34
The second guy will be minus
e to the minus t sin t.
• 5:34 - 5:37
Why do I do this?
• 5:37 - 5:41
Because I'm afraid that
this being on the final.
• 5:41 - 5:43
Well, it's good practice.
• 5:43 - 5:47
You may expect
something a little bit
• 5:47 - 5:53
similar to that, so why don't we
do this as part of our review,
• 5:53 - 5:55
which will be a very good idea.
• 5:55 - 5:58
We are gonna do lots of
review this week and next
• 5:58 - 6:02
the final is coming close
• 6:02 - 6:10
and you have to go over
everything that you've covered.
• 6:10 - 6:16
Let's square them,
• 6:16 - 6:23
• 6:23 - 6:24
OK.
• 6:24 - 6:32
this guy, the speed [INAUDIBLE]
• 6:32 - 6:34
is going to-- bless you.
• 6:34 - 6:37
It is going-- it's not going
to bless, it's going-- OK,
• 6:37 - 6:41
you are being blessed, and
now let's look at that.
• 6:41 - 6:46
You have e to the
minus 2t cosine squared
• 6:46 - 6:48
and e to the minus
2t sine squared,
• 6:48 - 6:51
those parts, the sine
• 6:51 - 6:53
squared plus cosine
squared stick together.
• 6:53 - 6:57
They form a block called 1.
• 6:57 - 6:59
Do you guys agree with me?
• 6:59 - 7:03
So what we have as the
first result of that
• 7:03 - 7:05
would be this guy.
• 7:05 - 7:10
But then, when you
take twice the product
• 7:10 - 7:13
of these guys in the
binomial formula,
• 7:13 - 7:16
and twice the product of these
guys, what do you notice?
• 7:16 - 7:20
We have exactly the
same individuals inside,
• 7:20 - 7:23
but when you do twice the
product of these two red ones,
• 7:23 - 7:27
you have minus, minus, plus.
• 7:27 - 7:29
But when you do twice the
product of these guys,
• 7:29 - 7:32
you have minus, plus, minus.
• 7:32 - 7:33
So they will cancel out.
• 7:33 - 7:37
The part in the middle
will cancel out.
• 7:37 - 7:41
And finally, when I square
this part and that part,
• 7:41 - 7:43
what's going to happen them?
• 7:43 - 7:47
And I'm gonna shut up because I
want you to give me the answer.
• 7:47 - 7:51
Square this animal, square
• 7:51 - 7:52
what do you have?
• 7:52 - 7:54
STUDENT: Squared, squared
total-- [INTERPOSING VOICES]
• 7:54 - 7:59
PROFESSOR: T to the minus 2t,
so exactly the same as this guy.
• 7:59 - 8:02
So all I know under
the square root,
• 8:02 - 8:12
I'm gonna get square root of
2 times e to the minus 2t.
• 8:12 - 8:15
Which is e to the minus
t square root of 2.
• 8:15 - 8:18
Am I right, [INAUDIBLE]
that's what we got last time?
• 8:18 - 8:19
All right.
• 8:19 - 8:26
So I know who this will be.
• 8:26 - 8:30
I don't know who this will be,
but I'm gonna need your help.
• 8:30 - 8:35
Here I write it, x squared
of t plus y squared of t
• 8:35 - 8:39
in terms of t, squaring them
• 8:39 - 8:43
It's gonna be again a piece of
cake, because you've got it.
• 8:43 - 8:46
How much is it?
• 8:46 - 8:48
I'm waiting for you to tell me.
• 8:48 - 8:51
This is this one.
• 8:51 - 8:52
[INAUDIBLE]
• 8:52 - 8:53
E to the?
• 8:53 - 8:54
STUDENT: Minus t
• 8:54 - 8:56
PROFESSOR: And anything else?
• 8:56 - 8:59
STUDENT: Was it 2?
• 8:59 - 9:00
[INAUDIBLE]
• 9:00 - 9:04
• 9:04 - 9:06
PROFESSOR: Why it times 2?
• 9:06 - 9:07
STUDENT: Times 2
in the last one.
• 9:07 - 9:10
Because we had an e to the minus
2t plus an e to the minus 2t.
• 9:10 - 9:14
PROFESSOR: So I took
this guy and squared it,
• 9:14 - 9:16
and I took this guy
and squared it--
• 9:16 - 9:17
STUDENT: No, we don't
have [INAUDIBLE].
• 9:17 - 9:19
PROFESSOR: And I sum them up.
• 9:19 - 9:22
And I close the issue.
• 9:22 - 9:28
Unless I have sine squared plus
cosine squared, which is 1,
• 9:28 - 9:31
so we adjust it to the minus 2t.
• 9:31 - 9:32
Agree with me?
• 9:32 - 9:34
All right, now we have
all the ingredients.
• 9:34 - 9:36
Do we have all the
ingredients we need?
• 9:36 - 9:38
We have this, we have
that, we have that.
• 9:38 - 9:41
And we should just go ahead
and solve the problem.
• 9:41 - 9:48
So, integral from 0 to pi over
2, this friend of yours, e
• 9:48 - 9:52
to the minus 2t
plus [INAUDIBLE].
• 9:52 - 9:59
The speed was over there, e to
the minus t times square root
• 9:59 - 10:00
of 2.
• 10:00 - 10:01
That was the speed.
• 10:01 - 10:05
[INAUDIBLE] magnitude, dt.
• 10:05 - 10:07
Is this what we got?
• 10:07 - 10:08
All right.
• 10:08 - 10:10
Now, we are almost
done, in the sense
• 10:10 - 10:14
that we should wrap things up.
• 10:14 - 10:17
Square root 2 gets out.
• 10:17 - 10:23
And then integral of it to the
minus 3t from zero to pi over 2
• 10:23 - 10:24
is our friend.
• 10:24 - 10:27
We know how to deal with him.
• 10:27 - 10:29
We have dt.
• 10:29 - 10:35
So when you integrate
that, what do you have?
• 10:35 - 10:37
Let me erase--
• 10:37 - 10:40
STUDENT: Negative square
root 2 over 3 [INAUDIBLE] 3t.
• 10:40 - 10:41
PROFESSOR: Right.
• 10:41 - 10:43
So let me erase this part.
• 10:43 - 10:48
• 10:48 - 10:51
So we have-- first we
have to copy this guy.
• 10:51 - 10:55
Then we have e to the
minus 3t divided by minus 3
• 10:55 - 10:58
because that is
the antiderivative.
• 10:58 - 11:05
And we take that into t equals
zero and t equals pi over 2.
• 11:05 - 11:11
Square root of 2 says I'm
going out, and actually minus 3
• 11:11 - 11:12
says also I'm going out.
• 11:12 - 11:16
So he doesn't want to be
involved in this discussion.
• 11:16 - 11:18
[INAUDIBLE]
• 11:18 - 11:24
Now, e to the minus 3 pi over
2 is the first thing we got.
• 11:24 - 11:27
And then minus e to the 0.
• 11:27 - 11:28
What's e to the 0?
• 11:28 - 11:29
STUDENT: 1.
• 11:29 - 11:30
PROFESSOR: 1.
• 11:30 - 11:35
So in the end, you have
to change the sign.
• 11:35 - 11:41
You have root 2 over 3
times bracket notation when
• 11:41 - 11:44
you type this in
WeBWorK because based
• 11:44 - 11:49
is bad, you are-- for example,
• 11:49 - 11:53
here we have to put
^ minus 3 pi over 2.
• 11:53 - 11:55
Are you guys with me?
• 11:55 - 11:57
Do you understand the words
coming out of my mouth?
• 11:57 - 12:02
So here you have to
type the right syntax,
• 12:02 - 12:04
and you did, and you got--
• 12:04 - 12:05
STUDENT: And I
didn't [INAUDIBLE]
• 12:05 - 12:08
but I need to type
• 12:08 - 12:10
In terms of decimal places.
• 12:10 - 12:12
PROFESSOR: This is a problem.
• 12:12 - 12:14
It shouldn't be like that.
• 12:14 - 12:18
Sometimes unfortunately--
well, fortunately it rarely
• 12:18 - 12:27
happens that WeBWorK program
• 12:27 - 12:28
in a certain format.
• 12:28 - 12:31
Maybe the pi screws
everything up.
• 12:31 - 12:32
I don't know.
• 12:32 - 12:35
But if you do this
• 12:35 - 12:38
eventually, you can What was
• 12:38 - 12:40
[INAUDIBLE]?
• 12:40 - 12:42
STUDENT: 0.467
• 12:42 - 12:44
PROFESSOR: And blah, blah, blah.
• 12:44 - 12:44
I don't know.
• 12:44 - 12:48
I think WeBWorK only cares
for the first two decimals
• 12:48 - 12:49
to be correct.
• 12:49 - 12:50
As I remember.
• 12:50 - 12:51
I don't know.
• 12:51 - 12:54
the programmer.
• 12:54 - 12:56
So this would be
approximately-- you
• 12:56 - 12:58
• 12:58 - 12:59
I solved the problem,
so I should give myself
• 12:59 - 13:03
the credit, plus
a piece of candy,
• 13:03 - 13:07
but I hope I was able to
save you from some grief
• 13:07 - 13:11
because you have so much review
going on that you shouldn't
• 13:11 - 13:14
spend time on problems
• 13:14 - 13:16
for computational reasons.
• 13:16 - 13:19
Actually, I have
computational reasons,
• 13:19 - 13:22
because we are not androids
and we are not computers.
• 13:22 - 13:27
What we can do is
think of a problem
• 13:27 - 13:31
and let the software
solve the problem for us.
• 13:31 - 13:36
So our strength does not consist
in how fast we can compute,
• 13:36 - 13:42
but on how well we can solve a
problem so that the calculator
• 13:42 - 13:46
or computer can carry on.
• 13:46 - 13:47
All right.
• 13:47 - 13:50
• 13:50 - 13:56
I know I covered
up to 13.6, and let
• 13:56 - 13:58
me remind you what we covered.
• 13:58 - 14:03
We covered some beautiful
sections that were called 13.4.
• 14:03 - 14:04
This was Green's theorem.
• 14:04 - 14:08
And now, I'm really proud
of you that all of you
• 14:08 - 14:10
know Green's theorem very well.
• 14:10 - 14:17
And the surface
integral, which was 13.5.
• 14:17 - 14:22
And then I promised
you that today we'd
• 14:22 - 14:26
move on to 13.6, which
is Stokes' theorem,
• 14:26 - 14:29
and I'm gonna do that.
• 14:29 - 14:34
But before I do that, I want
• 14:34 - 14:50
to a fact that this is a bigger
result that incorporates 13.4.
• 14:50 - 14:53
So Stokes' theorem
is a more general
• 14:53 - 14:58
result. So let me make a
diagram, like a Venn diagram.
• 14:58 - 15:02
This is all the cases
of Stokes' theorem,
• 15:02 - 15:04
and Green's is one of them.
• 15:04 - 15:07
• 15:07 - 15:11
And this is something you've
learned, and you did very well,
• 15:11 - 15:15
and we only considered
this theorem
• 15:15 - 15:18
on a domain that's
interconnected.
• 15:18 - 15:20
It has no holes in it.
• 15:20 - 15:22
Green's theorem
can be also taught
• 15:22 - 15:25
on something like a
doughnut, but that's not
• 15:25 - 15:27
the purpose of this course.
• 15:27 - 15:29
You have it in the book.
• 15:29 - 15:30
It's very sophisticated.
• 15:30 - 15:37
13.6 starts at-- oh, my
God, I don't know the pages.
• 15:37 - 15:40
And being a co-author
of this book
• 15:40 - 15:43
means that I should
remember the pages.
• 15:43 - 15:44
All right, there it is.
• 15:44 - 15:51
13.6 is that page 1075.
• 15:51 - 15:57
OK, and let's see what
• 15:57 - 16:01
• 16:01 - 16:05
I'm gonna state it as
first Stokes' theorem,
• 16:05 - 16:09
and then I will see why Green's
theorem is a particular case.
• 16:09 - 16:11
We don't know yet why that is.
• 16:11 - 16:14
Well, assume you
have a force field,
• 16:14 - 16:15
may the force be with you.
• 16:15 - 16:30
This is a big vector-valued
function over a domain in R 3
• 16:30 - 16:39
that includes a surface s.
• 16:39 - 16:42
We don't say much
• 16:42 - 16:46
because we try to
avoid the terminology,
• 16:46 - 16:48
but you guys should
assume that this
• 16:48 - 17:00
is a simply connected
surface patch with a boundary
• 17:00 - 17:11
c, such that c is
a Jordan curve.
• 17:11 - 17:14
• 17:14 - 17:19
We use the word Jordan curve
as a boundary of the surface,
• 17:19 - 17:21
but we don't say
simply connected.
• 17:21 - 17:23
you, what in the world
• 17:23 - 17:26
do we mean when we
simply connected?
• 17:26 - 17:28
I've used this before.
• 17:28 - 17:31
I just want to test your
memory and attention.
• 17:31 - 17:33
Do you remember what that meant?
• 17:33 - 17:39
I have some sort of little
hill, or s could be a flat disc,
• 17:39 - 17:41
or it could be a
patch of the plane,
• 17:41 - 17:46
or it could be just any
kind of surface that
• 17:46 - 17:49
is bounded by Jordan curve c.
• 17:49 - 17:50
What is a Jordan curve?
• 17:50 - 17:52
But what can we say about c?
• 17:52 - 17:56
• 17:56 - 18:00
So c would be nice
piecewise continuous-- we
• 18:00 - 18:02
assumed it continuous actually.
• 18:02 - 18:03
Most cases--
• 18:03 - 18:05
STUDENT: It has to connect
to itself, doesn't it?
• 18:05 - 18:08
PROFESSOR: No
self-intersections.
• 18:08 - 18:11
So we knew that from before,
but what does it mean,
• 18:11 - 18:15
simply connected for us?
• 18:15 - 18:16
I said it before.
• 18:16 - 18:20
I don't know how
attentive you were.
• 18:20 - 18:23
Connectedness makes
you think of something.
• 18:23 - 18:24
No holes in it.
• 18:24 - 18:26
So that means no holes.
• 18:26 - 18:27
No punctures.
• 18:27 - 18:31
No holes, no punctures.
• 18:31 - 18:33
So why-- don't draw it.
• 18:33 - 18:35
I will draw it so you can laugh.
• 18:35 - 18:37
Assume that the dog
came here and took
• 18:37 - 18:39
a bite of this surface.
• 18:39 - 18:42
And now you have a hole in it.
• 18:42 - 18:44
Well, you're not supposed
to have a hole in it,
• 18:44 - 18:46
so tell the dog to go away.
• 18:46 - 18:49
So you're not gonna have
any problems, any puncture,
• 18:49 - 18:53
any hole, any problem with this.
• 18:53 - 18:58
Now the surface is assumed
to be a regular surface,
• 18:58 - 19:00
and we've seen that before.
• 19:00 - 19:03
And since it's a regular
surface, that means
• 19:03 - 19:05
it's immersed in
the ambient space,
• 19:05 - 19:09
and you have an N orientation.
• 19:09 - 19:17
Orientation which is the
unit normal to the surface.
• 19:17 - 19:20
• 19:20 - 19:21
Can you draw it, Magdalena?
• 19:21 - 19:24
Yes, in a minute,
I will draw it.
• 19:24 - 19:30
At every point you
have an N unit normal.
• 19:30 - 19:33
What was the unit
normal for you when
• 19:33 - 19:36
you parametrize the surface?
• 19:36 - 19:40
That was the stick that
has length 1 perpendicular
• 19:40 - 19:42
to the tangent length, right?
• 19:42 - 19:45
So if you wanted to
do it for general R,
• 19:45 - 19:49
you would take those R sub u or
sub v the partial [INAUDIBLE].
• 19:49 - 19:53
And draw the cross
product, and this
• 19:53 - 19:55
is what I'm trying to do now.
• 19:55 - 19:58
And just make the length be 1.
• 19:58 - 20:02
So if the surface is
regular, I can parametrize it
• 20:02 - 20:07
as [INAUDIBLE] will exist
in that orientation.
• 20:07 - 20:09
I want something more.
• 20:09 - 20:17
I want N orientation to be
compatible to the direction
• 20:17 - 20:26
of travel on c-- along c.
• 20:26 - 20:30
Along, not on, but
• 20:30 - 20:32
So assume that this
is a hill, and I'm
• 20:32 - 20:34
running around the boundary.
• 20:34 - 20:37
Look, I'm just running around
the boundary, which is c.
• 20:37 - 20:40
Am I running in a
particular direction that
• 20:40 - 20:43
tells you I'm a mathematician?
• 20:43 - 20:45
It tells you that I'm a weirdo.
• 20:45 - 20:46
Yes.
• 20:46 - 20:49
So in what kind of
direction am I running?
• 20:49 - 20:51
Counterclockwise.
• 20:51 - 20:51
Why?
• 20:51 - 20:53
Because I'm a nerd.
• 20:53 - 20:55
Like Sheldon or something.
• 20:55 - 20:59
So let's go around,
and so what does
• 20:59 - 21:05
it mean I am compatible
with the orientation?
• 21:05 - 21:07
Think of the right hand rule.
• 21:07 - 21:10
rule, I hate that word.
• 21:10 - 21:13
Let's think faucet.
• 21:13 - 21:17
is along the c,
• 21:17 - 21:21
so that it's like you are
unscrewing the faucet,
• 21:21 - 21:23
it's going up.
• 21:23 - 21:25
That should mean
• 21:25 - 21:29
n should go up, or, not
down, in the other direction.
• 21:29 - 21:33
So if I take c to be my
orientation around the curve,
• 21:33 - 21:36
then the orientation of
the surface should go up.
• 21:36 - 21:41
Am I allowed to go around the
opposite direction on the c.
• 21:41 - 21:43
Yes I am.
• 21:43 - 21:46
That's, how it this called,
inverse trigonometric,
• 21:46 - 21:48
or how do we call such a thing.
• 21:48 - 21:49
STUDENT: Clockwise?
• 21:49 - 21:52
PROFESSOR: Clockwise,
you guessed it.
• 21:52 - 21:54
OK, clockwise.
• 21:54 - 21:57
If I would go
clockwise in plane,
• 21:57 - 22:01
then the N should
be pointing down.
• 22:01 - 22:04
So it should be oriented
just the opposite way
• 22:04 - 22:06
on the surface S.
• 22:06 - 22:11
All right, that's sort
of easy to understand now
• 22:11 - 22:13
because most of
you are engineers
• 22:13 - 22:17
and you deal with this
kind of stuff every day.
• 22:17 - 22:19
What is Stokes' theorem?
• 22:19 - 22:23
Stokes' theorem says well, in
that case, the path integral
• 22:23 - 22:30
over c of FdR, F dot dR.
• 22:30 - 22:32
What the heck is this?
• 22:32 - 22:35
I'm not gonna finish the
sentence, because I'm mean.
• 22:35 - 22:38
There is a sentence there,
an equation, but I'm mean.
• 22:38 - 22:43
what in the world is this?
• 22:43 - 22:46
F is the may the
force be with you.
• 22:46 - 22:48
R is the vector position.
• 22:48 - 22:50
What is this animal?
• 22:50 - 22:52
The book doesn't tell you.
• 22:52 - 22:54
This is the work that
you know so well.
• 22:54 - 22:56
All right.
• 22:56 - 23:00
So you may hear math majors
saying they don't care.
• 23:00 - 23:02
They don't care because they're
not engineers or physicists,
• 23:02 - 23:05
but work is very important.
• 23:05 - 23:11
The work along the curve
will be equal to-- now
• 23:11 - 23:14
comes the beauty--
the beautiful part.
• 23:14 - 23:17
This is a double integral
over the surface [INAUDIBLE]
• 23:17 - 23:22
with respect to the
area element dS.
• 23:22 - 23:27
Oh, guess what, you wouldn't
know unless somebody taught you
• 23:27 - 23:30
before coming to
class, this is going
• 23:30 - 23:34
to be curl F. What is curl?
• 23:34 - 23:35
It's a vector.
• 23:35 - 23:39
So I have to do dot product
with another vector.
• 23:39 - 23:41
And that vector is N.
• 23:41 - 23:45
ahead of time, which is great.
• 23:45 - 23:52
I would say 0.5% or less
• 23:52 - 23:53
in a textbook.
• 23:53 - 23:57
I used to do that
when I was young.
• 23:57 - 24:00
I didn't always have
the time to do it,
• 24:00 - 24:05
possibility I did it.
• 24:05 - 24:12
Now a quiz for you.
• 24:12 - 24:16
No, don't take any sheets
out, but a quiz for you.
• 24:16 - 24:19
Could you prove
to me, just based
• 24:19 - 24:25
on this thing that looks
weird, that Green's theorem is
• 24:25 - 24:28
a particular case of this?
• 24:28 - 24:32
So prove-- where
should I put it?
• 24:32 - 24:35
That was Stokes' theorem.
• 24:35 - 24:37
Stokes' theorem.
• 24:37 - 24:40
And I'll say
exercise number one,
• 24:40 - 24:43
sometimes I put this
in the final exam,
• 24:43 - 24:46
so I consider this
to be important.
• 24:46 - 25:07
Prove that Green's theorem is
nothing but a particular case
• 25:07 - 25:08
of Stokes' theorem.
• 25:08 - 25:12
• 25:12 - 25:17
And I make a face in the sense
that I'm trying to build trust.
• 25:17 - 25:19
Maybe you don't trust me.
• 25:19 - 25:22
But I-- let's do this together.
• 25:22 - 25:24
Let's prove together
that this is what it is.
• 25:24 - 25:31
Now, the thing is, if I were
to give you a test right now
• 25:31 - 25:34
on Green's theorem,
how many of you would
• 25:34 - 25:36
know what Green's theorem said?
• 25:36 - 25:40
So I'll put it here
in an-- open an icon.
• 25:40 - 25:44
Imagine this would be
an icon-- or a window,
• 25:44 - 25:48
a window on the computer screen.
• 25:48 - 25:53
Like a tutorial reminding
you what Green's theorem was.
• 25:53 - 25:55
So Green's theorem said what?
• 25:55 - 25:58
• 25:58 - 26:01
We have to-- bless you.
• 26:01 - 26:06
So Zander started
the theorem by-- we
• 26:06 - 26:13
have a domain D that was
also simply connected.
• 26:13 - 26:14
What does it mean?
• 26:14 - 26:16
No punctures, no holes.
• 26:16 - 26:16
No holes.
• 26:16 - 26:20
Even if you have a
puncture that's a point,
• 26:20 - 26:22
that's still a hole.
• 26:22 - 26:28
You may not see it, but if
anybody punctured the portion
• 26:28 - 26:30
of a plane, you are in trouble.
• 26:30 - 26:32
So there are no such things.
• 26:32 - 26:34
And c is a Jordan curve.
• 26:34 - 26:41
• 26:41 - 26:44
And then you say, OK,
how is it, how was F?
• 26:44 - 26:46
F was a c 1.
• 26:46 - 26:49
What does it mean
that if is a c 1?
• 26:49 - 26:53
F is a vector-valued function
that's differentiable,
• 26:53 - 26:56
and its derivatives
are continuous,
• 26:56 - 26:58
partial derivatives.
• 26:58 - 27:08
And so you think F of xy will
be M of xy I plus n of xyj
• 27:08 - 27:15
is a vector field, so
it's a multivariable,
• 27:15 - 27:17
so I have two variables.
• 27:17 - 27:18
OK?
• 27:18 - 27:21
So you think, OK, I
know what this is.
• 27:21 - 27:23
Like, this would be a force.
• 27:23 - 27:26
If this were a
force, I would get
• 27:26 - 27:34
the vector-- the work--
how can I write this again?
• 27:34 - 27:35
We didn't write it like that.
• 27:35 - 27:38
We wrote it as Mdx
plus Ndy, which
• 27:38 - 27:41
is the same thing as before.
• 27:41 - 27:41
Why?
• 27:41 - 27:46
Because Mr. F is MI plus NJ.
• 27:46 - 27:47
Not k, Magdalena.
• 27:47 - 27:50
You were too nice, but you
didn't want to shout at me.
• 27:50 - 27:57
And dR was what? dR was
dxI plus dyJ, right?
• 27:57 - 28:01
So when you do this, the
product which is called work,
• 28:01 - 28:04
the integral will
• 28:04 - 28:08
and this is what it
was in Green's theorem.
• 28:08 - 28:11
And what did we claim it was?
• 28:11 - 28:13
Now, you know it, because
you've done a lot of homework.
• 28:13 - 28:17
You're probably sick and tired
of Green's theorem and you say,
• 28:17 - 28:21
I understand that work-- a
path integral can be expressed
• 28:21 - 28:24
as a double integral some way.
• 28:24 - 28:26
Do you know this by heart?
• 28:26 - 28:29
You proved this to me last
time you know it by heart.
• 28:29 - 28:33
That was N sub x minus M sub y.
• 28:33 - 28:40
And we memorized it-- dA
over this is a planar domain.
• 28:40 - 28:44
It's a domain in
plane d, [INAUDIBLE].
• 28:44 - 28:46
I said it, but I
didn't write it down.
• 28:46 - 28:51
So double integral over d,
N sub x minus M sub y dA.
• 28:51 - 28:51
We've done that.
• 28:51 - 28:55
That was section-- what
section was that, guys?
• 28:55 - 28:57
13.4.
• 28:57 - 29:01
Yeah, for sure, you will have a
problem on the final like that.
• 29:01 - 29:04
Do not expect lots of problems.
• 29:04 - 29:05
Do not expect 25 problems.
• 29:05 - 29:07
You will not have the time.
• 29:07 - 29:09
So you will have
some 15, 16 problems.
• 29:09 - 29:12
This will be one of them.
• 29:12 - 29:15
You mastered this
Green's theorem.
• 29:15 - 29:19
When you sent me
questions from WeBWorK
• 29:19 - 29:23
I realized that you were able
to solve the problems where
• 29:23 - 29:26
these would be
easy to manipulate,
• 29:26 - 29:27
like constants and so on.
• 29:27 - 29:29
That's a beautiful case.
• 29:29 - 29:33
There was one that gave
• 29:33 - 29:35
and then I decided--
number 22, right?
• 29:35 - 29:40
Where this was more complicated
as an integrant in y, and guys,
• 29:40 - 29:43
• 29:43 - 29:47
And then normally to integrate
with respect to y and then x,
• 29:47 - 29:51
you would have had to split this
integral into two integrals--
• 29:51 - 29:53
one over a part of the
triangle, the other one
• 29:53 - 29:55
over part of the triangle.
• 29:55 - 30:00
So the easier way
was to do it how?
• 30:00 - 30:04
To do it like that, with
horizontal integrals.
• 30:04 - 30:05
And we've done that.
• 30:05 - 30:08
I told you-- I gave
you too much, actually,
• 30:08 - 30:13
I served it to you on a
plate, the proof-- solution
• 30:13 - 30:14
of that problem.
• 30:14 - 30:15
But you have many others.
• 30:15 - 30:21
Now, how do we prove that
this individual equation that
• 30:21 - 30:23
looks so sophisticated
is nothing
• 30:23 - 30:30
but that for the case
when S is a planar patch?
• 30:30 - 30:34
If S is like a hill,
yeah, then we believe it.
• 30:34 - 30:38
But what if S is the
domain d in plane Well,
• 30:38 - 30:43
then this S is exactly this d.
• 30:43 - 30:45
So it reduces to d.
• 30:45 - 30:47
So you say, wait a
minute, doesn't it
• 30:47 - 30:48
have to be curvilinear?
• 30:48 - 30:50
Nope.
• 30:50 - 30:56
Any surface that is bounded
by c verifies Stokes' theorem.
• 30:56 - 30:58
Say it again, Magdalena.
• 30:58 - 31:02
Any surface S that
is regular, so I'm
• 31:02 - 31:05
within the conditions
of the theorem, that
• 31:05 - 31:07
is bounded by a Jordan curve,
will satisfy the theorem.
• 31:07 - 31:09
So let's see what I've become.
• 31:09 - 31:12
That should became
a friend of yours,
• 31:12 - 31:15
who this guy is.
• 31:15 - 31:25
So the integral FdR is your
friend integral Mdx plus Ndy
• 31:25 - 31:27
that's staring at you over c.
• 31:27 - 31:31
It's an integral over one
form, and it says that's work.
• 31:31 - 31:35
And the right-hand side, it's
a little bit more complicated.
• 31:35 - 31:37
So we have to think.
• 31:37 - 31:38
We have to think.
• 31:38 - 31:42
• 31:42 - 31:46
are at identifying everybody.
• 31:46 - 31:50
If I go, for this particular
case, S is d, right?
• 31:50 - 31:52
Right.
• 31:52 - 31:57
So I have a double integral
over D. Sometimes you ask me,
• 31:57 - 32:02
but I saw that over a domain
that's a two-dimensional domain
• 32:02 - 32:05
people wrote only one
snake, and it looks fat,
• 32:05 - 32:08
like somebody fed
the snake too much.
• 32:08 - 32:11
Mathematicians are lazy people.
• 32:11 - 32:14
They don't want to write always
double snake, triple snake.
• 32:14 - 32:17
So sometimes they say,
I have an integral
• 32:17 - 32:18
over an n-dimensional domain.
• 32:18 - 32:20
I'll make it a fat snake.
• 32:20 - 32:23
And that should be enough.
• 32:23 - 32:26
Curl F N-- we have
to do this together.
• 32:26 - 32:26
Is it hard?
• 32:26 - 32:27
I don't know.
• 32:27 - 32:31
You have to help me.
• 32:31 - 32:33
So what in the world was that?
• 32:33 - 32:34
I pretend I forgot everything.
• 32:34 - 32:35
I have amnesia.
• 32:35 - 32:36
STUDENT: [INAUDIBLE].
• 32:36 - 32:39
• 32:39 - 32:41
PROFESSOR: Yeah, so
actually some of you
• 32:41 - 32:46
told me by email
that you prefer that.
• 32:46 - 32:49
I really like it
that you-- maybe I
• 32:49 - 32:52
should have started a
• 32:52 - 32:56
personal email interaction
• 32:56 - 33:00
between me and you,
everybody could see this.
• 33:00 - 33:04
So some of you tell me, I
like better this notation,
• 33:04 - 33:06
because I use it
in my engineering
• 33:06 - 33:12
course, curl F. OK, good, it's
up to you what you want to use.
• 33:12 - 33:16
d/dx, d/dy-- I mean it.
• 33:16 - 33:19
In principle, in r3,
but I'm really lucky.
• 33:19 - 33:23
Because in this case, F
is in r2, value in r2.
• 33:23 - 33:24
STUDENT: You mean d/dz?
• 33:24 - 33:25
PROFESSOR: Huh?
• 33:25 - 33:26
STUDENT: d/dx, d/dy, d/dz.
• 33:26 - 33:27
PROFESSOR: I'm sorry.
• 33:27 - 33:30
You are so on the ball.
• 33:30 - 33:31
Thank you, Alexander.
• 33:31 - 33:33
STUDENT: No, I thought I had
completely misunderstood--
• 33:33 - 33:36
PROFESSOR: No, no, no,
no, I wrote it twice.
• 33:36 - 33:42
So M and N and 0, M is a
function of x and y only.
• 33:42 - 33:44
N of course-- do I
have to write that?
• 33:44 - 33:47
No, I'm just being silly.
• 33:47 - 33:50
And what do I get in this case?
• 33:50 - 33:51
STUDENT: [INAUDIBLE].
• 33:51 - 33:56
• 33:56 - 33:59
PROFESSOR: I times this
guy-- how much is this guy?
• 33:59 - 34:00
STUDENT: 0.
• 34:00 - 34:00
PROFESSOR: 0.
• 34:00 - 34:02
Why is that 0?
• 34:02 - 34:05
Because this contains no z,
and I prime with respect to z.
• 34:05 - 34:12
So that is nonsense,
0i minus 0j.
• 34:12 - 34:12
Why is that?
• 34:12 - 34:17
Because 0 minus something
that doesn't depend on z.
• 34:17 - 34:23
So plus, finally-- the
only guy that matters there
• 34:23 - 34:28
is [INAUDIBLE], which
is this, which is that.
• 34:28 - 34:32
So because I have
derivative of N
• 34:32 - 34:35
with respect to h
minus derivative of M
• 34:35 - 34:36
with respect to y.
• 34:36 - 34:39
• 34:39 - 34:42
And now I stare at it,
and I say, times k.
• 34:42 - 34:46
That's the only guy that's
not 0, the only component.
• 34:46 - 34:49
Now I'm going to go
• 34:49 - 34:51
in the top product with him.
• 34:51 - 34:55
But we have to be smart
and think, N is what?
• 34:55 - 34:56
STUDENT: [INAUDIBLE].
• 34:56 - 34:58
PROFESSOR: It's
normal to the surface.
• 34:58 - 35:02
But the surface is
a patch of a plane.
• 35:02 - 35:03
The normal would be trivial.
• 35:03 - 35:05
What will the normal be?
• 35:05 - 35:11
The vector field of all
pencils that are k-- k.
• 35:11 - 35:13
It's all k, k everywhere.
• 35:13 - 35:15
All over the domain is k.
• 35:15 - 35:16
So N becomes k.
• 35:16 - 35:18
Where is it?
• 35:18 - 35:21
There, N becomes k.
• 35:21 - 35:24
So when you multiply
in the dot product
• 35:24 - 35:27
this guy with this
guy, what do you have?
• 35:27 - 35:28
STUDENT: [INAUDIBLE].
• 35:28 - 35:35
PROFESSOR: N sub x
minus M sub y dA.
• 35:35 - 35:38
• 35:38 - 35:42
QED-- what does it mean, QED?
• 35:42 - 35:44
\$1, which I don't
have, for the person
• 35:44 - 35:47
who will tell me what that is.
• 35:47 - 35:51
• 35:51 - 36:05
Latin-- quod erat demonstrandum,
which was to be proved, yes?
• 36:05 - 36:06
So I'm done.
• 36:06 - 36:10
When people put QED, that means
they are done with the proof.
• 36:10 - 36:14
But now since mathematicians
are a little bit illiterate,
• 36:14 - 36:17
philosophy or linguistics.
• 36:17 - 36:19
Now many of them,
• 36:19 - 36:23
they put a little square box.
• 36:23 - 36:25
And we do the same in our books.
• 36:25 - 36:28
So that means I'm
done with the proof.
• 36:28 - 36:31
Let's go home, but not that.
• 36:31 - 36:40
So we proved that for the
particular case of the planar
• 36:40 - 36:44
domains, Stokes' theorem
becomes Green's theorem.
• 36:44 - 36:46
And actually this is the curl.
• 36:46 - 36:50
And this-- well, not the curl.
• 36:50 - 36:55
But you have the curl of F
multiplied with dot product
• 36:55 - 36:58
with k and this green
fellow is exactly
• 36:58 - 37:02
the same as N sub
x minus N sub y
• 37:02 - 37:07
smooth function,
real value function.
• 37:07 - 37:10
All right, am I done?
• 37:10 - 37:13
Yes, with this Exercise 1,
which is a proof, I'm done.
• 37:13 - 37:18
You haven't seen many
proofs in calculus.
• 37:18 - 37:21
You've seen some from
me that we never cover.
• 37:21 - 37:25
We don't do epsilon delta in
regular classes of calculus,
• 37:25 - 37:26
only in honors.
• 37:26 - 37:29
And not in all the honors
you've seen some proofs
• 37:29 - 37:30
with epsilon delta.
• 37:30 - 37:36
You've seen one or two
proofs from me occasionally.
• 37:36 - 37:40
And this was one simple proof
that I wanted to work with you.
• 37:40 - 37:43
Now, do you know if
you're ever going
• 37:43 - 37:47
to see proofs in math
classes, out of curiosity?
• 37:47 - 37:51
US It depends how much
math you want to take.
• 37:51 - 37:56
If you're a math major, you
take a course called 3310.
• 37:56 - 37:59
That's called
Introduction to Proofs.
• 37:59 - 38:02
If you are not a math
major, but assume
• 38:02 - 38:06
you are in this
dual program-- we
• 38:06 - 38:11
have a beautiful and tough dual
major, mathematics and computer
• 38:11 - 38:15
science, 162 hours.
• 38:15 - 38:18
Then you see everything
you would normally
• 38:18 - 38:20
see for an engineering major.
• 38:20 - 38:22
see a few more courses
• 38:22 - 38:25
that have excellent proofs.
• 38:25 - 38:29
And one of them is linear
algebra, Linear Algebra 2360.
• 38:29 - 38:33
We do a few proofs--
depends who teaches that.
• 38:33 - 38:35
And in 3310 also
you see some proofs
• 38:35 - 38:38
let's prove this and that.
• 38:38 - 38:41
OK, so it's sort of fun.
• 38:41 - 38:46
But we don't attempt long
and nasty, complicated proofs
• 38:46 - 38:49
school, normally.
• 38:49 - 38:53
Some of you will do
• 38:53 - 38:54
Some of you-- I
know four of you--
• 38:54 - 38:58
want to go to medical school.
• 38:58 - 39:02
And then many of you hopefully
• 39:02 - 39:06
in engineering.
• 39:06 - 39:11
OK, let's see another
example for this section.
• 39:11 - 39:15
I don't particularly
like all the examples
• 39:15 - 39:18
we have in the book.
• 39:18 - 39:22
But I have my favorites.
• 39:22 - 39:31
And I'm going to go
• 39:31 - 39:35
• 39:35 - 39:38
There is one that's a
little bit complicated.
• 39:38 - 39:41
• 39:41 - 39:46
And I wanted to
• 39:46 - 39:51
Because it gave several
• 39:51 - 40:00
There is Example 1, which
says-- what does it say?
• 40:00 - 40:07
Evaluate fat integral
over C of 1 over 2 i
• 40:07 - 40:23
squared dx plus zdy plus xdz
where C is the intersection
• 40:23 - 40:47
curve between the plane x plus
z equals 1 and the ellipsoid x
• 40:47 - 40:50
squared plus 2y
squared plus z squared
• 40:50 - 41:01
equals 1 that's oriented
counterclockwise as viewed
• 41:01 - 41:03
from the above picture.
• 41:03 - 41:06
And I need to draw the picture.
• 41:06 - 41:10
The picture looks really ugly.
• 41:10 - 41:12
You have this ellipsoid.
• 41:12 - 41:17
• 41:17 - 41:24
And when you draw
this intersection
• 41:24 - 41:30
between this plane and the
ellipsoid, it looks horrible.
• 41:30 - 41:33
And the hint of
this problem-- well,
• 41:33 - 41:38
if you were to be given
such a thing on an exam,
• 41:38 - 41:42
the hint would be
that a projection--
• 41:42 - 41:43
look at the picture.
• 41:43 - 41:51
The projection of the curve of
intersection on the ground--
• 41:51 - 41:54
ground means the
plane on the equator.
• 41:54 - 41:55
How shall I say that?
• 41:55 - 42:01
The x, y plane is this.
• 42:01 - 42:03
It looks horrible.
• 42:03 - 42:10
• 42:10 - 42:12
And it looks like an egg.
• 42:12 - 42:14
It's not supposed
to be an egg, OK?
• 42:14 - 42:15
It's a circle.
• 42:15 - 42:17
I'm sorry if it
looks like an egg.
• 42:17 - 42:21
• 42:21 - 42:25
OK, and that would be the
only hint you would get.
• 42:25 - 42:29
figure out this circle
• 42:29 - 42:32
in polar coordinates.
• 42:32 - 42:37
And I'm not sure if all of
you would know how to do that.
• 42:37 - 42:41
And this is what worried me.
• 42:41 - 42:45
So before we do everything,
before everything,
• 42:45 - 42:50
can we express this
in polar coordinates?
• 42:50 - 42:54
How are you going to set
up something in r theta
• 42:54 - 42:59
for the same domain
inside this disc?
• 42:59 - 43:00
STUDENT: [INAUDIBLE].
• 43:00 - 43:28
• 43:28 - 43:29
PROFESSOR: So if we
were, for example,
• 43:29 - 43:33
to say x is r cosine
theta, can we do that?
• 43:33 - 43:36
And i to be r sine
theta, what would
• 43:36 - 43:38
we get instead of this equation?
• 43:38 - 43:41
Because it looks horrible.
• 43:41 - 43:45
We would get-- this equation,
let's brush it up a little bit
• 43:45 - 43:46
first.
• 43:46 - 43:48
It's x squared plus y squared.
• 43:48 - 43:49
And that's nice.
• 43:49 - 43:57
But then it's minus twice--
it's just x plus 1/4
• 43:57 - 44:01
equals 1/4, the heck with it.
• 44:01 - 44:02
My son says, don't say "heck."
• 44:02 - 44:03
• 44:03 - 44:05
I didn't know that.
• 44:05 - 44:08
But he says that he's being
told in school it's a bad word.
• 44:08 - 44:10
So he must know what
• 44:10 - 44:13
So this is r squared.
• 44:13 - 44:15
And x is r cosine theta.
• 44:15 - 44:21
Aha, so there we almost
did it in the sense
• 44:21 - 44:24
that r squared equals
r cosine theta is
• 44:24 - 44:27
the polar equation,
equation of the circle
• 44:27 - 44:29
in polar coordinates.
• 44:29 - 44:32
But we hate r.
• 44:32 - 44:33
Let's simplify by an r.
• 44:33 - 44:37
Because r is positive--
cannot be 0, right?
• 44:37 - 44:38
It would be a point.
• 44:38 - 44:44
So divide by r and get
r equals cosine theta.
• 44:44 - 44:46
So what is r equals
cosine theta?
• 44:46 - 44:50
r equals cosine theta
• 44:50 - 44:52
So I'm going to make a face.
• 44:52 - 44:55
nightmare in Calculus II.
• 44:55 - 44:59
And I was just talking to a
few colleagues in Calculus II
• 44:59 - 45:01
telling me that the
students don't know that,
• 45:01 - 45:06
and they have a big
hard time with that.
• 45:06 - 45:12
So the equation of this circle
is r equals cosine theta.
• 45:12 - 45:16
So if I were to
express this domain,
• 45:16 - 45:19
which in Cartesian
coordinates would be written--
• 45:19 - 45:22
I don't know if you want to--
as double integral, We'd?
• 45:22 - 45:25
Have to do the
vertical strip thingy.
• 45:25 - 45:29
But if I want to do it
in polar coordinates,
• 45:29 - 45:34
I'm going to say,
I start-- well,
• 45:34 - 45:36
you have to tell
me what you think.
• 45:36 - 45:42
• 45:42 - 45:45
We have an r that
starts with the origin.
• 45:45 - 45:50
And that's dr.
How far does r go?
• 45:50 - 45:56
For the domain inside, r goes
between 0 and cosine theta.
• 45:56 - 45:58
STUDENT: Why were you
able to divide by r
• 45:58 - 45:59
if it could have equaled 0?
• 45:59 - 46:00
• 46:00 - 46:03
STUDENT: Yes, but
then you just said
• 46:03 - 46:05
you could only do that
because it never equaled 0.
• 46:05 - 46:09
PROFESSOR: Right, and for
0 we pull out one point
• 46:09 - 46:13
where we take the
angle that we want.
• 46:13 - 46:16
We will still get
the same thing.
• 46:16 - 46:17
STUDENT: [INAUDIBLE].
• 46:17 - 46:20
• 46:20 - 46:22
PROFESSOR: No, r will be any--
• 46:22 - 46:25
• 46:25 - 46:26
STUDENT: Oh, I see.
• 46:26 - 46:28
PROFESSOR: Yeah,
so little r, what
• 46:28 - 46:30
is the r of any
little point inside?
• 46:30 - 46:34
The r of any little
point inside is
• 46:34 - 46:38
between 0 and N cosine theta.
• 46:38 - 46:42
Cosine theta would be the r
corresponding to the boundary.
• 46:42 - 46:45
Say it again-- so every
point on the boundary
• 46:45 - 46:49
will have that r
equals cosine theta.
• 46:49 - 46:56
The points inside the domain--
and this is on the circle,
• 46:56 - 46:59
on C. This is the circle.
• 46:59 - 47:03
Let's call it C ground.
• 47:03 - 47:08
That is the C.
• 47:08 - 47:14
So the r, the points
inside have one property,
• 47:14 - 47:16
that their r is between
0 and cosine theta.
• 47:16 - 47:19
If I take r theta
with this property,
• 47:19 - 47:21
I should be able to
get all the domain.
• 47:21 - 47:27
But theta, you have to be a
• 47:27 - 47:30
STUDENT: It goes from pi
over 2 to negative pi over 2.
• 47:30 - 47:31
PROFESSOR: Actually, yes.
• 47:31 - 47:38
So you have theta will be
between minus pi over 2
• 47:38 - 47:39
and pi over 2.
• 47:39 - 47:42
And you have to think
• 47:42 - 47:44
you set up the double integral.
• 47:44 - 47:45
But you're not there yet.
• 47:45 - 47:48
So when we'll be there
at the double integral
• 47:48 - 47:50
we will have to think about it.
• 47:50 - 47:54
• 47:54 - 47:56
What else did I want?
• 47:56 - 48:06
• 48:06 - 48:12
All right, so did I give
you the right form of F?
• 48:12 - 48:14
Yes.
• 48:14 - 48:18
I'd like you to compute curl
F and N all by yourselves.
• 48:18 - 48:19
So compute.
• 48:19 - 48:24
• 48:24 - 48:27
This is going to be F1.
• 48:27 - 48:29
This is going to be F2.
• 48:29 - 48:31
This is going to be F3.
• 48:31 - 48:35
And I'd like you to
realize that this
• 48:35 - 48:42
is nothing but integral over
C F dR. So who is this animal?
• 48:42 - 48:43
This is the work, guys.
• 48:43 - 48:46
• 48:46 - 48:50
All right, so I should
be able to set up
• 48:50 - 48:54
some integral, double
integral, over a surface
• 48:54 - 49:02
where I have curl F times N dS.
• 49:02 - 49:06
So what I want you to
do is simply-- maybe
• 49:06 - 49:08
I'm a little bit too lazy.
• 49:08 - 49:10
Take the curl of F and
tell me what it is.
• 49:10 - 49:15
Take the unit normal vector
field and tell me what it is.
• 49:15 - 49:23
• 49:23 - 49:24
And then we will
figure out the rest.
• 49:24 - 49:28
• 49:28 - 49:30
So you say, wait a
minute, Magdalena, now
• 49:30 - 49:34
you want me to look at this
Stokes' theorem over what
• 49:34 - 49:34
surface?
• 49:34 - 49:37
Because C is the red boundary.
• 49:37 - 49:44
So you want me to look at
this surface, right, the cap?
• 49:44 - 49:46
So the surface could be the cap.
• 49:46 - 49:49
But what did I tell you before?
• 49:49 - 49:57
I told you that Stokes' theorem
works for any kind of domain
• 49:57 - 50:02
that is bounded by the
curve C. So is this the way
• 50:02 - 50:06
you're going to do it-- take
the cap, put the normals,
• 50:06 - 50:10
find the normals, and do all
the horrible computation?
• 50:10 - 50:12
Or you will simplify
• 50:12 - 50:18
that this is exactly the same
as the integral evaluated
• 50:18 - 50:22
over any surface bounded by C.
• 50:22 - 50:25
Well, this horrible thing
is going to kill us.
• 50:25 - 50:27
So what's the simplest
way to do this?
• 50:27 - 50:32
• 50:32 - 50:35
It would be to do it
over another surface.
• 50:35 - 50:37
It doesn't matter
what surface you have.
• 50:37 - 50:41
This is the C. You can take any
surface that's bounded by C.
• 50:41 - 50:43
You can take this balloon.
• 50:43 - 50:45
You can take this one.
• 50:45 - 50:47
You can take the
disc bounded by C.
• 50:47 - 50:50
You can take any surface
that's bounded by C.
• 50:50 - 50:54
So in particular,
what if you take
• 50:54 - 51:00
the surface inside
this red disc,
• 51:00 - 51:04
the planar surface
inside that red disc?
• 51:04 - 51:08
OK, do you see it?
• 51:08 - 51:10
OK, that's going to
be part of a plane.
• 51:10 - 51:12
What is that plane?
• 51:12 - 51:15
x plus z equals 1.
• 51:15 - 51:18
So you guys have to
tell me who N will be
• 51:18 - 51:20
and who the curl will be.
• 51:20 - 51:25
• 51:25 - 51:30
And let me show you again
with my hands what you have.
• 51:30 - 51:34
You have a surface that's
curvilinear and round
• 51:34 - 51:35
and has boundary C.
• 51:35 - 51:40
The boundary is C. You
have another surface that's
• 51:40 - 51:45
an ellipse that has
C as a boundary.
• 51:45 - 51:47
And this is sitting in a plane.
• 51:47 - 51:51
And I want-- it's very hard
to model with my hands.
• 51:51 - 51:51
But this is it.
• 51:51 - 51:52
You see it?
• 51:52 - 51:53
You see it?
• 51:53 - 51:57
OK, when you project
this on the ground,
• 51:57 - 52:01
this is going to become that
circle that I just erased,
• 52:01 - 52:03
so this and that.
• 52:03 - 52:05
We have a surface integral.
• 52:05 - 52:09
Remember, you have dS here up,
and you have dA here down--
• 52:09 - 52:10
dS here up, dA here down.
• 52:10 - 52:13
So that shouldn't be
hard to do at all.
• 52:13 - 52:18
Now what is N?
• 52:18 - 52:22
N, for such an individual,
will be really nice and sassy.
• 52:22 - 52:27
x plus z equals 1.
• 52:27 - 52:34
So what is the normal
to the plane x plus z?
• 52:34 - 52:34
[INAUDIBLE]
• 52:34 - 52:39
• 52:39 - 52:47
So who is this normal for
D curl F times [INAUDIBLE]
• 52:47 - 52:55
but N d-- I don't know,
another S, S tilde.
• 52:55 - 53:01
So for this kind of surface,
I have another dS, dS tilde.
• 53:01 - 53:05
So who's going to
tell me who N is?
• 53:05 - 53:07
• 53:07 - 53:14
Well, it should be
x plus z equals 1.
• 53:14 - 53:15
What do we keep?
• 53:15 - 53:16
What do we throw away?
• 53:16 - 53:18
The plane is x plus z equals 1.
• 53:18 - 53:19
What's the normal?
• 53:19 - 53:23
• 53:23 - 53:28
So the plane is x plus
0y plus 1z equals 1.
• 53:28 - 53:30
What's the normal to the plane?
• 53:30 - 53:32
STUDENT: Is it i plus k
over square root of 2?
• 53:32 - 53:34
PROFESSOR: i plus k,
very good, but why
• 53:34 - 53:38
does Alexander say the
over square root of 2?
• 53:38 - 53:40
Because it says,
remember guys, that that
• 53:40 - 53:43
has to be a unit normal.
• 53:43 - 53:48
We cannot take i plus k based
on being perpendicular to the x
• 53:48 - 53:49
plus z.
• 53:49 - 53:51
Because you need
to normalize it.
• 53:51 - 53:52
So he did.
• 53:52 - 53:58
So he got i plus k
over square root of 2.
• 53:58 - 54:00
How much is curl F?
• 54:00 - 54:02
You have to do
this by yourselves.
• 54:02 - 54:04
I'll just give it to you.
• 54:04 - 54:06
I'll give you three
minutes, and then I'll
• 54:06 - 54:10
• 54:10 - 54:16
• 54:16 - 54:19
And in the end, I'll have to do
the dot product and keep going.
• 54:19 - 55:33
• 55:33 - 55:34
Is it hard?
• 55:34 - 55:37
I should do it
along with you guys.
• 55:37 - 55:42
I have i jk d/dx, d/dy, d/dz.
• 55:42 - 55:46
• 55:46 - 55:51
Who were the guys? y
squared over 2 was F1.
• 55:51 - 55:53
z was F2.
• 55:53 - 55:55
x was F3.
• 55:55 - 55:59
• 55:59 - 56:01
And let's see what you got.
• 56:01 - 56:05
• 56:05 - 56:09
I'm checking to see if
you get the same thing.
• 56:09 - 56:11
Minus psi is the first guy.
• 56:11 - 56:14
[INAUDIBLE] the next one?
• 56:14 - 56:15
STUDENT: Minus j.
• 56:15 - 56:17
PROFESSOR: Minus j.
• 56:17 - 56:19
STUDENT: Minus yk.
• 56:19 - 56:21
PROFESSOR: yk.
• 56:21 - 56:24
And I think that's
what it is, yes.
• 56:24 - 56:28
So when you do the integral,
what are you going to get?
• 56:28 - 56:30
I'm going to erase this here.
• 56:30 - 56:35
• 56:35 - 56:41
• 56:41 - 56:43
What was it again, Alexander?
• 56:43 - 56:47
i plus k over square
root of 2, right?
• 56:47 - 56:50
So let's write
down the integral W
• 56:50 - 56:56
will be-- double
integral over the domain.
• 56:56 - 57:03
Now, in our case, the domain
is this domain, this one here.
• 57:03 - 57:07
Let's call it-- do you want to
call it D or D star or D tilde?
• 57:07 - 57:08
I don't know what.
• 57:08 - 57:11
Because we use to call
the domain on the ground
• 57:11 - 57:15
D. Let's put here D star.
• 57:15 - 57:21
So over D star, and the cap
• 57:21 - 57:22
anymore.
• 57:22 - 57:24
You said, bye-bye bubble.
• 57:24 - 57:27
I can do the whole
computation on D star.
• 57:27 - 57:29
• 57:29 - 57:32
So you help me right?
• 57:32 - 57:38
I get minus 1 times
1 over root 2.
• 57:38 - 57:40
Am I right?
• 57:40 - 57:49
A 0 for the middle term, and
a minus y times 1 over root 2,
• 57:49 - 57:54
good-- this is the
whole thing over here.
• 57:54 - 57:57
My worry is about dS star.
• 57:57 - 58:00
What was dS star?
• 58:00 - 58:07
dS star is the area
limit for the plane-- are
• 58:07 - 58:11
limit for how can I call this?
• 58:11 - 58:15
For disc, for D star, not for D.
• 58:15 - 58:17
It's a little bit complicated.
• 58:17 - 58:19
D is a projection.
• 58:19 - 58:23
So who reminds me how we did it?
• 58:23 - 58:29
dS star was what times dA?
• 58:29 - 58:31
This is the surface area.
• 58:31 - 58:36
And if you have a surface that's
• 58:36 - 58:39
STUDENT: It's area, so r?
• 58:39 - 58:42
equation of this surface
• 58:42 - 58:43
up here?
• 58:43 - 58:48
This is the ellipse that goes
projected on the surface.
• 58:48 - 58:50
STUDENT: Cosine of theta.
• 58:50 - 58:52
PROFESSOR: The equation
of the plane, see?
• 58:52 - 58:54
The equation of the plane.
• 58:54 - 58:55
So I erased it.
• 58:55 - 58:57
So was it x plus z equals 1?
• 58:57 - 58:58
STUDENT: Yes.
• 58:58 - 59:02
PROFESSOR: So z
must be 1 minus x.
• 59:02 - 59:10
So this is going to be the
square root of 1 plus-- minus 1
• 59:10 - 59:11
is the first partial.
• 59:11 - 59:13
Are you guys with me?
• 59:13 - 59:19
Partial with respect to x of
this guy is minus 1 squared
• 59:19 - 59:23
plus the partial of
this with respect to y
• 59:23 - 59:26
is missing 0 squared.
• 59:26 - 59:28
And then comes
dA, and who is dA?
• 59:28 - 59:33
dA is dxdy in the floor plane.
• 59:33 - 59:40
This is the [INAUDIBLE] that
projects onto the floor.
• 59:40 - 59:44
Good, ds star is going to
be then square root 2dA.
• 59:44 - 59:46
Again, the old
trick that I taught
• 59:46 - 59:50
you guys is that
this will always
• 59:50 - 59:54
have to simplify with
[INAUDIBLE] on the bottom
• 59:54 - 59:56
of the N. Say what?
• 59:56 - 59:58
Magdalena, say it again.
• 59:58 - 60:03
Square root of 2DA, this
is that magic square root
• 60:03 - 60:06
of 1 plus [INAUDIBLE].
• 60:06 - 60:10
This guy, no matter what
exercise you are doing,
• 60:10 - 60:20
will always simplify with
the bottom of N [INAUDIBLE],
• 60:20 - 60:22
so you can do this
simplification
• 60:22 - 60:24
from the beginning.
• 60:24 - 60:26
And so in the end, what
are you going to have?
• 60:26 - 60:36
You're going to have
W is minus y minus 1
• 60:36 - 60:40
over the domain D in the
plane that this will claim.
• 60:40 - 60:43
• 60:43 - 60:46
The square root of
[INAUDIBLE], and then you'll
• 60:46 - 60:51
have dA, which is dxdy
• 60:51 - 60:56
OK, at this point suppose
that you are taking the 5.
• 60:56 - 60:59
And this is why I
got to this point
• 60:59 - 61:05
because I wanted
to emphasize this.
• 61:05 - 61:09
Whether you stop here
or you do one more step,
• 61:09 - 61:11
I would be happy.
• 61:11 - 61:13
Let's see what I mean.
• 61:13 - 61:22
So you would have minus who
is y r cosine theta minus 1.
• 61:22 - 61:25
of dxdy, you have--
• 61:25 - 61:25
STUDENT: [INAUDIBLE].
• 61:25 - 61:28
PROFESSOR: r, very
good. r dr is theta.
• 61:28 - 61:31
• 61:31 - 61:34
So you're thinking--
• 61:34 - 61:35
STUDENT: [INAUDIBLE].
• 61:35 - 61:40
PROFESSOR: --well,
so you're thinking--
• 61:40 - 61:48
I'm looking here what we have--
r was from 0 to cosine theta,
• 61:48 - 61:51
and theta is from
minus [INAUDIBLE].
• 61:51 - 61:54
• 61:54 - 61:59
So in the exam, we will not
expect-- on some integrals who
• 61:59 - 62:02
are not expected to
go on and do them,
• 62:02 - 62:05
which they set up the
integral and leave it.
• 62:05 - 62:05
Yes, sir?
• 62:05 - 62:08
STUDENT: Why did you throw
r cosine theta for y?
• 62:08 - 62:11
PROFESSOR: OK, because
let me remind you,
• 62:11 - 62:15
when you project the image
of this ellipse on the plane,
• 62:15 - 62:20
we got this fellow, which
is drawn in the book
• 62:20 - 62:23
as being this.
• 62:23 - 62:27
So we said, I want to
see how I set this up
• 62:27 - 62:29
in [INAUDIBLE] coordinates.
• 62:29 - 62:32
The equation of the
plane of the circle
• 62:32 - 62:34
was r equals cosine theta,
and this was calculus too.
• 62:34 - 62:38
That's why we
actually [INAUDIBLE].
• 62:38 - 62:40
So if somebody
• 62:40 - 62:46
area over the domain, what
• 62:46 - 62:49
if you compute for me the
linear area of the domain?
• 62:49 - 62:50
How would you do that?
• 62:50 - 62:58
Well, double integral of 1
or whatever-that-is integral
• 62:58 - 63:05
of r drd theta,
• 63:05 - 63:08
can have some other ugly
integral looking at you.
• 63:08 - 63:11
I put the stop here.
• 63:11 - 63:16
Theta is between minus
pie over 2n pi over 2
• 63:16 - 63:23
because I'm moving from here
to here, from here to here, OK?
• 63:23 - 63:28
Nr is between 0 and the margin.
• 63:28 - 63:29
Who is on the margin?
• 63:29 - 63:30
I started 0.
• 63:30 - 63:31
I ended cosine theta.
• 63:31 - 63:34
I started 0, ended cosine theta.
• 63:34 - 63:39
Cosine theta happens
online for the boundary,
• 63:39 - 63:40
so that's what you do.
• 63:40 - 63:41
Do we want you to do that?
• 63:41 - 63:43
No, we want you to leave it.
• 63:43 - 63:44
Yes?
• 63:44 - 63:50
STUDENT: He was asking why you
had a negative y minus 1 r sine
• 63:50 - 63:52
theta, not r cosine theta.
• 63:52 - 63:54
PROFESSOR: You are so right.
• 63:54 - 64:01
I forgot that x was r cosine
theta, and y was r sine theta.
• 64:01 - 64:02
You are correct.
• 64:02 - 64:05
And you have the group
good observation.
• 64:05 - 64:10
So r was [INAUDIBLE]
cosine theta.
• 64:10 - 64:14
And x was r cosine theta.
• 64:14 - 64:17
y was r sine theta.
• 64:17 - 64:19
Very good.
• 64:19 - 64:23
OK, so if you get something
like that, we will now
• 64:23 - 64:29
want you to go on, we
will want you to stop.
• 64:29 - 64:34
Let me show you one
where we wanted to go on,
• 64:34 - 64:38
and we indicate it
like this, example 3.
• 64:38 - 64:44
• 64:44 - 64:49
So here, we just dont'
want you to show some work,
• 64:49 - 64:53
we wanted to actually
• 64:53 - 64:57
And I'll draw the picture,
and don't be afraid of it.
• 64:57 - 64:59
It's going to look
a little bit ugly.
• 64:59 - 65:02
• 65:02 - 65:08
You have the surface
Z equals 1 minus x
• 65:08 - 65:12
squared minus 2y squared.
• 65:12 - 65:20
And you have to evaluate
over double the integral
• 65:20 - 65:24
of the surface S.
This is the surface.
• 65:24 - 65:25
Let me draw the surface.
• 65:25 - 65:30
We will have to understand
what kind of surface that is.
• 65:30 - 65:35
• 65:35 - 65:42
Double integral of curl F
[INAUDIBLE] dS evaluated
• 65:42 - 65:56
where F equals xI plus y squared
J plus-- this looks like a Z e
• 65:56 - 65:58
to the xy.
• 65:58 - 65:58
It's very tiny.
• 65:58 - 66:00
I bet you won't see it.
• 66:00 - 66:08
[INAUDIBLE] xy k and S. Is
that part of the surface?
• 66:08 - 66:14
• 66:14 - 66:17
Let me change the marker so
the video can see better.
• 66:17 - 66:20
• 66:20 - 66:22
Z-- this is a bad marker.
• 66:22 - 66:27
• 66:27 - 66:34
Z equals-- what was it, guys?
• 66:34 - 66:41
1 minus x squared minus 2y
squared with Z positive or 0.
• 66:41 - 66:44
• 66:44 - 66:48
And the [? thing ?] is
I think we may give you
• 66:48 - 66:50
this hint on the exam.
• 66:50 - 66:56
Think of the Stokes theorem
and the typical-- think
• 66:56 - 67:06
of the Stokes theorem
and the typical tools.
• 67:06 - 67:10
You have learned them.
• 67:10 - 67:11
OK, what does it mean?
• 67:11 - 67:14
We have like an
eggshell, which is coming
• 67:14 - 67:16
from the parabola [INAUDIBLE].
• 67:16 - 67:20
This parabola [INAUDIBLE]
is S minus x squared
• 67:20 - 67:24
minus 2y squared,
and we call that S,
• 67:24 - 67:27
but you see, we have
two surfaces that are
• 67:27 - 67:30
in this picture bounded by c.
• 67:30 - 67:34
The other one is the domain
D, and it's a simple problem
• 67:34 - 67:38
sitting on the xy plane.
• 67:38 - 67:43
So it's a blessing that you
already know what D will be.
• 67:43 - 67:48
D will be those pairs
xy with what property?
• 67:48 - 67:53
Can you guys tell
me what D will be?
• 67:53 - 67:55
Z should be 0, right?
• 67:55 - 67:58
If you impose it
to be 0, then this
• 67:58 - 68:02
has to satisfy x squared
plus 2y squared less than
• 68:02 - 68:03
or equal to 1.
• 68:03 - 68:06
Who is the C?
• 68:06 - 68:08
C are the points
on the boundary,
• 68:08 - 68:13
which means exactly x squared
plus 2y squared is equal to 1.
• 68:13 - 68:15
What in the world is this curve?
• 68:15 - 68:17
STUDENT: [INAUDIBLE].
• 68:17 - 68:18
PROFESSOR: It's an ellipse.
• 68:18 - 68:19
Is it an ugly ellipse?
• 68:19 - 68:22
Uh, not really.
• 68:22 - 68:22
It's a nice ellipse.
• 68:22 - 68:26
• 68:26 - 68:29
OK, what do they give us?
• 68:29 - 68:32
They give us xy
squared and Z times
• 68:32 - 68:38
e to the xy, so this
is F1, F2, and F3.
• 68:38 - 68:43
• 68:43 - 68:47
So the surface itself
is just the part
• 68:47 - 68:52
that corresponds to Z
positive, not all the surface
• 68:52 - 68:55
because the whole surface
will be infinitely large.
• 68:55 - 69:00
It's a paraboloid that keeps
going down to minus infinite,
• 69:00 - 69:02
so you only take this part.
• 69:02 - 69:08
It's a finite patch that I stop.
• 69:08 - 69:12
So this is a problem
that's amazingly simple
• 69:12 - 69:14
once you solve it one time.
• 69:14 - 69:21
You don't even have to show your
work much in the actual exam,
• 69:21 - 69:23
and I'll show you why.
• 69:23 - 69:27
So Stokes theorem tells
you what in this case?
• 69:27 - 69:29
Let's review what
Stokes theorem says.
• 69:29 - 69:34
Stokes theorem says, OK,
you have the work performed
• 69:34 - 69:38
by the four steps that's
given to you as a vector value
• 69:38 - 69:42
function along the path
C, which is given to you
• 69:42 - 69:44
as this wonderful ellipse.
• 69:44 - 69:49
Let me put C like I did it
before, C. This is not L,
• 69:49 - 69:53
it's C, which what is that?
• 69:53 - 69:58
It's the same as
double integral over S,
• 69:58 - 70:05
the round paraboloid [INAUDIBLE]
like church roof, S curl F
• 70:05 - 70:09
times N dS.
• 70:09 - 70:11
But what does it
say, this happens
• 70:11 - 70:15
for any-- for every, for
any, do you know the sign?
• 70:15 - 70:16
STUDENT: [INAUDIBLE].
• 70:16 - 70:23
PROFESSOR: Surface
is bounded by C.
• 70:23 - 70:28
And here is that winking
emoticon from-- how
• 70:28 - 70:30
• 70:30 - 70:32
Something like that?
• 70:32 - 70:35
A wink would be a good
hint on the final.
• 70:35 - 70:38
What are you going to do
when you see that wink?
• 70:38 - 70:41
If it's not on the
final, I will wink at you
• 70:41 - 70:43
until you understand
what I'm trying to say.
• 70:43 - 70:49
It means that you can change
the surface to any other surface
• 70:49 - 70:52
that has the boundary
C. What's the simplest
• 70:52 - 70:53
surface you may think of?
• 70:53 - 70:54
STUDENT: [INAUDIBLE].
• 70:54 - 70:57
PROFESSOR: The D.
So I'm going to say,
• 70:57 - 71:03
double integral over D. Curl
left, God knows what that is.
• 71:03 - 71:06
We still have to
do some work here.
• 71:06 - 71:09
face because I really
• 71:09 - 71:12
wanted no work whatsoever.
• 71:12 - 71:17
N becomes-- we've done this
argument three times today.
• 71:17 - 71:18
STUDENT: It's k.
• 71:18 - 71:19
PROFESSOR: It's a k.
• 71:19 - 71:21
• 71:21 - 71:24
That's what you have
to indicate on the exam
• 71:24 - 71:28
that N is k when I look
at the plane or domain.
• 71:28 - 71:29
STUDENT: And dS is DA.
• 71:29 - 71:32
PROFESSOR: And dS is dA.
• 71:32 - 71:35
It's much simpler than
before because you
• 71:35 - 71:36
don't have to project.
• 71:36 - 71:37
• 71:37 - 71:39
You are all to the floor.
• 71:39 - 71:40
You are on the ground.
• 71:40 - 71:42
What else do you have to do?
• 71:42 - 71:47
Not much, you just have to be
patient and compute with me
• 71:47 - 71:49
something I don't like to.
• 71:49 - 71:52
to do it by yourself,
• 71:52 - 71:56
but now I shouldn't be lazy.
• 71:56 - 71:58
• 71:58 - 72:00
You have to help me.
• 72:00 - 72:09
i j k z dx z dx z dz of what?
• 72:09 - 72:14
x y squared and
this horrible guy.
• 72:14 - 72:19
• 72:19 - 72:21
What do we get?
• 72:21 - 72:24
• 72:24 - 72:28
Well, it's not so
obvious anymore.
• 72:28 - 72:29
STUDENT: [INAUDIBLE].
• 72:29 - 72:35
PROFESSOR: It's Z prime this
guy with respect to y Zx,
• 72:35 - 72:36
very good.
• 72:36 - 72:44
The x into the xy times i, and
I don't care about the rest.
• 72:44 - 72:45
Why don't I care about the rest?
• 72:45 - 72:49
Because when I prime y squared
with respect to Z goes away.
• 72:49 - 72:52
So I'm done with the first term.
• 72:52 - 72:57
I'm going very slow as you
can see, but I don't care.
• 72:57 - 72:58
So I'm going to erase more.
• 72:58 - 73:01
• 73:01 - 73:07
Next guy, minus and then
we'll make an observation.
• 73:07 - 73:11
The same thing here,
I go [INAUDIBLE].
• 73:11 - 73:18
So I have x to the
Z Zy e to the xy.
• 73:18 - 73:22
Are you guys with me, or
am I talking nonsense?
• 73:22 - 73:26
So what am I saying?
• 73:26 - 73:32
I'm saying that I expand with
respect to the j element here.
• 73:32 - 73:34
I have a minus because
of that, and then I
• 73:34 - 73:36
have the derivative of
this animal with respect
• 73:36 - 73:43
to x, which is Zy into
the x y j, correct?
• 73:43 - 73:44
STUDENT: Yes.
• 73:44 - 73:48
PROFESSOR: Finally, last but
not least, and actually that's
• 73:48 - 73:51
the most important guy, and
I'll tell you in a second why.
• 73:51 - 73:52
What is the last guy?
• 73:52 - 73:53
STUDENT: [INAUDIBLE].
• 73:53 - 73:54
PROFESSOR: 0.
• 73:54 - 74:00
So one of you will
hopefully realize what
• 74:00 - 74:02
I'm going to ask you right now.
• 74:02 - 74:09
No matter what I got here,
this was-- what is that called?
• 74:09 - 74:15
Work that is not necessary,
it's some stupid word.
• 74:15 - 74:17
So why is that not necessary?
• 74:17 - 74:24
Why could I have said star
i plus start j-- God knows
• 74:24 - 74:27
what that is-- plus 0k.
• 74:27 - 74:33
Because in the end, I have to
multiply that product with k,
• 74:33 - 74:37
so no matter what we do
here, and we sweat a lot.
• 74:37 - 74:39
And so no matter
what we put here
• 74:39 - 74:42
it would not have made a
difference because I have
• 74:42 - 74:47
to take this whole curl and
multiply as a dot with k,
• 74:47 - 74:53
and what matters is
only what's left over.
• 74:53 - 74:57
So my observation is this
whole thing is how much?
• 74:57 - 74:58
STUDENT: 0.
• 74:58 - 74:59
PROFESSOR: 0, thank God.
• 74:59 - 75:01
• 75:01 - 75:03
And we've given
this problem where
• 75:03 - 75:09
times on four different finals.
• 75:09 - 75:12
The thing is that many
students won't study,
• 75:12 - 75:15
and they didn't know the trick.
• 75:15 - 75:19
When you have a surface like
that, that bounds the curve
• 75:19 - 75:21
Stokes over the surface,
• 75:21 - 75:26
you do Stokes over the domain
and plane, and you'll get zero.
• 75:26 - 75:29
So poor kids, they
• 75:29 - 75:32
to compute this from
scratch for the surface,
• 75:32 - 75:34
and they got nowhere.
• 75:34 - 75:37
And then I started the fights
with, of course, [INAUDIBLE],
• 75:37 - 75:39
but they don't want to
give them any credit.
• 75:39 - 75:42
And I wanted to give
them at least some credit
• 75:42 - 75:44
for knowing the
theorem, the statement,
• 75:44 - 75:51
and trying to do something for
the nasty surface, the roof
• 75:51 - 75:53
that is a paraboloid.
• 75:53 - 75:56
They've done something,
so in the end,
• 75:56 - 75:58
I said I want to
do whatever I want,
• 75:58 - 76:00
and I gave partial credit.
• 76:00 - 76:03
But normally, I was told
not to give partial credit
• 76:03 - 76:08
for this kind of a thing because
the whole key of the problem
• 76:08 - 76:12
is to be smart,
understand the idea,
• 76:12 - 76:17
and get 0 without doing any
work, and that was nice.
• 76:17 - 76:18
Yes, sir?
• 76:18 - 76:20
STUDENT: Does that mean
that all we would really
• 76:20 - 76:22
needed to do compute
the curl is the k part?
• 76:22 - 76:25
Because if k would
have been something,
• 76:25 - 76:26
then there would have
been a dot on it.
• 76:26 - 76:28
PROFESSOR: Exactly,
but only if-- guys,
• 76:28 - 76:35
no matter what, if we give you,
if your surface has a planer
• 76:35 - 76:37
boundary-- say it again?
• 76:37 - 76:39
no matter what it
• 76:39 - 76:45
is-- it could look like
• 76:45 - 76:50
a boundary in the plane xy
like it is in geography,
• 76:50 - 76:53
imagine you have a
hill or something,
• 76:53 - 76:54
and that's the sea level.
• 76:54 - 76:58
And around the hill you have
the rim of the [INAUDIBLE].
• 76:58 - 77:00
• 77:00 - 77:04
Then you can reduce
to the plane,
• 77:04 - 77:07
and all the arguments
will be like that.
• 77:07 - 77:15
So the thing is you get 0
when the curl has 0 here,
• 77:15 - 77:17
and there is [INAUDIBLE].
• 77:17 - 77:18
Say it again?
• 77:18 - 77:23
When the F is given to you
so that the last component
• 77:23 - 77:27
of the curl is zero, you
will get 0 for the work.
• 77:27 - 77:32
Otherwise, you can get something
else, but not bad at all.
• 77:32 - 77:36
You can get something
that-- let's do
• 77:36 - 77:40
another example like that where
you have a simplification.
• 77:40 - 77:43
and erase the whole--
• 77:43 - 77:50
STUDENT: So, let's say if I
knew the [INAUDIBLE] equal to 0,
• 77:50 - 77:51
so I--
• 77:51 - 77:55
PROFESSOR: Eh, you cannot know
unless you look at the F first.
• 77:55 - 77:56
You see--
• 77:56 - 77:58
STUDENT: Let's say that
I put the F on stop,
• 77:58 - 78:01
and I put the equation,
which is F d r,
• 78:01 - 78:05
and I put the curl
F [INAUDIBLE], so
• 78:05 - 78:07
and then I said--
I looked at it.
• 78:07 - 78:09
I said, oh, it's a 0.
• 78:09 - 78:11
PROFESSOR: If you
see that's a big 0,
• 78:11 - 78:13
you can go at them
to 0 at the end.
• 78:13 - 78:13
STUDENT: OK.
• 78:13 - 78:16
PROFESSOR: Because the
dot product between k,
• 78:16 - 78:17
that's what matters.
• 78:17 - 78:21
The dot product between k and
the last component of the curl.
• 78:21 - 78:24
And in the end,
integral of 0 is 0.
• 78:24 - 78:26
And that is the lesson.
• 78:26 - 78:29
STUDENT: We should
also have N equal to k
• 78:29 - 78:31
if we don't have that.
• 78:31 - 78:35
PROFESSOR: Yeah, so I'm saying
if-- um, that's a problem.
• 78:35 - 78:39
This is not going to happen, but
assume that somebody gives you
• 78:39 - 78:45
a hill that looks like that,
and this is not a planar curve.
• 78:45 - 78:48
This would be a really
nasty curve in space.
• 78:48 - 78:50
You cannot do that anymore.
• 78:50 - 78:54
You have to apply [INAUDIBLE]
for the general surface.
• 78:54 - 79:01
But if your boundary sees a
planar boundary [INAUDIBLE],
• 79:01 - 79:06
then you can do that,
• 79:06 - 79:07
So let me give you
another example.
• 79:07 - 79:11
• 79:11 - 79:16
This time it's not going to be--
OK, you will see the surprise.
• 79:16 - 79:50
• 79:50 - 79:54
And you have a sphere, and
you have a spherical cap,
• 79:54 - 80:00
the sphere of radius R, and
this is going to be, let's say,
• 80:00 - 80:02
R to be 5.
• 80:02 - 80:07
And this is z equals 3.
• 80:07 - 80:09
You have the surface.
• 80:09 - 80:15
Somebody gives
you the surface S.
• 80:15 - 80:23
That is the spherical cap
of the sphere x squared
• 80:23 - 80:31
plus y squared plus z squared
equals 25 above the plane z
• 80:31 - 80:31
equals 3.
• 80:31 - 80:38
• 80:38 - 80:53
Compute double
integral of F times--
• 80:53 - 80:56
how did we phrase this
if we phrase it as a--
• 80:56 - 80:58
STUDENT: Curl FN?
• 80:58 - 81:00
PROFESSOR: No, he said, curl FN.
• 81:00 - 81:10
I'm sorry, if we rephrase
it as work curl FN
• 81:10 - 81:15
over S, whereas this
is the spherical cap.
• 81:15 - 81:20
This is S.
• 81:20 - 81:23
So you're going to
have this on the final.
• 81:23 - 81:25
First thing is, stay calm.
• 81:25 - 81:26
Don't freak out.
• 81:26 - 81:30
This is a typical--
you have to say, OK.
• 81:30 - 81:31
She prepared me well.
• 81:31 - 81:33
I did review, [INAUDIBLE].
• 81:33 - 81:35
For God's sake, I'm
going to do fine.
• 81:35 - 81:38
Just keep in mind that
no matter what we do,
• 81:38 - 81:40
it's not going to involve
a heavy computation like we
• 81:40 - 81:42
saw in that horrible
first example
• 81:42 - 81:45
I gave you-- second
example I gave you.
• 81:45 - 81:49
So the whole idea is to make
• 81:49 - 81:49
harder.
• 81:49 - 81:51
So what's the
first thing you do?
• 81:51 - 81:56
You take curl F, and you want
to see what that will be.
• 81:56 - 82:05
i j k is going to
be d dx, d dy, d dz.
• 82:05 - 82:07
And you say, all
right, then I'll
• 82:07 - 82:16
have x squared yz xy
squared z and xy z squared.
• 82:16 - 82:20
And then you say, well,
this look ugly, right?
• 82:20 - 82:22
That's what you're going to say.
• 82:22 - 82:31
So what times i minus what times
j plus what times k remains up
• 82:31 - 82:35
to you to clue the computation,
and you say, wait a minute.
• 82:35 - 82:39
The first minor is it math?
• 82:39 - 82:43
No, the first minor-- minor is
the name of such a determinant
• 82:43 - 82:45
is just a silly path.
• 82:45 - 82:47
So you do x yz squared
with respect to y,
• 82:47 - 83:02
it's xz squared minus prime
with respect to z dz xy squared.
• 83:02 - 83:06
Next guy, what do we have?
• 83:06 - 83:07
Who tells me?
• 83:07 - 83:09
He's sort of significant
but not really--
• 83:09 - 83:09
STUDENT: yz squared?
• 83:09 - 83:11
PROFESSOR:yz squared, good.
• 83:11 - 83:13
It's symmetric in a way.
• 83:13 - 83:17
• 83:17 - 83:19
x squared y, right guys?
• 83:19 - 83:20
Are you with me?
• 83:20 - 83:21
STUDENT: Mh-hmm.
• 83:21 - 83:23
PROFESSOR: And for
the k, you will have?
• 83:23 - 83:24
STUDENT: y squared z.
• 83:24 - 83:26
PROFESSOR: y squared z.
• 83:26 - 83:30
STUDENT: And x squared z.
• 83:30 - 83:36
PROFESSOR: z squared z because
if you look at this guy--
• 83:36 - 83:38
so we [INAUDIBLE] again.
• 83:38 - 83:43
And you say, well, I have
derivative with respect
• 83:43 - 83:45
to x is y squared z.
• 83:45 - 83:49
The derivative with
respect to y is
• 83:49 - 83:56
x squared y squared z and
then minus x squared z.
• 83:56 - 84:01
Then I have-- [INAUDIBLE].
• 84:01 - 84:05
• 84:05 - 84:06
I did, right?
• 84:06 - 84:08
So this is squared.
• 84:08 - 84:12
What matters is that I
check what I'm going to do,
• 84:12 - 84:18
so now I say, my
c is a boundaries.
• 84:18 - 84:22
That's a circle, so the
meaning of this integral given
• 84:22 - 84:25
by Stokes is actually
a path integral
• 84:25 - 84:28
along the c at the
level z equals 3.
• 84:28 - 84:31
I'm at the third floor looking
at the world from up there.
• 84:31 - 84:35
I have the circle on the
third floor z equals 3.
• 84:35 - 84:40
And then I say, that's going
to be F dot dR God knows what.
• 84:40 - 84:42
That was originally the work.
• 84:42 - 84:45
And Stokes theorem
says, no matter
• 84:45 - 84:47
what surfaces
you're going to take
• 84:47 - 84:50
to have a regular surface
without controversy,
• 84:50 - 84:55
without holes that bounded
family by the circles c,
• 84:55 - 84:57
you're going to be in business.
• 84:57 - 85:03
So I say, the heck with
the S. I want the D,
• 85:03 - 85:08
and I want that D to be colorful
because life is great enough.
• 85:08 - 85:11
Let's make it D.
• 85:11 - 85:14
That D has what meaning?
• 85:14 - 85:15
z equals 3.
• 85:15 - 85:18
I'm at the level
three, but also x
• 85:18 - 85:23
squared plus y squared must
be less than or equal to sum.
• 85:23 - 85:25
Could anybody tell
me what that is?
• 85:25 - 85:27
STUDENT: 16.
• 85:27 - 85:29
PROFESSOR: So how do I know?
• 85:29 - 85:32
I will just plug in a 3 here.
• 85:32 - 85:34
3 squared is 9.
• 85:34 - 85:41
25 minus 9, so I get x squared
plus y squared equals 16.
• 85:41 - 85:46
So from here to here,
how much do I have?
• 85:46 - 85:46
STUDENT: 4.
• 85:46 - 85:48
PROFESSOR: 4, right?
• 85:48 - 85:53
So that little radius of that
yellow domain, [INAUDIBLE].
• 85:53 - 85:57
• 85:57 - 86:00
OK, so let's write
down the thing.
• 86:00 - 86:09
Let's go with D. This domain
is going to be called D.
• 86:09 - 86:12
And then I have this curl F.
• 86:12 - 86:15
And who is N?
• 86:15 - 86:16
N is k.
• 86:16 - 86:17
Why?
• 86:17 - 86:21
Because I'm in a
plane that's upstairs,
• 86:21 - 86:24
and I have dA because
whether the plane is
• 86:24 - 86:27
upstairs or downstairs
on the first floor,
• 86:27 - 86:29
dA will still be dxdy.
• 86:29 - 86:34
• 86:34 - 86:41
OK, so now let's compute
what we have backwards.
• 86:41 - 86:44
So this times k will
give me double integral
• 86:44 - 86:52
over D of y squared times.
• 86:52 - 86:56
Who is z?
• 86:56 - 86:58
I'm in a domain, d,
where z is fixed.
• 86:58 - 86:59
STUDENT: 3.
• 86:59 - 87:01
PROFESSOR: z is 3.
• 87:01 - 87:06
Minus x squared times 3.
• 87:06 - 87:07
And--
• 87:07 - 87:08
STUDENT: dx2.
• 87:08 - 87:10
PROFESSOR: I just came
up with this problem.
• 87:10 - 87:11
If I were to write
it for the final,
• 87:11 - 87:13
I would write it even simpler.
• 87:13 - 87:17
But let's see, 3 and
3, and then nothing.
• 87:17 - 87:20
And then da, dx dy, right?
• 87:20 - 87:23
Over the d, which
is x squared plus y
• 87:23 - 87:27
squared this is
[INAUDIBLE] over 16.
• 87:27 - 87:29
How do I solve such a integral?
• 87:29 - 87:31
I'm going to make it nicer.
• 87:31 - 87:33
OK.
• 87:33 - 87:35
How would I solve
such an integral?
• 87:35 - 87:40
Is it a painful thing?
• 87:40 - 87:42
STUDENT: [INAUDIBLE].
• 87:42 - 87:43
PROFESSOR: Well,
they're coordinates.
• 87:43 - 87:45
And somebody's going to help me.
• 87:45 - 87:47
And as soon as we are
done, we are done.
• 87:47 - 87:50
3 gets out.
• 87:50 - 87:53
plus y squared is then 16,
• 87:53 - 88:00
I have r between 0 and 4,
theta between 0 and 2pi.
• 88:00 - 88:03
I have to take
• 88:03 - 88:07
I've learned all the semester.
• 88:07 - 88:08
Knowledge is power.
• 88:08 - 88:09
What's missing?
• 88:09 - 88:10
r.
• 88:10 - 88:12
A 3 gets out.
• 88:12 - 88:15
And here I have to be just
smart and pay attention
• 88:15 - 88:16
to what you told me.
• 88:16 - 88:19
Because you told me, Magdalena,
why is our sine theta not
• 88:19 - 88:20
our cosine theta?
• 88:20 - 88:21
[INAUDIBLE]
• 88:21 - 88:27
This is r squared, sine
squared theta minus r
• 88:27 - 88:31
squared cosine squared theta.
• 88:31 - 88:35
So what have I taught
• 88:35 - 88:38
that can be expressed as
products of a function of theta
• 88:38 - 88:40
and function of r?
• 88:40 - 88:41
That they have a
blessing from God.
• 88:41 - 88:50
So you have 3 integral
from 0 to 2pi minus 1.
• 88:50 - 88:52
I have my plan when
it comes to this guy.
• 88:52 - 88:55
Because it goes on my nerves.
• 88:55 - 88:56
All right.
• 88:56 - 88:57
Do you see this?
• 88:57 - 88:58
[INAUDIBLE] theta.
• 88:58 - 88:59
STUDENT: That's the [INAUDIBLE].
• 88:59 - 88:59
[INTERPOSING VOICES]
• 88:59 - 89:02
PROFESSOR: Do you know
what I'm coming up with?
• 89:02 - 89:03
[INTERPOSING VOICES]
• 89:03 - 89:05
PROFESSOR: Cosine
of a double angle.
• 89:05 - 89:06
Very good.
• 89:06 - 89:08
I'm proud of you guys.
• 89:08 - 89:11
If I were to test--
oh, there was a test.
• 89:11 - 89:14
But [INAUDIBLE] next
for the whole nation.
• 89:14 - 89:19
students remembered that
• 89:19 - 89:22
by the end of the
calculus series.
• 89:22 - 89:26
But I think that's not--
that doesn't show weakness
• 89:26 - 89:27
of the [INAUDIBLE] programs.
• 89:27 - 89:33
It shows a weakness in
the trigonometry classes
• 89:33 - 89:37
that are either missing from
high school or whatever.
• 89:37 - 89:39
So you know that
you want in power.
• 89:39 - 89:45
Now, times what
integral from 0 to 4?
• 89:45 - 89:46
STUDENT: r squared.
• 89:46 - 89:47
Or r cubed.
• 89:47 - 89:51
PROFESSOR: r cubed, which again
is wonderful that we have.
• 89:51 - 89:55
And we should be able to
compute the whole thing easily.
• 89:55 - 89:58
Now if I'm smart,
• 89:58 - 89:59
STUDENT: [INAUDIBLE].
• 89:59 - 90:00
PROFESSOR: How can we see?
• 90:00 - 90:04
STUDENT: Because the cosine to
the integral is sine to theta.
• 90:04 - 90:05
And [INAUDIBLE].
• 90:05 - 90:05
PROFESSOR: Right.
• 90:05 - 90:10
So the sine to theta, whether
I put it here or here,
• 90:10 - 90:11
is still going to be 0.
• 90:11 - 90:13
The whole thing will be 0.
• 90:13 - 90:14
So I play the game.
• 90:14 - 90:18
Maybe I should've given
such a problem when we
• 90:18 - 90:21
wrote this edition of the book.
• 90:21 - 90:24
I think it's nicer than
the computational one
• 90:24 - 90:26
you saw before.
• 90:26 - 90:31
But I told you this trick so
you remember it for the final.
• 90:31 - 90:33
And you are to promise
that you'll remember it.
• 90:33 - 90:40
And that was the whole
essence of understanding
• 90:40 - 90:45
that the Stokes' theorem can
become Green's theorem very
• 90:45 - 90:47
easily when you work
with a surface that's
• 90:47 - 90:51
a domain in plane, a
planar domain. [INAUDIBLE].
• 90:51 - 90:53
Are you done with this?
• 90:53 - 90:54
OK.
• 90:54 - 91:02
• 91:02 - 91:05
So you say, OK, so what else?
• 91:05 - 91:08
This was something
that's sort of fun.
• 91:08 - 91:09
I understand it.
• 91:09 - 91:12
Is there anything left
in this whole chapter?
• 91:12 - 91:16
Fortunately or unfortunately,
there is only one section left.
• 91:16 - 91:19
And I'm going to
go over it today.
• 91:19 - 91:21
STUDENT: Can I ask you a quick
• 91:21 - 91:21
PROFESSOR: Yes, sire.
• 91:21 - 91:23
STUDENT: --before you move on?
• 91:23 - 91:23
PROFESSOR: Move on?
• 91:23 - 91:25
STUDENT: I was an idiot.
• 91:25 - 91:26
PROFESSOR: No, you are not.
• 91:26 - 91:27
STUDENT: And when I
was writing these down,
• 91:27 - 91:29
I missed the variable.
• 91:29 - 91:33
So I have the
integral of fdr over c
• 91:33 - 91:37
equals double integral
over f, curl f dot n.
• 91:37 - 91:37
PROFESSOR: ds.
• 91:37 - 91:40
STUDENT: I didn't
write down what c was.
• 91:40 - 91:41
I didn't write down
what this c was.
• 91:41 - 91:46
PROFESSOR: The c was
the whatever boundary
• 91:46 - 91:49
you had there of the surface s.
• 91:49 - 91:51
And that was in
the beginning when
• 91:51 - 91:56
we defined the sphere, when
we gave the general statement
• 91:56 - 91:59
for the function.
• 91:59 - 92:02
So I'm going to try
and draw a potato.
• 92:02 - 92:05
We don't do a very
good job in the book
• 92:05 - 92:08
drawing the solid body.
• 92:08 - 92:12
But I'll try and draw
a very nice solid body.
• 92:12 - 92:12
Let's see.
• 92:12 - 92:18
• 92:18 - 92:22
You have a solid body.
• 92:22 - 92:26
Imagine it as a potato,
topologically a sphere.
• 92:26 - 92:28
It's a balloon that you blow.
• 92:28 - 92:29
It's a closed surface.
• 92:29 - 92:32
It closes in itself.
• 92:32 - 92:36
And we call that r in the book.
• 92:36 - 92:43
It's a solid region enclosed
by the closed surfaces.
• 92:43 - 92:47
• 92:47 - 92:49
Sometimes we call
such a surface compact
• 92:49 - 92:52
for some topological reasons.
• 92:52 - 92:54
Let's put s.
• 92:54 - 92:56
s is the boundary of r.
• 92:56 - 93:01
• 93:01 - 93:05
We as you know our old friend
to be a vector value function.
• 93:05 - 93:10
• 93:10 - 93:14
And again, if you
like a force field,
• 93:14 - 93:16
think of it as a force field.
• 93:16 - 93:18
Now, I'm not going to
tell you what it is.
• 93:18 - 93:21
It's [INAUDIBLE] function
differential [INAUDIBLE]
• 93:21 - 93:25
the partial here is continuous.
• 93:25 - 93:29
The magic thing is that this
surface must be orientable.
• 93:29 - 93:34
And if we are going to immerse
it, it's a regular surface.
• 93:34 - 93:36
Then of course, n exists.
• 93:36 - 93:41
doesn't have to be outwards.
• 93:41 - 93:45
It could be inwards [INAUDIBLE].
• 93:45 - 93:52
Let's make the convention that n
will be outwards by convention.
• 93:52 - 93:55
So we have to have an agreement
like they do in politics,
• 93:55 - 93:58
between Fidel Castro and Obama.
• 93:58 - 94:01
By convention, whether
we like it or not,
• 94:01 - 94:07
let's assume the normal
will be pointing out.
• 94:07 - 94:10
Then something magic happens.
• 94:10 - 94:13
And that magic thing, I'm not
going to tell you what it is.
• 94:13 - 94:16
But you should tell
me if you remember
• 94:16 - 94:21
what the double integral was
in this case, intolerance
• 94:21 - 94:23
of physics.
• 94:23 - 94:24
Shut up, Magdalena.
• 94:24 - 94:27
Don't tell them everything.
• 94:27 - 94:29
Let people remember
what this was.
• 94:29 - 94:32
So what is the second term?
• 94:32 - 94:35
This is the so-called
famous divergence theorem.
• 94:35 - 94:38
• 94:38 - 94:40
So this is the divergence.
• 94:40 - 94:43
If you don't remember
that, we will review it.
• 94:43 - 94:45
dV is the volume integral.
• 94:45 - 94:48
I have a [INAUDIBLE] integral
over the solid potato,
• 94:48 - 94:49
of course.
• 94:49 - 94:53
What is this animal [INAUDIBLE]?
• 94:53 - 94:54
OK.
• 94:54 - 95:00
Take some milk and strain
it and make cheese.
• 95:00 - 95:03
And you have that kind
of piece of cloth.
• 95:03 - 95:05
And you hang it.
• 95:05 - 95:09
And the water goes through
that piece of cloth.
• 95:09 - 95:13
[INAUDIBLE] have this
kind of suggestive image
• 95:13 - 95:14
should make you
think of something we
• 95:14 - 95:16
• 95:16 - 95:21
Whether that was fluid
dynamics or electromagnetism,
• 95:21 - 95:25
[INAUDIBLE], this
has the same name.
• 95:25 - 95:32
f is some sort of field,
vector [INAUDIBLE] field.
• 95:32 - 95:35
N is the outer
normal in this case.
• 95:35 - 95:40
What is the meaning of that, for
a dollar, which I don't have?
• 95:40 - 95:41
It's a four-letter word.
• 95:41 - 95:43
It's an F word.
• 95:43 - 95:44
STUDENT: Flux.
• 95:44 - 95:44
PROFESSOR: Very good.
• 95:44 - 95:45
I'm proud of you.
• 95:45 - 95:48
Who said it first?
• 95:48 - 95:49
Aaron said it first?
• 95:49 - 95:50
I owe you a dollar.
• 95:50 - 95:51
You can stop by my office.
• 95:51 - 95:52
I'll give you a dollar.
• 95:52 - 95:54
STUDENT: He said it
five minutes ago.
• 95:54 - 95:55
PROFESSOR: So the flux-- He did?
• 95:55 - 95:56
STUDENT: Yeah, he did.
• 95:56 - 95:58
Silently.
• 95:58 - 96:00
PROFESSOR: Aaron
• 96:00 - 96:01
OK.
• 96:01 - 96:03
So the flux in the
left-hand side.
• 96:03 - 96:05
This thing you don't
know what it is.
• 96:05 - 96:07
But it's some sort of potato.
• 96:07 - 96:08
What is the divergence
of something?
• 96:08 - 96:11
• 96:11 - 96:20
So if somebody gives you the
vector field F1, [INAUDIBLE],
• 96:20 - 96:27
where these are functions
of xyz, [INAUDIBLE].
• 96:27 - 96:30
What is the divergence
of F by definition?
• 96:30 - 96:35
Remember section 13.1?
• 96:35 - 96:37
Keep it in mind for the final.
• 96:37 - 96:41
• 96:41 - 96:42
So what do we do?
• 96:42 - 96:45
Differentiate the
first component respect
• 96:45 - 96:50
to x plus differentiate
the second component
• 96:50 - 96:53
respect to y plus differentiate
the third component
• 96:53 - 96:55
respect to c, sum them up.
• 96:55 - 96:56
• 96:56 - 96:59
• 96:59 - 97:02
OK?
• 97:02 - 97:05
How do engineers
write divergence?
• 97:05 - 97:08
Not like a mathematician
or like a geometer.
• 97:08 - 97:10
I'm doing differential geometry.
• 97:10 - 97:11
How do they write?
• 97:11 - 97:13
STUDENT: Kinds of [INAUDIBLE].
• 97:13 - 97:17
PROFESSOR: [INAUDIBLE] dot if.
• 97:17 - 97:19
This is how engineers
write divergence.
• 97:19 - 97:22
And when they write curl,
how do they write it?
• 97:22 - 97:25
They write [INAUDIBLE]
cross product.
• 97:25 - 97:26
Because it has a meaning.
• 97:26 - 97:30
operator, you have ddx
• 97:30 - 97:35
applied to F1, ddy applied
to F2, ddz applied to F3.
• 97:35 - 97:39
So it's like having the dot
product between ddx, ddy,
• 97:39 - 97:44
ddz operators, which would
be the [INAUDIBLE] operator
• 97:44 - 97:47
acting on F1, F2, F3.
• 97:47 - 97:50
So you go first first,
plus second second,
• 97:50 - 97:52
plus third third, right?
• 97:52 - 97:58
It's exactly the same idea that
you inherited from dot product.
• 97:58 - 98:02
Now let's see the last two
problems of this semester.
• 98:02 - 98:04
except for step the review.
• 98:04 - 98:07
But the review's another story.
• 98:07 - 98:13
So I'm going to pick one
• 98:13 - 98:22
• 98:22 - 98:22
OK.
• 98:22 - 98:26
Example one, remember
• 98:26 - 98:28
I'm going to erase it.
• 98:28 - 98:32
• 98:32 - 98:36
Instead of the potato, you can
have something like a pyramid.
• 98:36 - 98:38
And you have example one.
• 98:38 - 98:40
• 98:40 - 98:44
Let's say, [INAUDIBLE]
we have that.
• 98:44 - 98:46
Somebody gives you the F.
• 98:46 - 98:49
I'm going to make
it nice and sassy.
• 98:49 - 98:53
Because the final is coming
and I want simple examples.
• 98:53 - 98:56
And don't expect anything
[INAUDIBLE] really nice
• 98:56 - 98:59
examples also on the final.
• 98:59 - 99:13
Apply divergence
theorem in order
• 99:13 - 99:22
to compute double integral
of F dot n ds over s,
• 99:22 - 99:38
where s is the surface of the
tetrahedron in the picture.
• 99:38 - 99:41
And that's your
favorite tetrahedron.
• 99:41 - 99:44
We've done that like
a million times.
• 99:44 - 99:51
Somebody gave you a--
shall I put 1 or a?
• 99:51 - 99:56
1, because [INAUDIBLE] is
[INAUDIBLE] is [INAUDIBLE].
• 99:56 - 99:59
So you have the plane
x plus y plus z.
• 99:59 - 100:03
Plus 1 you intersect
with the axis'.
• 100:03 - 100:06
The coordinates, you take
the place of coordinates
• 100:06 - 100:08
and you form a tetrahedron.
• 100:08 - 100:10
Next tetrahedron is a
little bit beautiful
• 100:10 - 100:15
that it has 90 degree
angles at the vertex.
• 100:15 - 100:18
And it has a name, OABC.
• 100:18 - 100:21
OABC is the tetrahedron.
• 100:21 - 100:25
And the surface of
the tetrahedron is s.
• 100:25 - 100:29
How are you going
to do this problem?
• 100:29 - 100:32
You're going to say, oh
my god, I don't know.
• 100:32 - 100:33
It's not hard.
• 100:33 - 100:34
STUDENT: It looks like
you're going to use
• 100:34 - 100:36
the formula you just gave us.
• 100:36 - 100:37
PROFESSOR: The
divergence theorem.
• 100:37 - 100:40
STUDENT: And the divergence
for that is really easy.
• 100:40 - 100:41
It's just a constant.
• 100:41 - 100:42
PROFESSOR: Right.
• 100:42 - 100:46
And we have to give a name to
the tetrahedron, [INAUDIBLE]
• 100:46 - 100:47
T, with the solid tetrahedron.
• 100:47 - 100:53
• 100:53 - 100:59
And its area, its
surface is this.
• 100:59 - 101:03
have the solid tetrahedron.
• 101:03 - 101:04
So what do you write?
• 101:04 - 101:10
Exactly what [INAUDIBLE] told
you, triple integral over T.
• 101:10 - 101:11
Of what?
• 101:11 - 101:16
The divergence of F, because
that's the divergence theorem,
• 101:16 - 101:17
dv.
• 101:17 - 101:20
• 101:20 - 101:21
Well, it should be easy.
• 101:21 - 101:25
Because just as you
said, divergence of F
• 101:25 - 101:26
would be a constant.
• 101:26 - 101:27
How come?
• 101:27 - 101:31
Differentiate this
with respect to x, 2.
• 101:31 - 101:33
This with respect to y, 3.
• 101:33 - 101:37
This with respect to z, 5.
• 101:37 - 101:41
Last time I checked this was
10 when I was [INAUDIBLE].
• 101:41 - 101:44
• 101:44 - 101:48
So 10 says I'm going for a walk.
• 101:48 - 101:54
And then triple integral of
the volume of 1dv over T,
• 101:54 - 101:55
what is this?
• 101:55 - 101:59
• 101:59 - 102:02
[INTERPOSING VOICES]
• 102:02 - 102:07
PROFESSOR: Well, because I
taught you how to cheat, yes.
• 102:07 - 102:14
But what if I were to ask
you to express this as--
• 102:14 - 102:15
STUDENT: [INAUDIBLE]?
• 102:15 - 102:16
PROFESSOR: Yeah.
• 102:16 - 102:17
Integrate one at a time.
• 102:17 - 102:22
So you have 1dz, dy, dx--
I'm doing review with you--
• 102:22 - 102:28
from 0 to 1 minus x
minus y from 0 to--
• 102:28 - 102:29
STUDENT: [INTERPOSING VOICES]
• 102:29 - 102:33
PROFESSOR: --1 minus
x from zero to 1.
• 102:33 - 102:36
And how did I teach
you how to cheat?
• 102:36 - 102:40
I taught you that in
this case you shouldn't
• 102:40 - 102:42
bother to compute that.
• 102:42 - 102:45
Remember that you
were in school and we
• 102:45 - 102:47
learned the volume
of a tetrahedron
• 102:47 - 102:54
was the area of the base
times the height divided by 3,
• 102:54 - 103:02
which was one half
times 1 divided by 3.
• 103:02 - 103:03
So you guys right.
• 103:03 - 103:06
The answer is 10 times 1 over 6.
• 103:06 - 103:08
Do I leave it like that?
• 103:08 - 103:08
No.
• 103:08 - 103:10
Because it's not nice.
• 103:10 - 103:11
• 103:11 - 103:14
• 103:14 - 103:19
Expect something like
that on the final,
• 103:19 - 103:21
something very similar.
• 103:21 - 103:25
So you'll have to apply
the divergence theorem
• 103:25 - 103:26
and do a good job.
• 103:26 - 103:29
And of course, you
have to be careful.
• 103:29 - 103:31
But it shouldn't be hard.
• 103:31 - 103:35
It's something that
should be easy to do.
• 103:35 - 103:39
Now, the last problem of the
semester that I want to do
• 103:39 - 103:41
is an application of
the divergence theorem
• 103:41 - 103:43
is over a cube.
• 103:43 - 103:49
So I'm going to erase
it, the whole thing.
• 103:49 - 103:53
And I'm going to
draw a cube, which is
• 103:53 - 103:56
an open-topped box upside down.
• 103:56 - 104:01
Say it again, an open-topped
box upside down, which
• 104:01 - 104:04
means somebody gives
you a cubic box
• 104:04 - 104:07
and tells you to
turn it upside down.
• 104:07 - 104:12
• 104:12 - 104:14
And you have from
here to here, 1.
• 104:14 - 104:17
All the dimensions
of the cube are 1.
• 104:17 - 104:23
The top is missing, so
there's faces missing.
• 104:23 - 104:25
The bottom face is missing.
• 104:25 - 104:28
Bottom face is missing.
• 104:28 - 104:31
Let's call it-- you know,
what shall we call it?
• 104:31 - 104:38
• 104:38 - 104:40
F1.
• 104:40 - 104:44
Because it was the top,
but now it's the bottom.
• 104:44 - 104:45
OK?
• 104:45 - 104:56
And the rest are F2, F3, F4,
F5, and F6, which is the top.
• 104:56 - 104:58
And I'm going to erase.
• 104:58 - 105:01
• 105:01 - 105:07
And the last thing before this
section is to do the following.
• 105:07 - 105:09
What do I want?
• 105:09 - 105:18
Evaluate the flux double
integral over s F dot n ds.
• 105:18 - 105:21
You have to evaluate that.
• 105:21 - 105:25
For the case when F-- I usually
don't take the exact data
• 105:25 - 105:26
from the book.
• 105:26 - 105:28
But in this case, I want to.
• 105:28 - 105:30
Because I know you'll read it.
• 105:30 - 105:34
And I don't want you to have any
difficulty with this problem.
• 105:34 - 105:35
I hate the data myself.
• 105:35 - 105:37
I didn't like it very much.
• 105:37 - 105:42
• 105:42 - 105:45
It's unit cube, OK.
• 105:45 - 105:48
So x must be between 0 and 1.
• 105:48 - 105:51
y must be between 0
and 1 including them.
• 105:51 - 105:58
But z-- attention guys-- must
be between 0 [INAUDIBLE],
• 105:58 - 106:00
without 0.
• 106:00 - 106:03
Because you remove the
face on the ground.
• 106:03 - 106:09
z is greater than 0 and
less than or equal to 1.
• 106:09 - 106:13
And do we want anything else?
• 106:13 - 106:13
No.
• 106:13 - 106:15
That is all.
• 106:15 - 106:21
So let's compute
the whole thing.
• 106:21 - 106:29
Now, assume the box
would be complete.
• 106:29 - 106:36
• 106:36 - 106:39
If the box were
complete, then I would
• 106:39 - 106:44
have the following, double
integral over all the
• 106:44 - 106:50
faces F2 union with
F3 union with F4 union
• 106:50 - 106:58
with F5 union with--
oh my God-- F6 of F
• 106:58 - 107:05
dot 10 ds plus double integral
over what's missing guys, F1?
• 107:05 - 107:10
• 107:10 - 107:15
Of n dot n ds-- F,
Magdalena, that's the flux.
• 107:15 - 107:18
F dot n ds.
• 107:18 - 107:21
If it were complete,
that would mean
• 107:21 - 107:25
I have the double integral
over all the six faces.
• 107:25 - 107:30
In that case, this sum would
be-- I can apply finally
• 107:30 - 107:32
the divergence theorem.
• 107:32 - 107:35
That would be triple
integral of-- God
• 107:35 - 107:43
knows what that is-- divergence
of F dv over the cube.
• 107:43 - 107:46
What do you want us
to call the cube?
• 107:46 - 107:47
STUDENT: C.
• 107:47 - 107:51
PROFESSOR: C is usually what
we denote for the curve.
• 107:51 - 107:53
• 107:53 - 107:54
PROFESSOR: Beautiful.
• 107:54 - 107:55
Sounds like.
• 107:55 - 107:56
Oh, I like that. q.
• 107:56 - 108:01
q is the cube inside
the whole thing.
• 108:01 - 108:03
Unfortunately, this
is not very nicely
• 108:03 - 108:10
picked just to make
• 108:10 - 108:14
So you have dv x over y.
• 108:14 - 108:17
There is no j, at least that.
• 108:17 - 108:20
ddz minus this way.
• 108:20 - 108:22
As soon as we are
done with this,
• 108:22 - 108:25
since I gave you no break,
I'm going to let you go.
• 108:25 - 108:26
So what do we have?
• 108:26 - 108:29
y minus 2z.
• 108:29 - 108:30
Does it look good?
• 108:30 - 108:32
No.
• 108:32 - 108:33
• 108:33 - 108:37
• 108:37 - 108:39
If I were to solve
the problem, I
• 108:39 - 108:46
would have to say triple,
triple, triple y minus 2z.
• 108:46 - 108:53
And now-- oh my
God-- dz, dy, dx.
• 108:53 - 108:59
I sort of hate when a little bit
of computation 0 to 1, 0 to 1,
• 108:59 - 109:00
0 to 1.
• 109:00 - 109:03
But this is for
[INAUDIBLE] theorem.
• 109:03 - 109:06
Is there anybody
missing from the picture
• 109:06 - 109:10
so I can reduce it
to a double integral?
• 109:10 - 109:10
STUDENT: x.
• 109:10 - 109:11
PROFESSOR: x is missing.
• 109:11 - 109:15
So I say there is no x inside.
• 109:15 - 109:19
I go what is integral
from 0 to 1 of 1dx?
• 109:19 - 109:19
STUDENT: 1
• 109:19 - 109:20
PROFESSOR: 1.
• 109:20 - 109:24
So I will rewrite it
as integral from 0 to1,
• 109:24 - 109:30
integral from 0 to1,
y minus 2z, dz, dy.
• 109:30 - 109:32
Is this hard?
• 109:32 - 109:33
Eh, no.
• 109:33 - 109:35
But it's a little bit obnoxious.
• 109:35 - 109:38
• 109:38 - 109:42
When I integrate with
respect to z, what do I get?
• 109:42 - 109:44
• 109:44 - 109:50
Yz minus z squared.
• 109:50 - 109:59
No, not that-- between z equals
0 1 down, z equals 1 up to z
• 109:59 - 110:03
equals 0 down dy.
• 110:03 - 110:07
So z goes from 0
to 1 [INAUDIBLE].
• 110:07 - 110:10
When z is 0 down,
I have nothing.
• 110:10 - 110:12
STUDENT: Yeah, 1
minus-- y minus 1.
• 110:12 - 110:16
PROFESSOR: 1. y
• 110:16 - 110:18
• 110:18 - 110:25
So I get y squared
over 2 minus y.
• 110:25 - 110:31
Between 0 and 1, what do I get?
• 110:31 - 110:31
STUDENT: Negative 1/2.
• 110:31 - 110:32
PROFESSOR: Negative 1/2.
• 110:32 - 110:37
• 110:37 - 110:37
All right.
• 110:37 - 110:39
Let's see what we've got here.
• 110:39 - 110:39
Yeah.
• 110:39 - 110:41
They got [INAUDIBLE].
• 110:41 - 110:43
what's going to happen.
• 110:43 - 110:47
Our contour is the
open-topped box upside down.
• 110:47 - 110:49
This is what we need.
• 110:49 - 110:54
This is what we--
• 110:54 - 110:57
STUDENT: Couldn't you
just the double integral?
• 110:57 - 111:01
PROFESSOR: We just have
to compute this fellow.
• 111:01 - 111:04
We need to compute that fellow.
• 111:04 - 111:07
So how do we do that?
• 111:07 - 111:10
How do we do that?
• 111:10 - 111:13
STUDENT: What is the
• 111:13 - 111:17
PROFESSOR: So the problem
• 111:17 - 111:22
but only over the box'
walls and the top.
• 111:22 - 111:25
The top, one, two, three,
four, without the bottom,
• 111:25 - 111:27
which is missing.
• 111:27 - 111:29
In order to apply
divergence theorem,
• 111:29 - 111:36
I have to put the bottom back
and have a closed surface that
• 111:36 - 111:38
is enclosing the whole cube.
• 111:38 - 111:41
So this is what I want.
• 111:41 - 111:43
This is what I know.
• 111:43 - 111:44
How much is it guys?
• 111:44 - 111:45
Minus 1/2.
• 111:45 - 111:48
And this is, again, what I need.
• 111:48 - 111:48
Right?
• 111:48 - 111:53
That's the last thing
I'm going to do today.
• 111:53 - 111:54
[INTERPOSING VOICES]
• 111:54 - 111:58
• 111:58 - 112:00
STUDENT: F times k da.
• 112:00 - 112:03
PROFESSOR: Let's compute it.
• 112:03 - 112:06
k is a blessing, as
you said, [INAUDIBLE].
• 112:06 - 112:07
It's actually minus k.
• 112:07 - 112:09
Why is it minus k?
• 112:09 - 112:11
Because it's upside down.
• 112:11 - 112:12
And it's an altered normal.
• 112:12 - 112:13
STUDENT: Oh, it is the
[INAUDIBLE] normal.
• 112:13 - 112:14
OK.
• 112:14 - 112:14
Yeah.
• 112:14 - 112:15
That's right.
• 112:15 - 112:15
PROFESSOR: [INAUDIBLE].
• 112:15 - 112:16
So minus k.
• 112:16 - 112:18
But it doesn't [INAUDIBLE].
• 112:18 - 112:19
The sign matters.
• 112:19 - 112:21
So I have to be careful.
• 112:21 - 112:26
F is-- z is 0, thank God.
• 112:26 - 112:27
So that does away.
• 112:27 - 112:36
So I have x y i dot
product with minus k.
• 112:36 - 112:38
What's the beauty of this?
• 112:38 - 112:38
0.
• 112:38 - 112:40
STUDENT: 0.
• 112:40 - 112:41
PROFESSOR: Yay.
• 112:41 - 112:42
0.
• 112:42 - 112:46
problem is minus 1/2.
• 112:46 - 112:51
So the answer is minus 1.
• 112:51 - 112:53
And we are done with
the last section
• 112:53 - 112:57
of the book, which is 13.7.
• 112:57 - 112:59
It was a long way.
• 112:59 - 113:01
We came a long way
to what I'm going
• 113:01 - 113:06
to do next time and
the times to come.
• 113:06 - 113:08
First of all, ask me from now
on you want a break or not.
• 113:08 - 113:11
Because I didn't give
you a break today.
• 113:11 - 113:13
We are not in a hurry.
• 113:13 - 113:15
I will pick up exams.
• 113:15 - 113:18
And I will go over
them together with you.
• 113:18 - 113:22
And by the time we
finish this review,
• 113:22 - 113:26
we will have solved two or
three finals completely.
• 113:26 - 113:30
We will be [INAUDIBLE].
• 113:30 - 113:39
And so the final is on the 11th,
May 11 at 10:30 in the morning.
• 113:39 - 113:39
I think.
• 113:39 - 113:40
STUDENT: It's the 11.
• 113:40 - 113:41
The 12 is [INAUDIBLE].
• 113:41 - 113:44
The 12th is the other class.
• 113:44 - 113:44
STUDENT: Yeah.
• 113:44 - 113:45
I'm positive.
• 113:45 - 113:46
PROFESSOR: We are
switching the two classes.
• 113:46 - 113:47
STUDENT: [INAUDIBLE].
• 113:47 - 113:51
PROFESSOR: And it's May 11
at 10:30 in the morning.
• 113:51 - 113:56
On May 12, there are other
math courses that have a final.
• 113:56 - 114:00
But fortunately for
them, they start at 4:30.
• 114:00 - 114:02
I'm really blessed that I
don't have that [INAUDIBLE].
• 114:02 - 114:06
They start at 4:30,
and they end at 7:00.
• 114:06 - 114:08
Can you imagine how
frisky you feel when you
• 114:08 - 114:09
take that final in the night?
• 114:09 - 114:12
• 114:12 - 114:13
Good luck with the homework.
• 114:13 - 114:17
homework if you have them.
• 114:17 - 114:20
Title:
TTU Math2450 Calculus3 Secs 13.6 - 13.7
Description:

Stokes' Theorem and Divergence Theorem

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Video Language:
English
 jackie.luft edited English subtitles for TTU Math2450 Calculus3 Secs 13.6 - 13.7