## TTU Math2450 Calculus3 Secs 12.1-12.2

• 0:00 - 0:04
PROFESSOR: I have
some assignments
• 0:04 - 0:06
that I want to give you back.
• 0:06 - 0:10
And I'm just going
to put them here,
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and I'll ask you to pick them
up as soon as we take a break.
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There are explanations there
how they were computed in red.
• 0:21 - 0:23
If you have questions,
you can as me
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Now, I promised you that
I would move on today,
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and that's what I'm going to do.
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I'm moving on to something
that you're gong to love.
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[? Practically ?] chapter 12
is integration of functions
• 0:48 - 0:49
of several variables.
• 0:49 - 0:59
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And to warn you
we're going to see
• 1:01 - 1:09
how we introduce introduction
to the double integral.
• 1:09 - 1:16
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But you will say, wait a minute.
• 1:17 - 1:22
I don't even know if I
remember the simple integral.
• 1:22 - 1:24
And that's why I'm here.
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I want to remind you what the
definite integral was both
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as a formal definition-- let's
do it as a formal definition
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first, then come up with a
geometric interpretation based
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on that.
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And finally write
down the definition
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and the fundamental
theorem of calculus.
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So assume you have a
function that's continuous.
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Continuous over a certain
integral of a, b interval in R.
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And you know that
in that case, you
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can "define the
definite integral of f
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of x from or between a and b."
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And as the notation is denoted,
by integral from a to b f of x
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dx.
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Well, how do we define this?
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This is just the notation.
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How do we define it?
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We have to have a set up, and
we are thinking of a x, y frame.
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You have a function,
f, that's continuous.
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And you are thinking,
oh, wait a minute.
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I would like to be
able to evaluate
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the area under the integral.
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when you are in fourth grade,
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I can give you some graphing
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paper.
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And with that
graphing paper, you
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can eventually approximate
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Sort of what you get here is
like you draw a horizontal
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so that the little part
above the horizontal
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cancels out with the little
part below the horizontal.
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So more or less,
the pink rectangle
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is a good approximation
of the first slice.
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But you say yeah, but the first
slice is a curvilinear slice.
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Yes, but we make it
like a stop function.
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So then you say, OK,
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I'm going to approximate
it in a similar way,
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and I'm going to have a bunch
of rectangles on this graphing
• 4:15 - 4:16
paper.
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And I'm going to
compute their areas,
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and I'm going to come up
with an approximation,
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and I'll give it to my
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And that's what we
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but this is not fourth grade.
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And actually, it's
very relevant to us
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that this has
applications to our life,
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to our digital world,
that people did not
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understand when Riemann
introduced the Riemann sum.
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They thought, OK, the idea
makes sense that practically we
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have a huge picture
here, and I'm
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taking a and b and a function
that's continuous over a and b.
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And then I say I'm
going to split this
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into a equidistant intervals.
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I don't know how
many I want, but let
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me make them eight of them.
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I don't know.
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They have to have
the same length.
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And I'll call this delta x.
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It has to be the same.
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me for the horrible picture.
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They don't look like
the same step, delta x,
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but it should be the same.
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In each of them I
arbitrarily, say it again,
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Magdalena, arbitrarily pick
x1 star, and another point,
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x2 star wherever I want inside.
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I'm just getting [INAUDIBLE].
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X4 star, and this is x8 star.
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But let's say that in general
I don't know they are 8.
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They could be n.
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xn star.
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And passing to the
limit with respect
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to n going to infinity,
what am I going to get?
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Well, in the first
cam I'm going up,
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and I'm hitting
at what altitude?
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I'm hitting at the altitude
called f of x1 star.
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And that's going to be the
height of this-- what is this?
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Strip?
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Right?
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Or rectangle.
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OK.
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And I'm going to do
the same with green
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for the second rectangle.
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I'll pick x2 star, and
then that doesn't work.
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And I'll take this.
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Let's see if I can do
the light green one,
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because spring is here.
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Let's see.
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That's beautiful.
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I go up.
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I hit here at x2 star.
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I get f of x2 star.
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And so on and so forth.
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Until I get to, let's say,
the last of the Mohicans.
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This will be xn minus
1, and this is going
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to be xn star, the purple guy.
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And this is going
to be the height
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of that last of the Mohicans.
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So when I compute the sum, I
call that approximating sum
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or Riemann approximating sum,
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better to do than invent it.
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He didn't even know
that we are going
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to get pixels that are in
larger and larger quantities.
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Like, we get 3,000 by 900.
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He didn't know we are going to
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But passing to the
limit practically should
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be easier to understand
for teenagers now
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age, because it's like
making the number of pixels
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larger and larger, and the
pixels practically invisible.
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Remember, I mean, I don't
know, those old TVs,
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color TVs where you could
still see the squares?
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STUDENT: Mm-hm.
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PROFESSOR: Well, yeah.
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When you were little.
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But I remember them
much better than you.
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And, yes, as the number
of pixels will increase,
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that means I'm taking the limit
and going larger and larger.
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That means
practically limitless.
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Infinity will give
me an ideal image.
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My eye will be as if I could see
the image that's a curvilinear
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image as a real person.
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And, of course, the
quality of our movies
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really increased a lot.
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And this is what I'm
trying to emphasize here.
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So you have f of x1 star delta
x plus the last rectangle
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area, f of xn star delta x.
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Well, as a mathematician,
I don't write it like that.
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How do I write it
as a mathematician?
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Well, we are funny people.
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We like Greek.
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It's all Greek to me.
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So we go sum and from-- no. k
from 1 to n, f of x sub k star.
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So I have k from 1 to n exactly
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And this is going to
be [INAUDIBLE], which
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is the same everywhere.
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the partition is equal.
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So practically I have
the same distance.
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And what is this
limit? [? Lim ?]
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is going to be exactly integral
from a to b of f of x dx.
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And I make a smile here,
and I say I'm very happy.
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This is as a meaning is
the area under the graph.
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If-- well, I didn't
say something.
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If I want it to be
positive, otherwise it's
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getting not to be the
area under the graph.
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The integral will still
be defined like that.
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But what's going to happen if
I have, for example, half of it
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above and half of it below?
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I'm going to get this,
and I'm going to get that.
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And when I add them, I'm going
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because this is a negative
area, and that's a positive area
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and they try to
annihilate each other.
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But this guy under
the water is stronger,
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like an iceberg that's
20% on tip of the water,
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80% of the iceberg
is under the water.
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So the same thing.
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I'm going to get a negative
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OK.
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Now, we remember that
very well, but now we
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have to generalize this
thingy to something else.
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And I will give you
a curvilinear domain.
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Where shall I erase?
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I don't know.
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Here.
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What if somebody gives you
the image of a potatoe-- well,
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I don't know.
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Something.
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A blob.
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Some nice curvilinear domain--
and says, you know what?
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I want to approximate the area
of this image, curvilinear
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image, to the best
of my abilities.
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And compute it, and eventually I
have some weighted sum of that.
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So if one would have
to compute the area,
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it wouldn't be so hard,
because we would say,
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OK, I have to
"partition this domain
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into small sections using
a rectangular partition
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or square partition."
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And how?
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Well, I'm going to--
you have to imagine
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that I have a bunch
of a grid, and I'm
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partitioning the whole thing.
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And you say, wait a minute.
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Wait a minute.
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It's not so easy.
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I mean, they are not all
the same area, Magdalena.
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Even if you tried to make these
equidistant in both directions,
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look at this guy.
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Look at that guy.
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He's much bigger than that.
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Look at this small
guy, and so on.
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So we have to imagine that we
look at the so-called normal
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the partition.
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And let's say in the normal,
or the length of the partition,
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is denoted like that.
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We have to give that a meaning.
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Well, let's say "this
is the highest diameter
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for all subdomains
in the picture."
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And you say, wait a minute.
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But these subdomains
should have names.
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Well, they don't have names,
but assume they have areas.
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This would be-- I have to
find a way to denote them
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and be orderly.
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A1, A2, A3, A4, A5, AN,
AM, AN, stuff like that.
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So practically I'm looking
at the highest diameter.
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When I have a domain, I
look at the largest instance
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inside that domain.
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So what would be the diameter?
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The largest distance between
two points in that domain.
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I'll call that the diameter.
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OK.
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I want that diameter to
go got 0 in the limit.
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So I want this partition
to go to 0 in the limit.
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And that means I'm
"shrinking" the pixels.
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"Shrinking" in
quotes, the pixels.
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How would I mimic
what I did here?
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Well, it would be
easier to get the area.
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In this case, I would have
some sort of A sum limit.
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I'm sorry.
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The curvilinear
area of the domain.
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Let's call it-- what
do you want to call it?
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D for domain--
inside the domain.
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OK?
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This whole thing would be what?
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Would be limit of summation of,
let's say, limit of what kind?
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k from 1 to n.
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Limit n goes to infinity.
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K from 1 to n of
these tiny A sub k's,
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areas of the subdomain.
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Wait a minute.
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But you say, but what if
I want something else?
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Like, I'm going to
build some geography.
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This is the domain.
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That's something like
on a map, and I'm
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going to build a
mountain on top of it.
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I'll take some Play-Do,
I'll take some Play-Do,
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and I'm going to
model some geography.
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And you say, wait a minute.
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Do you make mountains?
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I'm afraid to make Rocky
Mountains, because they
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may have points where the
function is not smooth.
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If I don't have
derivative at the peak,
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them I'm in trouble, in general.
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Although you say,
well, but the function
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has to be only continuous.
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I know.
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I know.
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But I don't want any kind
of really nasty singularity
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where I can have a
crack in the mountain
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or a well or
something like that.
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So I assume the
geography to be smooth,
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the function of
[INAUDIBLE] is continuous,
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and the picture
should look something
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like-- let's see
if I can do that.
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The projection, the
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would be the domain, [? D. ?]
And this is equal, f of x what?
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You say, what?
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Magdalena, I don't understand.
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The exact shadow of this fellow
where I have the sun on top
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here-- that's the sun.
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is the plain, or domain, x, y.
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I take all my points in x, y.
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I mean, I take really
all my points in x, y,
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and the value of the altitude
on this geography at the point
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x, y would be z
equals f of x, y.
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if this would be a can of Coke,
• 18:21 - 18:24
it would be easy to
compute the volume, right?
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Practically you have a
constant altitude everywhere,
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and you have the area of
the base times the height,
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• 18:33 - 18:39
But what if somebody asks you to
find the volume under the hat?
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"Find the volume
undo this graph."
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STUDENT: I would take it
more as two functions.
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So the top line would
be the one function,
• 18:53 - 18:55
and the bottom line would
be another function.
• 18:55 - 18:59
So if you take the volume of the
top function minus the volume
• 18:59 - 19:00
of the bottom
function, it'd give you
• 19:00 - 19:03
the total volume of the object.
• 19:03 - 19:06
PROFESSOR: And actually,
I want the total volume
• 19:06 - 19:08
above the sea level.
• 19:08 - 19:13
So I'm going to--
sometimes I can take it up
• 19:13 - 19:16
to a certain level where-- let's
say the mountain is up to here,
• 19:16 - 19:18
and I want it only up to here.
• 19:18 - 19:22
So I want everything,
including the-- the walls
• 19:22 - 19:24
would be cylindrical.
• 19:24 - 19:25
STUDENT: Yeah.
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PROFESSOR: If I
want all the volume,
• 19:26 - 19:28
that's going to be
a little bit easier.
• 19:28 - 19:29
Let's see why.
• 19:29 - 19:32
I will have limit.
• 19:32 - 19:36
The idea is, as you
said very well, limit.
• 19:36 - 19:37
n goes to infinity.
• 19:37 - 19:42
A sum k from 1 to n.
• 19:42 - 19:45
And what kind of
partition can I build?
• 19:45 - 19:48
I'll take the
line, and I'll say,
• 19:48 - 19:53
I'll build myself
a partition with a,
• 19:53 - 19:58
let's say, the
typical domain, AK.
• 19:58 - 20:01
I have A1, A2 A3, A4, AK, AN.
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How may of those little domains?
• 20:03 - 20:04
AN.
• 20:04 - 20:08
That will be all the
little subdomains
• 20:08 - 20:12
inside the green curve.
• 20:12 - 20:15
The green loop.
• 20:15 - 20:17
In that case, what do I do?
• 20:17 - 20:24
For each of these guys, I go
up, and I go, oh, my god, this
• 20:24 - 20:27
looks like a skyscraper,
but the corners,
• 20:27 - 20:29
when I go through
this surface, are
• 20:29 - 20:31
in the different dimensions.
• 20:31 - 20:32
What am I going to do?
• 20:32 - 20:35
That forces me to
build a skyscraper
• 20:35 - 20:39
by thinking I take a
point in the domain,
• 20:39 - 20:45
I go up until that hits the
surface, pinches the surface,
• 20:45 - 20:48
and this is the
altitude that I'm going
• 20:48 - 20:50
to select for my skyscraper.
• 20:50 - 20:54
And here I'm going to have
another skyscraper, and here
• 20:54 - 20:58
another one and another one,
so practically it's dense.
• 20:58 - 21:02
I have a skyscraper next
to the other or a less like
• 21:02 - 21:03
[INAUDIBLE].
• 21:03 - 21:07
Not so many gaps
in certain areas.
• 21:07 - 21:13
So I'm going to say
f of x kappa star.
• 21:13 - 21:19
Now those would be the
altitudes of the buildings.
• 21:19 - 21:21
Magdalena, you don't
know how to spell.
• 21:21 - 21:28
Altitudes of the buildings.
• 21:28 - 21:32
• 21:32 - 21:33
What are they?
• 21:33 - 21:34
Parallel [INAUDIBLE] by P's.
• 21:34 - 21:36
Can you say parallel by P?
• 21:36 - 21:37
OK.
• 21:37 - 21:41
[INAUDIBLE] what.
• 21:41 - 21:50
Ak where Ak will be the basis
of the area of the basis.
• 21:50 - 21:52
is of my building.
• 21:52 - 21:55
• 21:55 - 21:55
OK.
• 21:55 - 21:59
The green part will
be the flat area
• 21:59 - 22:04
of the floor of the skyscraper.
• 22:04 - 22:06
Is this hard?
• 22:06 - 22:07
Gosh, yes.
• 22:07 - 22:13
If you want to do it by
hand and take the limit
• 22:13 - 22:16
you would really kill
yourself in the process.
• 22:16 - 22:17
This is how you introduce it.
• 22:17 - 22:22
You can prove this limit exists,
and you can prove that limits
• 22:22 - 22:33
exist and will be the volume of
the region under the geography
• 22:33 - 22:39
z equals f of x,y and
above the sea level.
• 22:39 - 22:43
• 22:43 - 22:47
The seal level
meaning z equals z.
• 22:47 - 22:50
STUDENT: What's under a of k?
• 22:50 - 22:50
PROFESSOR: Ak.
• 22:50 - 22:51
STUDENT: What is [INAUDIBLE]
• 22:51 - 22:54
PROFESSOR: Volume of the region.
• 22:54 - 22:56
STUDENT: Oh, I know,
like what under it?
• 22:56 - 22:56
PROFESSOR: Here?
• 22:56 - 22:57
STUDENT: No, up.
• 22:57 - 22:58
PROFESSOR: Here?
• 22:58 - 22:59
STUDENT: Yes.
• 22:59 - 23:01
PROFESSOR: Area of the
basis of a building.
• 23:01 - 23:02
STUDENT: Oh, the basis.
• 23:02 - 23:04
PROFESSOR: So practically
this green thingy
• 23:04 - 23:11
is a basis like the base rate.
• 23:11 - 23:13
How large is the basement
of that building.
• 23:13 - 23:16
Ak.
• 23:16 - 23:18
Now how am I going
to write this?
• 23:18 - 23:19
This is something new.
• 23:19 - 23:27
We have to invent a notion
for it, and since it's Ak,
• 23:27 - 23:31
looks more or less like
a square or a rectangle.
• 23:31 - 23:35
You think, well, wouldn't--
OK, if it's a rectangle,
• 23:35 - 23:38
I know I'm going to get
delta x and delta y right?
• 23:38 - 23:42
The width times the height,
whatever those two dimensions.
• 23:42 - 23:42
It makes sense.
• 23:42 - 23:45
But what if I have
this domain that's
• 23:45 - 23:47
curvilinear or that
domain or that domain.
• 23:47 - 23:50
Of course, the diameter
of such a domain
• 23:50 - 23:54
is less than the diameter of the
partition, so I'm very happy.
• 23:54 - 23:56
The highest diameter,
say I can get it here,
• 23:56 - 23:59
and this is shrinking
to zero, and pixels
• 23:59 - 24:02
are shrinking to zero.
• 24:02 - 24:05
But what am I going to
• 24:05 - 24:10
Well, you can assume that
I am still approximating
• 24:10 - 24:15
with some squares and as
the pixels are getting
• 24:15 - 24:17
to be many, many,
many more, it doesn't
• 24:17 - 24:20
matter that I'm doing this.
• 24:20 - 24:22
Let me show you what I'm doing.
• 24:22 - 24:29
So on the floor, on the-- this
is the city floor, whatever.
• 24:29 - 24:32
What we do in practice,
we approximate that
• 24:32 - 24:42
like on the graphing paper
with tiny square domains,
• 24:42 - 24:49
and we call them delta Ak will
be delta Sk times delta Yk,
• 24:49 - 24:54
and I tried to make it a uniform
partition as much as I can.
• 24:54 - 24:56
Now as the number of
pixels goes to infinity
• 24:56 - 24:59
and those pixels will
become smaller and smaller,
• 24:59 - 25:04
it doesn't there that the actual
• 25:04 - 25:07
will look like graphing paper.
• 25:07 - 25:10
It will get refined, more
refined, more refined, smoother
• 25:10 - 25:13
and smoother, and
it's going to be
• 25:13 - 25:18
really close to the ideal
image, which is a curve.
• 25:18 - 25:20
So as that end goes
to infinity, you're
• 25:20 - 25:25
not going to see this-- what is
this called-- zig zag thingy.
• 25:25 - 25:26
Not anymore.
• 25:26 - 25:32
The zig zag thingy will go into
the limit to the green curve.
• 25:32 - 25:35
This is what the
• 25:35 - 25:38
This is how our
life changed a lot.
• 25:38 - 25:39
OK?
• 25:39 - 25:40
All right.
• 25:40 - 25:42
Now good.
• 25:42 - 25:45
How am I going
compute this thing?
• 25:45 - 25:48
• 25:48 - 25:52
Well, I don't know, but let
me give it a name first.
• 25:52 - 25:56
It's going to be double
integral over-- what
• 25:56 - 25:59
do want the floor to be called?
• 25:59 - 26:02
• 26:02 - 26:04
We called d domain before.
• 26:04 - 26:06
What should I call this?
• 26:06 - 26:10
Big D. Not round.
• 26:10 - 26:14
Over D. That's the
floor, the foundation
• 26:14 - 26:17
of the whole city-- of
the whole area of the city
• 26:17 - 26:18
that I'm looking at.
• 26:18 - 26:28
Then I have f of xy,
da, and what is this?
• 26:28 - 26:30
This is exactly that.
• 26:30 - 26:35
It's the limit of sum of the--
what is the difference here?
• 26:35 - 26:37
You say, wait a
minute, Magdalena,
• 26:37 - 26:40
but I think I don't
understand what you did.
• 26:40 - 26:44
You tried to copy the
concept from here,
• 26:44 - 26:48
but you forgot you have a
function of two variables.
• 26:48 - 26:53
In that case, this mister,
whoever it is that goes up
• 26:53 - 26:58
is not xk, it's XkYk.
• 26:58 - 27:02
So I have two variables--
doesn't change anything
• 27:02 - 27:04
for the couple.
• 27:04 - 27:08
This couple represents a
point on the skyscraper
• 27:08 - 27:16
so that when I go up, I hit the
roof with this exact altitude.
• 27:16 - 27:20
So what is the double integral
of a continuous function
• 27:20 - 27:26
f of x and y, two variables,
with respect to area level.
• 27:26 - 27:33
Well, it's going to be just
the limit of this huge thing.
• 27:33 - 27:38
In fact, it's how
do we compute it?
• 27:38 - 27:41
Let's see how we
compute it in practice.
• 27:41 - 27:43
It shouldn't be a big deal.
• 27:43 - 27:56
• 27:56 - 27:58
What if I have a
rectangular domain,
• 27:58 - 28:01
and that's going to
make my life easier.
• 28:01 - 28:06
I'm going to have a
rectangular domain in plane,
• 28:06 - 28:08
and which one is the x-axis?
• 28:08 - 28:10
This one.
• 28:10 - 28:15
From A to B, I have the x
moving between a and Mr. y
• 28:15 - 28:20
says, I'm going to
be between c and d.
• 28:20 - 28:23
C is here, and d is here.
• 28:23 - 28:28
So this is going to be
the so-called rectangle
• 28:28 - 28:37
a, b cross c, d meaning
the set of all the pairs--
• 28:37 - 28:42
or the couples xy-- inside
it, what does it mean?
• 28:42 - 28:46
x, y you playing with
the property there.
• 28:46 - 28:49
X is between a and b, thank god.
• 28:49 - 28:50
It's easy.
• 28:50 - 28:54
And y must be between
c and d, also easy.
• 28:54 - 28:58
A, b, c, d are fixed real
numbers in this order.
• 28:58 - 29:02
A is less than b, and c is less.
• 29:02 - 29:05
And we have this
geography on top,
• 29:05 - 29:09
and I will tell you
what it looks like.
• 29:09 - 29:14
I'm going to try and draw
some beautiful geography.
• 29:14 - 29:20
And now I'm thinking
of my son, who is 10.
• 29:20 - 29:24
He played with this kind of toy
that was exactly this color,
• 29:24 - 29:26
• 29:26 - 29:28
Do you guys remember that toy?
• 29:28 - 29:31
I am sure you're young
enough to remember that.
• 29:31 - 29:35
You have your palm like that,
and you see this square thingy,
• 29:35 - 29:37
of needles that
• 29:37 - 29:41
look like thin,
tiny skyscrapers,
• 29:41 - 29:47
and you push through and all
those needles go up and take
• 29:47 - 29:50
• 29:50 - 29:52
And of course, he would
put it on his face,
• 29:52 - 29:54
and you could see
his face and so on.
• 29:54 - 29:56
But what is that?
• 29:56 - 30:00
That's exactly the Riemann
sum, the Riemann approximation,
• 30:00 - 30:02
because if you think of
all those needles or tiny--
• 30:02 - 30:07
what are they, like
the tiny skyscrapers--
• 30:07 - 30:12
the sum of the them approximates
the curvilinear shape.
• 30:12 - 30:16
If you put that over your face,
your face is nice and smooth,
• 30:16 - 30:19
curvilinear except for
a few single areas,
• 30:19 - 30:25
but if you actually
look at that needle
• 30:25 - 30:28
thingy that is
giving the figure,
• 30:28 - 30:30
you recognize the figure.
• 30:30 - 30:34
It's like a pattern recognition,
• 30:34 - 30:35
I mean it is and it's not.
• 30:35 - 30:39
It's an approximation of
your face, a very rough face.
• 30:39 - 30:42
You have to take that
• 30:42 - 30:44
and smooth it out.
• 30:44 - 30:44
How?
• 30:44 - 30:50
By passing to the
limit, and this is what
• 30:50 - 30:52
animation is doing actually.
• 30:52 - 30:56
On top of that you want this
to have some other properties--
• 30:56 - 31:01
illumination of some sort--
light coming from what angle.
• 31:01 - 31:05
That is all rendering
techniques are actually
• 31:05 - 31:07
applied mathematics.
• 31:07 - 31:10
In animation, the
people who programmed
• 31:10 - 31:13
Toy Story-- that
was a long time ago,
• 31:13 - 31:17
but everything that
came after Toy Story 2
• 31:17 - 31:21
was based on mathematical
rendering techniques.
• 31:21 - 31:24
Everything based on
the notion of length.
• 31:24 - 31:25
All right.
• 31:25 - 31:28
So the way we compute
this in practice
• 31:28 - 31:31
is going to be very simple,
because you're going to think,
• 31:31 - 31:33
how am I going to do the
rectangle for the rectangles?
• 31:33 - 31:36
That'll be very easy.
• 31:36 - 31:44
I split the rectangle perfectly
into other tiny rectangle.
• 31:44 - 31:47
Every rectangle will
have the same dimension.
• 31:47 - 31:49
Delta x and delta y.
• 31:49 - 31:52
• 31:52 - 31:53
Does it makes sense?
• 31:53 - 31:56
So practically when
I go to the limit,
• 31:56 - 32:02
I have summation f
of xk star, yk star
• 32:02 - 32:07
inside the delta x
delta y delta Magdalena,
• 32:07 - 32:12
the same kind of displacement
when I take k from 1 to n,
• 32:12 - 32:16
and I pass to the limit
according to the partition,
• 32:16 - 32:18
what's going to happen?
• 32:18 - 32:21
These guys, according
to Mr. Linux,
• 32:21 - 32:26
will go to be infinitesimal
elements, dx, dy.
• 32:26 - 32:30
This whole thing will
go to double integral
• 32:30 - 32:37
of f of x and y,
and Mr. y says, OK
• 32:37 - 32:39
it's like you want him to
integrate him one at a time.
• 32:39 - 32:44
This is actually something that
we are going to see in a second
• 32:44 - 32:45
and verify it.
• 32:45 - 32:50
X goes between a and b,
and y goes between c and d,
• 32:50 - 32:55
and this is an application
of a big theorem called
• 32:55 - 33:01
Fubini's Theorem that
says, wait a minute,
• 33:01 - 33:07
if you do it like this over
a rectangle a,b cross c,d,
• 33:07 - 33:12
you're double integral can
be written as three things.
• 33:12 - 33:18
Double integral over your
square domain f of x,y dA,
• 33:18 - 33:22
or you integral from c to
d, integral from a to b,
• 33:22 - 33:28
f of x,y dx dy, or you
can also swap the order,
• 33:28 - 33:32
because you say, well, you can
do the integration with respect
• 33:32 - 33:35
to y first.
• 33:35 - 33:37
Nobody stops you
from doing that,
• 33:37 - 33:41
and y has to be
between what and what?
• 33:41 - 33:42
STUDENT: C and d.
• 33:42 - 33:43
PROFESSOR: C and d, thank you.
• 33:43 - 33:47
And then whatever you get,
you get to integrate that
• 33:47 - 33:52
with respect to x from a to b.
• 33:52 - 33:56
So no matter in what
order you do it,
• 33:56 - 33:59
you'll get the same thing.
• 33:59 - 34:03
Let's see an easy example,
and you'll say, well,
• 34:03 - 34:06
example, Magdalena,
• 34:06 - 34:09
because we are just
starting, and that's
• 34:09 - 34:10
exactly what I'm going to.
• 34:10 - 34:12
I will just misbehave.
• 34:12 - 34:15
I'm not going to go by the book.
• 34:15 - 34:19
And I will say I'm going
by whatever I want to go.
• 34:19 - 34:28
X is between 0, 2, and
y is between 0 and 2
• 34:28 - 34:33
and 3-- this is 2, this
is 3-- and my domain
• 34:33 - 34:39
will be the rectangle
0, 2 times 0, 3.
• 34:39 - 34:42
This is neat on the floor.
• 34:42 - 35:00
Compute the volume of the
box of basis d and height 5.
• 35:00 - 35:02
Can I draw that?
• 35:02 - 35:04
It gets out of the picture.
• 35:04 - 35:05
I'm just kidding.
• 35:05 - 35:08
This is 5, and that's
sort of the box.
• 35:08 - 35:11
• 35:11 - 35:14
And you say, wait a minute, I
• 35:14 - 35:16
• 35:16 - 35:17
How do we do that?
• 35:17 - 35:21
We go 2 units times
3 units that's
• 35:21 - 35:25
going to be 6 square inches
on the bottom of the box,
• 35:25 - 35:28
and then times 5.
• 35:28 - 35:31
So the volume has to be
2 times 3 times 5, which
• 35:31 - 35:35
is 30 square inches.
• 35:35 - 35:37
I don't care what it is.
• 35:37 - 35:39
I'm a mathematician, right?
• 35:39 - 35:40
OK.
• 35:40 - 35:44
How does somebody who just
learned Tonelli's-- Fubini
• 35:44 - 35:47
Tonelli's Theorem
do the problem.
• 35:47 - 35:49
That person will
say, wait a minute,
• 35:49 - 35:55
now I know that the
function is going to be z
• 35:55 - 36:00
equals f of xy, which in
this case happens to be cost.
• 36:00 - 36:05
According to what you told us,
the theorem you claim Magdalena
• 36:05 - 36:07
proved to this theorem,
but there is a sketch
• 36:07 - 36:09
of the proof in the book.
• 36:09 - 36:13
According to this,
the double integral
• 36:13 - 36:21
that you have over the
domain d, and this is dA.
• 36:21 - 36:31
DA will be called element of
area, which is also dx dy.
• 36:31 - 36:35
This can be solved in
two different ways.
• 36:35 - 36:38
You take integral
from-- where is x going?
• 36:38 - 36:42
Do we want to do it
first in x or in y?
• 36:42 - 36:45
If we put dy dx, that means
we integrate with respect
• 36:45 - 36:49
to y first, and y
goes between 0 and 3,
• 36:49 - 36:53
so I have to pay attention
to the limits of integration.
• 36:53 - 36:56
And then x between
0 and 2 and again
• 36:56 - 36:59
I have to pay attention to
the limits of integration
• 36:59 - 37:03
all the time and,
here, who is my f?
• 37:03 - 37:07
Is the altitude 5 that's
constant in my case?
• 37:07 - 37:08
• 37:08 - 37:11
Let me see if I get 30?
• 37:11 - 37:16
I'm just checking if this
theorem was true or is just
• 37:16 - 37:21
something that you cannot apply.
• 37:21 - 37:26
How do you integrate
5 with respect to y?
• 37:26 - 37:26
STUDENT: 5y.
• 37:26 - 37:28
PROFESSOR: 5y, very good.
• 37:28 - 37:34
So it's going to be 5y between
y equals 0 down and y equals 3
• 37:34 - 37:39
up, and how much
is that 5y, we're
• 37:39 - 37:41
doing y equals 0 down
and y equals 3 up,
• 37:41 - 37:43
what number is that?
• 37:43 - 37:45
STUDENT: 25.
• 37:45 - 37:46
PROFESSOR: What?
• 37:46 - 37:47
STUDENT: 25.
• 37:47 - 37:48
PROFESSOR: 25?
• 37:48 - 37:50
STUDENT: One [INAUDIBLE] 15.
• 37:50 - 37:54
PROFESSOR: No, you did--
• 37:54 - 38:00
So I go 5 times 3 minus
5 times 0 equals 15.
• 38:00 - 38:04
So when I compute this
variation of 5y between y
• 38:04 - 38:07
equals 3 and y equals
0, I just block in
• 38:07 - 38:08
and make the difference.
• 38:08 - 38:10
Why do I do that?
• 38:10 - 38:16
It's the simplest application
of that FT, fundamental theorem.
• 38:16 - 38:19
The one that I did not
specify in [INAUDIBLE].
• 38:19 - 38:23
I should have specified when
I have a g function that
• 38:23 - 38:28
is continuous between
alpha and beta, how do we
• 38:28 - 38:30
integrate with respect to x?
• 38:30 - 38:33
I get the antiderivative
of rule G. Let's call
• 38:33 - 38:37
that big G. Compute
it at the end points,
• 38:37 - 38:39
and I make the difference.
• 38:39 - 38:42
So I compute the
antiderivative at an endpoint--
• 38:42 - 38:44
at the other endpoint-- then I'm
going to make the difference.
• 38:44 - 38:50
That's the same thing I do
here, so 5 times 3 is 15,
• 38:50 - 38:54
5 times 0 is 0,
15 minus 0 is 15.
• 38:54 - 38:56
I can keep moving.
• 38:56 - 38:59
Everything in the
parentheses is the number 15.
• 38:59 - 39:04
I copy and paste, and that
should be a piece of cake.
• 39:04 - 39:07
What do I get?
• 39:07 - 39:10
STUDENT: 15.
• 39:10 - 39:16
PROFESSOR: I get 15
times x between 0 and 2.
• 39:16 - 39:17
Integral of 1 is x.
• 39:17 - 39:20
Integral of 1 is x
with respect to x,
• 39:20 - 39:24
so I get 15 times 2, which
is 30, and you go, duh,
• 39:24 - 39:27
[INAUDIBLE].
• 39:27 - 39:29
That was elementary mathematics.
• 39:29 - 39:32
Yes, you were lucky you
knew that volume of the box,
• 39:32 - 39:36
but what if somebody gave
you a curvilinear area?
• 39:36 - 39:39
What if somebody gave you
something quite complicated?
• 39:39 - 39:41
What would you do?
• 39:41 - 39:43
You have know calculus.
• 39:43 - 39:46
• 39:46 - 39:52
If you don't calculus,
• 39:52 - 40:01
So I'm saying, how
• 40:01 - 40:04
That look like it's
complicated, but calculus
• 40:04 - 40:08
is something
[INAUDIBLE] with that.
• 40:08 - 40:16
Suppose that I have a square
in the plane between-- this
• 40:16 - 40:20
is x and y-- do you
want square 0,1 0,1
• 40:20 - 40:23
or you want minus 1
to 1 minus 1 to 1.
• 40:23 - 40:26
• 40:26 - 40:28
It doesn't matter.
• 40:28 - 40:33
Well, let's take minus
1 to 1 and minus 1 to 1,
• 40:33 - 40:36
and I'll try to draw
as well as I can,
• 40:36 - 40:38
which I cannot but it's OK.
• 40:38 - 40:42
You will forgive me.
• 40:42 - 40:42
This is the floor.
• 40:42 - 40:46
• 40:46 - 40:48
If I were just a
little tiny square
• 40:48 - 40:52
in this room plus the
equivalent square in that room
• 40:52 - 40:54
and that room and that room.
• 40:54 - 40:56
This is the origin.
• 40:56 - 40:58
Are you guys with me?
• 40:58 - 41:00
So what you're
looking at right now
• 41:00 - 41:05
is this square foot
of carpet that I have,
• 41:05 - 41:12
but I have another one here and
another one behind the wall,
• 41:12 - 41:16
and so do I everything in mind?
• 41:16 - 41:21
X is between minus 1 and 1,
y is between minus 1 and 1.
• 41:21 - 41:25
• 41:25 - 41:30
And somebody gives you z
to be a positive function,
• 41:30 - 41:36
continuous function, which
is x squared plus y squared.
• 41:36 - 41:37
• 41:37 - 41:39
Oh, my god.
• 41:39 - 41:42
kind of hard function.
• 41:42 - 41:44
It's not a hard thing to do.
• 41:44 - 41:45
Let's draw that.
• 41:45 - 41:48
What are we going to get?
• 41:48 - 41:56
that goes like this.
• 41:56 - 42:00
And imagine what's
going to happen
• 42:00 - 42:03
with this is like a vase.
• 42:03 - 42:07
Inside, it has this
circular paraboloid.
• 42:07 - 42:18
But the walls of this vase are--
I cannot draw better than that.
• 42:18 - 42:25
So the walls of this
vase are squares.
• 42:25 - 42:30
And what you have inside is
the carved circular paraboloid.
• 42:30 - 42:33
• 42:33 - 42:45
you, how do I find
• 42:45 - 43:02
volume of the body under and
above D, which is minus 1,
• 43:02 - 43:04
1, minus 1, 1.
• 43:04 - 43:05
It's hard to draw that, right?
• 43:05 - 43:07
It's hard to draw.
• 43:07 - 43:10
So what do we do?
• 43:10 - 43:16
• 43:16 - 43:17
We start imagining things.
• 43:17 - 43:20
• 43:20 - 43:24
Actually, when you cut with
a plane that is y equals 1,
• 43:24 - 43:28
you would get a parabola.
• 43:28 - 43:36
And so when you look at what the
picture is going to look like,
• 43:36 - 43:40
you're going to have
a parabola like this,
• 43:40 - 43:42
a parabola like that,
exactly the same,
• 43:42 - 43:46
a parallel parabola like this
and a parabola like that.
• 43:46 - 43:49
Now I started drawing better.
• 43:49 - 43:52
And you say, how did you
start drawing better?
• 43:52 - 43:53
Well, with a little
bit of practice.
• 43:53 - 44:00
Where are the maxima
of this thing?
• 44:00 - 44:01
At the corners.
• 44:01 - 44:01
Why is that?
• 44:01 - 44:06
Because at the corners,
you get 1, 1 for both.
• 44:06 - 44:10
Of course, to do the absolute
extrema, minimum, maximum,
• 44:10 - 44:15
we would have to go back to
section 11.7 and do the thing.
• 44:15 - 44:19
But practically, it's easy
to see that at the corners,
• 44:19 - 44:23
you have the height 2 because
this is the point 1, 1.
• 44:23 - 44:29
And the same height, 2 and 2
and 2, are at every corner.
• 44:29 - 44:32
That would be the
maximum that you have.
• 44:32 - 44:39
So you have 1 minus 1 and so
on-- minus 1, 1, and minus 1,
• 44:39 - 44:42
minus 1, who is behind
me, minus 1, minus 1.
• 44:42 - 44:47
That goes all the way to 2.
• 44:47 - 44:51
So it's hard to do an
approximation with a three
• 44:51 - 44:53
dimensional model.
• 44:53 - 44:55
Thank god there is calculus.
• 44:55 - 44:59
So you say integral of x
squared plus y squared,
• 44:59 - 45:06
as simple as that, da over the
domain, D, which is minus 1,
• 45:06 - 45:08
1, minus 1, 1.
• 45:08 - 45:10
How do you write it
according to the theorem
• 45:10 - 45:13
that I told you
• 45:13 - 45:20
Then you have integral integral
x squared plus y squared dy dx.
• 45:20 - 45:22
• 45:22 - 45:25
Doesn't matter which
one I'm taking.
• 45:25 - 45:27
I can do dy dx.
• 45:27 - 45:28
I can do dx dy.
• 45:28 - 45:31
I just have to pay
attention to the endpoints.
• 45:31 - 45:33
Lucky for you the
endpoints are the same.
• 45:33 - 45:35
y is between minus 1 and 1.
• 45:35 - 45:37
x is between minus 1 and 1.
• 45:37 - 45:41
• 45:41 - 45:45
I wouldn't known how to compute
the volume of this vase made
• 45:45 - 45:46
of whatever you
• 45:46 - 45:54
want to make it unless I knew
to compute this integral.
• 45:54 - 45:59
Now you have to help me
because it's not hard
• 45:59 - 46:04
but it's not easy either, so we
need a little bit of attention.
• 46:04 - 46:06
We always start from the
inside to the outside.
• 46:06 - 46:11
The outer person has to be just
neglected for the time being
• 46:11 - 46:15
and I focus all my attention
to this integration.
• 46:15 - 46:18
And when I integrate
with respect to y,
• 46:18 - 46:20
y is the variable for me.
• 46:20 - 46:23
Nothing else exists
for the time being,
• 46:23 - 46:28
but y being a variable,
x being like a constant.
• 46:28 - 46:30
So when you integrate x
squared plus y squared
• 46:30 - 46:35
with respect to y, you have
to pay attention a little bit.
• 46:35 - 46:40
It's about the same if you
had 7 squared plus y squared.
• 46:40 - 46:43
So this x squared
is like a constant.
• 46:43 - 46:45
So what do you get inside?
• 46:45 - 46:47
Let's apply the fundamental
theorem of calculus.
• 46:47 - 46:48
STUDENT: x squared y.
• 46:48 - 46:49
PROFESSOR: x squared y.
• 46:49 - 46:50
Excellent.
• 46:50 - 46:52
I'm very proud of you.
• 46:52 - 46:53
Plus?
• 46:53 - 46:54
STUDENT: y cubed over 3.
• 46:54 - 46:55
PROFESSOR: y cubed over three.
• 46:55 - 46:57
Again, I'm proud of you.
• 46:57 - 47:03
Evaluated between y equals
minus 1 down, y equals 1 up.
• 47:03 - 47:07
And I will do the math later
because I'm getting tired.
• 47:07 - 47:10
• 47:10 - 47:12
Now let's do the math.
• 47:12 - 47:13
I don't know what
I'm going to get.
• 47:13 - 47:19
I get minus 1 to 1, a
big bracket, and dx.
• 47:19 - 47:22
And in this big bracket, I
have to do the difference
• 47:22 - 47:23
between two values.
• 47:23 - 47:27
So I put two parentheses.
• 47:27 - 47:30
When y equals 1, I
get x squared 1--
• 47:30 - 47:34
I'm not going to write
that down-- plus 1 cubed
• 47:34 - 47:37
over 3, 1/3.
• 47:37 - 47:42
I'm done with evaluating
this sausage thingy at 1.
• 47:42 - 47:44
It's an expression
that I evaluate.
• 47:44 - 47:47
It could be a lot longer.
• 47:47 - 47:49
I'm not planning to give you
long expressions in the midterm
• 47:49 - 47:52
because you're going to
make algebra mistakes,
• 47:52 - 47:55
and that's not what I want.
• 47:55 - 48:01
For minus 1, what do we
have Minus x squared.
• 48:01 - 48:04
What is y equals minus
1 plugged in here?
• 48:04 - 48:05
Minus 1/3.
• 48:05 - 48:09
• 48:09 - 48:11
I have to pay attention.
• 48:11 - 48:16
You realize that if I mess
up a sign, it's all done.
• 48:16 - 48:21
So in this case, I say, but
this I have minus, minus.
• 48:21 - 48:24
A minus in front of
a minus is a plus,
• 48:24 - 48:31
so I'm practically doubling
the x squared plus 1/3
• 48:31 - 48:34
and taking it
between minus 1 and 1
• 48:34 - 48:37
and just with respect to x.
• 48:37 - 48:38
So you say, wait a minute.
• 48:38 - 48:39
But that's easy.
• 48:39 - 48:41
I've done that when
I was in Calc 1.
• 48:41 - 48:42
Of course.
• 48:42 - 48:47
This is the nice part that
you get, a simple integral
• 48:47 - 48:52
from the ones in Calc 1.
• 48:52 - 48:57
Let's solve this one and find
out what the area will be.
• 48:57 - 48:59
What do we get?
• 48:59 - 49:00
Is it hard?
• 49:00 - 49:01
No.
• 49:01 - 49:02
Kick Mr. 2 out.
• 49:02 - 49:05
He's just messing
• 49:05 - 49:06
Kick him out.
• 49:06 - 49:08
2, out.
• 49:08 - 49:12
And then integral of
x squared plus 1/3
• 49:12 - 49:16
is going to be x
cubed over 3 plus--
• 49:16 - 49:17
STUDENT: x over 3.
• 49:17 - 49:19
PROFESSOR: x over 3, very good.
• 49:19 - 49:23
Evaluated between x equals
minus 1 down, x equals 1 up.
• 49:23 - 49:26
• 49:26 - 49:28
Let's see what we get.
• 49:28 - 49:31
2 times bracket.
• 49:31 - 49:34
I'll put a parentheses
for the first fractions,
• 49:34 - 49:37
and another minus, and
another parentheses.
• 49:37 - 49:42
What's the first edition
of fractions that I get?
• 49:42 - 49:44
1/3 plus 1/3.
• 49:44 - 49:47
I'll put 2/3 because I'm lazy.
• 49:47 - 49:49
Then minus what?
• 49:49 - 49:51
STUDENT: Minus 1/3.
• 49:51 - 49:56
PROFESSOR: Minus 1/3
minus 1/3, minus 2/3.
• 49:56 - 50:01
And now I should be able to
not beat around the bush.
• 50:01 - 50:04
will be in the end.
• 50:04 - 50:06
STUDENT: 8/3.
• 50:06 - 50:08
PROFESSOR: 8/3.
• 50:08 - 50:10
Does that make sense?
• 50:10 - 50:12
When you do that in
math, you should always
• 50:12 - 50:17
think-- one of the famous
professors at Harvard
• 50:17 - 50:22
was saying one time
• 50:22 - 50:24
how many hours of
life do we have have
• 50:24 - 50:26
in one day, blah, blah, blah?
• 50:26 - 50:30
And many students
came up with 36, 37.
• 50:30 - 50:36
So always make sure that the
• 50:36 - 50:38
This is part of a cube, right?
• 50:38 - 50:43
It's like carved in a
cube or a rectangle.
• 50:43 - 50:46
• 50:46 - 50:49
Now, what's the height?
• 50:49 - 50:53
If this were to go
up all the way to 2,
• 50:53 - 50:59
it would be 2, 2, and 2.
• 50:59 - 51:04
2 times 2 times 2 equals 8,
and what we got is 8 over 3.
• 51:04 - 51:09
Now, using our imagination,
it makes sense.
• 51:09 - 51:11
If I got a 16, I
would say, oh my god.
• 51:11 - 51:12
No, no, no, no.
• 51:12 - 51:14
What is that?
• 51:14 - 51:18
So a little bit, I would think,
does this make sense or not?
• 51:18 - 51:22
• 51:22 - 51:24
Let's do one more,
a similar one.
• 51:24 - 51:28
Now I'm going to count
on you a little bit more.
• 51:28 - 51:39
• 51:39 - 51:41
STUDENT: Professor,
did you calculate that
• 51:41 - 51:46
by just doing a quarter, and
then just multiplying it by 4?
• 51:46 - 51:47
Because then that
would just leave us
• 51:47 - 51:49
with zeroes [INAUDIBLE].
• 51:49 - 51:51
PROFESSOR: You mean in
that particular figure?
• 51:51 - 51:51
Yeah.
• 51:51 - 51:54
STUDENT: Yeah, because it
was perfectly [INAUDIBLE].
• 51:54 - 51:55
PROFESSOR: Yeah.
• 51:55 - 51:57
It's nice.
• 51:57 - 52:03
It's a little bit related
to some other problems that
• 52:03 - 52:04
come from pyramids.
• 52:04 - 52:07
• 52:07 - 52:16
By the way, how can you compute
the volume of a square pyramid?
• 52:16 - 52:21
• 52:21 - 52:26
Suppose that you have
the same problem.
• 52:26 - 52:31
Minus 1 to 1 for x and y.
• 52:31 - 52:35
Minus 1 to 1, minus 1 to 1.
• 52:35 - 53:05
Let's say the pyramid would
have the something like that.
• 53:05 - 53:07
What would be the volume
of such a pyramid?
• 53:07 - 53:10
• 53:10 - 53:13
STUDENT: [INAUDIBLE].
• 53:13 - 53:18
PROFESSOR: The height
is h for extra credit.
• 53:18 - 53:33
Can you compute the
volume of this pyramid
• 53:33 - 53:34
using double integrals?
• 53:34 - 53:41
• 53:41 - 53:49
Say the height is h and the
bases is the square minus 1,
• 53:49 - 53:52
1, minus 1, 1.
• 53:52 - 53:54
I'm sure it can be
done, but you know--
• 53:54 - 53:58
now I'm testing what you
remember in terms of geometry
• 53:58 - 54:01
because we will deal
with geometry a lot
• 54:01 - 54:03
in volumes and areas.
• 54:03 - 54:07
So how do you do that
in general, guys?
• 54:07 - 54:10
STUDENT: 1/3 [INAUDIBLE].
• 54:10 - 54:14
PROFESSOR: 1/3 the
height times the area
• 54:14 - 54:19
of the bases, which is what?
• 54:19 - 54:20
2 times 2.
• 54:20 - 54:28
2 times 2, 3, over 3, 4/3 h.
• 54:28 - 54:30
Can you prove that
with calculus?
• 54:30 - 54:31
That's all I'm saying.
• 54:31 - 54:34
One point extra credit.
• 54:34 - 54:36
Can you prove that
with calculus?
• 54:36 - 54:41
Actually, you would have
to use what you learned.
• 54:41 - 54:45
You can use Calc 2 as well.
• 54:45 - 54:47
Do you guys remember
that there were
• 54:47 - 54:53
some cross-sectional areas, like
this would be made of cheese,
• 54:53 - 54:57
and you come with a vertical
knife and cut cross sections.
• 54:57 - 54:58
They go like that.
• 54:58 - 54:59
But that's awfully hard.
• 54:59 - 55:03
Maybe you can do it differently
with Calc 3 instead of Calc 2.
• 55:03 - 55:08
• 55:08 - 55:10
Let's pick one from
the book as well.
• 55:10 - 55:31
• 55:31 - 55:33
OK.
• 55:33 - 55:39
So the same idea of using
the Fubini-Tonelli argument
• 55:39 - 55:46
and have an iterative-- evaluate
the following double integral
• 55:46 - 55:49
over the rectangle
of vertices 0, 0--
• 55:49 - 55:52
write it down-- 3,
0, 3, 2, and 0, 2.
• 55:52 - 56:02
So on the bases, you have a
rectangle of vertices 3, 0, 0,
• 56:02 - 56:14
0, 3, 2, and 0, 2.
• 56:14 - 56:19
And then somebody
tells you, find us
• 56:19 - 56:29
the double integral
of 2 minus y da
• 56:29 - 56:36
over r where r represents the
• 56:36 - 56:38
This is exactly [INAUDIBLE].
• 56:38 - 56:42
• 56:42 - 56:45
should get is 6.
• 56:45 - 56:49
And I'm saying on top of
what we said in the book,
• 56:49 - 56:53
can you give a geometric
interpretation?
• 56:53 - 56:55
Does this have a
geometric interpretation
• 56:55 - 56:57
you can think of or not?
• 56:57 - 57:01
• 57:01 - 57:04
Well, first of all,
what is this animal?
• 57:04 - 57:07
According to the Fubini
theorem, this animal
• 57:07 - 57:14
will have to be-- I have
it over a rectangle,
• 57:14 - 57:18
so assume x will be
between a and b, y
• 57:18 - 57:22
will be between c and d.
• 57:22 - 57:25
I have to figure
out who those are.
• 57:25 - 57:32
2 minus y and dy dx.
• 57:32 - 57:36
• 57:36 - 57:38
Where is y between?
• 57:38 - 57:40
I should draw the
picture for the rectangle
• 57:40 - 57:43
because otherwise, it's
not so easy to see.
• 57:43 - 57:51
I have 0, 0 here, 3, 0 here, 3,
2 over here, shouldn't be hard.
• 57:51 - 57:53
So this is going to be 0, 2.
• 57:53 - 57:57
That's the y-axis and
that's the x-axis.
• 57:57 - 58:01
Let's see if we can see it.
• 58:01 - 58:05
And what is the meaning
of the 6, I'm asking you?
• 58:05 - 58:07
I don't know.
• 58:07 - 58:11
x should be between
0 and 3, right?
• 58:11 - 58:15
y should be between
0 and 2, right?
• 58:15 - 58:17
Now you are experts in this.
• 58:17 - 58:21
We've done this twice, and
you already know how to do it.
• 58:21 - 58:23
Integral from 0 to 3.
• 58:23 - 58:27
Then I take that,
and that's going
• 58:27 - 58:39
to be 2y minus y
squared over 2 between y
• 58:39 - 58:43
equals 0 down and
y equals 2 up dx.
• 58:43 - 58:48
• 58:48 - 58:55
That means integral from 0
to 3, bracket minus bracket
• 58:55 - 58:59
to make my life easier, dx.
• 58:59 - 59:02
Now, there is no x, thank god.
• 59:02 - 59:05
So that means I'm going
to have a constant
• 59:05 - 59:10
minus another constant, which
means I go 4 minus 4 over 2.
• 59:10 - 59:13
2, right?
• 59:13 - 59:18
The other one, for 0, I get 0.
• 59:18 - 59:21
I'm very happy I get 0
because in that case,
• 59:21 - 59:25
it's obvious that I get
2 times 3, which is 6.
• 59:25 - 59:29
So I got what the book
said I'm going to get.
• 59:29 - 59:32
But do I have a geometric
interpretation of that?
• 59:32 - 59:37
I would like to see
if anybody can--
• 59:37 - 59:41
I'm going to give you a
break in a few minues--
• 59:41 - 59:46
if anybody can think of a
geometric interpretation.
• 59:46 - 59:53
What is this f of xy if I were
to interpret this as a graph?
• 59:53 - 59:55
x equals f of x and y.
• 59:55 - 59:56
Is this--
• 59:56 - 59:58
STUDENT: 2 minus y.
• 59:58 - 60:05
PROFESSOR: So z equals 2
minus y is a plane, right?
• 60:05 - 60:08
STUDENT: Yes, but then you have
the parabola is going down.
• 60:08 - 60:11
PROFESSOR: And how do I get
to draw this plane the best?
• 60:11 - 60:14
Because there are
many ways to do it.
• 60:14 - 60:17
I look at this wall.
• 60:17 - 60:19
The y-axis is this.
• 60:19 - 60:21
The z-axis is the vertical line.
• 60:21 - 60:23
So I'm looking at this plane.
• 60:23 - 60:28
y plus z must be equal to 2.
• 60:28 - 60:30
So when is y plus z equal to 2?
• 60:30 - 60:34
When I am on a
line in the plane.
• 60:34 - 60:39
I'm going to draw that line
with pink because I like pink.
• 60:39 - 60:41
This is y plus z equals 2.
• 60:41 - 60:44
• 60:44 - 60:50
And imagine this line will be
shifted by parallelism as it
• 60:50 - 60:55
comes towards you on all these
other parallel vertical planes
• 60:55 - 60:58
that are parallel to the board.
• 60:58 - 61:05
So I'm going to have an
entire plane like that,
• 61:05 - 61:09
and I'm going to stop here.
• 61:09 - 61:13
When I'm in the plane
that's called x equals 3--
• 61:13 - 61:15
this is the plane
called x equals
• 61:15 - 61:20
3-- I have exactly this
triangle, this [INAUDIBLE].
• 61:20 - 61:24
It's in the plane
that faces me here.
• 61:24 - 61:26
I don't know if
you realize that.
• 61:26 - 61:31
house or something nice.
• 61:31 - 61:33
I think I'm getting hungry.
• 61:33 - 61:36
I imagine this again as
being a piece of cheese,
• 61:36 - 61:40
or it looks even like a piece
of cake would be with layers.
• 61:40 - 61:43
• 61:43 - 61:48
So our question is, if
we didn't know calculus
• 61:48 - 61:51
but we knew how to draw
this, and somebody gave you
• 61:51 - 61:54
this at the GRE
or whatever exam,
• 61:54 - 61:56
how could you have done
it without calculus?
• 61:56 - 62:00
Just by cheating and
pretending, I know how to do it,
• 62:00 - 62:03
but you've never done a
• 62:03 - 62:06
So I know it's a volume.
• 62:06 - 62:09
How do I get the volume?
• 62:09 - 62:10
What kind of geometric
body is that?
• 62:10 - 62:12
STUDENT: A triangle.
• 62:12 - 62:14
STUDENT: It's a
triangular prism.
• 62:14 - 62:16
PROFESSOR: It's a
triangular prism.
• 62:16 - 62:16
Good.
• 62:16 - 62:20
And a triangular prism
has what volume formula?
• 62:20 - 62:21
STUDENT: Base times height.
• 62:21 - 62:22
PROFESSOR: Base
times the height.
• 62:22 - 62:26
And the height has what area?
• 62:26 - 62:27
Let's see.
• 62:27 - 62:30
The base would be that, right?
• 62:30 - 62:34
And the height would be 3.
• 62:34 - 62:36
Am I right or not?
• 62:36 - 62:38
The height would be 3.
• 62:38 - 62:38
This is not--
• 62:38 - 62:39
STUDENT: It's 2.
• 62:39 - 62:39
Yeah.
• 62:39 - 62:40
STUDENT: No, it's 3.
• 62:40 - 62:42
DR. MAGDALENA TODA:
From here to here?
• 62:42 - 62:42
STUDENT: 3.
• 62:42 - 62:44
DR. MAGDALENA TODA: It's 3.
• 62:44 - 62:47
So how much is that?
• 62:47 - 62:48
How much-- OK.
• 62:48 - 62:50
From here to here is 2.
• 62:50 - 62:54
From here to here,
it's how much?
• 62:54 - 62:56
STUDENT: The height
is only-- I see--
• 62:56 - 62:57
STUDENT: It's also 2.
• 62:57 - 62:59
DR. MAGDALENA TODA: It's
also 2 because look at that.
• 62:59 - 63:02
It's an isosceles triangle.
• 63:02 - 63:04
This is 45 to 45.
• 63:04 - 63:05
So this is also 2.
• 63:05 - 63:09
2 to-- that's 90
degrees, 45, 45.
• 63:09 - 63:09
OK.
• 63:09 - 63:13
So the area of the shaded purple
triangle-- how much is that?
• 63:13 - 63:14
STUDENT: 2.
• 63:14 - 63:15
DR. MAGDALENA TODA: 2.
• 63:15 - 63:17
2 times 2 over 2.
• 63:17 - 63:20
2 times 3 equals 6.
• 63:20 - 63:22
I don't need calculus.
• 63:22 - 63:24
In this case, I
don't need calculus.
• 63:24 - 63:27
But when I have those
nasty curvilinear
• 63:27 - 63:32
z equals f of x, y, complicated
expressions, I have no choice.
• 63:32 - 63:35
I have to do the
double integral.
• 63:35 - 63:38
But in this case, even if
I didn't know how to do it,
• 63:38 - 63:39
I would still get the 6.
• 63:39 - 63:40
Yes, sir?
• 63:40 - 63:43
STUDENT: What if we
did that on the exam?
• 63:43 - 63:44
DR. MAGDALENA TODA:
Well, that's good.
• 63:44 - 63:46
I will then keep it in mind.
• 63:46 - 63:46
Yes.
• 63:46 - 63:49
It doesn't matter to me.
• 63:49 - 63:51
I have other colleagues who
• 63:51 - 63:52
and start complaining.
• 63:52 - 63:56
I don't care how you
• 63:56 - 63:57
as long as you got
• 63:57 - 63:59
Let me tell you my logic.
• 63:59 - 64:04
Suppose somebody hired you
thinking you're a good worker,
• 64:04 - 64:05
and you're smart and so on.
• 64:05 - 64:10
Would they care how you got to
the solution of the problem?
• 64:10 - 64:14
As long as the problem
was solved correctly, no.
• 64:14 - 64:18
And actually, the elementary
way is the fastest
• 64:18 - 64:20
because it's just 10 seconds.
• 64:20 - 64:21
You draw.
• 64:21 - 64:22
You imagine.
• 64:22 - 64:23
You know what it is.
• 64:23 - 64:28
So your boss will want you to
find the fastest way to provide
• 64:28 - 64:29
the correct solution.
• 64:29 - 64:33
He's not going to
care how you got that.
• 64:33 - 64:36
So no matter how
you do it, as long
• 64:36 - 64:40
as you've got the right
answer, I'm going to be happy.
• 64:40 - 64:49
go to page 927 in the book
• 64:49 - 64:50
• 64:50 - 64:53
It's only one page.
• 64:53 - 64:55
That whole end section, 12.1.
• 64:55 - 65:00
It's called an informal
argument for Fubini's theorem.
• 65:00 - 65:06
Practically, it's a proof of
Fubini's theorem, page 927.
• 65:06 - 65:09
And then I'm going to go
• 65:09 - 65:12
four, if you don't mind.
• 65:12 - 65:16
I'm going to go into WeBWork
and give you homework four.
• 65:16 - 65:19
And the first few
problems that you
• 65:19 - 65:21
are going to be
expected to solve
• 65:21 - 65:27
will be out of 12.1,
which is really easy.
• 65:27 - 65:29
I'll give you a
few minutes back.
• 65:29 - 65:33
And we go on with 12.2,
and it's very similar.
• 65:33 - 65:35
You're going to like that.
• 65:35 - 65:40
And then we'll go home or
wherever we need to go.
• 65:40 - 65:42
So you have a few
minutes of a break.
• 65:42 - 65:46
• 65:46 - 65:47
I'll call the names.
• 65:47 - 65:49
Lily.
• 65:49 - 65:52
You got a lot of points.
• 65:52 - 65:55
And [INAUDIBLE].
• 65:55 - 65:57
And you have two separate ones.
• 65:57 - 65:58
Nathan.
• 65:58 - 65:59
Nathan?
• 65:59 - 66:02
• 66:02 - 66:03
Rachel Smith.
• 66:03 - 66:06
• 66:06 - 66:06
Austin.
• 66:06 - 66:09
• 66:09 - 66:10
Thank you.
• 66:10 - 66:13
• 66:13 - 66:14
Edgar.
• 66:14 - 66:16
[INAUDIBLE]
• 66:16 - 66:17
Aaron.
• 66:17 - 66:24
• 66:24 - 66:25
Andre.
• 66:25 - 66:32
• 66:32 - 66:35
Aaron.
• 66:35 - 66:36
Kasey.
• 66:36 - 66:40
• 66:40 - 66:43
Kasey came up with
a very good idea
• 66:43 - 66:48
that I will write
a review sample.
• 66:48 - 66:49
Did I promise that?
• 66:49 - 66:52
A review sample for the midterm.
• 66:52 - 66:54
And so I said yes.
• 66:54 - 66:57
• 66:57 - 67:01
Karen and Matthew.
• 67:01 - 67:08
• 67:08 - 67:08
Reagan.
• 67:08 - 67:16
• 67:16 - 67:18
Aaron.
• 67:18 - 67:21
When you submitted,
you submitted.
• 67:21 - 67:21
Yeah.
• 67:21 - 67:22
And [INAUDIBLE].
• 67:22 - 67:26
• 67:26 - 67:27
here.
• 67:27 - 67:28
And I'm done.
• 67:28 - 67:46
• 67:46 - 67:48
STUDENT: Did we
turn in [INAUDIBLE]?
• 67:48 - 67:50
DR. MAGDALENA TODA:
Yes, absolutely.
• 67:50 - 68:09
• 68:09 - 68:12
Now once we go over
12.2, you will say, oh,
• 68:12 - 68:14
but I understand
the Fubini theorem.
• 68:14 - 68:21
• 68:21 - 68:24
I didn't know whether
there's room for Fubini,
• 68:24 - 68:29
because once I cover the more
general case, which is in 12.2,
• 68:29 - 68:34
you are going to understand
Why Fubini-Tonelli
• 68:34 - 68:37
works for rectangles.
• 68:37 - 68:47
So if I think of a domain
that is of the following form,
• 68:47 - 68:54
in the x, y plane, I go x
is between and and b, right?
• 68:54 - 69:00
That's my favorite x.
• 69:00 - 69:02
So I take the pink
segment, and I
• 69:02 - 69:05
say, everything that
happens-- it's going
• 69:05 - 69:08
to happen on top of this world.
• 69:08 - 69:11
I have, let's say,
two functions.
• 69:11 - 69:14
To make my life easier, I'll
assume both of them [INAUDIBLE]
• 69:14 - 69:16
one bigger than the other.
• 69:16 - 69:24
But in case they are
not both positive,
• 69:24 - 69:28
I just need f to be bigger
than g for every point.
• 69:28 - 69:33
And the same argument
will function.
• 69:33 - 69:39
This is f, continuous positive.
• 69:39 - 69:42
Then g, continuous
positive but smaller
• 69:42 - 69:45
in values than this one.
• 69:45 - 69:48
• 69:48 - 69:49
Yes, sir?
• 69:49 - 69:51
STUDENT: [INAUDIBLE]
12.2 that we're starting?
• 69:51 - 69:52
DR. MAGDALENA TODA: 12.2.
• 69:52 - 69:56
And you are more organized
than I am, and I appreciate it.
• 69:56 - 70:02
So integration over a
non-rectangular domain.
• 70:02 - 70:07
• 70:07 - 70:10
And we call this a
type one because this
• 70:10 - 70:12
is what many books are using.
• 70:12 - 70:17
And this is that x is
between two fixed end points.
• 70:17 - 70:21
But y is between two
variable end points.
• 70:21 - 70:24
So what's going to happen to y?
• 70:24 - 70:29
y is going to take
values between the lower,
• 70:29 - 70:35
the bottom one, which is
g of x, and the upper one,
• 70:35 - 70:37
which is f of x.
• 70:37 - 70:40
So this is how we
define the domain that's
• 70:40 - 70:45
• 70:45 - 70:48
This is the domain.
• 70:48 - 70:56
Now you really do
not need to prove
• 70:56 - 71:09
that double integral over
1 dA over-- let's call
• 71:09 - 71:15
the domain D-- is what?
• 71:15 - 71:18
• 71:18 - 71:28
Integral between f of x
minus g of x from a to b dx.
• 71:28 - 71:31
• 71:31 - 71:32
And you say, what?
• 71:32 - 71:34
Magdalena, what are
you trying to say?
• 71:34 - 71:35
OK.
• 71:35 - 71:37
Let's go back and
say, what if somebody
• 71:37 - 71:41
same question in calculus 2?
• 71:41 - 71:45
Saying, guys I have a
• 71:45 - 71:49
vertical strip thing.
• 71:49 - 71:51
How are we going
to compute that?
• 71:51 - 71:54
And you would say,
oh, I have an idea.
• 71:54 - 72:04
I take the area under the graph
f, and I shade that in orange.
• 72:04 - 72:06
And I know what that is.
• 72:06 - 72:08
So you would say, I
know what that is.
• 72:08 - 72:09
That's going to be what?
• 72:09 - 72:13
Integral from a to be f of x dx.
• 72:13 - 72:17
Let's call that A1, right?
• 72:17 - 72:20
A1.
• 72:20 - 72:28
Then you go, minus the area
with-- I'm just going to shade
• 72:28 - 72:32
that, brown strips under g.
• 72:32 - 72:35
• 72:35 - 72:38
g of x dx.
• 72:38 - 72:39
And call that A2.
• 72:39 - 72:42
• 72:42 - 72:45
A1 minus A2.
• 72:45 - 72:49
We know both of these
formulas from where?
• 72:49 - 72:53
Calc 1 because that's where
• 72:53 - 72:55
under the graph of a curve.
• 72:55 - 72:58
This is the area under
the graph of a curve f.
• 72:58 - 73:01
This is the area under
the graph of the curve g.
• 73:01 - 73:05
The black striped area
is their difference.
• 73:05 - 73:05
All right.
• 73:05 - 73:07
And so how much is that?
• 73:07 - 73:09
I'm sorry I put the wrong thing.
• 73:09 - 73:12
a, b.
• 73:12 - 73:14
That's going to be
integral from a to b.
• 73:14 - 73:16
Now you say, wait,
wait, wait a minute.
• 73:16 - 73:17
Based on what?
• 73:17 - 73:20
Based on some sort of
• 73:20 - 73:24
of the integral of one
variable, which says integral
• 73:24 - 73:27
from a to b of f plus g.
• 73:27 - 73:29
You can have f plus, minus g.
• 73:29 - 73:31
It doesn't matter.
• 73:31 - 73:32
dx.
• 73:32 - 73:38
You have integral from a to b f
dx plus integral from a to b g
• 73:38 - 73:39
dx.
• 73:39 - 73:42
It doesn't matter what.
• 73:42 - 73:46
You can have a linear
combination of f and g.
• 73:46 - 73:47
Yes, Matthew?
• 73:47 - 73:49
MATTHEW: So this is
just for the domain?
• 73:49 - 73:53
So if you put it,
that would be down.
• 73:53 - 73:56
So there might be
another formula up here
• 73:56 - 73:57
that would be curved surface.
• 73:57 - 74:00
And this is the bottom,
so you're using integral
• 74:00 - 74:01
to find the base,
and then you're
• 74:01 - 74:04
going to plug that integral
into the other integral.
• 74:04 - 74:06
DR. MAGDALENA TODA: So I'm
just using the property that's
• 74:06 - 74:11
called linearity of
the simple integral,
• 74:11 - 74:15
meaning that if I have even
a linear combination like af
• 74:15 - 74:22
plus bg, then a-- I have not a.
• 74:22 - 74:27
Let me call it big A and
big B. Big A Af integral
• 74:27 - 74:29
of f plus big B integral of g.
• 74:29 - 74:31
You've learned that in Calc 2.
• 74:31 - 74:34
I'm doing this to apply it for
these areas that are subtracted
• 74:34 - 74:36
from one another.
• 74:36 - 74:39
If I were to add, as you
said, I would put something
• 74:39 - 74:40
on top of that.
• 74:40 - 74:45
And then it would be like
a superimposition onto it.
• 74:45 - 74:54
So I have integral from a to
b of f of x minus g of x dx.
• 74:54 - 74:57
And I claim that
this is the same
• 74:57 - 75:07
as double integral of the
1dA over the domain D.
• 75:07 - 75:10
How can you write
that differently?
• 75:10 - 75:12
I'll tell you how you
write that differently.
• 75:12 - 75:19
Integral from a to b of
integral from-- what's
• 75:19 - 75:21
the bottom value of Mr. Y?
• 75:21 - 75:24
• 75:24 - 75:27
So Mr. X knows what he's doing.
• 75:27 - 75:29
He goes all the way from a to b.
• 75:29 - 75:31
The bottom value of y is g of x.
• 75:31 - 75:36
You go from the bottom value
of y g of x to the upper value
• 75:36 - 75:38
f of x.
• 75:38 - 75:42
And then you here put 1 and dy.
• 75:42 - 75:45
Is this the same thing?
• 75:45 - 75:46
You say, OK, I know this one.
• 75:46 - 75:49
I know this one from calc 2.
• 75:49 - 75:54
But Magdalena, the one
you gave us is new.
• 75:54 - 75:55
It's new and not new, guys.
• 75:55 - 75:59
This is Fubini's
theorem but generalized
• 75:59 - 76:01
to something that depends on x.
• 76:01 - 76:03
So how do I do that?
• 76:03 - 76:05
Integral of 1dy.
• 76:05 - 76:07
That's what?
• 76:07 - 76:12
That's y measured between two
values that don't depend on y.
• 76:12 - 76:17
They depend only on x, g of x on
the bottom, f of x on the top.
• 76:17 - 76:20
So this is exactly the
integral from a to b.
• 76:20 - 76:22
In terms of the
round parentheses,
• 76:22 - 76:26
I put-- what is y between
f of x and g of x?
• 76:26 - 76:30
f of x minus g of x dx.
• 76:30 - 76:34
So it is exactly the
same thing from Calc 2
• 76:34 - 76:36
expressed as a double integral.
• 76:36 - 76:42
• 76:42 - 76:43
All right.
• 76:43 - 76:54
Now This is a type one
• 76:54 - 77:00
A type two region is a
similar region, practically.
• 77:00 - 77:03
What you have to keep
in mind is they're both
• 77:03 - 77:06
given here as examples.
• 77:06 - 77:09
But the technique is
absolutely the same.
• 77:09 - 77:13
taking this picture,
• 77:13 - 77:20
I would take y to move
between fixed values,
• 77:20 - 77:26
like y has to be between
c and d-- this is my y.
• 77:26 - 77:29
These are the fixed values.
• 77:29 - 77:33
And then give me
some nice colors.
• 77:33 - 77:42
This curve and
that curve-- OK, I
• 77:42 - 77:50
have to rotate my head because
then this is going to be x.
• 77:50 - 77:52
This is going to be y.
• 77:52 - 77:57
And the blue thingy has
to be a function of y.
• 77:57 - 77:59
x is a function of y.
• 77:59 - 78:01
So how do I call that?
• 78:01 - 78:10
I have x or whatever
equals big F of y.
• 78:10 - 78:17
And here in the red one, I
have x equals big G of y.
• 78:17 - 78:23
And how am I going to
evaluate the striped area?
• 78:23 - 78:31
Of course striped because I
have again y is between c and d.
• 78:31 - 78:34
And what's moving is Mr. X.
• 78:34 - 78:37
And Mr. X refuses to
have fixed variables.
• 78:37 - 78:42
Now he goes, I move from
the bottom, which is G of y,
• 78:42 - 78:47
to the top, which is F of y.
• 78:47 - 78:51
How am I going to write
the double integral
• 78:51 - 78:58
over this domain of
1dA, where dA is dxdy.
• 78:58 - 79:00
Who's going to tell me?
• 79:00 - 79:05
Similarly, the same
reasoning as for this one.
• 79:05 - 79:10
I'm going to have the
integral from what to what
• 79:10 - 79:12
of integral from what to what?
• 79:12 - 79:15
Who comes first, dx or dy?
• 79:15 - 79:16
STUDENT: dx.
• 79:16 - 79:17
DR. MAGDALENA TODA:
dx, very good.
• 79:17 - 79:19
And dy at the end.
• 79:19 - 79:23
So y will be between
c and d, and x
• 79:23 - 79:32
is going to be between
G of y and F of y.
• 79:32 - 79:32
And here is y.
• 79:32 - 79:35
• 79:35 - 79:39
How can I rewrite this integral?
• 79:39 - 79:40
Very easily.
• 79:40 - 79:46
The integral from c to
d of the guy on top,
• 79:46 - 79:54
the blue guy, F of y, minus the
guy on the bottom, G of y, dy.
• 79:54 - 80:00
Some people call the
vertical stip method
• 80:00 - 80:03
compared to the horizontal
strip method, where
• 80:03 - 80:05
in this kind of
horizontal strip method,
• 80:05 - 80:08
you just have to view
x as a function of y
• 80:08 - 80:12
the same reasoning as before.
• 80:12 - 80:13
It's not a big deal.
• 80:13 - 80:16
You just need a little
bit of imagination,
• 80:16 - 80:20
and the result is the same.
• 80:20 - 80:25
An example that's
not too hard-- I
• 80:25 - 80:26
want to give you
several examples.
• 80:26 - 80:29
• 80:29 - 80:31
We have plenty of time.
• 80:31 - 80:36
Now it says, we have
a triangular region.
• 80:36 - 80:41
And that is enclosed by lines
y equals 0, y equals 2x,
• 80:41 - 80:44
and x equals 1.
• 80:44 - 80:48
Let's see what that means
and be able to draw it.
• 80:48 - 80:51
It's very important to be
able to draw in this chapter.
• 80:51 - 80:55
If you're not, just
learn how to draw,
• 80:55 - 80:57
and that will give
you lots of ideas
• 80:57 - 80:58
on how to solve the problems.
• 80:58 - 81:18
• 81:18 - 81:23
Chapter 12 is included
completely on the midterm.
• 81:23 - 81:25
So the midterm is
on the 2nd of April.
• 81:25 - 81:30
For the midterm, we have chapter
10, those three sections.
• 81:30 - 81:32
Then we have chapter
11 completely,
• 81:32 - 81:40
and then we have chapter 12
not completely, up to 12.6.
• 81:40 - 81:41
All right.
• 81:41 - 81:44
So what did I say?
• 81:44 - 81:49
I have a triangular region that
is obtained by intersecting
• 81:49 - 81:51
the following lines.
• 81:51 - 81:59
y equals 0, x equals
1, and y equals 2x.
• 81:59 - 82:02
Can I draw them and
see how they intersect?
• 82:02 - 82:03
It shouldn't be a big problem.
• 82:03 - 82:06
This is a line that
passes through the origin
• 82:06 - 82:08
and has slope 2.
• 82:08 - 82:11
So it should be
very easy to draw.
• 82:11 - 82:18
At 1, x equals 1, the y will
be 2 for this line of slope 2.
• 82:18 - 82:21
So I'll try to draw.
• 82:21 - 82:24
Does this look double to you?
• 82:24 - 82:29
So this is 2.
• 82:29 - 82:32
This is the point 1, 2.
• 82:32 - 82:35
And that's the line y equals 2x.
• 82:35 - 82:38
And that's the line y equals 0.
• 82:38 - 82:40
And that's the line x equals 1.
• 82:40 - 82:43
So can I shade this triangle?
• 82:43 - 82:48
Yeah, I can eventually,
depending on what they ask me.
• 82:48 - 82:49
• 82:49 - 82:58
Find the double
integral of x plus y dA
• 82:58 - 83:06
with respect to the area element
over T, T being the triangle.
• 83:06 - 83:10
So now I'm going to ask,
did they say by what method?
• 83:10 - 83:13
Unfortunately, they say,
do it by both methods.
• 83:13 - 83:17
That means both by x
intregration first and then
• 83:17 - 83:20
y integration and
the other way around.
• 83:20 - 83:23
So they ask you to change
the order of the integration
• 83:23 - 83:25
or do what?
• 83:25 - 83:27
Switch from vertical
strip method
• 83:27 - 83:30
to horizontal strip method.
• 83:30 - 83:31
You should get the same answer.
• 83:31 - 83:34
That's a typical
final exam problem.
• 83:34 - 83:40
When we test you, if
you are able to do this
• 83:40 - 83:43
through the vertical
strip or horizontal
• 83:43 - 83:45
strip and change the
order of integration.
• 83:45 - 83:48
If I do it with the
vertical strip method,
• 83:48 - 83:52
who comes first,
the dy or the dx?
• 83:52 - 83:53
Think a little bit.
• 83:53 - 83:56
Where do I put d--
Fubini [INAUDIBLE]
• 83:56 - 83:59
comes dy dx or dx dy?
• 83:59 - 84:00
STUDENT: dy.
• 84:00 - 84:02
PROFESSOR: dy dx.
• 84:02 - 84:04
So VSM.
• 84:04 - 84:07
You're going to laugh.
• 84:07 - 84:08
It's not written in the book.
• 84:08 - 84:11
It's like a childish name,
Vertical Strip Method,
• 84:11 - 84:13
meeting integration
with respect to y
• 84:13 - 84:15
and then with respect to x.
• 84:15 - 84:18
It helped my students
• 84:18 - 84:20
the vertical strips.
• 84:20 - 84:25
And that's why I say something
that's not using the book, VSM.
• 84:25 - 84:36
Now, I have integral from-- so
who is Mr. X going from 0 to 1?
• 84:36 - 84:36
He's stable.
• 84:36 - 84:38
He's happy.
• 84:38 - 84:40
He's going between
two fixed values.
• 84:40 - 84:44
y goes between the
bottom line, which is 0.
• 84:44 - 84:45
We are lucky.
• 84:45 - 84:48
It's a really nice problem.
• 84:48 - 84:51
Going to y equals 2x.
• 84:51 - 84:54
So it's not hard at all.
• 84:54 - 84:59
And we have to integrate
the function x plus y.
• 84:59 - 85:02
It should be a piece of cake.
• 85:02 - 85:07
Let's do this together because
you've accumulated seniority
• 85:07 - 85:08
in this type of problem.
• 85:08 - 85:11
• 85:11 - 85:12
What do I put inside?
• 85:12 - 85:15
What's integral of x
plus y with respect to y?
• 85:15 - 85:16
Is it hard?
• 85:16 - 85:19
• 85:19 - 85:23
xy plus-- somebody tell me.
• 85:23 - 85:25
STUDENT: y squared.
• 85:25 - 85:29
PROFESSOR: y squared
over 2, between y
• 85:29 - 85:33
equals 0 on the bottom,
y equals 2x on top.
• 85:33 - 85:37
I have to be smart and
plug in the values y.
• 85:37 - 85:39
Otherwise, I'll never make it.
• 85:39 - 85:40
STUDENT: Professor?
• 85:40 - 85:41
PROFESSOR: Yes, sir?
• 85:41 - 85:43
STUDENT: Why did you take
2x as the final value
• 85:43 - 85:45
because you have a
specified triangle.
• 85:45 - 85:49
PROFESSOR: Because y
equals 2x is the expression
• 85:49 - 85:52
of the upper function.
• 85:52 - 85:55
The upper function is
the line y equals 2x.
• 85:55 - 85:56
They provided that.
• 85:56 - 86:00
So from the bottom function
to the upper function,
• 86:00 - 86:02
the vertical strips go
between two functions.
• 86:02 - 86:05
• 86:05 - 86:08
So when I plug in
here y equals 2x,
• 86:08 - 86:10
I have to pay attention
to my algebra.
• 86:10 - 86:14
If I forget the 2, it's all
over for me, zero points.
• 86:14 - 86:16
Well, not zero points,
but 10% credit.
• 86:16 - 86:20
I have no idea what I would
get, so I have to pay attention.
• 86:20 - 86:27
2x times x is 2x squared
plus 2x all squared-- guys,
• 86:27 - 86:31
keep an eye on me--
4x squared over 2.
• 86:31 - 86:36
I put the first value
in a pink parentheses,
• 86:36 - 86:41
and then I move on to
the line parentheses.
• 86:41 - 86:43
Evaluate it at 0.
• 86:43 - 86:45
That line is very lucky.
• 86:45 - 86:51
I get a 0 because y
equals 0 will give me 0.
• 86:51 - 86:54
What am I going to get here?
• 86:54 - 86:57
2x squared plus 2x squared.
• 86:57 - 86:57
Good.
• 86:57 - 86:59
What's 2x squared
plus 2x squared?
• 86:59 - 87:00
4x squared.
• 87:00 - 87:02
So a 4 goes out.
• 87:02 - 87:03
Kick him out.
• 87:03 - 87:06
Integral from 0
to 1 x squared dx.
• 87:06 - 87:08
Integral of x squared is?
• 87:08 - 87:12
• 87:12 - 87:14
Integral of x squared is?
• 87:14 - 87:15
STUDENT: x cubed over 3.
• 87:15 - 87:16
PROFESSOR: x cubed over 3.
• 87:16 - 87:19
And if you take it
between 1 and 0, you get?
• 87:19 - 87:21
STUDENT: 1.
• 87:21 - 87:21
PROFESSOR: 1/3.
• 87:21 - 87:24
1/3 times 4 is 4/3.
• 87:24 - 87:27
• 87:27 - 87:29
Suppose this is going to
happen on the midterm,
• 87:29 - 87:32
and I'm asking you to do it
reversing the integration
• 87:32 - 87:34
order.
• 87:34 - 87:38
Then you are going to check
• 87:38 - 87:42
in the sense that
you say, well, now
• 87:42 - 87:46
I'm going to see if I made
a mistake in this one.
• 87:46 - 87:47
What do I do?
• 87:47 - 87:50
I erase the whole thing, and
• 87:50 - 87:56
I'm going to put
horizontal strips.
• 87:56 - 88:01
And you say, well, life is a
little bit harder in this case
• 88:01 - 88:05
because in this
case, I have to look
• 88:05 - 88:11
at y between fixed
values, y between 0 and 1.
• 88:11 - 88:18
So y is between 0 and 1--
0 and 2, fixed values.
• 88:18 - 88:23
And Mr. X says, I'm going
between two functions of y.
• 88:23 - 88:26
I don't know what those
functions of y are.
• 88:26 - 88:28
I'm puzzled.
• 88:28 - 88:31
You have to help
Mr. X know where
• 88:31 - 88:35
he's going because his life
right now is a little bit hard.
• 88:35 - 88:39
So what is the
function for the blue?
• 88:39 - 88:42
• 88:42 - 88:44
Now he's not blue anymore.
• 88:44 - 88:45
He's brown.
• 88:45 - 88:48
x equals 1.
• 88:48 - 88:50
So he knows what
he's going to be.
• 88:50 - 88:53
What is the x function
for the red line
• 88:53 - 88:55
• 88:55 - 88:55
STUDENT: y over 2.
• 88:55 - 88:58
PROFESSOR: x must be y over 2.
• 88:58 - 89:01
It's the same thing, but I have
to express x in terms of y.
• 89:01 - 89:05
So I erase and I say
x equals y over 2.
• 89:05 - 89:07
Same thing.
• 89:07 - 89:11
So x has to be between what and
what, the bottom and the top?
• 89:11 - 89:14
• 89:14 - 89:19
The top must be x equals 1,
and the bottom one is y over 2.
• 89:19 - 89:25
That's the bottom one,
the bottom value for x.
• 89:25 - 89:27
Now wish me luck because I
have to get the same thing.
• 89:27 - 89:36
So integral from 0 to 2 of
integral from y over 2 to 1.
• 89:36 - 89:38
Changing the order
of integration
• 89:38 - 89:41
doesn't change the
integrand, which is exactly
• 89:41 - 89:43
the same function, f of xy.
• 89:43 - 89:47
This is the f function.
• 89:47 - 89:48
Then what changes?
• 89:48 - 89:49
The order of integration.
• 89:49 - 89:52
So I go dx first,
dy next and stop.
• 89:52 - 89:55
• 89:55 - 90:00
I copy and paste the outer
ones, and I focus my attention
• 90:00 - 90:06
to the red parentheses
inside, which I'm
• 90:06 - 90:08
going to copy and paste here.
• 90:08 - 90:12
I'll have to do some
math very carefully.
• 90:12 - 90:14
So what do I have?
• 90:14 - 90:17
I have x plus y integrated
with respect to x.
• 90:17 - 90:19
If I rush, it's a bad thing.
• 90:19 - 90:21
STUDENT: So that
would be x squared.
• 90:21 - 90:22
PROFESSOR: x squared.
• 90:22 - 90:23
STUDENT: Over 2.
• 90:23 - 90:24
PROFESSOR: Over 2.
• 90:24 - 90:25
STUDENT: Plus xy.
• 90:25 - 90:30
PROFESSOR: Plus xy taken
between the following.
• 90:30 - 90:33
When x equals 1,
I have it on top.
• 90:33 - 90:38
When x equals y over 2,
I have it on the bottom.
• 90:38 - 90:40
OK.
• 90:40 - 90:43
This red thing, I'm a
little bit too lazy.
• 90:43 - 90:48
I'll copy and paste
it separately.
• 90:48 - 90:52
For the upper part, it's
really easy to compute.
• 90:52 - 90:53
What do I get?
• 90:53 - 91:02
When x is 1, 1/2, 1/2
plus when x is 1, y.
• 91:02 - 91:07
Minus integral of--
when x is y over 2,
• 91:07 - 91:13
I get y squared over
4 up here over 2.
• 91:13 - 91:19
So I should get y
squared over 8 plus--
• 91:19 - 91:22
I've got an x equals y over 2.
• 91:22 - 91:24
What do I get?
• 91:24 - 91:26
y squared over 2.
• 91:26 - 91:29
Is this hard?
• 91:29 - 91:32
It's very easy to make an
algebra mistake on such
• 91:32 - 91:33
a problem, unfortunately.
• 91:33 - 91:37
I have y plus 1/2 plus what?
• 91:37 - 91:41
What is 1/2 plus 1/8?
• 91:41 - 91:42
STUDENT: 5/8.
• 91:42 - 91:49
PROFESSOR: 5 over 8
with a minus y squared.
• 91:49 - 91:52
• 91:52 - 91:54
So hopefully I did this right.
• 91:54 - 92:01
Now I'll go, OK, integral from
0 to 2 of all of this animal, y
• 92:01 - 92:06
plus 1/2 minus 5
over 8, y squared.
• 92:06 - 92:11
What happens if I don't
• 92:11 - 92:13
Then I go back and
check my work because I
• 92:13 - 92:15
know I'm supposed to get 4/3.
• 92:15 - 92:16
That was easy.
• 92:16 - 92:23
So what is integral of this
sausage, whatever it is?
• 92:23 - 92:30
y squared over 2 plus y
over 2 minus 5 over 8--
• 92:30 - 92:42
oh my god-- 5 over 8, y
cubed over 3, between 2 up
• 92:42 - 92:44
and 0 down.
• 92:44 - 92:47
When I have 0 down,
I plug y equals 0.
• 92:47 - 92:48
It's a piece of cake.
• 92:48 - 92:49
It's 0.
• 92:49 - 92:52
So what matters is
what I get when I plug
• 92:52 - 92:54
in the value 2 instead of y.
• 92:54 - 92:56
So what do I get?
• 92:56 - 93:07
4 over 2 is 2, plus 2 over 2
is 1, minus 2 cubed, thank god.
• 93:07 - 93:08
That's 8.
• 93:08 - 93:11
8 simplifies with 8 minus 5/3.
• 93:11 - 93:16
• 93:16 - 93:24
So I got 9/3 minus 5/3,
and I did it carefully.
• 93:24 - 93:25
I did a good job.
• 93:25 - 93:28
I got the same thing, 4/3.
• 93:28 - 93:31
So no matter which
method, the vertical strip
• 93:31 - 93:34
or the horizontal strip
method, I get the same thing.
• 93:34 - 93:37
And of course, you'll
• 93:37 - 93:43
because this is what the Fubini
theorem extended to this case
• 93:43 - 93:44
is telling you.
• 93:44 - 93:47
It doesn't matter the
order of integration.
• 93:47 - 93:51
• 93:51 - 93:55
I would advise you to go
through the theory in the book.
• 93:55 - 93:58
• 93:58 - 94:02
area and volume on page 934.
• 94:02 - 94:08
I'd like you to read that.
• 94:08 - 94:12
And let's see what I want to do.
• 94:12 - 94:14
Which one shall I do?
• 94:14 - 94:18
There are a few examples
that are worth it.
• 94:18 - 94:21
• 94:21 - 94:29
I'll pick the one that gives
people the most trouble.
• 94:29 - 94:30
• 94:30 - 94:33
I take the few examples that
give people the most trouble.
• 94:33 - 94:39
One example that popped up on
almost each and every final
• 94:39 - 94:44
in the past 13 years
that involves changing
• 94:44 - 94:46
the order of integration.
• 94:46 - 94:57
• 94:57 - 95:10
So example problem on changing
the order of integration.
• 95:10 - 95:15
• 95:15 - 95:20
A very tricky, smart
problem is the following.
• 95:20 - 95:31
Evaluate integral from 0
to 1, integral from x to 1,
• 95:31 - 95:34
e to the y squared dy dx.
• 95:34 - 95:42
• 95:42 - 95:44
I don't know if you've
seen anything like that
• 95:44 - 95:46
in AP Calculus or Calc 2.
• 95:46 - 95:52
Maybe you have, in which case
• 95:52 - 95:54
you that this is nasty.
• 95:54 - 95:57
• 95:57 - 96:00
You say, in what
sense is it nasty?
• 96:00 - 96:05
There is no expressible
anti-derivative.
• 96:05 - 96:22
So this cannot be expressed in
terms of elementary functions
• 96:22 - 96:22
explicitly.
• 96:22 - 96:29
• 96:29 - 96:31
It's not that there
is no anti-derivative.
• 96:31 - 96:35
There is an anti-derivative--
a whole family, actually--
• 96:35 - 96:39
but you cannot express them in
terms of elementary functions.
• 96:39 - 96:43
And actually, most functions
are not so bad in real world,
• 96:43 - 96:44
in real life.
• 96:44 - 96:49
Now, could you compute, for
example, integral from 1 to 3
• 96:49 - 96:51
of e to the t squared dt?
• 96:51 - 96:52
Yes.
• 96:52 - 96:54
How do you do that?
• 96:54 - 96:56
With a calculator.
• 96:56 - 96:58
And what if you don't have one?
• 96:58 - 96:59
You go to the lab over there.
• 96:59 - 97:00
There is MATLAB.
• 97:00 - 97:03
MATLAB will compute it for you.
• 97:03 - 97:05
How does MATLAB know
how to compute it
• 97:05 - 97:08
if there is no way to
express the anti-derivative
• 97:08 - 97:12
and take the value of the
anti-derivative between b
• 97:12 - 97:16
and a, like in the fundamental
theorem of calculus?
• 97:16 - 97:21
Well, the calculator or the
computer program is smart.
• 97:21 - 97:25
He uses numerical analysis
to approximate this type
• 97:25 - 97:27
of integral.
• 97:27 - 97:28
So he's fooling you.
• 97:28 - 97:30
He's just playing smarty pants.
• 97:30 - 97:33
He's smarter than
you at this point.
• 97:33 - 97:34
OK.
• 97:34 - 97:38
So you cannot do this by hand,
so this order of integration is
• 97:38 - 97:39
fruitless.
• 97:39 - 97:43
• 97:43 - 97:47
And there are people who
tried to do this on the final.
• 97:47 - 97:49
Of course, they
didn't get anywhere
• 97:49 - 97:51
because they couldn't
integrate it.
• 97:51 - 97:56
The whole idea of this one
is to-- some professors
• 97:56 - 97:59
are so mean they
don't even tell you,
• 97:59 - 98:01
hint, change the
order of integration
• 98:01 - 98:03
because it may work
the other way around.
• 98:03 - 98:06
They just give it to you, and
then people can spend an hour
• 98:06 - 98:08
and they don't get anywhere.
• 98:08 - 98:12
If you want to be mean to a
student, that's what you do.
• 98:12 - 98:17
So I will tell
you that one needs
• 98:17 - 98:20
to change the order of
integration for this.
• 98:20 - 98:21
This is the function.
• 98:21 - 98:26
We keep the function, but let's
see what happens if you draw.
• 98:26 - 98:31
The domain will be
x between 0 and 1.
• 98:31 - 98:34
• 98:34 - 98:37
y will be between x and 1.
• 98:37 - 98:40
So it's like you have a square.
• 98:40 - 98:44
y equals x is your
diagonal of the square.
• 98:44 - 98:49
And you go from--
• 98:49 - 98:55
You go from y equals x on the
bottom and y equals 1 on top.
• 98:55 - 98:57
And so the domain is
this beautiful triangle
• 98:57 - 99:03
that I make all in line
with vertical strips.
• 99:03 - 99:06
This is what it means,
vertical strips.
• 99:06 - 99:12
But if I do horizontal strips, I
have to change the color, blue.
• 99:12 - 99:15
And for horizontal
strips, I'm going
• 99:15 - 99:17
to have a different problem.
• 99:17 - 99:20
Integral, integral dx dy.
• 99:20 - 99:23
And I just hope to god
that what I'm going to get
• 99:23 - 99:27
is doable because if
not, then I'm in trouble.
• 99:27 - 99:30
So help me on this one.
• 99:30 - 99:34
If y is between what and what?
• 99:34 - 99:35
It's a square.
• 99:35 - 99:39
It's a square, so this will
be the same, 0 to 1, right?
• 99:39 - 99:39
STUDENT: Yep.
• 99:39 - 99:40
PROFESSOR: But Mr. X?
• 99:40 - 99:42
• 99:42 - 99:46
STUDENT: And then it
will be between 1 and y.
• 99:46 - 99:49
PROFESSOR: Between--
Mr. X is this guy.
• 99:49 - 99:52
And he doesn't go between 1.
• 99:52 - 99:55
He goes between the
sea level, which is
• 99:55 - 100:03
x equals 0, to x equals what?
• 100:03 - 100:03
STUDENT: [INAUDIBLE].
• 100:03 - 100:05
PROFESSOR: Right?
• 100:05 - 100:10
So from x equals 0
through x equals y.
• 100:10 - 100:15
And you have the same individual
e to the y squared that before
• 100:15 - 100:17
• 100:17 - 100:20
Now he's not so bad, actually.
• 100:20 - 100:22
Why is he not so bad?
• 100:22 - 100:25
Look what happens in
the first parentheses.
• 100:25 - 100:27
This is so beautiful
that it's something
• 100:27 - 100:29
you didn't even hope for.
• 100:29 - 100:34
So we copy and paste
it from 0 to 1 dy.
• 100:34 - 100:38
These guys stay
outside and they wait.
• 100:38 - 100:40
Inside, it's our
• 100:40 - 100:45
So Mr. X is independent
from e to the y squared.
• 100:45 - 100:47
So e to the y squared pulls out.
• 100:47 - 100:48
He's a constant.
• 100:48 - 100:53
And you have integral
of 1 dx between 0 and y.
• 100:53 - 100:57
How much is that?
• 100:57 - 100:58
1.
• 100:58 - 101:04
x between x equals
0 and x equals y.
• 101:04 - 101:05
So it's y.
• 101:05 - 101:06
So I'm being serious.
• 101:06 - 101:08
So I should have said y.
• 101:08 - 101:12
• 101:12 - 101:21
have given you, in Calc 2,
• 101:21 - 101:25
this, how would
you have done it?
• 101:25 - 101:27
STUDENT: U-substitution.
• 101:27 - 101:28
PROFESSOR: U-substitution.
• 101:28 - 101:29
Excellent.
• 101:29 - 101:32
What kind of
u-substitution [INAUDIBLE]?
• 101:32 - 101:35
STUDENT: y squared equals u.
• 101:35 - 101:40
PROFESSOR: y squared
equals u, du equals 2y dy.
• 101:40 - 101:44
So y dy together.
• 101:44 - 101:46
They stick together.
• 101:46 - 101:47
They stick together.
• 101:47 - 101:50
They attract each
other as magnets.
• 101:50 - 101:57
So y dy is going to be
1/2 du-- 1/2 pulls out--
• 101:57 - 102:00
integral e to the u du.
• 102:00 - 102:01
Attention.
• 102:01 - 102:03
When y is moving
between 0 and 1,
• 102:03 - 102:06
u is moving also
between 0 and 1.
• 102:06 - 102:12
So it really should
be a piece of cake.
• 102:12 - 102:13
Are you guys with me?
• 102:13 - 102:16
Do you understand what I did?
• 102:16 - 102:19
Do you understand the words
coming out of my mouth?
• 102:19 - 102:24
• 102:24 - 102:25
It's easy.
• 102:25 - 102:29
• 102:29 - 102:30
Good.
• 102:30 - 102:35
So what is integral
of e to the u du?
• 102:35 - 102:39
e to the u between
1 up and 0 down.
• 102:39 - 102:44
So e to the u de to the 1
minus e to the 0 over 2.
• 102:44 - 102:48
• 102:48 - 102:51
That is e minus 1 over 2.
• 102:51 - 102:54
• 102:54 - 102:59
I could not have solved this
if I tried it by integration
• 102:59 - 103:03
with y first and then x.
• 103:03 - 103:05
The only way I
could have done this
• 103:05 - 103:08
is by changing the
order of integration.
• 103:08 - 103:12
So how many times have I seen
this in the past 12 years
• 103:12 - 103:13
on the final?
• 103:13 - 103:15
At least six times.
• 103:15 - 103:18
It's a problem that
could be a little bit
• 103:18 - 103:21
hard if the student has
never seen it before
• 103:21 - 103:24
and doesn't know what to
do [? at that point. ?]
• 103:24 - 103:27
Let's do a few more
in the same category.
• 103:27 - 103:37
• 103:37 - 103:38
STUDENT: Professor?
• 103:38 - 103:38
PROFESSOR: Yes?
• 103:38 - 103:41
STUDENT: Where did this shape--
where did this graph come from?
• 103:41 - 103:44
Were we just saying
it was with the same--
• 103:44 - 103:45
PROFESSOR: OK.
• 103:45 - 103:47
• 103:47 - 103:50
So this and that are the key.
• 103:50 - 103:55
This is telling me x is between
0 and 1, and at the same,
• 103:55 - 103:59
time y is between x and 1.
• 103:59 - 104:03
information on the graph,
• 104:03 - 104:06
I say, well, x is
between 0 and 1.
• 104:06 - 104:09
Mr. Y has the freedom to go
between the first bisector,
• 104:09 - 104:14
which is that, and the
cap, his cap, y equals 1.
• 104:14 - 104:17
So that's how I got
to the line strips.
• 104:17 - 104:21
And from the line strips, I said
that I need horizontal strips.
• 104:21 - 104:24
So I changed the
color and I said
• 104:24 - 104:28
the blue strips go between x.
• 104:28 - 104:32
x will be x equals
0 and x equals y.
• 104:32 - 104:37
And then y between 0
and 1, just the same.
• 104:37 - 104:38
It's a little bit tricky.
• 104:38 - 104:42
That's why I want to do one or
two more problems like that,
• 104:42 - 104:47
because I know that I remember
20-something years ago,
• 104:47 - 104:52
I myself needed a little
bit of time understanding
• 104:52 - 104:57
the meaning of reversing
the order of integration.
• 104:57 - 104:59
STUDENT: Does it matter
which way you put it?
• 104:59 - 105:02
PROFESSOR: In this case, it's
important that you do reverse.
• 105:02 - 105:06
But in general, it's
doable both ways.
• 105:06 - 105:10
I mean, in the other problems
I'm going to give you today,
• 105:10 - 105:12
you should be able
to do either way.
• 105:12 - 105:19
So I'm looking for a problem
that you could eventually
• 105:19 - 105:21
do another one.
• 105:21 - 105:26
• 105:26 - 105:28
We don't have so many.
• 105:28 - 105:32
I'm going to go ahead and
look into the homework.
• 105:32 - 105:32
Yeah.
• 105:32 - 105:35
• 105:35 - 105:41
So it says, you
have this integral,
• 105:41 - 105:44
the integral from 0
to 4 of the integral
• 105:44 - 105:50
from x squared to 4y dy dx.
• 105:50 - 105:56
Draw, compute, and also
compute with reversing
• 105:56 - 105:59
the order of integration
• 105:59 - 106:01
When I say that,
it sounds horrible.
• 106:01 - 106:04
But in reality, the
more you work on
• 106:04 - 106:08
that one, the more familiar
you're going to feel.
• 106:08 - 106:10
So what did I just say?
• 106:10 - 106:13
Problem number 26.
• 106:13 - 106:18
You have integral
from 0 to 4, integral
• 106:18 - 106:25
from x squared to 4x dy dx.
• 106:25 - 106:27
• 106:27 - 106:31
Interpret geometrically,
whatever that means,
• 106:31 - 106:35
and then compute the
integral in two ways,
• 106:35 - 106:38
with this given order
integration, which
• 106:38 - 106:40
is what kind of strips, guys?
• 106:40 - 106:42
Vertical strips.
• 106:42 - 106:45
Or reversing the
order of integration.
• 106:45 - 106:50
And check that the answer is the
same just to check your work.
• 106:50 - 106:52
STUDENT: So first--
• 106:52 - 106:53
PROFESSOR: First you draw.
• 106:53 - 106:56
First you draw because
if you don't draw,
• 106:56 - 107:00
you don't understand what
• 107:00 - 107:02
And you say, wait a minute.
• 107:02 - 107:05
and do it without drawing?
• 107:05 - 107:08
Yeah, but you're not
going to get too far.
• 107:08 - 107:12
So let's see what kind
of problem you have.
• 107:12 - 107:13
y and x.
• 107:13 - 107:17
y equals x squared is a what?
• 107:17 - 107:19
It's a pa--
• 107:19 - 107:20
STUDENT: Parabola.
• 107:20 - 107:21
PROFESSOR: Parabola.
• 107:21 - 107:24
And this parabola should
be nice and sassy.
• 107:24 - 107:26
Is it fat enough?
• 107:26 - 107:27
I think it is.
• 107:27 - 107:34
And the other one will
be 4x, y equals 4x.
• 107:34 - 107:36
What does that look like?
• 107:36 - 107:39
It looks like a line passing
through the origin that
• 107:39 - 107:43
has slope 4, so the
slope is really high.
• 107:43 - 107:44
STUDENT: Just straight.
• 107:44 - 107:48
• 107:48 - 107:52
PROFESSOR: y equals 4x
versus y equals x squared.
• 107:52 - 107:54
Now, do they meet?
• 107:54 - 107:57
• 107:57 - 107:58
STUDENT: Yes.
• 107:58 - 107:59
PROFESSOR: Yes.
• 107:59 - 108:00
Exactly where do they meet?
• 108:00 - 108:00
Exactly here.
• 108:00 - 108:01
STUDENT: 4.
• 108:01 - 108:04
PROFESSOR: So 4x equals x
squared, where do they meet?
• 108:04 - 108:07
• 108:07 - 108:13
They meet at-- it has
two possible roots.
• 108:13 - 108:18
One is x equals
0, which is here,
• 108:18 - 108:21
and one is x equals
4, which is here.
• 108:21 - 108:27
So really, my graph looks
just the way it should look,
• 108:27 - 108:29
only my parabola is
a little bit too fat.
• 108:29 - 108:34
• 108:34 - 108:44
This is the point of
coordinates 4 and 16.
• 108:44 - 108:46
Are you guys with me?
• 108:46 - 108:52
And Mr. X is moving
between 0 and 4.
• 108:52 - 108:57
This is the maximum
level x can get.
• 108:57 - 109:02
And where he stops here
at 4, a miracle happens.
• 109:02 - 109:07
The two curves intersect each
other exactly at that point.
• 109:07 - 109:12
So this looks like a
leaf, a slice of orange.
• 109:12 - 109:12
Oh my god.
• 109:12 - 109:13
I don't know.
• 109:13 - 109:18
I'm already hungry so I cannot
wait to get out of here.
• 109:18 - 109:21
I bet you're hungry as well.
• 109:21 - 109:24
Let's do this problem
both ways and then go
• 109:24 - 109:27
home or to have
something to eat.
• 109:27 - 109:32
How are you going to advise
me to solve it first?
• 109:32 - 109:34
up to be solved.
• 109:34 - 109:35
So it's vertical strips.
• 109:35 - 109:38
And I will say
integral from 0 to 4,
• 109:38 - 109:41
copy and paste the outer part.
• 109:41 - 109:46
Take the inner part, and do the
inner part because it's easy.
• 109:46 - 109:50
And if it's easy, you tell
me how I'm going to do it.
• 109:50 - 109:54
Integral of 1 dy is y.
• 109:54 - 109:59
y measured at 4x is 4x,
and y measured at x squared
• 109:59 - 110:01
is x squared.
• 110:01 - 110:02
Oh thank god.
• 110:02 - 110:06
This is so beautiful
and so easy.
• 110:06 - 110:09
Let's integrate again.
• 110:09 - 110:16
4 x squared over 2 times x cubed
over 3 between x equals 0 down
• 110:16 - 110:18
and x equals 4 up.
• 110:18 - 110:22
• 110:22 - 110:24
What do I get?
• 110:24 - 110:30
I get 4 cubed over 2
minus 4 cubed over 3.
• 110:30 - 110:32
This 4 cubed is an obsession.
• 110:32 - 110:34
Kick him out.
• 110:34 - 110:36
1/2 minus 1/3.
• 110:36 - 110:40
• 110:40 - 110:41
How much is 1/2 minus 1/3?
• 110:41 - 110:42
My son knows that.
• 110:42 - 110:44
STUDENT: 1/6.
• 110:44 - 110:44
PROFESSOR: OK.
• 110:44 - 110:46
1/6, yes.
• 110:46 - 110:49
So we simply take it.
• 110:49 - 110:50
We can leave it like that.
• 110:50 - 110:56
If you leave it like that on
the exam, I don't mind at all.
• 110:56 - 110:59
But you could always put
64 over 6 and simplify it.
• 110:59 - 111:02
• 111:02 - 111:03
Are you guys with me?
• 111:03 - 111:07
You can simplify
it and get what?
• 111:07 - 111:08
32 over 3.
• 111:08 - 111:11
• 111:11 - 111:13
Don't give me decimals.
• 111:13 - 111:15
I'm not impressed.
• 111:15 - 111:16
You're not supposed
to use the calculator.
• 111:16 - 111:21
You are supposed to leave
this is exact fraction
• 111:21 - 111:25
form like that, irreducible.
• 111:25 - 111:26
Let's do it the
other way around,
• 111:26 - 111:30
and that will be the
last thing we do.
• 111:30 - 111:34
The other way around means
I'll take another color.
• 111:34 - 111:37
I'll do the horizontal stripes.
• 111:37 - 111:40
• 111:40 - 111:44
And I will have to rewrite
the meaning of these two
• 111:44 - 111:50
branches of functions with
x expressed in terms of y.
• 111:50 - 111:52
That's the only thing
I need to do, right?
• 111:52 - 111:56
So what is this?
• 111:56 - 111:59
If y is x squared, what is x?
• 111:59 - 112:00
STUDENT: Root y.
• 112:00 - 112:04
PROFESSOR: The inverse
function. x will be root of y.
• 112:04 - 112:06
You said very well.
• 112:06 - 112:07
So I have to write.
• 112:07 - 112:10
In [INAUDIBLE], I
have what I need
• 112:10 - 112:13
to have for the line
horizontal strip method.
• 112:13 - 112:16
• 112:16 - 112:20
And then for the other one,
x is going to be y over 4.
• 112:20 - 112:23
• 112:23 - 112:24
So what do I do?
• 112:24 - 112:32
So integral, integral, a
1 that was here hidden,
• 112:32 - 112:36
but I'll put it because
that's the integral.
• 112:36 - 112:39
And then I go dx dy.
• 112:39 - 112:45
All I have to care about is the
endpoints of the integration.
• 112:45 - 112:48
Now, pay attention a little
bit because Mr. Y is not
• 112:48 - 112:50
between 0 and 4.
• 112:50 - 112:53
students under stress
• 112:53 - 112:56
in the final putting 0 and 4.
• 112:56 - 112:57
Don't do that.
• 112:57 - 112:59
So pay attention to the
limits of integration.
• 112:59 - 113:01
What are the limits?
• 113:01 - 113:02
0 and--
• 113:02 - 113:02
STUDENT: 16.
• 113:02 - 113:03
PROFESSOR: 16.
• 113:03 - 113:05
Very good.
• 113:05 - 113:10
And x will be between root
y-- well, which one is on top?
• 113:10 - 113:12
Which one is on the bottom?
• 113:12 - 113:17
Because if I move my head,
I'll say that's on top
• 113:17 - 113:19
and that's on the bottom.
• 113:19 - 113:22
STUDENT: The right side
is always on the top.
• 113:22 - 113:26
PROFESSOR: So the one that
looks higher is this one.
• 113:26 - 113:29
This is more than
that in this frame.
• 113:29 - 113:37
So square of y is on top and
y over 4 is on the bottom.
• 113:37 - 113:39
I should get the same answer.
• 113:39 - 113:40
If I don't, then I'm in trouble.
• 113:40 - 113:43
So what do I get?
• 113:43 - 113:49
Integral from 0 to 16.
• 113:49 - 113:52
Tonight, when I
go home, I'm going
• 113:52 - 113:57
to cook up the homework
for 12.1 and 12.1 at least.
• 113:57 - 113:59
I'll put some problems
similar to that
• 113:59 - 114:03
because I want to emphasize
the same type of problem
• 114:03 - 114:05
in at least two or three
applications for the homework
• 114:05 - 114:07
for the midterm.
• 114:07 - 114:11
And maybe one like that will
be on the final as well.
• 114:11 - 114:13
It's very important for
you to understand how,
• 114:13 - 114:15
with this kind of
domain, you reverse
• 114:15 - 114:17
the order of integration.
• 114:17 - 114:20
Who's helping me here?
• 114:20 - 114:22
Root y.
• 114:22 - 114:26
What is root y
when-- y to the 1/2.
• 114:26 - 114:28
I need to integrate.
• 114:28 - 114:34
So I need minus y over 4 and dy.
• 114:34 - 114:39
• 114:39 - 114:42
Can you help me integrate?
• 114:42 - 114:44
STUDENT: [INAUDIBLE].
• 114:44 - 114:50
PROFESSOR: 2/3 y
to the 3/2 minus--
• 114:50 - 114:51
STUDENT: y squared.
• 114:51 - 114:56
PROFESSOR: y squared
over 8, y equals 0
• 114:56 - 114:58
on the bottom, piece of cake.
• 114:58 - 115:00
That will give me 0.
• 115:00 - 115:01
I'm so happy.
• 115:01 - 115:05
And y equals 16 on top.
• 115:05 - 115:10
So for 16, I have 2/3.
• 115:10 - 115:12
And who's telling me what else?
• 115:12 - 115:13
STUDENT: 64.
• 115:13 - 115:14
PROFESSOR: 64.
• 115:14 - 115:14
4 cubed.
• 115:14 - 115:23
I can leave it 4 cubed if I want
to minus another-- well here,
• 115:23 - 115:25
I have to pay attention.
• 115:25 - 115:27
So I have 16 here.
• 115:27 - 115:31
I got square root of
16, which is 4, cubed.
• 115:31 - 115:39
Here, I put minus 4
squared, which was there.
• 115:39 - 115:41
How do you want me to
do this simplification?
• 115:41 - 115:42
STUDENT: [INAUDIBLE].
• 115:42 - 115:45
PROFESSOR: I can
do 4 to the fourth.
• 115:45 - 115:47
Are you guys with me?
• 115:47 - 115:52
I can put, like you
prefer, 16 squared over 8.
• 115:52 - 115:58
• 115:58 - 115:59
• 115:59 - 116:00
I don't know.
• 116:00 - 116:02
Let's see.
• 116:02 - 116:09
This is really 4 to the 4,
so I have 4 times 4 cubed.
• 116:09 - 116:20
4 cubed gets out and
I have 2/3 minus 1/2.
• 116:20 - 116:24
• 116:24 - 116:28
And how much is that?
• 116:28 - 116:29
Again 1/6.
• 116:29 - 116:31
Are you guys with me?
• 116:31 - 116:32
1/6.
• 116:32 - 116:37
So again, I get 4 cubed
over 6, so I'm done.
• 116:37 - 116:40
4 cubed over 6 equals 32 over 3.
• 116:40 - 116:43
I am happy that
I checked my work
• 116:43 - 116:44
through two different methods.
• 116:44 - 116:46
• 116:46 - 116:49
• 116:49 - 116:52
Now, let me tell you something.
• 116:52 - 116:55
There were also times
when on the midterm
• 116:55 - 117:00
or on the final, due to
lack of time and everything,
• 117:00 - 117:03
we put the following
kind of problem.
• 117:03 - 117:11
Without solving this integral--
without solving-- indicate
• 117:11 - 117:16
the corresponding integral
with the order reversed.
• 117:16 - 117:20
So all you have to
do-- don't do that.
• 117:20 - 117:25
Just from here,
write this and stop.
• 117:25 - 117:28
• 117:28 - 117:30
If you do the whole thing,
it's going to take you
• 117:30 - 117:31
10 minutes, 15 minutes.
• 117:31 - 117:34
If you do just reversing
the order of integration,
• 117:34 - 117:38
I don't know what it takes, a
minute and a half, two minutes.
• 117:38 - 117:42
So in order to save
time, at times,
• 117:42 - 117:46
we gave you just don't
solve the problem. reverse
• 117:46 - 117:48
the order of integration.
• 117:48 - 117:54
• 117:54 - 117:56
One last one.
• 117:56 - 117:58
One last one.
• 117:58 - 118:00
But I don't want to finish it.
• 118:00 - 118:03
I want to give you
• 118:03 - 118:06
or maybe you can finish it.
• 118:06 - 118:08
It should be shorter.
• 118:08 - 118:14
You have a circular parabola,
• 118:14 - 118:16
• 118:16 - 118:19
So x is positive.
• 118:19 - 118:20
STUDENT: Question.
• 118:20 - 118:21
PROFESSOR: I don't know.
• 118:21 - 118:22
I have to find it.
• 118:22 - 118:24
Find the volume.
• 118:24 - 118:25
Example 4, page 934.
• 118:25 - 118:29
Find the volume
of the solid bound
• 118:29 - 118:33
in the above-- this is a
little tricky-- by the plane z
• 118:33 - 118:38
equals y and below
in the xy plane
• 118:38 - 118:42
by the part of the disk
• 118:42 - 118:48
So z equals y means this
is your f of x and y.
• 118:48 - 118:51
So they gave it to you.
• 118:51 - 118:54
But then they say, but
also, in the xy plane,
• 118:54 - 119:00
you have to have the part of
the disk in the first quadrant.
• 119:00 - 119:02
This is not so easy.
• 119:02 - 119:05
They draw it for you to
• 119:05 - 119:08
• 119:08 - 119:14
How do you write the unit
circle, x squared equals 1,
• 119:14 - 119:17
x squared plus y squared
less than or equal to 1,
• 119:17 - 119:19
and x and y are both positive.
• 119:19 - 119:21
• 119:21 - 119:23
How do you compute?
• 119:23 - 119:27
So they say compute the
volume, and I say just
• 119:27 - 119:28
set up the volume.
• 119:28 - 119:30
• 119:30 - 119:33
I could put it in the
midterm just like that.
• 119:33 - 119:36
Set up an integral
without solving it
• 119:36 - 119:46
that indicates the volume
under z equals f of xy-- that's
• 119:46 - 119:51
the geography of z-- and above
a certain domain in plane,
• 119:51 - 119:56
above D in plane.
• 119:56 - 119:58
So you have, OK, what
this should teach you?
• 119:58 - 120:09
Should teach you that double
integral over d f of xy da
• 120:09 - 120:11
can be solved.
• 120:11 - 120:13
Do I ask to be solved?
• 120:13 - 120:14
No.
• 120:14 - 120:14
Why?
• 120:14 - 120:18
Because you can finish
it later, finish at home.
• 120:18 - 120:27
Or maybe, I don't even want
you to compute on the final.
• 120:27 - 120:29
So how do we do that?
• 120:29 - 120:33
f is y.
• 120:33 - 120:37
Would I be able to choose
whichever order integration I
• 120:37 - 120:38
want?
• 120:38 - 120:40
It shouldn't matter which order.
• 120:40 - 120:43
It should be more
or less the same.
• 120:43 - 120:45
What if I do dy dx?
• 120:45 - 120:48
• 120:48 - 120:52
Then I have to do the Fubini.
• 120:52 - 120:54
But it's not a
rectangular domain.
• 120:54 - 120:55
Aha.
• 120:55 - 120:57
So Magdalena, be a
little bit careful
• 120:57 - 121:00
because this is going to
be two finite numbers,
• 121:00 - 121:02
but these are functions.
• 121:02 - 121:04
STUDENT: It will
be an x function.
• 121:04 - 121:08
PROFESSOR: So the x
is between 0 and 1,
• 121:08 - 121:10
and that's going to be z.
• 121:10 - 121:11
You do vertical strips.
• 121:11 - 121:14
That's a piece of cake.
• 121:14 - 121:18
But if you do the
vertical strips,
• 121:18 - 121:22
you have to pay attention to
the endpoints for x and y,
• 121:22 - 121:23
and one is easy.
• 121:23 - 121:24
Which one is trivial?
• 121:24 - 121:25
STUDENT: Zero.
• 121:25 - 121:26
PROFESSOR: The bottom one, zero.
• 121:26 - 121:29
The one that's nontrivial
is the upper one.
• 121:29 - 121:31
STUDENT: There will be 1 minus--
• 121:31 - 121:34
STUDENT: Square root
of 1 minus y squared.
• 121:34 - 121:34
PROFESSOR: Very good.
• 121:34 - 121:36
Square root of 1
minus y squared.
• 121:36 - 121:41
• 121:41 - 121:47
So if I were to go one more step
further without solving this,
• 121:47 - 121:51
I'm going to ask you, could
this be solved by hand?
• 121:51 - 121:58
Well, so you have
it in the book--
• 121:58 - 122:00
STUDENT: Professor, should be
a [INAUDIBLE] minus x squared?
• 122:00 - 122:03
• 122:03 - 122:04
PROFESSOR: Oh yeah.
• 122:04 - 122:05
1 minus x squared.
• 122:05 - 122:07
Excuse me.
• 122:07 - 122:08
Didn't I write it?
• 122:08 - 122:12
Yeah, here I should have written
y equals square root of 1
• 122:12 - 122:14
minus x squared.
• 122:14 - 122:21
So when you do it-- thank you
so much-- you go integrate,
• 122:21 - 122:27
and you have y squared over 2.
• 122:27 - 122:30
And you evaluate
between y equals 0
• 122:30 - 122:33
and y equals square
root 1 minus x squared,
• 122:33 - 122:35
and then you do the [INAUDIBLE].
• 122:35 - 122:41
• 122:41 - 122:45
In the book, they
do it differently.
• 122:45 - 122:50
They do it with respect to
dx and dy and integrate.
• 122:50 - 122:53
But it doesn't
matter how you do it.
• 122:53 - 122:55
You should get the same answer.
• 122:55 - 122:58
• 122:58 - 123:00
All right?
• 123:00 - 123:01
[INAUDIBLE]?
• 123:01 - 123:04
STUDENT: [INAUDIBLE]
in that way,
• 123:04 - 123:07
doesn't the square root work out
• 123:07 - 123:08
a y there?
• 123:08 - 123:09
PROFESSOR: In the other case--
• 123:09 - 123:11
STUDENT: Doing dy dx.
• 123:11 - 123:13
PROFESSOR: Yeah,
in the other way,
• 123:13 - 123:14
it works a little
bit differently.
• 123:14 - 123:17
You can do
u-substitution, I think.
• 123:17 - 123:20
So if you do it the other
way, it will be what?
• 123:20 - 123:24
Integral from 0 to
1, integral form 0
• 123:24 - 123:32
to square root of 1
minus y squared, y dx dy.
• 123:32 - 123:35
And what do you do in this case?
• 123:35 - 123:37
You have integral from 0 to 1.
• 123:37 - 123:43
Integral of y dx is going
to be y is a constant.
• 123:43 - 123:49
x between the two values will
be simply 1 minus y squared dy.
• 123:49 - 123:50
So you're right.
• 123:50 - 123:53
Matthew saw that,
because he's a prophet,
• 123:53 - 123:56
and he could see
• 123:56 - 123:58
This is very nice
what you observed.
• 123:58 - 123:59
What do you do?
• 123:59 - 124:03
You take a u-substitution
when you go home.
• 124:03 - 124:06
You get u equals
1 minus y squared.
• 124:06 - 124:13
du will be minus 2y
dy, and you go on.
• 124:13 - 124:17
So in the book, we got 1/3.
• 124:17 - 124:20
If you continue
with this method,
• 124:20 - 124:21
I think it's the same answer.
• 124:21 - 124:21
STUDENT: Yeah.
• 124:21 - 124:22
I got 1/3.
• 124:22 - 124:23
PROFESSOR: You got 1/3.
• 124:23 - 124:26
So sounds good.
• 124:26 - 124:28
We will stop here.
• 124:28 - 124:30
You will get homework.
• 124:30 - 124:33
How long should I
leave that homework on?
• 124:33 - 124:36
Because I'm thinking maybe
• 124:36 - 124:38
don't procrastinate.
• 124:38 - 124:41
So let's say until
the end of March.
• 124:41 - 124:44
And keep in mind
that we have included
• 124:44 - 124:48
one week of spring
break here, which you
• 124:48 - 124:51
can do whatever you want with.
• 124:51 - 124:58
Some of you may be in Florida
swimming and working on a tan,
• 124:58 - 124:59
and not working on homework.
• 124:59 - 125:02
So no matter how, plan ahead.
• 125:02 - 125:03
Plan ahead and you will do well.
• 125:03 - 125:10
31st of March for
the whole chapter.
• 125:10 - 125:11
Title:
TTU Math2450 Calculus3 Secs 12.1-12.2
Description:

Double Integral over a rectangular region and Double Integral over a non rectangular region

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Video Language:
English
 jackie.luft edited English subtitles for TTU Math2450 Calculus3 Secs 12.1-12.2