## TTU Math2450 Calculus3 Sec 12.4

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DR. MAGDALENA TODA: Sorry.
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I really don't mind if you
walk in a little bit late.
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I know that you guys come
from other buildings,
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and some professors
keep you overtime.
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So as long as you
quietly enter the room,
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I have no problem with
walking in a little bit late.
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Would anybody want to
start an attendance sheet?
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Who wants to be the one?
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Thank you so much.
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All right.
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We went through chapter
12 on Monday fast.
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a review of 12.1, 12.2, 12.3.
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So two thing we will do today.
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Part one will be review of
chapter 12, sections to 12.1,
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12.3 from the book and
starting chapter 12,
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section 12.4 today later.
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integrals, surface area,
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and [INAUDIBLE].
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All right.
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What have you seen
in 12.1, 12.3?
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Let's review quickly
what you've learned.
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how to interpret
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an integral with a positive
function that is smooth.
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Well, we said
continuous-- that would
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be enough-- over a
rectangular region.
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And the geometric meaning
of such a problem,
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integrate f of x, y positive
over a domain was what?
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The volume of a body under the
graph and above that domain,
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so projected down, protecting
down on the domain.
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Evaluate that body.
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How did we do it?
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Double integral of f
of x, y, dxdy or dA.
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But then we said, OK, if you
have a rectangular region
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on the ground, then it's easy.
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You apply the Fubini theorem.
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And then you'll have
integral from A to B,
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integral from C to
D, fixed end points.
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When you didn't have
a rectangular region
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to integrate over, you
would have such a type one,
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type two regions, who
are easy to deal with,
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which were the case of regions
like the ones between two
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straight lines
and two functions.
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And then you had the type
two, two straight lines
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and two functions,
where the functions
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were assumed differentiable
actually in our examples.
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Type one, type two.
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What did we do after that?
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After that, we said, well,
what if you're not so lucky
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and have such nice domains?
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Or maybe you have
something with a corner.
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What do you do if
you have a corner?
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Well, you'd still be able to
divide the surface into two,
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where you have two
separate areas.
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And then you integrate on them
separately at the same time.
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And you have an
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Those are easy to deal with.
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Well, what if you had something
that is more sophisticated,
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like a disk or an annulus?
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And in that case, it's
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considering how to do this
using one of the previous steps.
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polar coordinates.
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And we have to
think, what change
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do I have from x, y to r,
theta, polar coordinates
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back and forth?
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And when we did the double
integral over a domain f of x,
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y function positive dA in
the Cartesian coordinates.
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When we switched to
polar coordinates,
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happen, which was what?
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Some f of x of r,
theta, y of r, theta.
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I say theta.
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I put phi.
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It doesn't matter.
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Let me put theta if
you prefer theta.
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A change of
coordinates, a Jacobian.
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That was what?
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Do you guys remember that?
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r.
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Very good.
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I'm proud of you, r.
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And then drd theta.
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kind of homework, integrals,
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double integrals in
polar coordinates.
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dr will be between
certain values,
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hopefully fixed
values because that
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will make the Fubini-Tonelli
a piece of cake.
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Theta, also fixed values.
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But not always will you have
fixed values, especially
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in the first part.
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You may have some function
of r, function of r.
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And here, theta 1 and theta 2.
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So I want to see a
few more examples
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before I move on to section 12.4
because, as the Romans said,
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review is the
mother of studying,
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which is [LATIN], which means
go ahead and do a lot of review
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if you really want to
master the concepts.
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OK.
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I'm going to take the plunge
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I've been pondering
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We've done problems that I made
up, like the ones in the book.
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And I also took problems
straight out of the book.
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But I would like to go over
some homework type problems
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in order to assist you in more
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In chapter 12, homework
four-- am I right,
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homework number four?
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You have a big array of
problems, all sorts of problems
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because mathematicians
have all sorts of problems.
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For example, an easy
one that you're not
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going to have a problem with--
and I'm using my own end
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points.
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different in the homework.
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It would be homework four,
chapter 12, number four.
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And you say-- most of
you should say, oh,
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that's a piece of cake.
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I don't know why she even talks
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right?
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Many of you have said that.
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Well, I am willing
to review everything
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so that you have a better
grasp of the material.
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On this one, since it's so
easy, I want you to help me.
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What kind of problem is that?
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As I said, mathematicians have
all sorts of problems, right?
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So a problem where you
have a product inside
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as an integrand, where the
variables are completely
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separated-- what does it mean?
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The function
underneath is a product
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of two functions, one
function of x only,
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the other function
of y only, which
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is a blessing in disguise.
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Why is that a blessing?
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I told you last time that you
can go ahead and write this
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as product of integrals.
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what those integrals will be?
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Let's see how much you
mastered the material.
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STUDENT: x over 2y times--
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DR. MAGDALENA TODA:
From 1 to 2, you said?
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STUDENT: Yeah, from 1 to 2.
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I'm sorry.
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DR. MAGDALENA TODA: Of what?
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X, dx times the integral
from 0 to pi of what?
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STUDENT: Cosine y.
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DR. MAGDALENA TODA: Cosine y.
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Do we need to
re-prove this result?
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No, we proved it last time.
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But practically, if
you forget, the idea
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is a very simple thing.
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When you integrate
with respect to y,
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Mr. X said, I'm
not married to y.
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I'm out of here.
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I'm out of the picture.
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I'm going for a walk.
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So the integral of cosine is
in itself to be treated first,
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independently.
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And it's inside,
and it's a constant.
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And it pulls out in the end.
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And since it pulls
out, what you're
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going to be left with afterwards
will be that integral of 1
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to 2x dx.
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So we've done that
last time as well.
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Yes, sir?
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STUDENT: So you
would-- would you
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not be able to do that
if it was cosine x, y?
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DR. MAGDALENA TODA: Absolutely.
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x, y, it's bye, bye.
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STUDENT: So it's only when
they're completely separate--
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DR. MAGDALENA TODA: When
you are lucky enough
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to have a functional of only
that's a function of y only.
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example, sine of x plus y,
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anything that mixes them up--
that would be a bad thing.
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Do I have to compute this?
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Not if I'm smart.
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At the blink of an eye,
I can sense that maybe I
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should do this one first.
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Why?
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Integral of cosine is sine.
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And sine is 0 at both 0 and pi.
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So it's a piece of pie.
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So if I have 0, and
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So you say, OK,
give us something
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like that on the midterm
because this problem is
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a piece of cake.
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Uh, yeah.
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I can do that.
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Probably you will have something
like that on the midterm,
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on the April 2 midterm.
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So since Alex just
entered, I'm not
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going to erase this for a while
until you are able to copy it.
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I announced starting the
surface area integral today.
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Section 12.4, we'll
do that later on.
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And I will move on to
another example right now.
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Oh, now they learn.
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that I have lots of needs.
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And I don't complain.
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But they noticed that these
were disappearing really fast.
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Everybody else told me that
I write a lot on the board,
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compared to other professors.
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So I don't know if that is true.
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But I really need
this big bottle.
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OK.
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So you can actually
solve this by yourself.
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You just don't realize it.
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I'm not going to take
any credit for that.
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And I'm going to go ahead
and give you something
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more challenging, see if
you are ready for the review
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and for the midterm.
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OK.
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That's number nine
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that may have again
the data changed.
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But it's the same
type of problem.
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number nine anymore directly
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from WeBWork, because I'll
say, I did that in class.
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And if you have
difficulty with it,
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that means you did
not cover the notes.
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This is pretty.
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You've seen that one before.
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And I would suspect
that you're not
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going to even let me
talk, because look at it.
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Evaluate the following integral.
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And it doesn't
matter what numbers
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we are going to put on that
and what funny polynomial I'm
• 12:08 - 12:08
going to put here.
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You are going to have
all sorts of numbers.
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Maybe these are not
the most inspired ones,
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but this is WeBWork.
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It creates problems at
random, and every student
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may have a different
problem, that is,
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in order to minimize cheating.
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And that's OK.
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The type of the problem
is what matters.
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So if we were in
Calc 1 right now,
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and somebody would say, go
ahead and take an integral of e
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to the x squared dx and compute
it by hand, see what you get,
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They don't know, poor people.
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They don't know.
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But you know because I told
you that this is a headache.
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You need another way out.
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You cannot do that in Calc 2.
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And you cannot do that in
an elementary way by hand.
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This is something that MATLAB
would solve numerically for you
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in no time if you gave
certain values and so on.
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But to find an explicit
form of that anti-derivative
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would be a hassle.
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The same thing would happen
if I had the minus here.
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In that case, I wouldn't be able
to express the anti-derivative
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as an elementary
function at all.
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OK.
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So this is giving
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I'm going to make a face.
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And I'll say, oh, my god.
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Unless you help me get out of
trouble, I cannot solve that.
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MATLAB can do that for me.
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On Maple, I can go in and
plug in the endpoints and hope
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and pray that I'm
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numerical approximation
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But what if I want a precise
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Then I better put my
mind, my own mind,
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my own processor to work and
not rely on MATLAB or Maple.
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OK.
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Hmm.
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And I leave it unsimplified
hopefully, yes.
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We need to think of what
technique in this case?
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STUDENT: Changing the order.
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DR. MAGDALENA TODA: Change
the order of integration.
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OK.
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All right.
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And in that case, the
integrand stays the same.
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These two guys are
swapped, and the end points
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are changing
completely because I
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will have to switch from one
domain to the other domain.
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The domain that's given here by
this problem is the following.
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x is between 7y 7, and
y is between 0 and 1.
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So do they give you horizontal
strip or vertical strip domain?
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Horizontal.
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Very good.
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I wasn't sure if
I heard it right.
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But anyway, what is this
function and that function?
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So x equals 7 would be what?
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x equals 7 will be far away.
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I have to do one, two, three,
four-- well, five, six, seven.
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Then is a vertical
line. x equals 7.
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That's the x-axis.
• 15:42 - 15:42
That's the y-axis.
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I'm trying to draw the domain.
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And what is x equals 7y?
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X equals 7y is the
same as y equals 1/7x.
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Uh-huh.
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That should be a
friendlier function
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to draw because I'm smart
enough to even imagine
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what it looks like.
• 16:05 - 16:11
y equals mx is a line that
passes through the origin.
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It's part of a pencil of planes.
• 16:13 - 16:18
A pencil of planes is infinitely
many-- pencil of lines,
• 16:18 - 16:19
I'm sorry.
• 16:19 - 16:21
Infinitely many lines that all
pass through the same point.
• 16:21 - 16:23
So they all pass
through the origin.
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For 7, x equals 7 is going
to give me y, 1, y equals 1.
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So I'm going to erase this
dotted line and draw the line.
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This is y equals x/7,
and we look at it,
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and we think how nice
it is and how ugly it
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is because it's [? fat ?].
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It's not a straight line.
• 16:44 - 16:47
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Now it looks straighter.
• 16:49 - 16:55
So simply, I get to 1, y equals
1 here, which is good for me
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because that's
exactly what I wanted.
• 16:58 - 17:05
I wanted to draw the horizontal
strips for y between 0 and 1.
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I know I'm going
very slow, but that's
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kind of the idea
because-- do you
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mind that I'm going so slow?
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OK.
• 17:12 - 17:17
This is review for the
midterm slowly, a little bit.
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So y between 0 and 1.
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I'm drawing the
horizontal strips,
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and this is exactly
what you guys have.
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This is the red domain.
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Let's call it d.
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It's the same domain but
with horizontal strips.
• 17:31 - 17:34
And I'm going to
draw the same domain.
• 17:34 - 17:35
What color do you like?
• 17:35 - 17:37
I like green because it's
in contrast with red.
• 17:37 - 17:46
I'm going to use green to
draw the vertical strip domain
• 17:46 - 17:47
and say, all right.
• 17:47 - 17:49
Now I know what I'm
supposed to say,
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that d with vertical strips is
going to be x between-- what?
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Yes.
• 17:59 - 18:04
First the fixed
numbers, 0 and 7.
• 18:04 - 18:05
And y between--
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STUDENT: 0 and x plus 7.
• 18:13 - 18:13
STUDENT: And 1.
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DR. MAGDALENA TODA: This one.
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x/7, 1/7x.
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Right?
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Is it x/7?
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x/7y equals x is the same
thing as y equals x/7.
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So y equals x/7 is
this problem, which was
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the same as x equals 7y before.
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OK.
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So how do I set up
the new integral?
• 18:46 - 18:54
I'm going to say dydx, and then
y will be between 0 and x/7.
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And x will be between 0 and 7.
• 18:57 - 19:03
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Is it solved?
• 19:04 - 19:04
No.
• 19:04 - 19:09
But I promise from my heart that
if you do that on the midterm,
• 19:09 - 19:13
you'll get 75% on this
problem, even if doesn't say
• 19:13 - 19:14
don't compute it.
• 19:14 - 19:16
If it says, don't compute
it or anything like that,
• 19:16 - 19:18
you got 100%.
• 19:18 - 19:19
OK?
• 19:19 - 19:21
So this is the most
important step.
• 19:21 - 19:24
From this on, I know
you can do it with what
• 19:24 - 19:26
you've learned in Calc 1 and 2.
• 19:26 - 19:29
It's a piece of cake, and you
should do it with no problem.
• 19:29 - 19:35
Now how are we going
to handle this fellow?
• 19:35 - 19:41
This fellow says, I have nothing
to do with you, Mr. Y. I'm out,
• 19:41 - 19:43
and you're alone.
• 19:43 - 19:44
I don't need you as my friend.
• 19:44 - 19:45
I'm out.
• 19:45 - 19:46
I'm independent.
• 19:46 - 19:49
So Mr. Y starts sulking.
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And say I have an integral
of 1dy between 0 and x/7.
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I'm x/7.
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So you are reduced to
a very simple integral.
• 19:59 - 20:02
That is the integral that you
learned in-- was it Calc 1?
• 20:02 - 20:06
Calc 1, yes, the end of Calc 1.
• 20:06 - 20:07
All right.
• 20:07 - 20:09
So you don't need
the picture anymore.
• 20:09 - 20:12
You've done most of
the work, and you say,
• 20:12 - 20:25
I have an integral from 0 to
7, x over-- so this guy-- which
• 20:25 - 20:26
one shall I put first?
• 20:26 - 20:27
It doesn't matter.
• 20:27 - 20:30
e to the x squared
got out first.
• 20:30 - 20:32
He said, I'm out.
• 20:32 - 20:35
And then the integral of
1dy was y between these two,
• 20:35 - 20:40
so it's x/7 dx.
• 20:40 - 20:41
And this is a 7.
• 20:41 - 20:44
• 20:44 - 20:45
All right.
• 20:45 - 20:46
We are happy.
• 20:46 - 20:48
So what happens?
• 20:48 - 20:51
1/7 also goes for a walk.
• 20:51 - 20:55
And xdx says, OK, I need
to think about who I am.
• 20:55 - 20:58
I have to find my own
identity because I
• 20:58 - 20:59
don't know who I am anymore.
• 20:59 - 21:02
So he says, I need
a u substitution.
• 21:02 - 21:06
u substitution is
u equals x squared.
• 21:06 - 21:09
du equals 2xdx.
• 21:09 - 21:13
So xdx says, I know at least
that I am a differential
• 21:13 - 21:18
form, a 1 form, which is du/2.
• 21:18 - 21:23
And that's exactly what you
guys need to change the inputs.
• 21:23 - 21:24
1/7 was a [? custom ?].
• 21:24 - 21:26
He got out of here.
• 21:26 - 21:31
But you have to think,
when x is 0, what is u?
• 21:31 - 21:32
0.
• 21:32 - 21:37
When x is 7, what is u?
• 21:37 - 21:38
49.
• 21:38 - 21:39
Even my son would know this one.
• 21:39 - 21:40
He would know more.
• 21:40 - 21:42
He would know
fractions and stuff.
• 21:42 - 21:43
OK.
• 21:43 - 21:45
So e to the u.
• 21:45 - 21:48
• 21:48 - 21:50
And the 1/7 was out.
• 21:50 - 21:52
But what is xdx?
• 21:52 - 21:53
du/2.
• 21:53 - 21:55
So I'll say 1/2 du.
• 21:55 - 21:59
• 21:59 - 22:00
Are you guys with me?
• 22:00 - 22:01
• 22:01 - 22:01
Yes.
• 22:01 - 22:03
It shouldn't be a problem.
• 22:03 - 22:05
Now 1/7 got out.
• 22:05 - 22:07
1/2 gets out.
• 22:07 - 22:10
Everybody gets out.
• 22:10 - 22:14
And the guy in the middle who is
left alone, the integral from e
• 22:14 - 22:17
to the u du-- what is he?
• 22:17 - 22:18
e to the u.
• 22:18 - 22:20
Between what values?
• 22:20 - 22:23
Between 49 and 0.
• 22:23 - 22:25
So I'm going to--
shall I write it again?
• 22:25 - 22:26
I'm too lazy for that.
• 22:26 - 22:29
e to the u-- OK, I'll write it.
• 22:29 - 22:33
e to the u between 49 and 0.
• 22:33 - 22:40
So I have 1/14, parentheses,
e to the 49 minus e to the 0.
• 22:40 - 22:42
That's a piece of cake.
• 22:42 - 22:43
1.
• 22:43 - 22:47
OK, so presumably if you
• 22:47 - 22:49
• 22:49 - 22:53
Finding it correctly, you
• 22:53 - 22:56
Of course, you could do
that with the calculator.
• 22:56 - 22:57
MATLAB could do it for you.
• 22:57 - 22:58
Maple could do it for you.
• 22:58 - 23:00
Mathematica could do it for you.
• 23:00 - 23:04
But they will come up
• 23:04 - 23:06
an approximation.
• 23:06 - 23:08
And you haven't learned
anything in the process.
• 23:08 - 23:12
Somebody just served you
• 23:12 - 23:14
and that's not the idea.
• 23:14 - 23:18
• 23:18 - 23:20
Is this hard?
• 23:20 - 23:27
I'm saying on the midterm that's
based on the double integral
• 23:27 - 23:30
with switching order integrals,
this is as hard as it can get.
• 23:30 - 23:33
It cannot get worse than that.
• 23:33 - 23:38
So that will tell you about
the level of the midterm that's
• 23:38 - 23:42
coming up on the 2nd of April,
not something to be worried
• 23:42 - 23:43
• 23:43 - 23:47
Do you need to learn a little
bit during the spring break?
• 23:47 - 23:48
Maybe a few hours.
• 23:48 - 23:52
But I would not worry my
• 23:52 - 23:53
there is this witch.
• 23:53 - 23:56
And I'm going back
to [? Lubbock ?],
• 23:56 - 23:59
and I have to take
her stinking midterm.
• 23:59 - 24:03
And that stresses me out, so I
cannot enjoy my spring break.
• 24:03 - 24:05
By all means, enjoy
• 24:05 - 24:09
And just devote a few
• 24:09 - 24:11
But don't fret.
• 24:11 - 24:15
Don't be worried
• 24:15 - 24:17
because you will be prepared.
• 24:17 - 24:18
And I'm going to do
more review so that you
• 24:18 - 24:23
• 24:23 - 24:23
Another one.
• 24:23 - 24:25
Well, they're all easy.
• 24:25 - 24:28
to the best of my extent.
• 24:28 - 24:32
• 24:32 - 24:33
One more.
• 24:33 - 24:35
Here also is-- I don't-- OK.
• 24:35 - 24:41
Let's take this one because
it's not computational.
• 24:41 - 24:42
And I love it.
• 24:42 - 24:43
It's number 14.
• 24:43 - 24:47
Number 14 and number 15
are so much the same type.
• 24:47 - 24:48
And 16.
• 24:48 - 24:49
It's a theoretical problem.
• 24:49 - 24:53
It practically tests if
you understood the idea.
• 24:53 - 24:55
That's why I love this problem.
• 24:55 - 24:58
And it appears
obsessively, this problem.
• 24:58 - 25:01
I saw it in-- I've
been here for 14 years.
• 25:01 - 25:05
I've seen it at least on 10
different finals, the same type
• 25:05 - 25:08
of theoretical problem.
• 25:08 - 25:15
So it's number 14
over homework four.
• 25:15 - 25:19
Find an equivalent integral
with the order of integration
• 25:19 - 25:20
reversed.
• 25:20 - 25:24
So you need to
reverse some integral.
• 25:24 - 25:31
And since you are so savvy about
reversing the ordered integral,
• 25:31 - 25:35
you should not have
a problem with it.
• 25:35 - 25:48
• 25:48 - 25:51
And WeBWork is
• 25:51 - 25:53
in the following expressions.
• 25:53 - 25:56
You know the type.
• 25:56 - 26:00
f of y, you have to
• 26:00 - 26:03
And g of y, to type
• 26:03 - 26:06
• 26:06 - 26:07
OK.
• 26:07 - 26:12
So you're thinking, I know
how to do this problem.
• 26:12 - 26:16
It must be the idea as before.
• 26:16 - 26:20
This integral should
be-- according
• 26:20 - 26:23
to the order of
integration, it should
• 26:23 - 26:28
be a vertical strip thing
switching to a horizontal strip
• 26:28 - 26:29
thing.
• 26:29 - 26:33
And once I draw the domain,
I'm going to know everything.
• 26:33 - 26:35
you can do this problem
• 26:35 - 26:37
• 26:37 - 26:40
The moment you've learned
it and understood it,
• 26:40 - 26:42
it's going to go very smoothly.
• 26:42 - 26:45
And to convince
you, I'm just going
• 26:45 - 26:49
to go ahead and say, 0 and 1.
• 26:49 - 26:51
And draw, Magdalena.
• 26:51 - 26:52
You know how to draw.
• 26:52 - 26:53
Come on.
• 26:53 - 26:54
OK.
• 26:54 - 26:56
From 1-- 1, 1, right?
• 26:56 - 26:59
Is this the corner-- does
it look like a square?
• 26:59 - 27:00
Yes.
• 27:00 - 27:06
So the parabola y equals x
squared is the bottom one.
• 27:06 - 27:07
Am I right?
• 27:07 - 27:09
That is the bottom one, guys?
• 27:09 - 27:12
But when you see-- when
you are between 0 and 1,
• 27:12 - 27:16
x squared is a lot less
than the square root of x.
• 27:16 - 27:20
The square root of x is the
top, is the function on top.
• 27:20 - 27:25
And then you say, OK, I
got-- somebody gave me
• 27:25 - 27:27
the vertical strips.
• 27:27 - 27:30
I'll put the [INAUDIBLE],
but I don't need them.
• 27:30 - 27:37
and take the purple,
• 27:37 - 27:43
and I'll draw the
horizontal strips.
• 27:43 - 27:47
there because I
• 27:47 - 27:50
see the light in your eyes.
• 27:50 - 27:53
So tell me what you
have. y between n--
• 27:53 - 27:53
STUDENT: 0 and 1.
• 27:53 - 27:55
DR. MAGDALENA TODA: 0 and 1.
• 27:55 - 27:56
Excellent.
• 27:56 - 27:59
And y between what and x?
• 27:59 - 28:00
Oh, sorry, guys.
• 28:00 - 28:02
I need to protect my hand.
• 28:02 - 28:04
That's the secret recipe.
• 28:04 - 28:07
x is between a function of y.
• 28:07 - 28:10
Now what's the
highest function of y?
• 28:10 - 28:11
STUDENT: Square root of y.
• 28:11 - 28:12
DR. MAGDALENA TODA:
Square root of y.
• 28:12 - 28:14
And who is that fellow?
• 28:14 - 28:15
This one.
• 28:15 - 28:18
x equals square root
of y, the green fellow.
• 28:18 - 28:21
I should have written in
green, but I was too lazy.
• 28:21 - 28:26
And this one is going to
be just x equals y squared.
• 28:26 - 28:30
So between y square
down, down, down, down.
• 28:30 - 28:31
Who is down? f is down.
• 28:31 - 28:33
Right, guys?
• 28:33 - 28:35
The bottom one is f.
• 28:35 - 28:37
The bottom one is y squared.
• 28:37 - 28:44
The upper one is the
square root of y.
• 28:44 - 28:46
You cannot type that
in WeBWork, right?
• 28:46 - 28:49
You type sqrt, what?
• 28:49 - 28:51
y, caret, 2.
• 28:51 - 28:52
And here, what do you have?
• 28:52 - 28:53
0 and 1.
• 28:53 - 28:55
So I talk too much.
• 28:55 - 28:58
But if you were on your
own doing this in WeBWork,
• 28:58 - 29:02
it would take you no
more than-- I don't
• 29:02 - 29:05
know-- 60 seconds to type in.
• 29:05 - 29:07
Remember this problem
for the midterm.
• 29:07 - 29:09
It's an important idea.
• 29:09 - 29:11
And you've seen it emphasized.
• 29:11 - 29:17
You will see it emphasized
in problems 14, 15, 16.
• 29:17 - 29:20
It's embedded in this
type of exchange,
• 29:20 - 29:22
change the order of
integration type problem.
• 29:22 - 29:25
• 29:25 - 29:27
OK?
• 29:27 - 29:32
Anything else I would like
to show you from-- there
• 29:32 - 29:34
are many things I
would like to show you.
• 29:34 - 29:37
But I better let you
• 29:37 - 29:42
How about 17, which is a
similar type of problem,
• 29:42 - 29:44
theoretical, just like this one?
• 29:44 - 29:49
But it's testing
if you know the--
• 29:49 - 29:53
if you understood the idea
behind polar integration,
• 29:53 - 29:56
integration in
polar coordinates.
• 29:56 - 29:57
Can I erase?
• 29:57 - 30:00
OK.
• 30:00 - 30:03
So let's switch to number
• 30:03 - 30:06
• 30:06 - 30:09
Write down the problems
we are going over,
• 30:09 - 30:10
so when you do
• 30:10 - 30:13
• 30:13 - 30:14
This is not a lecture.
• 30:14 - 30:16
What is this, what
you're doing now?
• 30:16 - 30:19
It's like-- what is this?
• 30:19 - 30:23
An application session,
a problem session.
• 30:23 - 30:25
OK.
• 30:25 - 30:30
Number 17, homework four.
• 30:30 - 30:36
On this one, unfortunately
• 30:36 - 30:39
because there is no data.
• 30:39 - 30:46
So when-- it's the unique
problem you're going to get.
• 30:46 - 30:53
You have a picture, and that
picture looks like that.
• 30:53 - 30:58
From here, [INAUDIBLE]
a half of an annulus.
• 30:58 - 31:01
• 31:01 - 31:03
You have half of a ring.
• 31:03 - 31:07
And it says, suppose that
• 31:07 - 31:10
in the figure.
• 31:10 - 31:13
As an iterated integral
in polar coordinates,
• 31:13 - 31:20
the double integral
over R f of x, y dA
• 31:20 - 31:24
is the integral from A to
B of the integral from C
• 31:24 - 31:38
to B of f of r, theta times r
drd theta with the following
• 31:38 - 31:40
limits of integration.
• 31:40 - 31:45
A. And WeBWork says, you say it.
• 31:45 - 31:45
You say.
• 31:45 - 31:47
It's playing games with you.
• 31:47 - 31:48
B, you say.
• 31:48 - 31:50
It's a guessing game.
• 31:50 - 31:51
C, you say.
• 31:51 - 31:55
Then D, you say it.
• 31:55 - 31:58
And let's see what you say.
• 31:58 - 32:02
• 32:02 - 32:06
Well, we say, well,
how am I going to go?
• 32:06 - 32:09
I have to disclose
the graphing paper.
• 32:09 - 32:11
They are so mean.
• 32:11 - 32:15
They don't show you
the actual numbers.
• 32:15 - 32:17
They only give you
graphing paper.
• 32:17 - 32:19
I'm not good at graphing, OK?
• 32:19 - 32:23
So you will have to
guess what this says.
• 32:23 - 32:24
That should be good enough.
• 32:24 - 32:25
Perfect.
• 32:25 - 32:29
So the unit supposedly
is this much.
• 32:29 - 32:32
1 inch, whatever.
• 32:32 - 32:33
I don't care.
• 32:33 - 32:36
So is it hard?
• 32:36 - 32:37
It's a piece of cake.
• 32:37 - 32:39
It's a 10 second problem.
• 32:39 - 32:41
It's a good problem for the
midterm because it's fast.
• 32:41 - 32:49
• 32:49 - 32:53
Theta is a wonderful angle.
• 32:53 - 32:56
• 32:56 - 32:58
It is nice to look at.
• 32:58 - 33:02
And they really don't
put numbers here?
• 33:02 - 33:04
They do.
• 33:04 - 33:07
They do on the margin
of the graphing paper.
• 33:07 - 33:09
They have a scale.
• 33:09 - 33:09
OK.
• 33:09 - 33:11
So come on.
• 33:11 - 33:12
This is easy.
• 33:12 - 33:15
You guys are too smart
for this problem.
• 33:15 - 33:18
From what to what?
• 33:18 - 33:18
STUDENT: [INAUDIBLE].
• 33:18 - 33:20
DR. MAGDALENA TODA: Nope.
• 33:20 - 33:21
No, that's a problem.
• 33:21 - 33:23
So when we measure
the angle theta,
• 33:23 - 33:25
where do we start measuring?
• 33:25 - 33:25
STUDENT: 0.
• 33:25 - 33:27
DR. MAGDALENA TODA: Over here.
• 33:27 - 33:29
So we go down there
clockwise because that's
• 33:29 - 33:31
how we mix in the bowl,
counter-clockwise.
• 33:31 - 33:38
So 0-- so this is
going to be pi.
• 33:38 - 33:39
Pi.
• 33:39 - 33:43
• 33:43 - 33:45
And what is the end?
• 33:45 - 33:46
STUDENT: 2 pi.
• 33:46 - 33:47
DR. MAGDALENA TODA: 2 pi.
• 33:47 - 33:47
2 pi.
• 33:47 - 33:50
• 33:50 - 33:53
Don't type-- oh, I mean, you
cannot type the symbol part,
• 33:53 - 33:54
right?
• 33:54 - 33:58
And then what do you
type, in terms of C and D?
• 33:58 - 33:59
STUDENT: [INAUDIBLE]
• 33:59 - 34:00
DR. MAGDALENA TODA: Nope.
• 34:00 - 34:00
No, no.
• 34:00 - 34:03
• 34:03 - 34:06
STUDENT: 0 to 1.
• 34:06 - 34:07
DR. MAGDALENA TODA: 1 to 2.
• 34:07 - 34:08
Why 1 to 2?
• 34:08 - 34:09
Excellent.
• 34:09 - 34:13
represents the half of a donut.
• 34:13 - 34:14
You have nothing inside.
• 34:14 - 34:18
There is a whole in
here, in the donut.
• 34:18 - 34:20
So between 0 and 1,
you have nothing.
• 34:20 - 34:23
• 34:23 - 34:24
It's here.
• 34:24 - 34:26
• 34:26 - 34:32
you guys see on the picture is
a value that's between 1 and 2,
• 34:32 - 34:35
between 1 and 2.
• 34:35 - 34:36
And that's it.
• 34:36 - 34:38
That was a 10 second problem.
• 34:38 - 34:38
So promise me.
• 34:38 - 34:41
You are going to do
the homework and stuff.
• 34:41 - 34:44
You have two or three like that.
• 34:44 - 34:46
If you see this on
the midterm, are you
• 34:46 - 34:50
going to remember the procedure,
the idea of the problem?
• 34:50 - 34:53
OK.
• 34:53 - 34:55
I'm going to also think
of writing a sample.
• 34:55 - 34:58
I promised Stacy I'm
going to do that.
• 34:58 - 34:59
And I did not forget.
• 34:59 - 35:00
It's going to happen.
• 35:00 - 35:02
After spring break, you're
going to get a review
• 35:02 - 35:05
sheet for the midterm.
• 35:05 - 35:09
I promised you a sample, right?
• 35:09 - 35:09
OK.
• 35:09 - 35:15
• 35:15 - 35:18
Shall I do more or not?
• 35:18 - 35:21
Yes?
• 35:21 - 35:23
You know what I'm
afraid of, really?
• 35:23 - 35:29
I think you will be able to do
fine with most of the problems
• 35:29 - 35:32
you have here.
• 35:32 - 35:39
geometric representations
• 35:39 - 35:44
you guys became familiar
• 35:44 - 35:48
with only now,
only this semester.
• 35:48 - 35:51
And you have a grasp of them.
• 35:51 - 35:52
You've seen them.
• 35:52 - 35:57
But you're still not
very friendly with them,
• 35:57 - 35:59
and you don't
quite like to draw.
• 35:59 - 36:03
So let's see if we can learn
how to draw one of them together
• 36:03 - 36:08
and see if it's a big
deal or not because it's
• 36:08 - 36:10
pretty as a picture.
• 36:10 - 36:12
And when we set it
up as an integral,
• 36:12 - 36:17
it should be done wisely.
• 36:17 - 36:19
It shouldn't be hard.
• 36:19 - 36:22
• 36:22 - 36:25
We have to do a good job
from the moment we draw.
• 36:25 - 36:30
And if we don't do that,
we don't have much chance.
• 36:30 - 36:36
The problem is going to
change the data a little bit
• 36:36 - 36:39
to numbers that I like.
• 36:39 - 36:40
29.
• 36:40 - 36:45
• 36:45 - 36:47
You have a solid.
• 36:47 - 36:49
And I say solid gold, 24 k.
• 36:49 - 36:51
I don't know what.
• 36:51 - 36:54
That is between two paraboloids.
• 36:54 - 36:57
And those paraboloids
are given, and I'd
• 36:57 - 37:00
like you to tell me
what they look like.
• 37:00 - 37:06
One paraboloid is y-- no.
• 37:06 - 37:09
• 37:09 - 37:09
Yeah.
• 37:09 - 37:14
One paraboloid is
y-- I'll change it.
• 37:14 - 37:15
z.
• 37:15 - 37:17
So I can change your
problem, and then you
• 37:17 - 37:19
will figure it out by yourself.
• 37:19 - 37:21
z equals x squared
plus y squared.
• 37:21 - 37:24
They give you y equals x
squared plus d squared.
• 37:24 - 37:27
So you have to change
completely the configuration
• 37:27 - 37:29
• 37:29 - 37:33
And then z equals 8 minus
x squared minus y squared.
• 37:33 - 37:39
I'm I'm changing problem 29,
but it's practically the same.
• 37:39 - 37:44
Find the volume of the solid
enclosed by the two paraboloids
• 37:44 - 37:45
• 37:45 - 37:49
• 37:49 - 37:56
Find the volume of
the solid enclosed
• 37:56 - 37:58
by the two paraboloids.
• 37:58 - 37:59
You go, oh, my god.
• 37:59 - 38:03
How am I going to do that?
• 38:03 - 38:04
STUDENT: Draw the pictures.
• 38:04 - 38:05
DR. MAGDALENA TODA:
Draw the pictures.
• 38:05 - 38:06
Very good.
• 38:06 - 38:09
So he's teaching
my sensing to me
• 38:09 - 38:11
and draw the picture.
• 38:11 - 38:13
Don't be lazy, because
if you don't, it's
• 38:13 - 38:15
never going to happen.
• 38:15 - 38:17
You're never going
to see the domain
• 38:17 - 38:20
if you don't draw the pictures.
• 38:20 - 38:24
So the first one will
be the shell of the egg.
• 38:24 - 38:25
Easter is coming.
• 38:25 - 38:30
So that's something
like the shell.
• 38:30 - 38:34
It's a terrible shell,
a paraboloid, circular
• 38:34 - 38:35
paraboloid.
• 38:35 - 38:41
And that is called z equals
x squared plus y squared.
• 38:41 - 38:50
• 38:50 - 38:50
OK.
• 38:50 - 38:53
• 38:53 - 38:56
This guy keeps going.
• 38:56 - 38:58
But there will be
another paraboloid
• 38:58 - 39:05
that has the shape of exactly
the same thing upside down.
• 39:05 - 39:07
STUDENT: Where's 8?
• 39:07 - 39:08
DR. MAGDALENA TODA: Where is 8?
• 39:08 - 39:10
The 8 is far away.
• 39:10 - 39:11
STUDENT: It's on--
• 39:11 - 39:12
DR. MAGDALENA TODA: I'll try.
• 39:12 - 39:17
• 39:17 - 39:19
STUDENT: Did they tell you
that a had to be positive?
• 39:19 - 39:20
DR. MAGDALENA TODA: Huh?
• 39:20 - 39:22
STUDENT: Did they tell
you a had to be positive?
• 39:22 - 39:24
DR. MAGDALENA TODA: Which a?
• 39:24 - 39:26
STUDENT: That a or whatever.
• 39:26 - 39:27
DR. MAGDALENA TODA: 8.
• 39:27 - 39:28
STUDENT: Oh, 8.
• 39:28 - 39:29
DR. MAGDALENA TODA: 8.
• 39:29 - 39:29
STUDENT: Oh, that's
why I'm confused.
• 39:29 - 39:31
DR. MAGDALENA TODA:
How do I know it's 8?
• 39:31 - 39:34
Because when I put
x equals 7 equals 0,
• 39:34 - 39:37
I get z equals 8
for this paraboloid.
• 39:37 - 39:41
This is the red paraboloid.
• 39:41 - 39:43
The problem-- my
question is, OK, it's
• 39:43 - 39:46
like it is two
eggshells that are
• 39:46 - 39:49
connecting, exactly this egg.
• 39:49 - 39:53
But the bound-- the--
how do you call that?
• 39:53 - 39:56
Boundary, the thing where
they glue it together.
• 39:56 - 39:58
What is the equation
of this circle?
• 39:58 - 39:59
This is the question.
• 39:59 - 40:01
Where do they intersect?
• 40:01 - 40:05
How do you find out where
two surfaces intersect?
• 40:05 - 40:06
STUDENT: [INAUDIBLE]
• 40:06 - 40:08
DR. MAGDALENA TODA:
Solve a system.
• 40:08 - 40:13
Make a system of two equations
and solve the system.
• 40:13 - 40:15
You have to intersect them.
• 40:15 - 40:21
So whoever x, y, z will be, they
have to satisfy both equations.
• 40:21 - 40:22
Oh, my god.
• 40:22 - 40:25
So we have to look for the
solutions of both equations
• 40:25 - 40:32
at the same time, which
means that I'm going
• 40:32 - 40:36
to say these are equal, right?
• 40:36 - 40:38
Let's write that down.
• 40:38 - 40:41
x squared plus y
squared equals 8 minus x
• 40:41 - 40:43
squared minus y squared.
• 40:43 - 40:48
Then z is whatever.
• 40:48 - 40:52
What is this equation?
• 40:52 - 40:54
We'll find out who
z is in a second.
• 40:54 - 40:57
• 40:57 - 41:00
z has to be x squared
plus y squared.
• 41:00 - 41:04
If we find out who the sum
of the squares will be,
• 41:04 - 41:07
we'll find out the altitude z.
• 41:07 - 41:09
z equals what number?
• 41:09 - 41:10
This is the whole idea.
• 41:10 - 41:11
So x squared.
• 41:11 - 41:14
I move everything to
the left hand side.
• 41:14 - 41:17
So I have 2x squared
plus 2y squared equals 8.
• 41:17 - 41:23
• 41:23 - 41:29
And then I have z equals
x squared plus y squared.
• 41:29 - 41:33
And then that's if and only
if x squared plus y squared
• 41:33 - 41:36
equals 4.
• 41:36 - 41:37
STUDENT: Then z equals 4.
• 41:37 - 41:39
DR. MAGDALENA TODA:
So z equals 4.
• 41:39 - 41:45
So z equals 4 is exactly what
I guessed because come on.
• 41:45 - 41:48
The two eggshells
have to be equal.
• 41:48 - 41:52
So this should be in the
middle between 0 and 8.
• 41:52 - 41:54
So I knew it was z equals 4.
• 41:54 - 41:56
it mathematically.
• 41:56 - 41:59
So z equals 4, and x
squared plus y squared
• 41:59 - 42:02
equals 4 is the boundary.
• 42:02 - 42:05
Let's make it purple
because it's the same
• 42:05 - 42:07
as the purple equation there.
• 42:07 - 42:09
• 42:09 - 42:16
So the domain has to be the
projection of this purple--
• 42:16 - 42:20
it looks like a sci-fi thing.
• 42:20 - 42:22
You have some hologram.
• 42:22 - 42:23
I don't know what it is.
• 42:23 - 42:25
• 42:25 - 42:28
You want to know the domain
D. Could somebody tell me
• 42:28 - 42:31
what the domain D will be?
• 42:31 - 42:37
It will be those x's and y's
on the floor with the quality
• 42:37 - 42:42
that x squared plus y squared
will be between 0 and--
• 42:42 - 42:43
STUDENT: 4
• 42:43 - 42:44
DR. MAGDALENA TODA: --4.
• 42:44 - 42:47
So I can do everything
in polar coordinates.
• 42:47 - 42:51
This is the same thing as
saying rho, theta-- r, theta.
• 42:51 - 42:52
Not rho.
• 42:52 - 42:53
Rho is Greek.
• 42:53 - 42:54
It's all Greek to me.
• 42:54 - 42:57
So rho is sometimes
used by people
• 42:57 - 43:02
for the polar coordinates,
rho and theta.
• 43:02 - 43:05
But we use r.
• 43:05 - 43:09
r squared between 0 and 4.
• 43:09 - 43:10
You'll say, Magdalena, come on.
• 43:10 - 43:11
That's silly.
• 43:11 - 43:14
Why didn't you write
r between 0 and 2?
• 43:14 - 43:16
I will.
• 43:16 - 43:16
I will.
• 43:16 - 43:17
I will.
• 43:17 - 43:18
This is 2, right?
• 43:18 - 43:21
So r between 0 and 2.
• 43:21 - 43:23
I erase this.
• 43:23 - 43:27
And theta is between 0 and 2 pi.
• 43:27 - 43:28
And I'm done.
• 43:28 - 43:29
Why 0 and 2 pi?
• 43:29 - 43:30
Because we have the whole egg.
• 43:30 - 43:33
I mean, I could
cut the egg in half
• 43:33 - 43:37
and say 0 to pi or something,
invent a different problem.
• 43:37 - 43:40
But for the time
being, I'm rotating
• 43:40 - 43:45
a full rotation of 2 pi to
create the egg all around.
• 43:45 - 43:49
So finally, what is
the volume of-- suppose
• 43:49 - 43:56
this is like in the story
with the golden eggs.
• 43:56 - 43:58
They are solid gold eggs.
• 43:58 - 44:00
Wouldn't that be wonderful?
• 44:00 - 44:03
We want to know the
volume of this golden egg.
• 44:03 - 44:08
What's inside the solid egg, not
the shell, not just the shell
• 44:08 - 44:09
• 44:09 - 44:12
The whole thing is made of gold.
• 44:12 - 44:15
And who's coming tomorrow
to the-- sorry, guys.
• 44:15 - 44:16
Mathematician talking.
• 44:16 - 44:18
Switching from
another-- who's coming
• 44:18 - 44:20
tomorrow to the honors society?
• 44:20 - 44:23
Do you-- did you decide?
• 44:23 - 44:24
You have.
• 44:24 - 44:26
And Rachel comes.
• 44:26 - 44:27
Are you coming?
• 44:27 - 44:28
No, no, no, no.
• 44:28 - 44:30
Tomorrow night.
• 44:30 - 44:30
Tomorrow night.
• 44:30 - 44:32
Tomorrow.
• 44:32 - 44:34
What time does that--
• 44:34 - 44:34
STUDENT: 3:00.
• 44:34 - 44:36
DR. MAGDALENA
TODA: At 3 o'clock.
• 44:36 - 44:37
At 3 o'clock.
• 44:37 - 44:38
OK.
• 44:38 - 44:41
So if you want, I can
• 44:41 - 44:45
And then you'll be members.
• 44:45 - 44:47
I saw one of the certificates.
• 44:47 - 44:48
It was really beautiful.
• 44:48 - 44:50
That one [INAUDIBLE].
• 44:50 - 44:53
It was really-- some
parents frame these things.
• 44:53 - 44:55
My parents don't care.
• 44:55 - 44:56
But I wish they cared.
• 44:56 - 45:01
So the more certificates you
get, and the older you get,
• 45:01 - 45:04
the nicer it is to put them,
frame them and put them
• 45:04 - 45:06
on the wall of
fame of the family.
• 45:06 - 45:09
This certificate, the KME
one, looks so much better
• 45:09 - 45:15
than my own diplomas, the PhD
diplomas, the math diplomas.
• 45:15 - 45:20
And it's huge, and it
has a golden silver seal
• 45:20 - 45:22
will all the stuff.
• 45:22 - 45:25
And it's really nice.
• 45:25 - 45:26
OK.
• 45:26 - 45:29
Now coming back to this thing.
• 45:29 - 45:32
• 45:32 - 45:34
STUDENT: Can we multiply by 2?
• 45:34 - 45:35
Just find the--
• 45:35 - 45:36
DR. MAGDALENA TODA: Exactly.
• 45:36 - 45:37
That's what we will do.
• 45:37 - 45:44
We could set up the integral
from whatever it is.
• 45:44 - 45:48
My one function to
another function.
• 45:48 - 45:51
But the simplest way
to compute the volume
• 45:51 - 45:54
would be to say
there are two types.
• 45:54 - 45:58
And set up the
integral for this one,
• 45:58 - 46:02
for example or the other one.
• 46:02 - 46:05
It doesn't matter which one.
• 46:05 - 46:06
It doesn't really
matter which one.
• 46:06 - 46:07
Which one we would prefer?
• 46:07 - 46:08
I don't know.
• 46:08 - 46:11
Maybe you like the bottom
part of the [INAUDIBLE].
• 46:11 - 46:12
I don't know.
• 46:12 - 46:14
Do you guys understand
• 46:14 - 46:16
STUDENT: If we
just-- I don't know
• 46:16 - 46:25
where to find B. Find the
area left, like indented?
• 46:25 - 46:28
Because if you did it at the
bottom, the domain is zero.
• 46:28 - 46:31
Then you have-- wouldn't
it find the stuff that
• 46:31 - 46:34
was not cupped out, the edges?
• 46:34 - 46:36
DR. MAGDALENA TODA:
Isn't it exactly
• 46:36 - 46:39
the same volume up and down?
• 46:39 - 46:40
STUDENT: Yes.
• 46:40 - 46:42
DR. MAGDALENA TODA: It's
the same volume up and down.
• 46:42 - 46:45
So it's enough for
me to take the volume
• 46:45 - 46:48
of the lower part and w.
• 46:48 - 46:52
Can you help me set
up the lower part?
• 46:52 - 46:53
So I'm going to have two types.
• 46:53 - 46:56
Can I do that directly
in polar coordinates?
• 46:56 - 46:57
That's the thing.
• 46:57 - 46:59
1 is 1.
• 46:59 - 47:01
r is r.
• 47:01 - 47:03
r is going to be-- this is
the Jacobian r drd theta.
• 47:03 - 47:10
• 47:10 - 47:13
OK?
• 47:13 - 47:21
But now let me ask you, how
do we compute-- I'm sorry.
• 47:21 - 47:32
This is the function f of x, y.
• 47:32 - 47:34
• 47:34 - 47:38
Yeah, it's a little
bit more complicated.
• 47:38 - 47:41
So you have to subtract
from one the other one.
• 47:41 - 47:54
• 47:54 - 47:57
So I'm referring to the domain
as being only the planar
• 47:57 - 47:58
domain.
• 47:58 - 48:02
• 48:02 - 48:07
And I have first a graph
and then another graph.
• 48:07 - 48:12
So when I want to compute,
• 48:12 - 48:18
I want to compute the
volume of this, the volume
• 48:18 - 48:22
of this egg, the inside.
• 48:22 - 48:26
I have to say, OK, integral over
the d of the function that's
• 48:26 - 48:27
on top.
• 48:27 - 48:32
The function that's on top
is the z equals f of x, y.
• 48:32 - 48:35
And the function that's
on the bottom for this egg
• 48:35 - 48:37
is just this.
• 48:37 - 48:41
So this is just
a flat altitude g
• 48:41 - 48:43
of x, y equals-- what is that?
• 48:43 - 48:46
4.
• 48:46 - 48:54
So I have to subtract the two
because I have first this body.
• 48:54 - 48:58
If this would not exist, how
would I get the purple part?
• 48:58 - 49:03
I would say for the function f,
the protection on the ground,
• 49:03 - 49:09
I have this whole body
that looks like a crayon.
• 49:09 - 49:11
A whole body that
looks like crayon.
• 49:11 - 49:13
This is the first integral.
• 49:13 - 49:18
I minus the cylinder
that's dotted with floating
• 49:18 - 49:21
points, which is this part.
• 49:21 - 49:23
So it's V1 minus V2.
• 49:23 - 49:27
V1 is the volume of the whole
body that looks like a crayon.
• 49:27 - 49:32
V2 is just the volume of the
cylinder under the crayon.
• 49:32 - 49:36
We want-- minus
V2 is exactly half
• 49:36 - 49:42
of the egg, the volume of the
half of the egg, give or take.
• 49:42 - 49:43
So is this hard?
• 49:43 - 49:45
It shouldn't be hard.
• 49:45 - 49:50
f of x, y-- can you guys
tell me who that is?
• 49:50 - 49:54
A minus x squared
minus y squared.
• 49:54 - 49:55
And who is g?
• 49:55 - 49:59
• 49:59 - 49:59
4.
• 49:59 - 50:02
Just the altitude, 4.
• 50:02 - 50:03
OK.
• 50:03 - 50:05
So I'm going to go
• 50:05 - 50:11
I have to integrate 2 double
integral over D. 8 minus 4
• 50:11 - 50:18
is 4 minus x squared
minus y squared dA.
• 50:18 - 50:24
dA is the area element dxdy.
• 50:24 - 50:28
Now switch to polar.
• 50:28 - 50:29
How do you switch to polar?
• 50:29 - 50:33
• 50:33 - 50:35
You can also set this
up as a triple integral.
• 50:35 - 50:37
And that's what I
wanted to do at first.
• 50:37 - 50:40
But then I realized that you
don't know triple integrals,
• 50:40 - 50:42
so I set it up as
a double integral.
• 50:42 - 50:44
For a triple integral,
you have three snakes.
• 50:44 - 50:47
And you integrate the
element 1, and that's
• 50:47 - 50:48
going to be the volume.
• 50:48 - 50:52
And I'll teach you in
the next two sessions.
• 50:52 - 50:56
2 times the double integral.
• 50:56 - 50:57
Who is this nice fellow?
• 50:57 - 50:59
Look how nice and sassy he is.
• 50:59 - 51:09
4 minus r squared times--
never forget the r drd theta.
• 51:09 - 51:15
Theta goes between 0
and 2 pi and r between--
• 51:15 - 51:17
STUDENT: 0 and 2.
• 51:17 - 51:18
DR. MAGDALENA TODA: 0 and--
• 51:18 - 51:18
STUDENT: 2.
• 51:18 - 51:19
DR. MAGDALENA TODA: 2.
• 51:19 - 51:20
Excellent.
• 51:20 - 51:25
Because when I had 4 here,
• 51:25 - 51:27
So r is 2.
• 51:27 - 51:28
Look at this integral.
• 51:28 - 51:31
Is it hard?
• 51:31 - 51:33
Not so hard.
• 51:33 - 51:34
Not so hard at all.
• 51:34 - 51:37
So what would you
do if you were me?
• 51:37 - 51:38
Would you do a u substitution?
• 51:38 - 51:41
Do you need a u
substitution necessarily?
• 51:41 - 51:42
You don't need it.
• 51:42 - 51:48
So just say 4r minus r cubed.
• 51:48 - 51:50
Now what do you see again?
• 51:50 - 51:52
Theta is missing
from the picture.
• 51:52 - 51:54
Theta says, I'm out of here.
• 51:54 - 51:55
I don't care.
• 51:55 - 52:00
So you get 2 times the integral
from 0 to 2 pi of nothing--
• 52:00 - 52:04
well, of 1d theta, not of
nothing-- times the integral
• 52:04 - 52:16
from 0 to 2 of 4r
minus r cubed dr.
• 52:16 - 52:22
4r minus r cubed dr, the
integral from 0 to 2.
• 52:22 - 52:22
Good.
• 52:22 - 52:32
• 52:32 - 52:33
Who's going to help me?
• 52:33 - 52:35
• 52:35 - 52:39
I give you how
much money-- money.
• 52:39 - 52:43
Time shall I give
you to do this one?
• 52:43 - 52:46
And I need three people
to respond and get
• 52:46 - 52:48
• 52:48 - 52:54
So [INAUDIBLE] 2r squared
minus r to the 4 over 4
• 52:54 - 52:56
between 0 and 2.
• 52:56 - 53:00
• 53:00 - 53:02
STUDENT: 16 [INAUDIBLE].
• 53:02 - 53:04
DR. MAGDALENA TODA:
How much did you get?
• 53:04 - 53:04
STUDENT: 4.
• 53:04 - 53:05
DR. MAGDALENA TODA:
How much did you get?
• 53:05 - 53:06
STUDENT: For just this one?
• 53:06 - 53:08
DR. MAGDALENA
TODA: For all this.
• 53:08 - 53:10
• 53:10 - 53:11
STUDENT: 4.
• 53:11 - 53:18
• 53:18 - 53:20
DR. MAGDALENA TODA:
Yes, it is, right?
• 53:20 - 53:21
Are you with me?
• 53:21 - 53:23
You have 2r squared
when you integrate
• 53:23 - 53:26
minus r to the 4 over
4 between 0 and 2.
• 53:26 - 53:31
That means 2 times 4 minus 16/4.
• 53:31 - 53:33
8 minus 4 is 4.
• 53:33 - 53:38
So with 4 for this guy, 2 pi
for this guy, and one 2 outside,
• 53:38 - 53:43
you have 16 pi.
• 53:43 - 53:47
And that was-- I remember
it as if it was yesterday.
• 53:47 - 53:55
That was on a final
two or three years ago.
• 53:55 - 53:58
OK.
• 53:58 - 54:04
So you've seen many
of these problems now.
• 54:04 - 54:07
It shouldn't be complicated
• 54:07 - 54:08
• 54:08 - 54:11
If you want, go ahead and
• 54:11 - 54:13
that we did today.
• 54:13 - 54:16
And when you see numbers
changed or something,
• 54:16 - 54:18
problem the same way.
• 54:18 - 54:20
Make sure you understood it.
• 54:20 - 54:22
I'm going to do more.
• 54:22 - 54:23
Is this useful for you?
• 54:23 - 54:25
I mean-- OK.
• 54:25 - 54:27
So you agree that
every now and then,
• 54:27 - 54:31
we do homework in the classroom?
• 54:31 - 54:33
Homework like problems
in the classroom.
• 54:33 - 54:36
In the homework, you
may have different data,
• 54:36 - 54:39
but it's the same
type of problem.
• 54:39 - 54:39
OK.
• 54:39 - 54:45
• 54:45 - 54:50
I'm going to remind you
of some Calc 2 notions
• 54:50 - 54:57
because today I will
cover the surface area.
• 54:57 - 54:58
STUDENT: Dr. Toda?
• 54:58 - 54:58
DR. MAGDALENA TODA: Yes, sir?
• 54:58 - 55:00
STUDENT: I have a question
on the last problem.
• 55:00 - 55:01
DR. MAGDALENA TODA: Yes, sir?
• 55:01 - 55:03
something like that on the exam
• 55:03 - 55:07
and had done it using the fact
that it's a solid revolution--
• 55:07 - 55:09
DR. MAGDALENA TODA:
Yeah, you can do that.
• 55:09 - 55:11
There are at least four
methods to do this problem.
• 55:11 - 55:12
One would be with
triple integral.
• 55:12 - 55:14
One would be with
a double integral
• 55:14 - 55:17
of a function on top
minus the function below.
• 55:17 - 55:21
One would be with solid of
revolution like in Calc 2,
• 55:21 - 55:24
where your axis is the z axis.
• 55:24 - 55:26
I don't care how you
solve the problem.
• 55:26 - 55:29
Again, if I were
the CEO of a company
• 55:29 - 55:31
or the boss of a
firm or something,
• 55:31 - 55:37
I would care for my employees to
be solving problems the fastest
• 55:37 - 55:38
possible way.
• 55:38 - 55:40
As long as the
• 55:40 - 55:41
I don't care how you do it.
• 55:41 - 55:42
STUDENT: Thank you, Doctor.
• 55:42 - 55:44
DR. MAGDALENA TODA: So go ahead.
• 55:44 - 55:44
All right.
• 55:44 - 55:47
• 55:47 - 56:00
Oh, and by the way, I want to
give you another example where
• 56:00 - 56:03
the students were able
to very beautifully cheat
• 56:03 - 56:06
• 56:06 - 56:09
That was funny.
• 56:09 - 56:12
But that is again
a Calc 3 problem
• 56:12 - 56:17
in an elementary way
that can be solved
• 56:17 - 56:23
with the notions you have from
K-12, if you mastered them
• 56:23 - 56:25
[INAUDIBLE].
• 56:25 - 56:29
So you are given x
plus y plus z equals 1.
• 56:29 - 56:31
Before I do the
surface integral--
• 56:31 - 56:34
I could do the surface
integral for such a problem.
• 56:34 - 56:49
This is a plane that intersects
the coordinate planes
• 56:49 - 56:55
and forms a
tetrahedron with them.
• 56:55 - 57:05
• 57:05 - 57:07
Find the volume of
that tetrahedron.
• 57:07 - 57:19
• 57:19 - 57:26
Now I say, with Calc 3,
because the course coordinator
• 57:26 - 57:30
several years ago did not
specify with what you learned.
• 57:30 - 57:33
Set up a double integral
or set up-- he simply
• 57:33 - 57:36
said, find the volume.
• 57:36 - 57:40
So the students-- what's
the simplest way to do it?
• 57:40 - 57:41
STUDENT: That's
just half a cube.
• 57:41 - 57:45
DR. MAGDALENA TODA:
Just draw the thingy.
• 57:45 - 57:46
And they were smart.
• 57:46 - 57:47
They knew how to draw it.
• 57:47 - 57:51
The knew what the vertices were.
• 57:51 - 57:53
The plane looks like this.
• 57:53 - 57:57
If you shade it, you see
that it's x plus y plus z.
• 57:57 - 58:00
And I'm going to try
and write with my hands.
• 58:00 - 58:01
It's very hard.
• 58:01 - 58:04
But it comes from 0, 0, 1 point.
• 58:04 - 58:06
This is the 0, 0, 1.
• 58:06 - 58:08
And it comes like that.
• 58:08 - 58:11
And it hits the floor over here.
• 58:11 - 58:16
And these points are 1,
0, 0; 0, 1, 0; and 0, 0,
• 58:16 - 58:20
1 on the vertices
of a tetrahedron,
• 58:20 - 58:22
including the origin.
• 58:22 - 58:25
How do I know those are
exactly the vertices
• 58:25 - 58:27
of the tetrahedron?
• 58:27 - 58:30
Because they verify x
plus y plus z equals 1.
• 58:30 - 58:33
As long as the point
verifies the equation,
• 58:33 - 58:35
it is in the plane.
• 58:35 - 58:38
For example, another point
that's not in the picture
• 58:38 - 58:40
would be 1/3 plus 1/3 plus 1/3.
• 58:40 - 58:42
1/3 and 1/3 is 1/3.
• 58:42 - 58:47
Anything that verifies the
equation is in the plane.
• 58:47 - 58:49
So the tetrahedron has a name.
• 58:49 - 58:57
It's called-- let's call
this A, B, C, and O. OABC.
• 58:57 - 58:58
It's a tetrahedron.
• 58:58 - 59:00
It's a pyramid.
• 59:00 - 59:07
So how does the smart
student who was not given
• 59:07 - 59:09
a specific method solve that?
• 59:09 - 59:11
They did that on the final.
• 59:11 - 59:12
I'm so proud of them.
• 59:12 - 59:13
I said, come on now.
• 59:13 - 59:14
The final is two
hours and a half.
• 59:14 - 59:16
You don't know what to do first.
• 59:16 - 59:22
So they said-- they
did the base multiplied
• 59:22 - 59:25
by the height divided by 3.
• 59:25 - 59:30
So you get 1 times 1.
• 59:30 - 59:32
So practically, divided by 2.
• 59:32 - 59:33
1/2.
• 59:33 - 59:36
You don't even have
to do the-- even
• 59:36 - 59:41
my son would know that this
is half of a square, a 1
• 59:41 - 59:42
by 1 square.
• 59:42 - 59:46
So it's half the area of the
base times the height, which
• 59:46 - 59:49
is 1, divided by 3 is 1/6.
• 59:49 - 59:52
And goodbye and see you later.
• 59:52 - 59:56
But if you wanted-- if
the author of the problem
• 59:56 - 59:58
would indicate, do
that with Calculus 3,
• 59:58 - 60:01
then that's another
story because you
• 60:01 - 60:06
have to realize what the domain
would be, the planar domain.
• 60:06 - 60:09
You practically have a surface.
• 60:09 - 60:12
equilateral triangle
• 60:12 - 60:16
call it c of f from surface.
• 60:16 - 60:20
But this would be
z equals f of x, y.
• 60:20 - 60:22
How do you get to that?
• 60:22 - 60:24
You get it from here.
• 60:24 - 60:28
The explicit equation
is-- [INAUDIBLE].
• 60:28 - 60:29
1 minus x minus y.
• 60:29 - 60:33
That is the surface,
the green surface.
• 60:33 - 60:35
And the domain--
let's draw that in.
• 60:35 - 60:39
Do you prefer red or purple?
• 60:39 - 60:40
You don't care?
• 60:40 - 60:44
• 60:44 - 60:47
OK, I'll take red.
• 60:47 - 60:47
Red.
• 60:47 - 60:51
• 60:51 - 60:52
Red.
• 60:52 - 60:57
That's the domain D. So you'll
have to set up I, integral.
• 60:57 - 61:05
I for an I. And volume, double
integral over D of f of x, y,
• 61:05 - 61:08
whatever that is, dA.
• 61:08 - 61:11
That's going to be-- who is D?
• 61:11 - 61:14
Somebody help me, OK?
• 61:14 - 61:15
That's not easy.
• 61:15 - 61:20
So to draw the domain D, I have
to have a little bit of skill,
• 61:20 - 61:23
if I don't have any skill, I
don't belong in this class.
• 61:23 - 61:25
What do I have to draw?
• 61:25 - 61:26
Guys, tell me what to do.
• 61:26 - 61:29
0, x, and y.
• 61:29 - 61:37
To draw z, 0, z equals 0 gives
me x plus y equals 1, right?
• 61:37 - 61:39
So this is the floor.
• 61:39 - 61:42
Guys, this is the floor.
• 61:42 - 61:44
So why don't I shade it?
• 61:44 - 61:46
Because I'm not sure
which one I want.
• 61:46 - 61:48
Do I want vertical strips
or horizontal strips?
• 61:48 - 61:49
You're the boss.
• 61:49 - 61:52
You tell me what I want.
• 61:52 - 61:55
So do you want vertical strips?
• 61:55 - 61:58
• 61:58 - 62:01
Let's draw vertical strips.
• 62:01 - 62:03
So how do I
represent this domain
• 62:03 - 62:06
from the vertical strips?
• 62:06 - 62:09
x is between 0 and 1.
• 62:09 - 62:14
These are fixed
variable values of x
• 62:14 - 62:15
between fixed values 0 and 1.
• 62:15 - 62:21
For any such blue choice
of a point, I have a strip,
• 62:21 - 62:28
a vertical strip that goes
from y equals 0 down to--
• 62:28 - 62:29
STUDENT: 1 minus x.
• 62:29 - 62:31
DR. MAGDALENA TODA:
--1 minus x up.
• 62:31 - 62:33
Excellent, excellent.
• 62:33 - 62:37
This is exactly-- Roberto, you
were the one who said that?
• 62:37 - 62:38
OK.
• 62:38 - 62:39
So this is the domain.
• 62:39 - 62:42
So how do we write it down?
• 62:42 - 62:43
0 to 1.
• 62:43 - 62:46
0 to 1 minus x.
• 62:46 - 62:47
That is what I want to write.
• 62:47 - 62:49
No polar coordinates.
• 62:49 - 62:49
Goodbye.
• 62:49 - 62:51
There is no problem.
• 62:51 - 62:53
This is all a typical
Cartesian problem.
• 62:53 - 62:56
• 62:56 - 62:59
f-- f.
• 62:59 - 63:04
f is 1 minus x minus
y, thank you very much.
• 63:04 - 63:08
This is f dydx.
• 63:08 - 63:12
• 63:12 - 63:17
Homework, get 1/6.
• 63:17 - 63:20
• 63:20 - 63:26
So trying to do
that and get a 1/6.
• 63:26 - 63:29
And of course in the
exam-- oh, in the exam,
• 63:29 - 63:31
you will cheat big time, right?
• 63:31 - 63:32
What would you do?
• 63:32 - 63:37
You would set it up and forget
• 63:37 - 63:40
one at a time, doing this.
• 63:40 - 63:42
And you would put equals 1/6.
• 63:42 - 63:44
Thank you very much.
• 63:44 - 63:45
Right?
• 63:45 - 63:47
Can I check that you
didn't do the work?
• 63:47 - 63:50
No.
• 63:50 - 63:50
You trapped me.
• 63:50 - 63:51
You got me.
• 63:51 - 63:55
I have no-- I mean, I need
to say, is this correct?
• 63:55 - 63:56
Yes.
• 63:56 - 63:57
• 63:57 - 63:58
Yes.
• 63:58 - 64:00
Do they get full credit?
• 64:00 - 64:01
Yes.
• 64:01 - 64:05
So it's a sneaky problem.
• 64:05 - 64:06
OK.
• 64:06 - 64:12
Now finally, let's plunge
into 12.4, which is-- can you
• 64:12 - 64:14
remember this problem for 12.4?
• 64:14 - 64:16
I want to draw this again.
• 64:16 - 64:20
So I'll try not to
erase the picture.
• 64:20 - 64:23
I'll erase all the
data I have here.
• 64:23 - 64:27
And I'll keep the future because
I don't want to draw it again.
• 64:27 - 64:30
• 64:30 - 64:40
When we were small-- I
mean small in Calc 1 and 2,
• 64:40 - 64:47
they gave us a function
y equals f of x.
• 64:47 - 64:50
That is smooth, at least C1.
• 64:50 - 64:58
C1 means differentiable, and
the derivative is continuous.
• 64:58 - 65:01
• 65:01 - 65:05
And we said, OK, between
x equals a and x equals b,
• 65:05 - 65:10
I want you-- you, any
student-- to compute
• 65:10 - 65:13
the length of the arch.
• 65:13 - 65:15
Length of the arch.
• 65:15 - 65:17
And how did we do
that in Calc 2?
• 65:17 - 65:21
I have colleagues who
drive me crazy by refusing
• 65:21 - 65:24
to teach that in Calc 2.
• 65:24 - 65:26
Well, I disagree.
• 65:26 - 65:29
Of course, I can teach
it only in Calc 3,
• 65:29 - 65:32
and I can do it with
parametrization, which we did,
• 65:32 - 65:36
and then come back to the case
you have, y equals f of x,
• 65:36 - 65:39
and get the formula.
• 65:39 - 65:42
But it should be taught at
both levels, in both courses.
• 65:42 - 65:47
So when you have a
general parametrization
• 65:47 - 65:51
rt equals x of ty
of t [INAUDIBLE],
• 65:51 - 65:57
this is a parametrized
curve that's in C1, in time.
• 65:57 - 66:01
What is the length of the
arch between time t0 and time
• 66:01 - 66:02
t1 [INAUDIBLE]?
• 66:02 - 66:05
• 66:05 - 66:09
The integral from t0 to t1 or
the speed because the space
• 66:09 - 66:11
is the integral
of speed in time.
• 66:11 - 66:14
• 66:14 - 66:17
So in terms of speed,
remember that we
• 66:17 - 66:24
put square root of x prime of
t squared plus y prime of t
• 66:24 - 66:26
squared dt.
• 66:26 - 66:27
Why is that?
• 66:27 - 66:28
Somebody tell me.
• 66:28 - 66:31
That was the speed.
• 66:31 - 66:36
That was the magnitude
of the velocity vector.
• 66:36 - 66:39
And we've done that, and we
discovered that in Calculus 3.
• 66:39 - 66:44
In Calculus 2, they
taught this for what case?
• 66:44 - 66:48
The case when x is t--
say it again, I will now.
• 66:48 - 66:53
The case when x is
t, and y is f of t,
• 66:53 - 66:57
which is f of x--
and in that case,
• 66:57 - 66:59
as I told you before,
the length will
• 66:59 - 67:05
be the integral from A to B.
Whatever, it's the same thing.
• 67:05 - 67:07
A to B.
• 67:07 - 67:12
Square root-- since x
is t, x prime of t is 1.
• 67:12 - 67:16
So you get 1 plus-- y is just f.
• 67:16 - 67:17
y is f.
• 67:17 - 67:23
So you have f prime
of x squared dx.
• 67:23 - 67:30
So the length of this arch--
let me draw the arch in green,
• 67:30 - 67:32
so it's beautiful.
• 67:32 - 67:37
The length of this green
arch will be the length of r.
• 67:37 - 67:42
The integral from A to B square
root of 1 plus f prime 1x
• 67:42 - 67:44
squared dx.
• 67:44 - 67:46
Now there is a beautiful,
beautiful generalization
• 67:46 - 67:57
of that for-- generalization
for extension gives you
• 67:57 - 68:11
the surface area of a graph z
equals f of x, y over domain D.
• 68:11 - 68:12
Say what?
• 68:12 - 68:16
OK, it's exactly the
same formula generalized.
• 68:16 - 68:19
And I would like you to guess.
• 68:19 - 68:21
So I'd like you to
experimentally get
• 68:21 - 68:22
to the formula.
• 68:22 - 68:24
It can be proved.
• 68:24 - 68:29
It can be proved by taking
the equivalence of some sort
• 68:29 - 68:33
of Riemann summation
and passing to the limit
• 68:33 - 68:34
and get the formula.
• 68:34 - 68:42
But I would like you
to imagine you have--
• 68:42 - 68:45
so you have z equals f of x, y.
• 68:45 - 68:55
That projects exactly over
D. The area of the surface--
• 68:55 - 68:56
let's call it A of s.
• 68:56 - 69:03
• 69:03 - 69:09
The area of the
surface will be--
• 69:09 - 69:10
what do you think it will be?
• 69:10 - 69:12
You are smart people.
• 69:12 - 69:17
It will be double integral
• 69:17 - 69:19
over-- what do you think?
• 69:19 - 69:20
Over the domain.
• 69:20 - 69:23
• 69:23 - 69:26
What's the simplest
way to generalize this
• 69:26 - 69:30
through probably [INAUDIBLE]?
• 69:30 - 69:31
Another square root.
• 69:31 - 69:34
• 69:34 - 69:37
We don't have just one
derivative, f prime of x.
• 69:37 - 69:40
We are going to have two
derivatives, f sub x and f sub
• 69:40 - 69:41
y.
• 69:41 - 69:44
So what do you think the
simplest generalization
• 69:44 - 69:45
will look like?
• 69:45 - 69:46
STUDENT: 1 plus [INAUDIBLE].
• 69:46 - 69:52
DR. MAGDALENA TODA: 1 plus
f sub x squared plus f sub y
• 69:52 - 69:54
squared dx.
• 69:54 - 69:55
That's it.
• 69:55 - 69:57
So you say, oh, I'm a genius.
• 69:57 - 69:58
I discovered it.
• 69:58 - 69:59
Yes, you are.
• 69:59 - 70:01
I mean, in a sense that-- no.
• 70:01 - 70:03
In the sense that we
all have that kind
• 70:03 - 70:07
of mathematical intuition,
creativity that you come up
• 70:07 - 70:08
with something.
• 70:08 - 70:10
And you say, OK, can I verify?
• 70:10 - 70:11
Can I prove it?
• 70:11 - 70:11
Yes.
• 70:11 - 70:13
Can you discover
• 70:13 - 70:15
Yes, you can.
• 70:15 - 70:18
Actually, that's how all
the mathematical minds came.
• 70:18 - 70:21
They came up to it
with a conjecture based
• 70:21 - 70:25
on some prior experiences, some
prior observations and said,
• 70:25 - 70:26
I think it's going
to look like that.
• 70:26 - 70:30
I bet you like 90% that it's
going to look like that.
• 70:30 - 70:32
But then it took them
time to prove it.
• 70:32 - 70:37
But they were convinced that
this is what it's going to be.
• 70:37 - 70:37
OK.
• 70:37 - 70:44
So if you want the area
of the patch of a surface,
• 70:44 - 70:48
that's going to be 4.1, and
that's page-- god knows.
• 70:48 - 70:49
Wait a second.
• 70:49 - 70:49
Wait.
• 70:49 - 70:50
Bare with me.
• 70:50 - 71:02
It starts at page 951,
and it ends at page 957.
• 71:02 - 71:05
It's only seven pages, OK?
• 71:05 - 71:07
So it's not hard.
• 71:07 - 71:08
You have several examples.
• 71:08 - 71:10
I'm going to work
on an example that
• 71:10 - 71:12
is straight out of the book.
• 71:12 - 71:14
And guess what?
• 71:14 - 71:16
You see, that's why I like
this problem, because it's
• 71:16 - 71:22
in-- example one is exactly the
one that I came up with today
• 71:22 - 71:26
and says, find the
same tetrahedron thing.
• 71:26 - 71:29
Find the surface area of
the portion of the plane
• 71:29 - 71:39
x plus y plus z equals
1, which was that, which
• 71:39 - 71:42
lies in the first octant
where-- what does it mean,
• 71:42 - 71:43
first octant?
• 71:43 - 71:46
It means that x is
positive. y is positive.
• 71:46 - 71:49
z is positive for z.
• 71:49 - 71:52
• 71:52 - 71:53
OK.
• 71:53 - 71:54
Is this hard?
• 71:54 - 71:55
I don't know.
• 71:55 - 71:56
Let's find out.
• 71:56 - 71:59
• 71:59 - 72:05
So if we were to apply this
formula, how would we do that?
• 72:05 - 72:07
Is it hard?
• 72:07 - 72:10
I don't know.
• 72:10 - 72:12
We have to recollect
who everybody
• 72:12 - 72:16
is from scratch, one at a time.
• 72:16 - 72:20
• 72:20 - 72:20
Hmm?
• 72:20 - 72:24
STUDENT: Could we just
use our K-12 knowledge?
• 72:24 - 72:26
DR. MAGDALENA TODA: Well,
you can do that very well.
• 72:26 - 72:29
But let's do it
first-- you're sneaky.
• 72:29 - 72:31
Let's do it first as Calc 3.
• 72:31 - 72:37
And then let's see who comes
up with the fastest solution
• 72:37 - 72:38
in terms of surface area.
• 72:38 - 72:43
By the way, this individual--
this whole fat, sausage
• 72:43 - 72:46
kind of thing is ds.
• 72:46 - 72:51
This expression is called
the surface element.
• 72:51 - 72:54
Make a distinction
between dA, which is
• 72:54 - 72:56
called area element in plane.
• 72:56 - 73:00
• 73:00 - 73:08
ds is the surface element on the
surface, on the surface on top.
• 73:08 - 73:18
So practically, guys, you have
some [? healy ?] part, which
• 73:18 - 73:23
projects on a domain in plane.
• 73:23 - 73:27
The dA is the infinite
decimal area of this thingy.
• 73:27 - 73:30
And ds is the infinite
decimal area of that.
• 73:30 - 73:32
What do you mean by that?
• 73:32 - 73:33
OK.
• 73:33 - 73:37
Imagine this grid of pixels
that becomes smaller and smaller
• 73:37 - 73:37
and smaller.
• 73:37 - 73:38
OK?
• 73:38 - 73:41
make it infinitesimally small.
• 73:41 - 73:46
That's going to be
da dxdy, dx times dy.
• 73:46 - 73:52
What is the corresponding
pixel on the round surface?
• 73:52 - 73:53
I don't know.
• 73:53 - 73:56
It's still going to
be given by two lines,
• 73:56 - 74:02
and two lines form a
curvilinear domain.
• 74:02 - 74:06
And that curvilinear tiny-- do
you see how small it is that?
• 74:06 - 74:08
I bet the video cannot see it.
• 74:08 - 74:09
But you can see it.
• 74:09 - 74:12
So this tiny infinitesimally
small element
• 74:12 - 74:16
on the surface-- this is ds.
• 74:16 - 74:17
This is ds.
• 74:17 - 74:18
OK?
• 74:18 - 74:28
So if it were between a plane
and a tiny square, dxdy dA
• 74:28 - 74:32
and the ds here, it would
be easy between a plane
• 74:32 - 74:35
and a floor because
you can do some trick,
• 74:35 - 74:38
like a projection
with cosine and stuff.
• 74:38 - 74:39
But in general,
it's not so easy,
• 74:39 - 74:42
because you can have
a round patch that's
• 74:42 - 74:44
sitting above a domain.
• 74:44 - 74:48
And it's just-- you
have to do integration.
• 74:48 - 74:51
You have no other choice.
• 74:51 - 74:52
Let's compute it both ways.
• 74:52 - 74:54
Let's see.
• 74:54 - 75:00
A of s will be integral over
domain D. What in the world
• 75:00 - 75:02
was the domain D?
• 75:02 - 75:05
The domain D was the
domain on the floor.
• 75:05 - 75:08
And you told me what that
is, but I forgot, guys.
• 75:08 - 75:13
x is between 0 and 1.
• 75:13 - 75:14
Did you say so?
• 75:14 - 75:16
And y was between what and what?
• 75:16 - 75:19
Can you remind me?
• 75:19 - 75:21
STUDENT: [INAUDIBLE]
• 75:21 - 75:23
DR. MAGDALENA TODA:
Between 0 and--
• 75:23 - 75:23
STUDENT: 1 minus x.
• 75:23 - 75:25
DR. MAGDALENA TODA:
1 minus x, excellent.
• 75:25 - 75:29
So this is the
meaning of domain D.
• 75:29 - 75:35
And the square root
of-- who is f of x, y?
• 75:35 - 75:37
It's 1 minus x minus what?
• 75:37 - 75:38
Oh, right.
• 75:38 - 75:42
So you guys have to help
me compute this animal.
• 75:42 - 75:46
f sub x is negative 1.
• 75:46 - 75:47
• 75:47 - 75:51
f sub y is negative 1.
• 75:51 - 75:51
OK.
• 75:51 - 75:57
So I have to say 1 plus negative
1 squared plus negative 1
• 75:57 - 76:00
squared dA.
• 76:00 - 76:01
Gosh, I'm lucky.
• 76:01 - 76:01
Look.
• 76:01 - 76:03
I mean, I'm not
just lucky, but that
• 76:03 - 76:06
has to be-- it has to
be something beautiful
• 76:06 - 76:09
because otherwise the
elementary formula will not
• 76:09 - 76:10
be so beautiful.
• 76:10 - 76:14
This is root 3, and
it brings it back.
• 76:14 - 76:19
Root 3 pulls out
of the whole thing.
• 76:19 - 76:21
So you have root 3.
• 76:21 - 76:25
What is double integral--
OK, let's compute.
• 76:25 - 76:30
So first you go dy and dx.
• 76:30 - 76:32
x, again, you gave
it to me, guys.
• 76:32 - 76:33
0 to 1.
• 76:33 - 76:38
y between 0 and 1 minus x.
• 76:38 - 76:39
Great.
• 76:39 - 76:41
We are almost there.
• 76:41 - 76:41
We are almost there.
• 76:41 - 76:43
I just need your
help a little bit.
• 76:43 - 76:45
The square root of 3 goes out.
• 76:45 - 76:47
The integral from 0 to 1.
• 76:47 - 76:50
What is the integral of 1dy?
• 76:50 - 76:57
It's y, y between 1 minus x
on top and 0 on the bottom.
• 76:57 - 76:58
That means 1 minus x.
• 76:58 - 77:02
If I make a mistake, just shout.
• 77:02 - 77:05
dx.
• 77:05 - 77:11
The square root of 3 times
the integral of 1 minus x.
• 77:11 - 77:13
STUDENT: x minus
the square root.
• 77:13 - 77:15
DR. MAGDALENA TODA: x
minus the square root of 2.
• 77:15 - 77:18
Let me write it separately
because I should do that fast,
• 77:18 - 77:19
right?
• 77:19 - 77:20
Between 0 and 1.
• 77:20 - 77:21
What is that?
• 77:21 - 77:24
• 77:24 - 77:25
1/2.
• 77:25 - 77:26
That's a piece of cake.
• 77:26 - 77:27
This is 1/2.
• 77:27 - 77:30
So 1/2 is this thing.
• 77:30 - 77:32
And root 3 over 2.
• 77:32 - 77:34
And now Alex says,
congratulations
• 77:34 - 77:36
over 2, but I could
• 77:36 - 77:39
have told you that much faster.
• 77:39 - 77:42
So the question
is, how could Alex
• 77:42 - 77:47
have shown us this root
3 over 2 much faster?
• 77:47 - 77:49
STUDENT: Well, it's
just a triangle.
• 77:49 - 77:50
DR. MAGDALENA TODA:
It's just a triangle.
• 77:50 - 77:52
It's not just a triangle.
• 77:52 - 77:57
It's a beautiful triangle
that's an equilateral triangle.
• 77:57 - 78:01
And in school, they used
to teach more trigonometry.
• 78:01 - 78:03
And they don't.
• 78:03 - 78:05
choice-- I'm not
• 78:05 - 78:10
involved in the K-12 curriculum
standards for any state.
• 78:10 - 78:14
But if I had the choice, I
would say, teach the kids
• 78:14 - 78:16
a little bit more
geometry in school
• 78:16 - 78:19
because they don't know
anything in terms of geometry.
• 78:19 - 78:22
So there were
triumphs in the past,
• 78:22 - 78:26
know those better.
• 78:26 - 78:30
But If somebody gave you
an equilateral triangle
• 78:30 - 78:35
of a certain side, you
would be able to tell them,
• 78:35 - 78:37
• 78:37 - 78:38
I know the area.
• 78:38 - 78:45
I know the area being l squared,
the square root of 3 over 4.
• 78:45 - 78:47
• 78:47 - 78:48
• 78:48 - 78:49
He will know.
• 78:49 - 78:53
But we don't teach
that in school anymore.
• 78:53 - 78:56
The smart kids do this
by themselves how?
• 78:56 - 78:58
Can you show me how?
• 78:58 - 79:02
They draw the
perpendicular bisector.
• 79:02 - 79:05
And there is a
theorem actually--
• 79:05 - 79:07
but we never prove
that in school--
• 79:07 - 79:11
that if we draw that
perpendicular bisector,
• 79:11 - 79:15
then the two triangles
are congruent.
• 79:15 - 79:23
And as a consequence,
well, that is l/2, l/2.
• 79:23 - 79:24
OK?
• 79:24 - 79:29
So if you draw just
the perpendicular,
• 79:29 - 79:37
you can prove it using some
congruence of triangles
• 79:37 - 79:39
that what you get here
is also the median.
• 79:39 - 79:41
So it's going to keep
right in the middle
• 79:41 - 79:43
of the opposite side.
• 79:43 - 79:45
So you l/2, l/2.
• 79:45 - 79:46
OK.
• 79:46 - 79:47
That's what I wanted to say.
• 79:47 - 79:49
And then using the
Pythagorean theorem,
• 79:49 - 79:50
you're going to get the height.
• 79:50 - 79:55
So the height will be the square
root of l squared minus l/2
• 79:55 - 80:00
squared, which is the
square root of l squared
• 80:00 - 80:06
minus l squared over 4, which
is the square root of 3l squared
• 80:06 - 80:12
over 4, which simplified
will be l root 3 over 2.
• 80:12 - 80:19
l root 3 over 2 is
exactly the height.
• 80:19 - 80:25
And then the area will be
height times the base over 2
• 80:25 - 80:26
for any triangle.
• 80:26 - 80:34
So I have the height times
the base over 2, which
• 80:34 - 80:39
is root 3l squared over 4.
• 80:39 - 80:42
So many people when
they were young
• 80:42 - 80:45
• 80:45 - 80:48
Now in our case, it
should be a piece of cake.
• 80:48 - 80:49
Why?
• 80:49 - 80:54
Because we know who l is.
• 80:54 - 80:57
l is going to be the hypotenuse.
• 80:57 - 81:03
We have here a 1 and
a 1, a 1 and a 1.
• 81:03 - 81:07
So this is going to be the
hypotenuse, square root of 2.
• 81:07 - 81:10
So if I apply this
formula, which
• 81:10 - 81:13
is the area of the
equilateral triangle,
• 81:13 - 81:23
that says root 2 squared root
3 over 4 equals 2 root 3 over 4
• 81:23 - 81:25
equals root 3 over 2.
• 81:25 - 81:28
• 81:28 - 81:33
So can you do that?
• 81:33 - 81:35
Are you allowed to do that?
• 81:35 - 81:39
Well, we never formulated
it actually saying
• 81:39 - 81:46
compute the surface of
this patch of a plane using
• 81:46 - 81:48
the surface integral.
• 81:48 - 81:49
We didn't say that.
• 81:49 - 81:53
We said, compute it,
period We didn't care how.
• 81:53 - 81:56
If you can do it
by another method,
• 81:56 - 81:59
whether to stick to that
method, elementary method
• 81:59 - 82:01
or to just check
• 82:01 - 82:03
is it really a square
root of 3 over 2?
• 82:03 - 82:07
You are allowed to do that.
• 82:07 - 82:08
Questions?
• 82:08 - 82:10
STUDENT: So would the
length be square root
• 82:10 - 82:13
of 2 squared, which is 2.
• 82:13 - 82:17
2 divided by 4 is [INAUDIBLE]
square root of 3 over 2.
• 82:17 - 82:18
I'm just talking-- oh, yeah.
• 82:18 - 82:20
DR. MAGDALENA TODA: You are
just repeating what I have.
• 82:20 - 82:23
like this is elementary.
• 82:23 - 82:28
like this is with Calc 3.
• 82:28 - 82:31
gives me the reassurance
• 82:31 - 82:32
I wasn't speaking nonsense.
• 82:32 - 82:38
I did it in two different ways,
and I got the same answer.
• 82:38 - 82:42
Let's do one or
two more examples
• 82:42 - 82:49
of surface integrals, surface
areas and surface integrals.
• 82:49 - 82:50
It's not hard.
• 82:50 - 82:52
It's actually quite fun.
• 82:52 - 82:55
Some of them are
harder than others.
• 82:55 - 82:56
Let's see what we can do.
• 82:56 - 83:02
• 83:02 - 83:02
Oh, yeah.
• 83:02 - 83:04
I like this one very much.
• 83:04 - 83:09
• 83:09 - 83:14
I remember we gave it several
times on the final exams.
• 83:14 - 83:16
and do one like that
• 83:16 - 83:20
because you've
seen-- why don't we
• 83:20 - 83:23
pick the one I picked before
with the eggshells for Easter,
• 83:23 - 83:24
like Easter eggs?
• 83:24 - 83:30
What was the paraboloid I
had on top, the one on top?
• 83:30 - 83:31
STUDENT: 8 minus x--
• 83:31 - 83:33
DR. MAGDALENA TODA: 8 minus
x squared [INAUDIBLE].
• 83:33 - 83:34
That's what I'm going to pick.
• 83:34 - 83:40
And I'll say, as Easter
is coming, a word problem.
• 83:40 - 83:47
We want to compute the
surface of an egg that
• 83:47 - 83:56
is created by intersecting
the two paraboloids 8 minus x
• 83:56 - 83:59
squared minus y squared and
x squared plus y squared.
• 83:59 - 84:00
So let's see.
• 84:00 - 84:04
• 84:04 - 84:05
No.
• 84:05 - 84:07
Not y, Magdalena.
• 84:07 - 84:17
• 84:17 - 84:21
Intersect, make
the egg intersect.
• 84:21 - 84:21
Create the eggshells.
• 84:21 - 84:25
• 84:25 - 84:27
Shells.
• 84:27 - 84:29
Compute the area.
• 84:29 - 84:32
• 84:32 - 84:34
And you say, wait a minute.
• 84:34 - 84:37
The two eggshells were equal.
• 84:37 - 84:39
Yes, I know.
• 84:39 - 84:42
I know that the two
eggshells were equal.
• 84:42 - 84:45
But they don't look
equal in my picture.
• 84:45 - 84:45
I'll try better.
• 84:45 - 84:49
• 84:49 - 84:51
Assume they are parabolas.
• 84:51 - 84:56
• 84:56 - 84:58
Assume this was z equals 4.
• 84:58 - 85:01
This was 8 minus x
squared minus y squared.
• 85:01 - 85:05
This was x squared
plus y squared.
• 85:05 - 85:09
How do we compute-- just
like Matthew observed, 8
• 85:09 - 85:10
for the volume.
• 85:10 - 85:14
I only need half of the
8 multiplied by the 2.
• 85:14 - 85:15
The same thing.
• 85:15 - 85:22
I'm going to take one of
the two shells, this one.
• 85:22 - 85:29
And the surface of the egg will
be twice times the surface S1.
• 85:29 - 85:32
All I have to
compute is S1, right?
• 85:32 - 85:34
It shouldn't be a big problem.
• 85:34 - 85:36
I mean, what do I
need for that S1?
• 85:36 - 85:43
I only need the shadow of
it and the expression of it.
• 85:43 - 85:49
The shadow of it and the--
the shadow of it is this.
• 85:49 - 85:52
The shadow of this is this.
• 85:52 - 85:54
And the expression-- hmm.
• 85:54 - 85:55
It shouldn't be hard.
• 85:55 - 85:58
• 85:58 - 86:02
So I'm going to-- I'm
• 86:02 - 86:05
you to tell me what to write.
• 86:05 - 86:09
• 86:09 - 86:11
What?
• 86:11 - 86:13
STUDENT: Square root of 1 plus--
• 86:13 - 86:14
DR. MAGDALENA TODA: No.
• 86:14 - 86:16
First I will write
the definition.
• 86:16 - 86:20
Double integral over D,
square root of, as you said--
• 86:20 - 86:20
say it again.
• 86:20 - 86:26
STUDENT: 1 plus f
of x squared plus f
• 86:26 - 86:32
of y squared [INAUDIBLE] dA.
• 86:32 - 86:34
• 86:34 - 86:36
DR. MAGDALENA TODA: All right.
• 86:36 - 86:40
So this is ds, and I'm
integrating over the domain.
• 86:40 - 86:42
Should this be hard?
• 86:42 - 86:44
No, it shouldn't be hard.
• 86:44 - 86:48
But I'm going to get
something a little bit ugly.
• 86:48 - 86:53
And it doesn't matter, because
we will do it with no problem.
• 86:53 - 86:58
I'm going to say, integral
over-- now the domain
• 86:58 - 87:00
D-- I know what it is
because the domain D will
• 87:00 - 87:07
be given by x squared
plus y squared less than
• 87:07 - 87:08
or equal to 4.
• 87:08 - 87:14
So I would know how to
deal with that later on.
• 87:14 - 87:18
Now what scares me off
a little bit-- and look
• 87:18 - 87:20
what's going to happen.
• 87:20 - 87:30
When I compute f sub x and f sub
y, those will be really easy.
• 87:30 - 87:35
But when I plug
everything in here,
• 87:35 - 87:39
it's going to be
a little bit hard.
• 87:39 - 87:41
Never mind, I'm
going to have to have
• 87:41 - 87:46
to battle the problem
with polar coordinates.
• 87:46 - 87:51
That's why polar coordinates
exist, to help us.
• 87:51 - 87:55
So f sub x is minus 2x, right?
• 87:55 - 87:56
f sub y is minus 2y.
• 87:56 - 88:00
• 88:00 - 88:00
OK.
• 88:00 - 88:04
So what am I going
to write over here?
• 88:04 - 88:14
Minus 2x squared plus
minus 2y squared dx.
• 88:14 - 88:16
I don't have much room.
• 88:16 - 88:18
But that would mean dxdy.
• 88:18 - 88:20
Am I happy with that?
• 88:20 - 88:21
No.
• 88:21 - 88:24
I'm not happy with
it, because here it's
• 88:24 - 88:32
going to be x squared plus
y squared between 0 and 4.
• 88:32 - 88:36
• 88:36 - 88:39
And I'm not happy with it,
because it looks like a mess.
• 88:39 - 88:47
And I have to find this area
integral with a simple method,
• 88:47 - 88:49
something nicer.
• 88:49 - 88:54
Now the question is,
does my elementary math
• 88:54 - 88:57
help me find the
area of the egg?
• 88:57 - 88:59
Unfortunately, no.
• 88:59 - 89:03
So from this point on, it's
goodbye elementary geometry.
• 89:03 - 89:05
STUDENT: Unless you
• 89:05 - 89:07
DR. MAGDALENA TODA: But they
are not spheres or anything.
• 89:07 - 89:09
I can approximate the
eggs with spheres,
• 89:09 - 89:14
but I cannot do anything with
those paraboloids [INAUDIBLE].
• 89:14 - 89:17
STUDENT: I know the
function of the top.
• 89:17 - 89:19
DR. MAGDALENA TODA:
Yeah, yeah, yeah.
• 89:19 - 89:20
You can.
• 89:20 - 89:22
STUDENT: [INAUDIBLE]
the integration f prime.
• 89:22 - 89:24
DR. MAGDALENA TODA: But
it's still integration.
• 89:24 - 89:28
So can I pretend like I'm a
smart sixth grader, and I can--
• 89:28 - 89:32
how can I measure that if I'm
• 89:32 - 89:36
With some sort of graphic paper,
do some sort of approximation
• 89:36 - 89:38
of the area of the egg.
• 89:38 - 89:41
It's a school project that's
not worth anything because I
• 89:41 - 89:44
think not even at a science
fair, could I do it.
• 89:44 - 89:48
STUDENT: Unless--
• 89:48 - 89:51
I can draw the sphere in.
• 89:51 - 89:54
Then if I apply the
distance between the sphere
• 89:54 - 89:57
and the [INAUDIBLE] the
distance between [INAUDIBLE]
• 89:57 - 89:59
and take it all from there.
• 89:59 - 90:03
But then the function
actually will look easier
• 90:03 - 90:07
because it will go from the
y axis up to the A axis,
• 90:07 - 90:08
and they meet each other.
• 90:08 - 90:10
So I took up the
area and took up
• 90:10 - 90:11
the other area to [INAUDIBLE].
• 90:11 - 90:13
• 90:13 - 90:14
DR. MAGDALENA TODA: Yeah.
• 90:14 - 90:17
Well, wouldn't that
surface of the egg still
• 90:17 - 90:21
be an approximation
• 90:21 - 90:24
Anyway, let's come
back to the egg.
• 90:24 - 90:25
The egg, the egg.
• 90:25 - 90:27
The egg is [INAUDIBLE].
• 90:27 - 90:28
It's nice.
• 90:28 - 90:31
1 plus 4 x squared
plus y squared.
• 90:31 - 90:34
Look at the beauty of the
symmetry of polynomials.
• 90:34 - 90:37
x squared plus y squared says,
I'm a symmetric polynomial.
• 90:37 - 90:40
You're my friend
because I'm r squared,
• 90:40 - 90:42
and I know what I'm going to do.
• 90:42 - 90:44
So how do we compute?
• 90:44 - 90:47
What kind of integral
do we need to compute?
• 90:47 - 90:54
So S1 will be the integral of
integral of the square root
• 90:54 - 90:58
of 1 plus 4r squared.
• 90:58 - 91:02
Don't forget the dA
contains the Jacobian.
• 91:02 - 91:05
So don't write drd theta.
• 91:05 - 91:07
I had a student who wrote that.
• 91:07 - 91:10
That is worth
exactly zero points.
• 91:10 - 91:13
So say, times r.
• 91:13 - 91:19
r between-- oh, my god,
the poor egg-- 0 to 2.
• 91:19 - 91:24
And theta between 0 to 2 pi.
• 91:24 - 91:25
And come on.
• 91:25 - 91:28
We've done that in Calc 2.
• 91:28 - 91:33
I mean, it's not so hard.
• 91:33 - 91:36
So u substitution.
• 91:36 - 91:40
u is 4r squared plus 1.
• 91:40 - 91:41
That's our only hope.
• 91:41 - 91:43
We have no other hope.
• 91:43 - 91:48
du is going to be 8rdr.
• 91:48 - 91:51
And rdr is a married couple.
• 91:51 - 91:52
They stick together.
• 91:52 - 91:53
Where is the purple?
• 91:53 - 91:55
The purple is here.
• 91:55 - 91:59
rdr, rdr.
• 91:59 - 91:59
rdr is du/8.
• 91:59 - 92:03
• 92:03 - 92:06
This fellow's name is u.
• 92:06 - 92:07
He is u.
• 92:07 - 92:09
He is not u, but he's like u.
• 92:09 - 92:11
OK, not necessary.
• 92:11 - 92:12
OK.
• 92:12 - 92:16
So you go 2 pi-- because
there is no theta.
• 92:16 - 92:21
So no theta means-- let me
write it one more time for you.
• 92:21 - 92:24
The integral from
0 to 2 pi 1d theta.
• 92:24 - 92:26
And he goes out and has fun.
• 92:26 - 92:28
This is 2 pi.
• 92:28 - 92:34
But then all you have left
inside is the integral of u.
• 92:34 - 92:44
Square root of u times
1/8 du, close the bracket,
• 92:44 - 92:53
where u is between 1 and 17.
• 92:53 - 92:55
Isn't that beautiful?
• 92:55 - 92:56
That's 17.
• 92:56 - 93:00
So you have 2 squared
times 416 plus 117.
• 93:00 - 93:04
But believe me that from this
viewpoint, from this point on,
• 93:04 - 93:06
it's not really hard.
• 93:06 - 93:08
It just looks like the
surface of that egg
• 93:08 - 93:13
is-- whenever it was produced,
in what factory, in whatever
• 93:13 - 93:19
country is the toy factory, they
must have done this area stage.
• 93:19 - 93:22
So you have 2 pi.
• 93:22 - 93:27
1/8 comes out, whether
he wants out or not.
• 93:27 - 93:28
Integral of square root of u.
• 93:28 - 93:30
Do you like that?
• 93:30 - 93:32
I don't.
• 93:32 - 93:33
You have--
• 93:33 - 93:34
STUDENT: 2/3.
• 93:34 - 93:38
DR. MAGDALENA TODA:
2/3 u to the 3/2
• 93:38 - 93:41
between-- down is u equals 1.
• 93:41 - 93:43
Up is u equals 17.
• 93:43 - 93:45
because we've done
• 93:45 - 93:48
this in the past
reviews for the finals--
• 93:48 - 93:50
and several finals
are like that.
• 93:50 - 93:54
do I do in such a case?
• 93:54 - 93:54
Nothing.
• 93:54 - 93:56
I mean, you do nothing.
• 93:56 - 93:58
You just plug it in
and leave it as is.
• 93:58 - 94:02
So you have-- to simplify
your life a little bit, what
• 94:02 - 94:05
you can do is 2, 2, and 8.
• 94:05 - 94:06
What is 2 times 2?
• 94:06 - 94:06
4.
• 94:06 - 94:13
Divided by 8-- so you
have pi/6 overall.
• 94:13 - 94:17
• 94:17 - 94:27
pi/6 times 17 to
the 3/2 and minus 1.
• 94:27 - 94:30
• 94:30 - 94:34
One of my students, after he
got such an answer last time
• 94:34 - 94:37
we did the review, he
said, I don't like it.
• 94:37 - 94:41
I want to write this as
square root of 17 cubed.
• 94:41 - 94:43
You can write it
whatever you want.
• 94:43 - 94:45
It can be-- it
has to be correct.
• 94:45 - 94:48
I don't care how you write it.
• 94:48 - 94:49
What if you mess up?
• 94:49 - 94:52
You say, well, this
woman is killing me
• 94:52 - 94:54
with her algebra over here.
• 94:54 - 94:55
OK.
• 94:55 - 94:57
If you understood--
suppose that you
• 94:57 - 94:59
are taking the final right now.
• 94:59 - 95:01
You drew the
picture beautifully.
• 95:01 - 95:02
You remember the problem.
• 95:02 - 95:03
You remember the formula.
• 95:03 - 95:04
You write it down.
• 95:04 - 95:05
You wrote it down.
• 95:05 - 95:06
You got to this point.
• 95:06 - 95:11
have 50% of the problem.
• 95:11 - 95:11
Yup.
• 95:11 - 95:16
And then from this point on,
you do the polar coordinates,
• 95:16 - 95:19
and you still get another 25%.
• 95:19 - 95:20
You messed it up.
• 95:20 - 95:22
You lose some partial credit.
• 95:22 - 95:25
But everything you
write correctly
• 95:25 - 95:29
earns and earns
and earns points.
• 95:29 - 95:30
OK?
• 95:30 - 95:32
So don't freak out
thinking, I'm going
• 95:32 - 95:34
to mess up my algebra for sure.
• 95:34 - 95:39
If you do, it doesn't matter,
because even if this would
• 95:39 - 95:41
be a multiple choice--
some problems will
• 95:41 - 95:43
be show work completely,
and some problems
• 95:43 - 95:45
may be multiple
choice questions.
• 95:45 - 95:50
Even if this is going
to be a multiple choice,
• 95:50 - 95:54
I will still go over the entire
computation for everybody
• 95:54 - 95:55
and give partial credit.
• 95:55 - 96:00
This is my policy.
• 96:00 - 96:03
We are allowed to choose
our policies as instructors.
• 96:03 - 96:08
So you earn partial credit
for everything you write down.
• 96:08 - 96:09
OK.
• 96:09 - 96:10
• 96:10 - 96:12
It's one of the harder
problems in the book.
• 96:12 - 96:17
It is he similar to
example number-- well,
• 96:17 - 96:22
this is exactly like
example 2 in the section.
• 96:22 - 96:24
• 96:24 - 96:28
So we did these two
examples from the section.
• 96:28 - 96:31
And I want to give you one
more piece of information
• 96:31 - 96:36
that I saw, that unfortunately
my colleagues don't teach that.
• 96:36 - 96:39
And it sort of bothers me.
• 96:39 - 96:40
I wish they did.
• 96:40 - 96:43
• 96:43 - 96:46
Once upon a time,
a long time ago,
• 96:46 - 96:54
I taught you a little bit
• 96:54 - 96:56
of a surface.
• 96:56 - 96:59
And I want to give you yet
another formula, not just
• 96:59 - 97:02
this one but one more.
• 97:02 - 97:04
So what if you have a
generalized surface that
• 97:04 - 97:10
is parametrized, meaning
• 97:10 - 97:13
given as explicitly
z equals f of x, y?
• 97:13 - 97:16
That's the lucky case.
• 97:16 - 97:17
That's a graph.
• 97:17 - 97:20
We call that a graph,
z equals f of x and y.
• 97:20 - 97:22
And we call ourselves lucky.
• 97:22 - 97:25
But life is not always so easy.
• 97:25 - 97:35
Sometimes all you can get
is a parametrization r
• 97:35 - 97:38
of v, v for a surface.
• 97:38 - 97:42
And from that, you
have to deal with that.
• 97:42 - 97:47
So suppose somebody says,
I don't give you f of x, y,
• 97:47 - 97:51
although locally every
surface looks like the graph.
• 97:51 - 97:53
But a surface doesn't have
to be a graph in general.
• 97:53 - 97:57
Locally, it does look like
a graph on a small length.
• 97:57 - 98:02
But in general, it's
given by r, v, v equals--
• 98:02 - 98:04
and that was what?
• 98:04 - 98:12
I gave you something like x of
u, v I plus y of u, v J plus z
• 98:12 - 98:23
of u, v-- let's not put
things in alphabetical order.
• 98:23 - 98:29
• 98:29 - 98:34
z of u, v J and K.
• 98:34 - 98:38
And we said that
we have a point.
• 98:38 - 98:44
P is our coordinate u0, v0.
• 98:44 - 98:46
And we said we
look at that point,
• 98:46 - 98:49
and we try to draw the partials.
• 98:49 - 98:53
What are the partials from
a geometric viewpoint?
• 98:53 - 98:56
Well, if I want to
write the partials,
• 98:56 - 98:57
they would be various.
• 98:57 - 99:01
It's going to be the vector
x sub u, y sub u, z sub
• 99:01 - 99:11
u, and the vector x sub v, y
sub v, z sub v, two vectors.
• 99:11 - 99:14
Do you remember when I
drew them, what they were?
• 99:14 - 99:17
• 99:17 - 99:20
We said the following.
• 99:20 - 99:23
We said, let's assume
v will be a constant.
• 99:23 - 99:27
• 99:27 - 99:29
So we say, v is a constant.
• 99:29 - 99:30
And then v equals v0.
• 99:30 - 99:33
And then you have P of u0, v0.
• 99:33 - 99:38
• 99:38 - 99:44
And then we have another,
and we have u equals u0.
• 99:44 - 99:49
• 99:49 - 99:54
This guy is going to
be who of the two guys?
• 99:54 - 99:55
r sub u.
• 99:55 - 99:59
When we measure out
the point P, r sub u
• 99:59 - 100:07
is this guy, who is tangent
to the line r of u, v zero.
• 100:07 - 100:11
• 100:11 - 100:12
Does it look tangent?
• 100:12 - 100:14
I hope it looks tangent.
• 100:14 - 100:19
And this guy will be r of
u-- because u0 means what?
• 100:19 - 100:24
u0 and v. So who
is free to move?
• 100:24 - 100:33
v. So this guy, this r sub
v-- they are both tangents.
• 100:33 - 100:36
So do you have a surface?
• 100:36 - 100:38
This is the surface.
• 100:38 - 100:39
This is the surface.
• 100:39 - 100:43
And these two horns or
whatever they are-- those
• 100:43 - 100:47
are the tangents r sub u, r
sub v, the two tangent vectors,
• 100:47 - 100:48
the partial velocities.
• 100:48 - 100:51
And I told you before, they
form the tangent plane.
• 100:51 - 100:52
They are partial velocities.
• 100:52 - 100:56
They are both tangent to
the surface at that point.
• 100:56 - 100:57
They form a basis.
• 100:57 - 100:59
They are linearly independent.
• 100:59 - 100:59
Always?
• 100:59 - 101:00
No.
• 101:00 - 101:06
But we assume that r sub u
and r sub v are non-zero,
• 101:06 - 101:08
and they are not co-linear.
• 101:08 - 101:09
How do I write that?
• 101:09 - 101:11
They are not parallel.
• 101:11 - 101:13
So guys, what does it mean?
• 101:13 - 101:15
It means-- we talked
• 101:15 - 101:18
The velocities cannot be 0.
• 101:18 - 101:20
And r sub u, r sub v
cannot be parallel,
• 101:20 - 101:23
because if they are parallel,
there is no area element.
• 101:23 - 101:26
There is no tangent
plane between them.
• 101:26 - 101:31
What they form is
the area element.
• 101:31 - 101:36
So what do you think the
area element will look like?
• 101:36 - 101:39
It's a magic thing.
• 101:39 - 101:44
The surface element
actually will
• 101:44 - 101:59
be exactly the area between ru
and rv times the u derivative.
• 101:59 - 102:01
Say it again, Magdalena.
• 102:01 - 102:04
What the heck is the area
between the vectors r sub
• 102:04 - 102:06
u, r sub v?
• 102:06 - 102:10
You know it better than
me because you're younger,
• 102:10 - 102:11
• 102:11 - 102:16
And you just covered
this in chapter nine.
• 102:16 - 102:19
When you have a vector A
and a vector B that are not
• 102:19 - 102:23
co-linear, what was the
area of the parallelogram
• 102:23 - 102:24
that they form?
• 102:24 - 102:25
STUDENT: The
magnitude [INAUDIBLE].
• 102:25 - 102:26
DR. MAGDALENA TODA:
Magnitude of--
• 102:26 - 102:27
STUDENT: The cross product.
• 102:27 - 102:27
DR. MAGDALENA TODA:
The cross product.
• 102:27 - 102:28
Excellent.
• 102:28 - 102:31
This is exactly what
I was hoping for.
• 102:31 - 102:35
The magnitude of the
cross product is the area.
• 102:35 - 102:42
So you have ds, infinitesimal
of an answer plus area, surface
• 102:42 - 102:47
element will be
exactly the magnitude
• 102:47 - 102:52
of the cross product of the
two velocity vectors, dudv.
• 102:52 - 102:56
dudv can also be
written dA in this case
• 102:56 - 102:59
because it's a flat
area on the floor.
• 102:59 - 103:03
It's the area of a tiny
square on the floor,
• 103:03 - 103:05
infinitesimally small square.
• 103:05 - 103:07
So remember that.
• 103:07 - 103:10
And you say, well, Magdalena,
you are just feeding us
• 103:10 - 103:11
formula after formula.
• 103:11 - 103:13
But we don't even know.
• 103:13 - 103:14
OK, this makes sense.
• 103:14 - 103:18
This looks like I have some
sort of tiny parallelogram,
• 103:18 - 103:23
and I approximate the
actual curvilinear
• 103:23 - 103:25
patch, curvilinear
patch on-- I'm
• 103:25 - 103:27
going to draw it on my hand.
• 103:27 - 103:29
So this is-- oh, my god.
• 103:29 - 103:31
My son would make fun of me.
• 103:31 - 103:35
So this curvilinear patch
between two curves on my hand
• 103:35 - 103:39
will be actually
approximated by this.
• 103:39 - 103:40
What is this rectangle?
• 103:40 - 103:41
No, it's a--
• 103:41 - 103:42
STUDENT: Parallelogram.
• 103:42 - 103:42
DR. MAGDALENA TODA:
Parallelogram.
• 103:42 - 103:44
Thank you so much.
• 103:44 - 103:48
So this is an
approximation, again.
• 103:48 - 103:50
So this is the area
of the parallelogram.
• 103:50 - 103:55
• 103:55 - 104:00
And that's what we defined
as being the surface element.
• 104:00 - 104:02
It has to do with
the tangent plane.
• 104:02 - 104:05
but shouldn't this
• 104:05 - 104:09
be the same as the formula
root of 1 plus f sub x squared
• 104:09 - 104:12
plus f sub y squared dxdy?
• 104:12 - 104:13
Yes.
• 104:13 - 104:14
Let's prove it.
• 104:14 - 104:17
Let's finally prove that
the meaning of this area
• 104:17 - 104:22
will provide you
with the surface
• 104:22 - 104:26
element the terms of x
and y, just the way you--
• 104:26 - 104:28
you did not prove it.
• 104:28 - 104:29
You discovered it.
• 104:29 - 104:30
Remember, guys?
• 104:30 - 104:33
You came up with a
formula as a conjecture.
• 104:33 - 104:37
You said, if we generalize
the arch length,
• 104:37 - 104:39
it should look like that.
• 104:39 - 104:40
You sort of smelled it.
• 104:40 - 104:41
You said, I think.
• 104:41 - 104:42
I feel.
• 104:42 - 104:43
I'm almost sure.
• 104:43 - 104:45
But did you prove it?
• 104:45 - 104:46
No.
• 104:46 - 104:50
So starting from the
idea of the area element
• 104:50 - 104:53
that I gave before,
do you remember
• 104:53 - 104:57
that we also had that signed
area between the dx and dy,
• 104:57 - 105:00
and we used the area of
the parallelogram before?
• 105:00 - 105:06
We also allowed it to
go oriented plus, minus.
• 105:06 - 105:06
OK.
• 105:06 - 105:07
All right.
• 105:07 - 105:11
So this makes more sense
than what you gave me.
• 105:11 - 105:14
Can I prove what you
gave me based on this
• 105:14 - 105:16
and show it's one
and the same thing?
• 105:16 - 105:18
So hopefully, yes.
• 105:18 - 105:24
If I have my explicit form
z equals f of x and y,
• 105:24 - 105:27
I should be able to
parametrize this surface.
• 105:27 - 105:29
How do I parametrize
this surface
• 105:29 - 105:31
in the simplest possible way?
• 105:31 - 105:34
• 105:34 - 105:37
x is u.
• 105:37 - 105:45
y is v. z is f of
u, v. And that's it.
• 105:45 - 105:55
Then it's r of u, v as a vector
will be angular bracket, u, v,
• 105:55 - 106:02
f of u, v. Now you have to
help me compute r sub u and r
• 106:02 - 106:04
sub v. They have to
be these blue vectors,
• 106:04 - 106:06
the partial velocities.
• 106:06 - 106:08
r sub u, r sub v.
• 106:08 - 106:09
Is it hard?
• 106:09 - 106:10
Come on.
• 106:10 - 106:11
It shouldn't be hard.
• 106:11 - 106:13
I need to change colors.
• 106:13 - 106:15
So can you tell
me what they are?
• 106:15 - 106:20
• 106:20 - 106:22
What's the first-- 1?
• 106:22 - 106:23
Good.
• 106:23 - 106:23
What's next?
• 106:23 - 106:24
0.
• 106:24 - 106:25
Thank you.
• 106:25 - 106:26
STUDENT: F sub u.
• 106:26 - 106:27
DR. MAGDALENA TODA: f sub u.
• 106:27 - 106:29
Very good.
• 106:29 - 106:33
f sub u or f sub x is the same
because x and u are the same.
• 106:33 - 106:37
So let me rewrite
it 1, 0, f sub x.
• 106:37 - 106:41
Now the next batch.
• 106:41 - 106:50
0, 1, f sub v, which
is 0, 1, f sub y.
• 106:50 - 106:52
0, 1, f sub y.
• 106:52 - 106:57
• 106:57 - 106:59
Now I need to cross them.
• 106:59 - 107:02
And I need to cross them, and
I'm too lazy because it's 2:20.
• 107:02 - 107:03
But I'll do it.
• 107:03 - 107:04
I'll do it.
• 107:04 - 107:05
I'll cross.
• 107:05 - 107:08
and everything,
• 107:08 - 107:10
I'm going to get to
where I need to get.
• 107:10 - 107:23
• 107:23 - 107:24
You can start.
• 107:24 - 107:27
I mean, don't wait for me.
• 107:27 - 107:29
Try it yourselves
and see what you get.
• 107:29 - 107:34
And how hard do you think
it is to compute the thing?
• 107:34 - 107:35
STUDENT: [INAUDIBLE]
• 107:35 - 107:38
• 107:38 - 107:40
DR. MAGDALENA TODA: I
will do the normality
• 107:40 - 107:42
at the magnitude later.
• 107:42 - 107:58
r sub u, r sub v's cross product
will be I, J, K. 1, 0, f sub x,
• 107:58 - 107:59
0, 1, f sub y.
• 107:59 - 108:00
Is this hard?
• 108:00 - 108:02
It shouldn't be hard.
• 108:02 - 108:10
So I have minus f sub x, what?
• 108:10 - 108:11
I. I'm sorry.
• 108:11 - 108:13
I for an I. OK.
• 108:13 - 108:19
J, again minus because
I need to change.
• 108:19 - 108:23
When I expand along the row, I
have plus, minus, plus, minus,
• 108:23 - 108:25
plus, alternating.
• 108:25 - 108:27
So I need to have minus.
• 108:27 - 108:38
The determinant is f sub y times
J plus K times this fellow.
• 108:38 - 108:42
But that fellow is 1, is the
minor 1, for god's sakes.
• 108:42 - 108:45
So this is so easy.
• 108:45 - 108:49
I got the vector,
but I need the norm.
• 108:49 - 108:50
But so what?
• 108:50 - 108:51
Do you have it?
• 108:51 - 108:52
I'm there, guys.
• 108:52 - 108:54
I'm really there.
• 108:54 - 108:56
It's a piece of cake.
• 108:56 - 108:58
I take the components.
• 108:58 - 108:59
I squeeze them a little bit.
• 108:59 - 109:00
No.
• 109:00 - 109:04
I square them,
and I sum them up.
• 109:04 - 109:09
And I get the square
root of 1 plus-- exactly.
• 109:09 - 109:14
1 plus f sub x squared
plus f sub y squared d.
• 109:14 - 109:20
This is u and v, and this
is dxdy, which is dA.
• 109:20 - 109:26
This is the tiny floor square
of an infinitesimally square
• 109:26 - 109:28
on the floor.
• 109:28 - 109:29
OK?
• 109:29 - 109:31
And what is this again?
• 109:31 - 109:36
This is the area of
a tiny curvilinear
• 109:36 - 109:41
patch on the surface that's
projected on that tiny square
• 109:41 - 109:43
on the floor.
• 109:43 - 109:45
All right?
• 109:45 - 109:47
OK.
• 109:47 - 109:54
So Now you know why
you get what you get.
• 109:54 - 109:57
One the last problem
because time is up.
• 109:57 - 109:58
No, I'm just kidding.
• 109:58 - 110:00
We still have plenty of time.
• 110:00 - 110:09
• 110:09 - 110:11
This is a beautiful,
beautiful problem.
• 110:11 - 110:13
But I don't want to finish it.
• 110:13 - 110:21
I want to give you
the problem at home.
• 110:21 - 110:23
It's like the one
in the book, but I
• 110:23 - 110:25
don't want to give you
exactly the one in the book.
• 110:25 - 110:29
• 110:29 - 110:31
I want to cover
something special today.
• 110:31 - 110:37
• 110:37 - 110:44
We are all familiar with their
notion of a spiral staircase.
• 110:44 - 110:47
But spiral staircases
are everywhere,
• 110:47 - 110:56
in elegant buildings, official
buildings, palaces, theaters,
• 110:56 - 111:00
houses of multimillionaires
in California.
• 111:00 - 111:02
And even people who
are not millionaires
• 111:02 - 111:05
have some spiral
staircases in their houses,
• 111:05 - 111:09
or even marble.
• 111:09 - 111:14
Did you ever wonder why
the spiral staircases
• 111:14 - 111:16
were invented?
• 111:16 - 111:19
If you go to most of the
castles on the Loire Valley,
• 111:19 - 111:22
or many European castles
have spiral staircases.
• 111:22 - 111:30
Many mosques, many churches
have these spiral staircases.
• 111:30 - 111:35
I think it was about a few
thousand years ago that it
• 111:35 - 111:38
was documented
for the first time
• 111:38 - 111:43
that the spiral staircases
consumed the least
• 111:43 - 111:46
amount of materials to build.
• 111:46 - 111:50
• 111:50 - 111:55
is that for confined spaces--
• 111:55 - 112:01
you have something like a
cylinder tower like that--
• 112:01 - 112:05
that's the only shape you
can build that minimizes
• 112:05 - 112:09
the area of the staircase
because if you start building
• 112:09 - 112:11
a staircase like
ours here, it's not
• 112:11 - 112:14
efficient at all in
terms of construction,
• 112:14 - 112:16
in terms of materials.
• 112:16 - 112:22
So you get a struggle
at actually making
• 112:22 - 112:24
these stairs that are not even.
• 112:24 - 112:27
You know, they're not even even.
• 112:27 - 112:33
Each of them will have a
triangle, or what is this?
• 112:33 - 112:37
Not a triangle, but
more like a trapezoid.
• 112:37 - 112:40
And it keeps going up.
• 112:40 - 112:42
This comes from a
helix, obviously.
• 112:42 - 112:46
And we have to understand
why this happens.
• 112:46 - 112:48
And I will introduce the
surface called helicoid.
• 112:48 - 112:51
• 112:51 - 112:57
And the helicoid will have
the following parametrization
• 112:57 - 112:58
by definition.
• 112:58 - 113:10
u cosine v, u sine v, and v.
• 113:10 - 113:21
Assume u is between 0 and 1 and
assume v is between 0 and 2 pi.
• 113:21 - 113:22
Draw the surface.
• 113:22 - 113:25
• 113:25 - 113:39
Also find the surface area of
the patch u between 0 and 1,
• 113:39 - 113:41
v between 0 and pi/2.
• 113:41 - 113:45
• 113:45 - 113:46
So I go, uh oh.
• 113:46 - 113:47
I'm in trouble.
• 113:47 - 113:49
Now how in the world am I
going to do this problem?
• 113:49 - 113:51
It looks horrible.
• 113:51 - 113:53
And it looks hard.
• 113:53 - 113:55
And it even looks hard to draw.
• 113:55 - 113:58
It's not that hard.
• 113:58 - 114:00
It's not hard at
all, because you
• 114:00 - 114:04
have to think of these
extreme points, the limit
• 114:04 - 114:11
points of u and v and see what
they really represent for you.
• 114:11 - 114:14
to work and say,
• 114:14 - 114:17
this is the frame
I'm starting with.
• 114:17 - 114:22
This is the x and y and
z frame with origin 0.
• 114:22 - 114:27
And I better draw this helicoid
because it shouldn't be hard.
• 114:27 - 114:30
So for u equals
0, what do I have?
• 114:30 - 114:33
• 114:33 - 114:34
I don't know.
• 114:34 - 114:35
It looks weird.
• 114:35 - 114:37
But Alex said it.
• 114:37 - 114:41
0, 0, v. v is my
parameter in real life.
• 114:41 - 114:45
So I have the whole z-axis.
• 114:45 - 114:50
So one edge is going to
be the z-axis itself.
• 114:50 - 114:52
Does it have to be
only the positive one?
• 114:52 - 114:52
No.
• 114:52 - 114:55
Who said so?
• 114:55 - 114:59
My problem said so,
that v only takes
• 114:59 - 115:00
values between 0 and 2 pi.
• 115:00 - 115:04
Unfortunately, I'm limiting
v between 0 and 2 pi.
• 115:04 - 115:07
But in general, v could
be any real number.
• 115:07 - 115:12
So I'll take it from
0 to 2 pi, and this
• 115:12 - 115:15
is going to be what I'm thinking
of, one edge of the staircase.
• 115:15 - 115:18
• 115:18 - 115:20
It's the interior age, the axis.
• 115:20 - 115:24
Let's see what happens
when u equals 1.
• 115:24 - 115:27
That's another curve
of the surface.
• 115:27 - 115:28
Let's see what I get.
• 115:28 - 115:36
Cosine v, sine v, and v. And it
looks like a friend of yours.
• 115:36 - 115:38
And you have to
tell me who this is.
• 115:38 - 115:43
If v were bt, what
is sine, cosine tt?
• 115:43 - 115:44
STUDENT: A sphere.
• 115:44 - 115:45
DR. MAGDALENA TODA: It's your--
• 115:45 - 115:46
STUDENT: Helicoid.
• 115:46 - 115:49
DR. MAGDALENA TODA: --helix
that you loved in chapter 10
• 115:49 - 115:50
• 115:50 - 115:55
And it was a curve that had
constant curvature and constant
• 115:55 - 115:57
portion, and it
• 115:57 - 115:58
And the speed of
this, for example,
• 115:58 - 116:00
would be square root of 2,
if you have the curiosity
• 116:00 - 116:02
to compute it.
• 116:02 - 116:04
It will have square root of 2.
• 116:04 - 116:06
And can I draw it?
• 116:06 - 116:08
I better draw it,
but I don't know how.
• 116:08 - 116:12
• 116:12 - 116:20
So I have to think of drawing
this for v between 0 and 2 pi.
• 116:20 - 116:23
• 116:23 - 116:28
When I'm at 0, when I
have time v equals 0,
• 116:28 - 116:31
I have the point 1, 0, and 0.
• 116:31 - 116:32
And where am I?
• 116:32 - 116:33
Here.
• 116:33 - 116:36
1, 0, 0.
• 116:36 - 116:39
And from here, I start moving
on the helix and going up.
• 116:39 - 116:48
And see, my hand should be
on-- this is the stairs.
• 116:48 - 116:50
It's obviously a smooth surface.
• 116:50 - 116:53
This is a smooth
surface, but the stairs
• 116:53 - 116:55
are a discretization
• 116:55 - 116:58
of the smooth surface.
• 116:58 - 117:01
I have a step, another step,
another step, another step.
• 117:01 - 117:07
So it's like a smooth helicoid
but discretized step functions.
• 117:07 - 117:08
• 117:08 - 117:12
staircase in the church--
• 117:12 - 117:13
you don't want to go to church.
• 117:13 - 117:16
You want to go to
the water park.
• 117:16 - 117:17
You want to go to Six Flags.
• 117:17 - 117:21
You want to go to whatever,
Disney World, San Antonio,
• 117:21 - 117:22
somewhere.
• 117:22 - 117:24
This is a slide.
• 117:24 - 117:25
You let yourself go.
• 117:25 - 117:28
This is you going
down, swimming--
• 117:28 - 117:30
I don't know-- upside down.
• 117:30 - 117:32
I don't know how.
• 117:32 - 117:36
So this is a smooth
slide in a water park.
• 117:36 - 117:38
That's how you should
be imagining it.
• 117:38 - 117:40
And it keeps going.
• 117:40 - 117:44
If I start here-- if I
start here, again, look,
• 117:44 - 117:46
this is what I'm describing.
• 117:46 - 117:48
• 117:48 - 117:49
A helicoid.
• 117:49 - 117:53
My arm moved on this.
• 117:53 - 117:56
Again, I draw the same motion.
• 117:56 - 118:01
My elbow should not
do something crazy.
• 118:01 - 118:04
It should keep
moving on the z-axis.
• 118:04 - 118:14
And I perform the pi/2 motion
when the stair-- not the stair.
• 118:14 - 118:15
I don't know what to call it.
• 118:15 - 118:23
This line becomes horizontal
when v equals pi/2.
• 118:23 - 118:30
So for v equals pi/2, I moved
from here straight to here.
• 118:30 - 118:31
STUDENT: Doesn't it go around?
• 118:31 - 118:32
DR. MAGDALENA TODA:
It goes around.
• 118:32 - 118:36
But see, what I asked-- I
• 118:36 - 118:38
First of all, I
said it goes around.
• 118:38 - 118:39
So I'll try to go around.
• 118:39 - 118:41
But it's hard.
• 118:41 - 118:44
Oh, wish me luck.
• 118:44 - 118:47
One, two, three.
• 118:47 - 118:49
I cannot go higher.
• 118:49 - 118:50
It goes to pi.
• 118:50 - 118:52
STUDENT: If it went to
2 pi, it would actually
• 118:52 - 118:53
wrap completely around?
• 118:53 - 118:54
DR. MAGDALENA TODA: It
would wrap like that.
• 118:54 - 118:56
STUDENT: Right above
where it started?
• 118:56 - 118:57
DR. MAGDALENA TODA: Exactly.
• 118:57 - 119:03
So if I started here, I end
up parallel to that, up.
• 119:03 - 119:05
But 2 pi is too high for me.
• 119:05 - 119:07
So I should go slowly.
• 119:07 - 119:13
• 119:13 - 119:17
I'm up after 2 pi in
the same position.
• 119:17 - 119:20
STUDENT: But since it's pi/2,
it's just kind of like--
• 119:20 - 119:21
DR. MAGDALENA TODA:
When I'm pi/2,
• 119:21 - 119:25
I just performed
from here to here.
• 119:25 - 119:27
STUDENT: So that's it's
• 119:27 - 119:29
DR. MAGDALENA TODA:
• 119:29 - 119:36
what area does my arm from here
to here sweep at this point?
• 119:36 - 119:39
From this point to this point.
• 119:39 - 119:41
It's a smooth surface.
• 119:41 - 119:46
So it's generated by my motion.
• 119:46 - 119:48
And stop.
• 119:48 - 119:49
It's a root surface.
• 119:49 - 119:51
It's a root patch of a surface.
• 119:51 - 119:54
Somebody tell me how I'm going
to do this because this is not
• 119:54 - 119:58
with square root of 1 plus f sub
x squared plus f sub y squared.
• 119:58 - 119:59
That is for normal people.
• 119:59 - 120:01
You are not normal people.
• 120:01 - 120:05
They never teach this in honors.
• 120:05 - 120:09
In honors, we don't
cover this formula.
• 120:09 - 120:10
But you're honors.
• 120:10 - 120:15
So do I want you to finish
it at home with a calculator?
• 120:15 - 120:18
All I want is for you to be
able to set up the integral.
• 120:18 - 120:22
And I think, knowing you
better and working with you--
• 120:22 - 120:25
I think you have the potential
to do that without my help,
• 120:25 - 120:29
with all the elements
I gave you until now.
• 120:29 - 120:39
So the area of s-- it
will be the blue slide.
• 120:39 - 120:41
These are all--
when you slide down,
• 120:41 - 120:43
you slide down along helices.
• 120:43 - 120:47
going down along helices.
• 120:47 - 120:48
OK?
• 120:48 - 120:50
So that's what you have.
• 120:50 - 120:56
[INAUDIBLE] double integral
for a certain domain D.
• 120:56 - 120:57
Which is that domain D?
• 120:57 - 121:03
That domain D is u between 0
and 1 and v between 0 and pi/2
• 121:03 - 121:06
because that's what
I said I want here.
• 121:06 - 121:10
• 121:10 - 121:12
Of what?
• 121:12 - 121:19
Of magnitude of r sub u times
r sub v cross product dudv.
• 121:19 - 121:23
• 121:23 - 121:25
You need to help me
though because I don't
• 121:25 - 121:28
know what I'm going to do.
• 121:28 - 121:33
So who starts?
• 121:33 - 121:42
r sub u is-- I'm not doing
anything without you, I swear.
• 121:42 - 121:43
OK?
• 121:43 - 121:44
STUDENT: Cosine.
• 121:44 - 121:46
DR. MAGDALENA TODA: Cosine v.
• 121:46 - 121:47
STUDENT: Sine v.
• 121:47 - 121:48
DR. MAGDALENA TODA: Sine v.
• 121:48 - 121:48
STUDENT: And 0.
• 121:48 - 121:49
DR. MAGDALENA TODA: 0.
• 121:49 - 121:52
r sub v equals--
• 121:52 - 121:54
STUDENT: Negative u sine v.
• 121:54 - 121:56
DR. MAGDALENA TODA: Very good.
• 121:56 - 122:01
u cosine v. And 1, and
this goes on my nerves.
• 122:01 - 122:02
But what can I do?
• 122:02 - 122:04
Nothing.
• 122:04 - 122:05
OK?
• 122:05 - 122:06
All right.
• 122:06 - 122:12
So I have to compute the what?
• 122:12 - 122:13
STUDENT: The cross product.
• 122:13 - 122:15
DR. MAGDALENA TODA:
The cross product.
• 122:15 - 122:22
• 122:22 - 122:24
I, J, K, of course, right?
• 122:24 - 122:33
I, J, K, cosine v, sine v,
0, minus u sine-- oh, my god.
• 122:33 - 122:34
Where was it?
• 122:34 - 122:34
It's there.
• 122:34 - 122:36
You gave it to me.
• 122:36 - 122:39
u cosine v and 1.
• 122:39 - 122:44
Minus u sine v, u
cosine v, and 1.
• 122:44 - 122:46
OK.
• 122:46 - 122:49
I times that.
• 122:49 - 122:59
Sine v, I. sine v,
I. J. J has a friend.
• 122:59 - 123:02
Is this mine or--
cosine v times 1.
• 123:02 - 123:04
But I have to change the sine.
• 123:04 - 123:06
Are you guys with me?
• 123:06 - 123:11
It's really serious that I have
to think of changing the sine.
• 123:11 - 123:16
Minus cosine v times
J. Are you with me?
• 123:16 - 123:19
• 123:19 - 123:22
This times that
with a sign change.
• 123:22 - 123:28
And then plus-- what is--
well, this is not so obvious.
• 123:28 - 123:31
But you have so much practice.
• 123:31 - 123:32
Make me proud.
• 123:32 - 123:34
What is the minor
that multiplies K?
• 123:34 - 123:36
This red fellow.
• 123:36 - 123:39
You need to compute
it and simplify it.
• 123:39 - 123:41
So I don't talk too much.
• 123:41 - 123:43
• 123:43 - 123:44
u, excellent.
• 123:44 - 123:47
How did you do it, [INAUDIBLE]?
• 123:47 - 123:48
STUDENT: [INAUDIBLE].
• 123:48 - 123:50
DR. MAGDALENA TODA: So
you group together cosine
• 123:50 - 123:52
squared plus sine squared.
• 123:52 - 123:53
Minus, minus is a plus.
• 123:53 - 123:55
And you said u, it's u.
• 123:55 - 123:56
Good.
• 123:56 - 123:59
Plus u times K. Good.
• 123:59 - 124:01
• 124:01 - 124:03
Now that you look at it,
• 124:03 - 124:06
You have to set up the integral.
• 124:06 - 124:09
And that's going to be what?
• 124:09 - 124:13
The square root of
this fellow squared
• 124:13 - 124:16
plus that fellow squared
plus this fellow squared.
• 124:16 - 124:18
Let's take our time.
• 124:18 - 124:25
You take these three fellows,
• 124:25 - 124:27
and put them under
a square root.
• 124:27 - 124:29
Is it hard?
• 124:29 - 124:30
STUDENT: 1 plus u squared.
• 124:30 - 124:33
DR. MAGDALENA TODA:
1 plus u squared.
• 124:33 - 124:40
Now the thing is I
don't have a Jacobian.
• 124:40 - 124:42
This is dudv.
• 124:42 - 124:44
The Jacobian is what?
• 124:44 - 124:46
So this is what I have.
• 124:46 - 124:50
Now between the end
points, I have to think.
• 124:50 - 124:52
v has to be between 0 and pi/2.
• 124:52 - 124:56
• 124:56 - 124:58
And u has to be between 0 and 1.
• 124:58 - 125:05
• 125:05 - 125:06
Do you notice anything?
• 125:06 - 125:09
And that's exactly what
I wanted to tell you.
• 125:09 - 125:10
v is not inside.
• 125:10 - 125:12
v says, I'm independent.
• 125:12 - 125:14
• 125:14 - 125:17
I'll go off, take a break. pi/2.
• 125:17 - 125:20
But then you have
integral from 0
• 125:20 - 125:27
to 1 square root of
1 plus u squared du.
• 125:27 - 125:28
I want to say a remark.
• 125:28 - 125:32
• 125:32 - 125:35
Happy or not happy, shall I be?
• 125:35 - 125:40
Here, you need either
the calculator,
• 125:40 - 125:42
which is the simplest
way to do it, just
• 125:42 - 125:44
compute the simple integral.
• 125:44 - 125:49
Integral from 0 to 1 square
root of 1 plus u squared du.
• 125:49 - 125:54
Or what do you have in this
• 125:54 - 125:55
if you don't have a calculator?
• 125:55 - 125:56
STUDENT: [INAUDIBLE]
• 125:56 - 126:01
DR. MAGDALENA TODA: A table of
integration, integration table.
• 126:01 - 126:02
• 126:02 - 126:05
I mean, you cannot
give me an exact value.
• 126:05 - 126:12
But give me an approximate
value by Thursday.
• 126:12 - 126:13
Is today Tuesday?
• 126:13 - 126:14
Yes.
• 126:14 - 126:15
By Thursday.
• 126:15 - 126:18
So please let me know how much
you got from the calculator
• 126:18 - 126:21
or from integration tables.
• 126:21 - 126:27
So we have this
result. And then I
• 126:27 - 126:31
would like to interpret
this result geometrically.
• 126:31 - 126:37
What we can say about the
helicoid that I didn't tell you
• 126:37 - 126:41
but I'm going to tell you
just to finish-- have you
• 126:41 - 126:45
been to the OMNIMAX
Science Spectrum, the one
• 126:45 - 126:47
in Lubbock or any other?
• 126:47 - 126:48
I think they are
everywhere, right?
• 126:48 - 126:49
I mean-- everywhere.
• 126:49 - 126:51
We are a large city.
• 126:51 - 126:56
Only in the big cities can
you come to a Science Spectrum
• 126:56 - 126:58
museum kind of like that.
• 126:58 - 127:01
Have you played
with the soap films?
• 127:01 - 127:02
OK.
• 127:02 - 127:05
Do you remember what kind
• 127:05 - 127:11
They had the big tub with soapy
water, with soap solution.
• 127:11 - 127:14
And then there were all sorts
of [INAUDIBLE] in there.
• 127:14 - 127:20
They had the wire with pig
rods that looked like a prism.
• 127:20 - 127:25
They had a cube that they wanted
you to dip into the soap tub.
• 127:25 - 127:27
• 127:27 - 127:29
And there comes the
beautiful thing.
• 127:29 - 127:33
spring that they took
• 127:33 - 127:36
• 127:36 - 127:36
No.
• 127:36 - 127:39
I don't think it
was flexible at all.
• 127:39 - 127:44
It was a helix made with
a rod inside, a metal rod
• 127:44 - 127:49
inside and attached
to the frame of that.
• 127:49 - 127:55
There was the metal rod and
this helix made of hard iron,
• 127:55 - 127:56
and they were sticking together.
• 127:56 - 128:00
Have you dipped this
into the soap solution?
• 128:00 - 128:02
And what did you get?
• 128:02 - 128:03
STUDENT: [INAUDIBLE]
• 128:03 - 128:05
DR. MAGDALENA TODA: That's
exactly what you get.
• 128:05 - 128:06
You can get several surfaces.
• 128:06 - 128:10
You can even get the one on
the outside that's unstable.
• 128:10 - 128:11
It broke in my case.
• 128:11 - 128:14
It's almost like a cylinder.
• 128:14 - 128:19
The one that was pretty
• 128:19 - 128:31
The helicoid is a so-called
soap film, soap film
• 128:31 - 128:33
or minimal surface.
• 128:33 - 128:38
Minimal surface.
• 128:38 - 128:40
So what is a minimal surface?
• 128:40 - 128:42
A minimal surface
is a soap film.
• 128:42 - 128:47
A minimal surface
is a surface that
• 128:47 - 128:53
tends to minimize the area
enclosed in a certain frame.
• 128:53 - 129:00
You take a wire that looks like
a loop, its own skew curve.
• 129:00 - 129:03
You dip that into
the soap solution.
• 129:03 - 129:04
You pull it out.
• 129:04 - 129:05
You get a soap film.
• 129:05 - 129:08
That a minimal surface.
• 129:08 - 129:10
So all the things
that you created
• 129:10 - 129:15
by taking wires and dipping
them into the soap tub
• 129:15 - 129:18
and pulling them out-- they
are not just called soap films.
• 129:18 - 129:20
They are called
minimal surfaces.
• 129:20 - 129:23
Somewhere on the wall
of the Science Spectrum,
• 129:23 - 129:25
they wrote that.
• 129:25 - 129:27
the theory of minimal surfaces.
• 129:27 - 129:31
But there are people-- there
are famous mathematicians who
• 129:31 - 129:34
all their life studied
just minimal surfaces, just
• 129:34 - 129:35
soap films.
• 129:35 - 129:37
They came up with the results.
• 129:37 - 129:41
Some of them got very
prestigious awards
• 129:41 - 129:44
for that kind of thing
theory for minimal surfaces.
• 129:44 - 129:51
And these have been known
since approximately the middle
• 129:51 - 129:54
of the 19th century.
• 129:54 - 129:56
There were several
mathematicians
• 129:56 - 130:00
who discovered the most
important minimal surfaces.
• 130:00 - 130:04
There are several that
you may be familiar with
• 130:04 - 130:07
and several you may
not be familiar with.
• 130:07 - 130:12
But another one that you may
have known is the catenoid.
• 130:12 - 130:15
And that's the last thing I'm
• 130:15 - 130:19
Have you heard of a catenary?
• 130:19 - 130:22
Have you ever been to St. Louis?
• 130:22 - 130:25
St. Louis, St. Louis.
• 130:25 - 130:27
The city St. Louis
with the arch.
• 130:27 - 130:29
OK.
• 130:29 - 130:31
Did you go to the arch?
• 130:31 - 130:33
No?
• 130:33 - 130:34
You should go to the arch.
• 130:34 - 130:35
It looks like that.
• 130:35 - 130:37
It has a big base.
• 130:37 - 130:39
So it looks so beautiful.
• 130:39 - 130:42
It's thicker at the base.
• 130:42 - 130:46
This was based on a
mathematical equation.
• 130:46 - 130:55
The mathematical equation
it was based on was cosh x.
• 130:55 - 130:56
What is cosh as a function?
• 130:56 - 131:00
Now I'm testing you,
but I'm not judging you.
• 131:00 - 131:01
If you forgot--
• 131:01 - 131:02
STUDENT: e to the x.
• 131:02 - 131:05
DR. MAGDALENA TODA:
e to the x plus e
• 131:05 - 131:06
to the negative x over 3.
• 131:06 - 131:08
If it's minus, it's called sinh.
• 131:08 - 131:12
So the one with [? parts. ?]
You can have x/a,
• 131:12 - 131:14
and you can have 1/a in front.
• 131:14 - 131:17
It's still called a catenary.
• 131:17 - 131:20
Now what is-- this
is a catenary.
• 131:20 - 131:24
The shape of the arch of
St. Louis is a catenary.
• 131:24 - 131:29
But you are more used to the
catenary upside down, which
• 131:29 - 131:30
is any necklace.
• 131:30 - 131:31
Do you have a necklace?
• 131:31 - 131:34
If you take any
necklace-- that is,
• 131:34 - 131:36
it has to be homogeneous,
not one of those,
• 131:36 - 131:40
like you have a pearl hanging,
• 131:40 - 131:41
No.
• 131:41 - 131:45
It has to be a
homogeneous metal.
• 131:45 - 131:48
Think gold, solid gold, but
that kind of liquid gold.
• 131:48 - 131:50
Do you know what
• 131:50 - 131:54
Those beautiful bracelets
or necklaces that are fluid,
• 131:54 - 131:58
and you cannot even see
• 131:58 - 132:04
So you hang it at
the same height.
• 132:04 - 132:08
What you get-- Galileo
proved it was not
• 132:08 - 132:11
a parabola because people at
that time were really stupid.
• 132:11 - 132:16
So they thought, hang a
chain from a woman's neck
• 132:16 - 132:19
or some sort of
beautiful jewelry,
• 132:19 - 132:22
it must be a parabola because
it looks like a parabola.
• 132:22 - 132:25
And Galileo Galilei says,
these guys are nuts.
• 132:25 - 132:28
They don't know any mathematics,
any physics, any astronomy.
• 132:28 - 132:33
So he proved in no time
that thing, the chain,
• 132:33 - 132:34
cannot be a parabola.
• 132:34 - 132:38
And he actually
came up with that.
• 132:38 - 132:40
If a is 1, you just
have the cosh x.
• 132:40 - 132:42
So this is the chain.
• 132:42 - 132:46
If you take the chain--
if you take the chain--
• 132:46 - 132:48
that's a chain upside down.
• 132:48 - 132:51
If you take a chain--
let's say y equals cosh
• 132:51 - 132:56
x to make it easier--
and revolve that chain,
• 132:56 - 132:58
you get a surface of revolution.
• 132:58 - 133:03
• 133:03 - 133:05
And this surface of
revolution is called catenoid.
• 133:05 - 133:09
• 133:09 - 133:12
How can you get the catenoid
as a minimal surface, expressed
• 133:12 - 133:15
as the minimum surface?
• 133:15 - 133:16
There are people
who can do that.
• 133:16 - 133:18
They have the
ability to do that.
• 133:18 - 133:21
And they tried to have
an experimental thing
• 133:21 - 133:23
at the Science Spectrum as well.
• 133:23 - 133:25
And they did a beautiful job.
• 133:25 - 133:26
I was there.
• 133:26 - 133:30
So they took two
• 133:30 - 133:33
You can have them be
• 133:33 - 133:37
steel or anything.
• 133:37 - 133:41
But they have to be
equal, equal circles.
• 133:41 - 133:43
And you touch them,
and you dip them
• 133:43 - 133:46
both at the same time
into the soap solution.
• 133:46 - 133:48
And then you pull
it out very gently
• 133:48 - 133:50
because you have
to be very smart
• 133:50 - 133:52
and very-- be like a surgeon.
• 133:52 - 133:55
it's goodbye minimal surfaces
• 133:55 - 133:57
because they break.
• 133:57 - 133:58
They collapse.
• 133:58 - 134:02
So you have to pull those
circles with the same force,
• 134:02 - 134:05
gently, one away from the other.
• 134:05 - 134:06
What you're going
to get is going
• 134:06 - 134:09
to be a film that looks
exactly like that.
• 134:09 - 134:12
These are the circles.
• 134:12 - 134:16
After a certain distance
of moving them apart,
• 134:16 - 134:19
the soap film will
collapse and will burst.
• 134:19 - 134:22
There's no more surface inside.
• 134:22 - 134:25
But up to that moment,
you have a catenoid.
• 134:25 - 134:27
And this catenoid is
a minimal surface.
• 134:27 - 134:30
It's trying to-- of
the frame you gave it,
• 134:30 - 134:34
which is the wire frame--
minimize the area.
• 134:34 - 134:35
It's not going to be a cylinder.
• 134:35 - 134:38
It's way too much area.
• 134:38 - 134:41
It's going to be something
smaller than that,
• 134:41 - 134:44
so something that says,
I'm an elastic surface.
• 134:44 - 134:46
I'm occupying as
little area as I
• 134:46 - 134:51
can because I live in
a world of scarcity,
• 134:51 - 134:54
and I try to occupy
as little as I can.
• 134:54 - 134:57
So it's based on a
principle of physics.
• 134:57 - 135:01
The surface tension
of the soap films
• 135:01 - 135:04
will create this
minimization of the area.
• 135:04 - 135:08
So all you need to do is
remember you know the helicoid
• 135:08 - 135:12
and catenoid only because
you're honors students.
• 135:12 - 135:15
So thank [INAUDIBLE] college
for giving you this opportunity.
• 135:15 - 135:16
All right?
• 135:16 - 135:17
I'm not kidding.
• 135:17 - 135:21
It may sound like a joke, but
it's half joke, half truth.
• 135:21 - 135:24
We learn learn a little
bit more interesting stuff
• 135:24 - 135:26
than other kids.
• 135:26 - 135:27
• 135:27 - 135:29
Good luck with homework.
• 135:29 - 135:32
Come bug me abut any kind
of homework [INAUDIBLE].
• 135:32 - 135:43
• 135:43 - 135:44
STUDENT: Do you
know if I can talk
• 135:44 - 135:45
• 135:45 - 135:49
• 135:49 - 135:52
DR. MAGDALENA TODA:
Actually, my [INAUDIBLE].
• 135:52 - 135:54
She is the one who
does [INAUDIBLE].
• 135:54 - 136:02
But I can take you to her so
you can start [INAUDIBLE].
• 136:02 - 136:04
I think it would be a very
interesting [INAUDIBLE].
• 136:04 - 136:05
Maybe.
• 136:05 - 136:05
[INAUDIBLE]
• 136:05 - 136:15
• 136:15 - 136:16
STUDENT: OK.
• 136:16 - 136:19
DR. MAGDALENA TODA:
Second floor, [INAUDIBLE].
• 136:19 - 136:22
This is the [INAUDIBLE].
• 136:22 - 136:24
You have to go all
the way behind.
• 136:24 - 136:25
STUDENT: OK.
• 136:25 - 136:26
OK.
• 136:26 - 136:28
STUDENT: [INAUDIBLE]
• 136:28 - 136:32
• 136:32 - 136:35
DR. MAGDALENA TODA:
[INAUDIBLE] because I don't
• 136:35 - 136:37
want to give you anything new.
• 136:37 - 136:40
I don't want to get you
in any kind of trouble.
• 136:40 - 136:44
The problems that I solved
on the board are primarily,
• 136:44 - 136:47
I would say, 60% of what
you'll see on the midterm.
• 136:47 - 136:50
It's something that
we covered in class.
• 136:50 - 136:53
And the other 40% will be
something not too hard,
• 136:53 - 136:57
but something standard out of
• 136:57 - 136:58
that you studied.
• 136:58 - 137:00
It shouldn't be hard.
• 137:00 - 137:01
STUDENT: Thank you.
• 137:01 - 137:02
DR. MAGDALENA TODA:
You're welcome.
• 137:02 - 137:04
STUDENT: We're trying to
join the Honors Society.
• 137:04 - 137:05
But we can't make
that thing tomorrow.
• 137:05 - 137:06
Can we still join?
• 137:06 - 137:08
DR. MAGDALENA TODA:
You can still join.
• 137:08 - 137:11
Remind me to give you the
golden pin, the brochures, all
• 137:11 - 137:12
the information.
• 137:12 - 137:16
And then when you get those,
you'll give me the \$35.
• 137:16 - 137:17
• 137:17 - 137:18
STUDENT: Awesome.
• 137:18 - 137:18
Thank you.
• 137:18 - 137:20
DR. MAGDALENA TODA:
You're welcome.
• 137:20 - 137:21
Both of you want to--
• 137:21 - 137:22
STUDENT: Yeah.
• 137:22 - 137:23
DR. MAGDALENA TODA: And
you cannot come tomorrow?
• 137:23 - 137:23
STUDENT: Yeah.
• 137:23 - 137:24
I have my--
• 137:24 - 137:28
DR. MAGDALENA TODA: I wanted to
bring something, some snacks.
• 137:28 - 137:29
But I don't know.
• 137:29 - 137:32
I need to count and see
how many people can come.
• 137:32 - 137:34
And it's going to
be in my office.
• 137:34 - 137:34
STUDENT: OK.
• 137:34 - 137:35
DR. MAGDALENA TODA: All right.
• 137:35 - 137:36
STUDENT: Were you in
the tennis tournament?
• 137:36 - 137:37
STUDENT: Yeah.
• 137:37 - 137:38
STUDENT: You're the guy who won?
• 137:38 - 137:39
STUDENT: Yeah.
• 137:39 - 137:40
STUDENT: Congratulations.
• 137:40 - 137:42
I was like, I know that name.
• 137:42 - 137:43
He's in my [INAUDIBLE].
• 137:43 - 137:45
DR. MAGDALENA TODA: You won it?
• 137:45 - 137:45
STUDENT: Yeah.
• 137:45 - 137:46
DR. MAGDALENA TODA:
The tennis tournament?
• 137:46 - 137:47
STUDENT: Yeah.
• 137:47 - 137:48
It was [INAUDIBLE].
• 137:48 - 137:49
DR. MAGDALENA TODA:
Congratulations.
• 137:49 - 137:51
Why don't you blab a
• 137:51 - 137:53
You're so modest.
• 137:53 - 137:54
You never say anything.
• 137:54 - 137:56
STUDENT: It's all right.
• 137:56 - 137:57
STUDENT: I'm sorry.
• 137:57 - 137:57
STUDENT: No.
• 137:57 - 137:58
STUDENT: It's not a big deal.
• 137:58 - 137:59
STUDENT: Were people good?
• 137:59 - 138:02
Yeah?
• 138:02 - 138:03
DR. MAGDALENA TODA: All right.
• 138:03 - 138:05
STUDENT: I have my extra credit.
• 138:05 - 138:06
Title:
TTU Math2450 Calculus3 Sec 12.4
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 jackie.luft edited English subtitles for TTU Math2450 Calculus3 Sec 12.4