## TTU Math2450 Calculus3 Sec 12.5

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MAGDALENA TODA:
According to my watch,
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we are right on time to start.
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I may be one minute
early, or something.
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Do you have questions out of the
material we covered last time?
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What I'm planning on
doing-- let me tell you
• 0:20 - 0:21
what I'm planning to do.
• 0:21 - 0:24
I will cover triple
integrals today.
• 0:24 - 0:27
And this way, you
would have accumulated
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enough to deal with most of
the problems in homework four.
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You have mastered the
double integration by now,
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in all sorts of coordinates,
which is a good thing.
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Triple integrals
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If you have understood
the double integration,
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you will have no
problem understanding
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the triple integrals.
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The idea is the same.
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You look at different
domains, and then you
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realize that there are
Fubini-Tunelli type of results.
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I'm going to present
one right now.
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And there are also regions
of a certain type, that
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can be treated
differentially, and then you
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have cases in which reversing
the order of integration
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for those triple integrals
is going to help you a lot.
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OK, 12.5 is the name of the
section, triple integrals.
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So what should you imagine?
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You should imagine
that somebody gives you
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a function of three variables.
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Let's call that-- it doesn't
often have a name as a letter,
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but let's call it w.
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Being a function of three
coordinates, x, y, and z,
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where x, y, z is in R3.
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And we have some assumptions
about the domain D
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that you are working
over, and you
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have D as a closed-bounded
domain in R3.
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Examples that you're going
to do use frequently.
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Frequently used.
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A sphere, a ball,
actually, because in here,
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if a sphere is together with
a shell, it is the ball.
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Then you have some types of
polyhedra in r3 of all types.
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And by that, I mean
the classical polyhedra
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whose sides are just polygons.
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But you will also have some
curvilinear polyhedra, as well.
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What do I mean?
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I mean, we've seen that already.
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For example, somebody
give you a graph
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of a function, g of x and y, a
continuous function, and says,
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OK, can you estimate the
volume under the graph?
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Right?
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And until now, we treated
this volume under the graph
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as double integral of g of x, y,
continuous function over d, a,
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where d, a was dx dy.
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And we said double integral
over the projected domain
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in the plane.
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That's what I have.
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But can I treat it, this
volume, can I treat it
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as a triple integral?
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This is the question,
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so can I make three snakes?
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The answer is yes.
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And the way I'm
going to define that
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would be a triple
integral over a 3D domain.
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Let's call it curvilinear d in
r3, which is the volume-- which
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is the body under the graph
of this positive function,
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and above the projected
area in plane.
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So it's going to be a
cylinder in this case.
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And I'll put here 1 dv.
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And dv is a mysterious element.
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That's the volume element.
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And I will talk a little
bit about it right now.
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So what can you imagine?
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They give you a way to
look at it in the book.
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I mean, we give you a way
to look at it in the book.
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It's not very thorough
in explanations,
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but it certainly gives you the
general idea of what you want,
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what you need.
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So somebody gives you a potato.
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It doesn't have to
be this cylinder.
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It's something beautiful, a
body inside a compact surface.
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Let's say there are
no self-intersections.
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You have some compact surface,
like a sphere or a polyhedron,
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assume it's simply
connected, and it doesn't
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have any self-intersections.
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So a beautiful
potato that's smooth.
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If you imagine a potato
that has singularities,
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like most potatoes have
singularities, boo-boos,
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and cuts, so that's bad.
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So think about some
regular surface
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that's closed, no
self-intersections,
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and that is a potato
that is [INAUDIBLE].
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Oh let's call it--
p for potato, no,
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because I have got to
use p for the partition.
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So let me call it
D from 3D domain,
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because it's a three-dimensional
domain, enclosed
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by a curve, enclosed
by a compact surface.
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So think potato.
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What do we do in
terms of partitions?
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Those pixels were pixels for
the 2D world in flat line.
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But now, we don't
have pixels anymore.
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Yes we do.
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I was watching lots of
sci-fi, and the holograms
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have the
three-dimensional pixels.
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I'm going to try and
make a partition.
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It's going to be a hard way
to partition this potato.
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But you have to imagine you
have a rectangular partition,
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so every little pixel
will be a-- is not cube.
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It has to be a little
tiny parallelepiped.
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So a 3D pixel, let me
put pixel in quotes,
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because this is
kind of the idea.
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Well have what
kind of dimensions?
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We'll have three
dimensions, right?
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Three dimensions, a delta xk,
a delta yk, and a delta zk
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for the pixel number k.
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That's pixel number k.
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We have to number them,
see how many they are.
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Where k is from 1 to n, and
is the total number of 3D
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pixels that I'm covering
the whole thing.
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So don't think graphing
paper, anymore,
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because that's outdated.
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Don't even think of 2D image.
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Think of some hologram,
where you cover everything
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with tiny, tiny,
tiny 3D pixels so
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that in the limit, when you pass
to the limit, with respect to n
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and going to infinity,
the discrete image,
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you're going to have something
like a diamond shaped thingy,
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will convert to
the smooth potato.
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So as n goes to
infinity, that surface
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made of tiny, tiny squares
will convert to the data.
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So how do you actually find
a triple snake integral f
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of x, y, z over the
domain D, dx dy dz.
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OK, this theoretically
should be what?
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Think pixels.
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Limit as n goes to infinity
of the sum of the--
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what do I need to do?
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Think the whole
partition into pixels.
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How many pixels? n pixels
total is called a p.
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Script p.
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And the normal p will be the
highest diameter of-- highest
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diameter among all pixels.
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And you're going to
say, Oh, what the heck?
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I don't understand it.
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I have these three-dimensional
cubes, or three-dimensional
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barely by p that get tinier,
and tinier, and tinier.
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What in the world is going to
be a diameter of such a pixel?
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Well, you have to
take this pixel
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and magnify it so we can look
at it a little bit better.
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What do we mean by
diameter of this pixel?
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Let's call this pixel k.
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The maximum of all the distances
you can compute between two
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arbitrary points inside.
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Between two arbitrary
points inside,
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you have many [INAUDIBLE].
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So the maximum of the
distance between points,
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let's call them r and
q inside the pixel.
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So if it were for me
to ask you to find
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that diameter in this
case, what would that be?
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STUDENT: It's the diagonal.
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MAGDALENA TODA: It's the
diagonal between this corner
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and the opposite corner.
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So this would be the highest
distance inside this pixel.
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If it's a cube-- you see
that I wanted a cube.
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If it's a parallelepiped,
it's the same idea.
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So I have that opposite
corner distance kind of thing.
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OK.
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So I know what I want.
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I want n to go to infinity.
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That means I'm going to
have the p going to 0.
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The length of the
highest diameter
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will go shrinking to 0.
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And then I'm going to say
here, what do I have inside?
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F of some intermediate point.
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In every pixel, I take a point.
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Another pixel, another
point, and so on.
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So how many such
points do I have?
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n, because it's the
number of pixels.
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So inside, let's call
this as pixel p k.
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What is the little point that
I took out of the [INAUDIBLE]
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inside the cube, or
inside the pixel?
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Let's call that mister x k
star, y k star, x k star.
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Why do we put a star?
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Because he is a star.
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He wants to be number one
in these little domains,
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and says I'm a star.
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So we take that intermediate
point, x k star,
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y k star, x k star.
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Then we have this function
multiplied by the delta v k.
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somebody tell me what
this delta v k will mean,
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because ir really looks weird.
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And then k will be from 1 to n.
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So what do you do?
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You sum up all these
weighted volumes.
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This is a weight.
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So this is a volume.
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All these weighted volumes.
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We sum them up for all
the pixels k from 1 to n.
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We are going to get
something like this cover
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with tiny parallelepipeds
in the limit,
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as the partitions' [? norm ?]
go to 0, or the number of pixels
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goes to infinity.
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This discrete
surface will converge
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to the beautiful smooth
potato, and give you
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a perfect linear image.
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Actually, if we saw a
hologram, this is what it is.
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Our eyes actually
see a bunch of I
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tiny-- many, many, many,
many, millions of pixels
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that are cubes in 3D.
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But it's an optical illusion.
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We see, OK, it's a curvilinear,
it's a smooth body of a person.
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It's not smooth at all.
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If you get closer and
closer to that diagram
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and put your eye
glasses on, you are
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going to see, oh, this
is not a real person.
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It's made of pixels
that are all cubes.
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just the same, you see your
digital image of your picture
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on Facebook, whatever it is.
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If you would be able
to be enlarge it,
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you would see the pixels, being
little tiny squares there.
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The graphical
imaging has improved.
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The quality of our digital
imaging has improved a lot.
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But of course, 20 years ago,
when you weren't even born,
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we could still see the pixels
in the photographic images
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in a digital camera.
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Those tiny first cameras,
what was that, '98?
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STUDENT: Kodak.
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MAGDALENA TODA: Like AOL
cameras that were so cheap.
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The cheaper the camera,
the worse the resolution.
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I remember some resolutions
like 400 by 600.
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STUDENT: Black and white.
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MAGDALENA TODA: Not
black and white.
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Black and white
would have been neat.
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But really nasty in the sense
that you had the feeling
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that the colors
were not even-- they
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were blending into each
other, because the resolution
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was so small.
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So it was not at all
pleasing to the eye.
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What was good is that
any kind of defects you
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would have, something
like a pimple
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could not be seen in that,
because the resolution was
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so slow that you couldn't
see the boo-boos,
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the pimples, the defects
of a face or something.
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Now, you can see.
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With the digital cameras
we have now, we can do,
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Photoshop, and all of us
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will look great if we
photoshop our pictures.
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OK.
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So this is what it
is in the limit.
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But in reality, in
the everyday reality,
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you cannot take Riemann
sums like that--
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this is a Riemann
approximating sum--
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and then cast to the limit, and
get ideal curvilinear domains.
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No.
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You don't do that.
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You have to deal
with the equivalent
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of the fundamental
theorem of calculus
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from Calc I, which is called the
Fubini-Tonelli type of theorem
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in Calc III.
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So say it again.
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So the Fubini-Tunelli
theorem that you
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learned for double
integrals over a rectangle
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can be generalized to the
Fubini-Tonelli theorem
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over a parallelepiped.
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And it's the same
thing, practically,
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as applying the fundamental
theorem of calculus in Calc I.
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So somebody says, well, let me
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I give you a-- you
will say, Magdalena,
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you are offending us.
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This is way too easy.
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What do you think, that we
cannot understand the concept?
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the simplest possible example
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that I think of.
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So x is between a and b.
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In my case, they will
be positive numbers,
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because I want everything
to be in the first octant.
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First octant means x positive,
y positive, and z positive
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all together.
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To make my life easier,
I take that example,
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and I say, I know the
numbers for x, y, z.
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I would like you to
compute two integrals.
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One would be the
volume of this object.
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Let's call it body.
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It's not a dead body,
it's just body in 3D.
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The volume of the
body, we say it
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like a mathematician, V of B.
What is that by definition?
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Who's going to tell me?
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Triple snake.
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Don't say triple
snake to other people,
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because other professors
are more orthodox than me.
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They will laugh-- they
will not joke about it.
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So triple integral over the
body of-- to get the volume,
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the weight must be 1.
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f integral must be 1, and
then you have exactly dV.
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How can I convince
you what we have here,
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in terms of Fubini-Tonelli?
• 19:03 - 19:05
It's really beautiful.
• 19:05 - 19:12
B is going to be a, b segment
cross product, c, d segment,
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cross product.
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What is the altitude?
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E, f, e to f.
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Interval e to f
means the height.
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So length, width, height.
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This is the box, or carry-on,
or USPS parcel, or whatever
• 19:28 - 19:32
box you want to measure.
• 19:32 - 19:38
So how am I going to set up
the Fubini-Tonelli integral?
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a to b, c to d, e to f, 1.
• 19:44 - 19:46
And now, who counts first?
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dz, dy, dx.
• 19:48 - 19:55
So it is like the equivalent
of the vertical strip
• 19:55 - 19:59
thingy in double corners,
double integrals.
• 19:59 - 19:59
Yes, sir?
• 19:59 - 20:02
STUDENT: Professor,
why did you use 1 dV?
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Why 1?
• 20:03 - 20:04
MAGDALENA TODA: OK.
• 20:04 - 20:06
You'll see in a second.
• 20:06 - 20:09
This is the same
thing we do for areas.
• 20:09 - 20:14
So when you compute an
area-- very good question.
• 20:14 - 20:18
If you use 1 here, and
you put delta [? ak ?]
• 20:18 - 20:21
that is the graphing paper area.
• 20:21 - 20:25
It's going to be
all the tiny areas,
• 20:25 - 20:28
summed up, sum of all
the delta [? ak ?] which
• 20:28 - 20:31
means this little pixel,
plus this little pixel,
• 20:31 - 20:33
plus this little pixel,
plus this little pixel,
• 20:33 - 20:38
plus 1,000 pixels all together
will cover up the area.
• 20:38 - 20:42
If you have the
volume of a potato,
• 20:42 - 20:44
a body that is alive,
but shouldn't move.
• 20:44 - 20:47
OK, it should stay in one place.
• 20:47 - 20:50
Then, to compute the
volume of the potato,
• 20:50 - 20:54
you have to say, the
potato, the smooth potato,
• 20:54 - 20:59
is the limit of the sum of
all the tiny cubes of potato,
• 20:59 - 21:03
if you cut the potato in many
cubes, like you cut cheese.
• 21:03 - 21:06
They got a bunch of cheddar
cheese into small cubes,
• 21:06 - 21:11
and they feed us with crackers
and wine-- OK, no comments.
• 21:11 - 21:15
So you have delta vk, you
have 1,000 little cubes,
• 21:15 - 21:18
tiny, tiny, tiny,
like that Lego.
• 21:18 - 21:20
OK, forget about the cheese.
• 21:20 - 21:23
The cheese cubes
are way too big.
• 21:23 - 21:26
So imagine Legos that
are really performing
• 21:26 - 21:29
with millions of little pieces.
• 21:29 - 21:36
Have you seen the exhibit, Lego
exhibit with almost invisible
• 21:36 - 21:40
Legos at the Civic Center?
• 21:40 - 21:43
They have that art festival.
• 21:43 - 21:46
How many of you go
to the art festival?
• 21:46 - 21:49
Is it every April?
• 21:49 - 21:51
Something like that.
• 21:51 - 21:54
So imagine those little
tiny Legos, but being cubes
• 21:54 - 21:55
and put together.
• 21:55 - 21:56
This is what it is.
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So f Vy.
• 21:58 - 22:02
Now, can we verify
the volume of a box?
• 22:02 - 22:02
It's very easy.
• 22:02 - 22:04
What do we do?
• 22:04 - 22:08
Well, first of all, I
would do it in a slow way,
• 22:08 - 22:10
and you are going to
shout at me, I know.
• 22:10 - 22:14
But I'll tell you why
you need to bear with me.
• 22:14 - 22:17
So integral of 1 dz goes first.
• 22:17 - 22:19
That's z between f and e.
• 22:19 - 22:21
So it's f minus e, am I right?
• 22:21 - 22:23
You say, duh, that's
to easy for me.
• 22:23 - 22:26
I'm know it's too
easy for me, but I'm
• 22:26 - 22:27
going somewhere with it.
• 22:27 - 22:30
dy dx.
• 22:30 - 22:35
The one inside, f minus e
is a constant, pulls out,
• 22:35 - 22:37
completely out of the product.
• 22:37 - 22:43
And then I have integral from
a to b of-- what is that left?
• 22:43 - 22:47
1 dy, y between d and c.
• 22:47 - 22:48
So d minus c, right?
• 22:48 - 22:53
d minus c dx.
• 22:53 - 22:57
And so on and so
forth, until I get it.
• 22:57 - 23:02
If minus c times d
minus c times b minus a,
• 23:02 - 23:06
and goodbye, because this
is the volume of the box.
• 23:06 - 23:08
It's the height.
• 23:08 - 23:10
This is the height.
• 23:10 - 23:13
No, excuse me, guys.
• 23:13 - 23:16
The height is-- this
one is the height.
• 23:16 - 23:21
This is the width, and this is
the length, whatever you want.
• 23:21 - 23:22
All right.
• 23:22 - 23:25
How could I have done it if
I were a little bit smarter?
• 23:25 - 23:28
STUDENT: You could have just
put it in three integrals.
• 23:28 - 23:30
MAGDALENA TODA:
Right Hey, I have
• 23:30 - 23:36
a theorem, just like
before, which says
• 23:36 - 23:38
three integrals in a product.
• 23:38 - 23:40
This is what Matt
immediately remembered.
• 23:40 - 23:44
We had two integrals
in a product last time.
• 23:44 - 23:47
So what have we proved
in double integrals
• 23:47 - 23:50
remains valid in
triple integrals
• 23:50 - 23:53
if we have something like that.
• 23:53 - 23:55
So I'm going the same theorem.
• 23:55 - 23:56
It's in the book.
• 23:56 - 23:57
We have a proof.
• 23:57 - 24:02
So you have integral from
a to b, c to d, e to f.
• 24:02 - 24:08
And then, some guys that you
like, f of x, times g of y,
• 24:08 - 24:11
times h of z.
• 24:11 - 24:16
Functions of x, y, z,
separated variables.
• 24:16 - 24:19
So f, a function of
x only, g a function
• 24:19 - 24:21
of y only, h a
function of z only.
• 24:21 - 24:23
This is the complicated case.
• 24:23 - 24:26
And then I have
• 24:26 - 24:28
STUDENT: dz, dy, dx.
• 24:28 - 24:29
MAGDALENA TODA: dz, dy, dx.
• 24:29 - 24:30
Excellent.
• 24:30 - 24:33
Thanks for whispering,
because I was a little bit
• 24:33 - 24:36
confused for a second.
• 24:36 - 24:39
So, just as Matt said,
go ahead and observe
• 24:39 - 24:42
that you can treat them one
at a time like you did here,
• 24:42 - 24:47
and integrate one at a time, and
integrate again, and pull out
• 24:47 - 24:50
a constant, integrate
again, pull out a constant.
• 24:50 - 24:53
But practically this is
exactly the same as integral
• 24:53 - 25:02
from a to b of f of x
alone, dx, times integral
• 25:02 - 25:12
from c to d, g of y alone,
dy, and times integral from e
• 25:12 - 25:18
to f of h of z, dz, and close.
• 25:18 - 25:22
So you've seen the version
of the double integral,
• 25:22 - 25:27
and this is the same result
for triple integrals.
• 25:27 - 25:30
And it's practically--
what is the proof?
• 25:30 - 25:34
You just pull out one
at a time, so the proof
• 25:34 - 25:38
is that you start working and
say, mister z counts here,
• 25:38 - 25:40
and he's the only
one that counts.
• 25:40 - 25:44
These guys get out for a
walk one at a time outside
• 25:44 - 25:46
of the first integral inside.
• 25:46 - 25:49
And then, integral
of h of z, dz,
• 25:49 - 25:53
over the corresponding domain,
will be just a constant, c1,
• 25:53 - 25:55
that pulls out.
• 25:55 - 26:00
And that is that--
c1 that pulls out.
• 26:00 - 26:03
Ans since you pull them out
in this product one at a time,
• 26:03 - 26:05
that's what you get.
• 26:05 - 26:09
I'm not going to give you this
as an exercise in the midterm
• 26:09 - 26:12
with a proof, but this is
one of the first exercises
• 26:12 - 26:16
I had as a freshman
in my multi--
• 26:16 - 26:20
I took it as a freshman,
as multivariable calculus.
• 26:20 - 26:23
And it was a pop quiz.
• 26:23 - 26:27
My professor just came
one day, and said, guys,
• 26:27 - 26:30
you have to try to do this
[? before ?] by yourself.
• 26:30 - 26:33
And some of us did,
some of us didn't.
• 26:33 - 26:35
To me, it really
looked very easy.
• 26:35 - 26:41
I was very happy to prove it,
in an elementary way, of course.
• 26:41 - 26:41
OK.
• 26:41 - 26:46
So how hard is it
to generalize, to go
• 26:46 - 26:48
to non-rectangular domains?
• 26:48 - 26:49
Of course it's a pain.
• 26:49 - 26:54
It's really a pain,
like it was before.
• 26:54 - 27:00
But you will be able to
figure out what's going on.
• 27:00 - 27:03
In most cases,
you're going to have
• 27:03 - 27:07
a domain that's really
not bad, a domain that
• 27:07 - 27:09
has x between fixed values.
• 27:09 - 27:14
For example y between
• 27:14 - 27:20
something like f of x and
g of x, top and bottom.
• 27:20 - 27:22
That's what you had
for double integral.
• 27:22 - 27:26
in this case, you
• 27:26 - 27:31
will have z between--
let's make this guy
• 27:31 - 27:36
big F and big G,
other functions.
• 27:36 - 27:38
This is going to be
a function of x, y.
• 27:38 - 27:41
This is going to be
a function of x, y,
• 27:41 - 27:44
and that's the
upper and the lower.
• 27:44 - 27:50
And find the triple integral
of, let's say 1 over d dV
• 27:50 - 27:53
will be a volume of the potato.
• 27:53 - 27:56
Now, I'm sick of potatoes,
because they're not
• 27:56 - 27:59
my favorite food.
• 27:59 - 28:03
Let me imagine I'm making a
tetrahedron, a lot of cheese.
• 28:03 - 28:09
• 28:09 - 28:14
I'm going to draw this same
tetrahedron from last time.
• 28:14 - 28:15
So what did we do last time?
• 28:15 - 28:18
We took a plane
that was beautiful,
• 28:18 - 28:22
and we said let's
cut with that plane.
• 28:22 - 28:25
This is the plane we are
cutting the cheese with.
• 28:25 - 28:26
It's a knife.
• 28:26 - 28:28
x plus y plus z equals 1.
• 28:28 - 28:30
Imagine that there's
an infinite knife that
• 28:30 - 28:32
comes into the frame.
• 28:32 - 28:33
Everything is cheese.
• 28:33 - 28:37
The space, the universe is
covered in solid cheese.
• 28:37 - 28:42
So the whole thing,
the Euclidean space
• 28:42 - 28:44
is covered in cheddar cheese.
• 28:44 - 28:45
That's all there.
• 28:45 - 28:48
From everywhere, you
come with this knife,
• 28:48 - 28:53
and you cut along
this plane-- hi
• 28:53 - 28:55
let's call this
[? high plane. ?]
• 28:55 - 29:00
And then you cut the x
plane along the x, y plane,
• 29:00 - 29:03
y, z plane and x, z plane.
• 29:03 - 29:04
What are these called?
• 29:04 - 29:07
Planes of coordinates.
• 29:07 - 29:07
And what do you obtain?
• 29:07 - 29:09
Then, you throw
everything away, and you
• 29:09 - 29:15
maintain only the
tetrahedron made of cheese.
• 29:15 - 29:18
Now, you remember
what the corners were.
• 29:18 - 29:19
This is 0, 0, 0.
• 29:19 - 29:20
It's a piece of cake.
• 29:20 - 29:23
But I want to know the vertices.
• 29:23 - 29:28
And you know them, and I
don't want to spend time
• 29:28 - 29:30
discussing why you know them.
• 29:30 - 29:31
So
• 29:31 - 29:31
STUDENT: 0--
• 29:31 - 29:32
MAGDALENA TODA: 1, 0, 0.
• 29:32 - 29:33
Thank you.
• 29:33 - 29:34
Huh?
• 29:34 - 29:35
STUDENT: 0, 1, 0.
• 29:35 - 29:36
MAGDALENA TODA: Yes.
• 29:36 - 29:38
And 0, 0, 1.
• 29:38 - 29:40
All right.
• 29:40 - 29:40
Great.
• 29:40 - 29:44
The only thing is, if we see
the cheese being a solid,
• 29:44 - 29:50
we don't see this part, the
three axes of corners behind.
• 29:50 - 29:56
so I'm going to make them
dotted, and you see the slice,
• 29:56 - 29:59
here, it has to
be really planar.
• 29:59 - 30:01
And you ask yourself,
how do you set up
• 30:01 - 30:04
the triple integral
that represents
• 30:04 - 30:06
the volume of this object?
• 30:06 - 30:07
Is it hard?
• 30:07 - 30:09
It shouldn't be hard.
• 30:09 - 30:13
You just have to think what
the domain will be like,
• 30:13 - 30:17
and you say the domain is
inside the tetrahedron.
• 30:17 - 30:18
Do you want d or t?
• 30:18 - 30:21
T from tetrahedron.
• 30:21 - 30:22
It doesn't matter.
• 30:22 - 30:23
We have a new name.
• 30:23 - 30:27
We get bored of all sorts
of names and notations.
• 30:27 - 30:28
We change them.
• 30:28 - 30:34
Mathematicians have imagination,
so we change our notations.
• 30:34 - 30:35
Like we cannot change
our identities,
• 30:35 - 30:38
and we suffer because of that.
• 30:38 - 30:41
So you can be a nerd
mathematician imagining
• 30:41 - 30:44
you're Spiderman,
and you can take,
• 30:44 - 30:48
give any name you want, and
you can adopt a new name,
• 30:48 - 30:52
and this is behind
our motivation
• 30:52 - 30:56
why we like to change names
and change notation so much.
• 30:56 - 30:57
OK?
• 30:57 - 31:03
So we have triple integral of
this T. All right, of what?
• 31:03 - 31:06
1 dV.
• 31:06 - 31:07
Good.
• 31:07 - 31:10
Now we understand
what we need to do,
• 31:10 - 31:12
just like [? Miteish ?]
• 31:12 - 31:16
OK, now we know this is going
to be a limit of little cubes.
• 31:16 - 31:18
If were to cover
this piece of cheese
• 31:18 - 31:22
in tiny, tiny,
infinitesimally small cubes.
• 31:22 - 31:25
But now we know a
method to do it.
• 31:25 - 31:30
So according to--
Fubini-Tonelli type of result.
• 31:30 - 31:39
We would have a between--
no, x-- is first, dz.
• 31:39 - 31:43
z is first, y is moving
next, x is moving last.
• 31:43 - 31:47
z is constrained to
move between a and b.
• 31:47 - 31:51
But in this case, a and b should
be prescribed by you guys,
• 31:51 - 31:56
because you should think
where everybody lives.
• 31:56 - 32:00
Not you, I mean the coordinates
in their imaginary world.
• 32:00 - 32:03
The coordinates
represent somebodies.
• 32:03 - 32:04
STUDENT: 0.
• 32:04 - 32:08
MAGDALENA TODA: x, 0 to 1.
• 32:08 - 32:11
How should I give you
a feeling for that?
• 32:11 - 32:12
Just draw this line.
• 32:12 - 32:15
This red segment between 0 to 1.
• 32:15 - 32:18
That expresses everything
• 32:18 - 32:23
into pictures, because every
picture is worth 1,000 words.
• 32:23 - 32:27
y is married to
x, unfortunately.
• 32:27 - 32:31
y cannot say, oh, I am y,
I'm going wherever I want.
• 32:31 - 32:34
He hits his head against
this purple line.
• 32:34 - 32:37
He cannot go beyond
that purple line.
• 32:37 - 32:40
He's constrained, poor y.
• 32:40 - 32:41
So he says, I'm moving.
• 32:41 - 32:42
I'm mister y.
• 32:42 - 32:46
I'm moving in this direction,
but I cannot go past the purple
• 32:46 - 32:49
line in plane here.
• 32:49 - 32:52
• 32:52 - 32:57
I need you, because
if you go, I'm lost.
• 32:57 - 32:59
y is between 0 and--
• 32:59 - 33:00
STUDENT: 1 minus x.
• 33:00 - 33:01
MAGDALENA TODA: 1 minus x.
• 33:01 - 33:02
Excellent Roberto.
• 33:02 - 33:05
• 33:05 - 33:08
The purple line has
equation-- how do you
• 33:08 - 33:11
get to the equation of the
purple line, first of all?
• 33:11 - 33:15
your plug in z equals 0.
• 33:15 - 33:19
So the purple line would
be x plus y equals 1.
• 33:19 - 33:25
And so mister y will
be 1 minus x here.
• 33:25 - 33:28
That's how you got it.
• 33:28 - 33:32
And finally, z is that--
mister z foes from the floor
• 33:32 - 33:34
all the way--
imagine somebody who
• 33:34 - 33:40
is like-- z is a
helium balloon, and he
• 33:40 - 33:43
is left-- you let him
go from the floor,
• 33:43 - 33:45
and he goes all the
way to the ceiling.
• 33:45 - 33:49
And the ceiling is not
flat like our ceiling.
• 33:49 - 33:55
The ceiling is
this oblique plane.
• 33:55 - 33:59
So z is going to hit his head
against the roof at some point,
• 33:59 - 34:01
and he doesn't know
where he is going
• 34:01 - 34:06
to hit his head, unless you
tell him where that happens.
• 34:06 - 34:11
So he knows he leaves at 0,
and he's going to end up where?
• 34:11 - 34:13
STUDENT: 1 minus y minus x.
• 34:13 - 34:14
MAGDALENA TODA: Excellent.
• 34:14 - 34:16
1 minus x minus y.
• 34:16 - 34:18
How do we do that?
• 34:18 - 34:23
We pull z out of that, and
say, 1 minus x minus y.
• 34:23 - 34:27
So that is the equation of
the shaded purple plane,
• 34:27 - 34:30
and this is as
far as you can go.
• 34:30 - 34:34
You cannot go past the
roof of your house,
• 34:34 - 34:38
which is the purple plane,
the purple shaded plane.
• 34:38 - 34:40
So here you are.
• 34:40 - 34:40
Is this hard?
• 34:40 - 34:41
No.
• 34:41 - 34:44
In many problems on the
final and on the midterm,
• 34:44 - 34:48
we tell you, don't even
think about solving that,
• 34:48 - 34:53
because we believe you.
• 34:53 - 34:56
Just set up the integral.
• 34:56 - 34:59
I might give you something
like that again, just
• 34:59 - 35:04
set up the integral and
you have to do that.
• 35:04 - 35:08
But now, I would like
to actually work it out,
• 35:08 - 35:12
see how hard it is.
• 35:12 - 35:15
So is this hard
to work this out?
• 35:15 - 35:18
• 35:18 - 35:21
I have to do it one at
a time, because you see,
• 35:21 - 35:24
I don't have fixed endpoints.
• 35:24 - 35:28
I cannot say, I'm applying the
problem with the integral if f
• 35:28 - 35:32
times the integral of g, times--
so I have to integrate one
• 35:32 - 35:38
at a time, because I don't
have fixed endpoints.
• 35:38 - 35:42
And the integral of 1dz is
z between that and that.
• 35:42 - 35:47
So z, 1 minus x minus y
will be what's left over,
• 35:47 - 35:50
and then I have dy,
and then I have dx.
• 35:50 - 35:54
And at this point it
looks horrible enough,
• 35:54 - 35:56
but we have to pray
that in the end
• 35:56 - 36:02
it's not going to be so hard,
and I'm going to keep going.
• 36:02 - 36:05
So we have integral from 0 to 1.
• 36:05 - 36:12
We have integral
from 0 to 1 minus x..
• 36:12 - 36:17
I'll just copy and paste it.
• 36:17 - 36:20
Which is integral from 0 to 1.
• 36:20 - 36:25
Now I have to think,
and that's dangerous.
• 36:25 - 36:27
I have 1 minux x
with respect to y.
• 36:27 - 36:29
This is going to be ugly.
• 36:29 - 36:34
That's a constant with
respect to y, and times y,
• 36:34 - 36:39
minus-- integrate with
respect to y, y is [? what? ?]
• 36:39 - 36:40
y squared over 2.
• 36:40 - 36:46
• 36:46 - 36:49
Between y equals 0 down.
• 36:49 - 36:52
That's going to save my
life, because for y equals 0,
• 36:52 - 36:55
0 is going to be a
great simplification.
• 36:55 - 37:00
And for y equals
1 minus x on top,
• 37:00 - 37:02
hopefully it's not going
to be the end of the world.
• 37:02 - 37:07
It looks ugly now, but
I'm an optimistic person,
• 37:07 - 37:11
so I hope that this is
going to get better.
• 37:11 - 37:14
And I can see it's
going to get better.
• 37:14 - 37:16
So I have integral
from here to 1.
• 37:16 - 37:18
And now I say, OK, let me think.
• 37:18 - 37:20
Life is not so bad.
• 37:20 - 37:21
Why?
• 37:21 - 37:25
1 minus x, 1 minus x
is 1 minus x squared.
• 37:25 - 37:28
I could think faster, you
could think faster than me,
• 37:28 - 37:30
but I don't want to rush.
• 37:30 - 37:36
1 minus x squared over 2.
• 37:36 - 37:37
So it's not bad at all.
• 37:37 - 37:43
Look, I'm getting this guy who
is beautiful in the end, when
• 37:43 - 37:45
I'm going to
integrate, and you have
• 37:45 - 37:48
to keep your fingers
crossed for me,
• 37:48 - 37:54
because I don't know
what I'm going to get.
• 37:54 - 38:03
So I get integral from 0 to 1,
1/2 out, 1 minus x squared dx.
• 38:03 - 38:06
• 38:06 - 38:08
Can you do this by
yourself without my help?
• 38:08 - 38:11
What are you going to do?
• 38:11 - 38:16
x squared minus 2x plus 1.
• 38:16 - 38:17
That's the square.
• 38:17 - 38:19
STUDENT: Why not just change it?
• 38:19 - 38:20
MAGDALENA TODA: Huh?
• 38:20 - 38:22
STUDENT: Why not just change it?
• 38:22 - 38:24
MAGDALENA TODA: You
can do it in many ways.
• 38:24 - 38:26
You can do whatever you want.
• 38:26 - 38:28
I don't care.
• 38:28 - 38:32
I want you to the right
answer one way or another.
• 38:32 - 38:35
So I'm going to clean a
little bit around here.
• 38:35 - 38:40
• 38:40 - 38:41
It's dirty.
• 38:41 - 38:42
You do it.
• 38:42 - 38:47
You have one minute
and a half to finish.
• 38:47 - 38:50
And tell me what you get.
• 38:50 - 38:52
STUDENT: 1 minus x
cubed over six negative.
• 38:52 - 38:53
MAGDALENA TODA: No, no.
• 38:53 - 38:55
In the end is the number.
• 38:55 - 38:57
What number?
• 38:57 - 38:59
But you have to go slow.
• 38:59 - 39:02
I need three people to
give me the same answer.
• 39:02 - 39:04
Because then it's like in that
proverb, if two people tell
• 39:04 - 39:06
you drunk, you go to bed.
• 39:06 - 39:10
I need three people to tell
me what the answer is in order
• 39:10 - 39:13
to believe them.
• 39:13 - 39:14
Three witnesses.
• 39:14 - 39:16
STUDENT: 1 [? by ?] 6.
• 39:16 - 39:19
MAGDALENA TODA: Who got
1 over 6, raise hand?
• 39:19 - 39:20
Wow, guys, you're fast.
• 39:20 - 39:23
Can you raise hands again?
• 39:23 - 39:26
OK, being fast doesn't
mean you're the best,
• 39:26 - 39:30
but I agree you do a very good
job, all of you in general.
• 39:30 - 39:35
So I believe there were
eight people or nine people.
• 39:35 - 39:36
1 over 6.
• 39:36 - 39:41
Now, how could I have cheated
on this problem on the final?
• 39:41 - 39:43
STUDENT: It's a
[? junction ?] from this--
• 39:43 - 39:44
MAGDALENA TODA: Right.
• 39:44 - 39:48
In this case, being a volume,
I would have been lucky enough,
• 39:48 - 39:51
and say, it is the
volume of a tetrahedron.
• 39:51 - 39:56
I go, the tetrahedron
has area of the base 1/2,
• 39:56 - 39:57
the height is 1.
• 39:57 - 40:01
1/2 times 1 divided by 3 is 1/6.
• 40:01 - 40:06
And just pretend on the
final that I actually
• 40:06 - 40:07
computed everything.
• 40:07 - 40:12
I could have done that, from
here jump to here, or from here
• 40:12 - 40:13
jump straight to here.
• 40:13 - 40:16
And ask you, how did you
get from here to here?
• 40:16 - 40:19
And you say, I'm a genius.
• 40:19 - 40:20
Could I not believe you?
• 40:20 - 40:23
I have to give you full credit.
• 40:23 - 40:28
However, what would you
have done if I said compute,
• 40:28 - 40:31
I don't know, something
worse, something
• 40:31 - 40:37
like triple integral of x,
y, z over the tetrahedron 2.
• 40:37 - 40:40
In that case, you cannot cheat.
• 40:40 - 40:42
You're not lucky
enough to cheat.
• 40:42 - 40:44
You're lucky enough
to cheat when
• 40:44 - 40:47
you have a volume
of a prism, you
• 40:47 - 40:50
have a volume of-- and volume
means this should be the number
• 40:50 - 40:52
1 here, number 1.
• 40:52 - 40:56
So if you have number 1, here,
or I ask you for the volume,
• 40:56 - 40:59
and it's a prism, or
tetrahedron, or sphere,
• 40:59 - 41:02
or something, go
• 41:02 - 41:04
and pretend that you're
actually solving the integral.
• 41:04 - 41:05
Yes, sir.
• 41:05 - 41:07
STUDENT: What would that
represent, geometrically,
• 41:07 - 41:09
the triple integral of x, y, z?
• 41:09 - 41:12
MAGDALENA TODA: It's a
weighted triple integral.
• 41:12 - 41:16
I'm going to give
you examples later.
• 41:16 - 41:20
When you have mass and momentum,
when you compute the center
• 41:20 - 41:25
map, or you compute the
mass, and somebody give you
• 41:25 - 41:26
densities.
• 41:26 - 41:33
Let me get -- If you have a
triple integral over row at x,
• 41:33 - 41:37
y, z, this could be it,
but I [? recall ?] it row
• 41:37 - 41:40
for a reason, not just for fun.
• 41:40 - 41:43
And here, dx, dy, dz.
• 41:43 - 41:45
Very good question, and
it's very insightful.
• 41:45 - 41:48
For a physicist or
engineer, the guy
• 41:48 - 41:52
needs to know why we take
this weighted [? integral. ?]
• 41:52 - 41:55
If row is the
density of an object,
• 41:55 - 41:59
if it's everywhere the same, if
row is a homogeneous density,
• 41:59 - 42:02
for that piece of cheddar
cheese-- Oh my God
• 42:02 - 42:05
I'm so hungry-- row
would be constant.
• 42:05 - 42:08
If it's a quality cheddar
made in Vermont in the best
• 42:08 - 42:13
factory, whatever, row would
be considered to be a constant,
• 42:13 - 42:14
right?
• 42:14 - 42:15
And in that case, what happens?
• 42:15 - 42:18
If it's a constant,
it's a gets out,
• 42:18 - 42:21
and then you have row
times triple integral 1
• 42:21 - 42:23
dV, which is what?
• 42:23 - 42:25
The volume.
• 42:25 - 42:28
And then the volume times the
density of the piece of cheese
• 42:28 - 42:29
will be?
• 42:29 - 42:30
STUDENT: [INAUDIBLE]
• 42:30 - 42:33
MAGDALENA TODA: The mass
of the piece of cheese,
• 42:33 - 42:37
in kilograms, because I
think in kilograms because I
• 42:37 - 42:38
can eat more.
• 42:38 - 42:39
OK?
• 42:39 - 42:42
Actually, no, I'm just kidding.
• 42:42 - 42:47
You guys have really-- I
mean, 2 pounds and 1 kilogram
• 42:47 - 42:48
is not the same thin.
• 42:48 - 42:50
Can somebody tell me why?
• 42:50 - 42:52
I mean, you know it's not
the same thing because,
• 42:52 - 42:54
the approximation.
• 42:54 - 42:58
But I'm claiming you cannot
compare pounds with kilograms
• 42:58 - 42:58
at all.
• 42:58 - 43:00
STUDENT: Pounds is
a measure of weight,
• 43:00 - 43:02
whereas kilograms is
a measure of mass.
• 43:02 - 43:03
MAGDALENA TODA: Excellent.
• 43:03 - 43:05
Kilogram is a measure
of mass, pound
• 43:05 - 43:08
is a measure of the
gravitational force.
• 43:08 - 43:11
It's a force measure.
• 43:11 - 43:16
So OK.
• 43:16 - 43:20
• 43:20 - 43:23
Which reminds me,
there was-- I don't
• 43:23 - 43:25
know if you saw this short
movie for 15 minutes that
• 43:25 - 43:29
got an award the
previous Oscar last year,
• 43:29 - 43:35
and there was an old lady
telling another old lady
• 43:35 - 43:42
in Great Britain, get
2 pounds of sausage.
• 43:42 - 43:45
And the other one says,
I thought we got metric,
• 43:45 - 43:47
because we are in
the European Union.
• 43:47 - 43:51
And she said, then get me
just the one meter of sausage,
• 43:51 - 43:53
or something.
• 43:53 - 43:55
So it was funny.
• 43:55 - 43:56
So it can be mass.
• 43:56 - 44:00
But what if this
density is not the same?
• 44:00 - 44:04
This is exactly why we
need to do the integral.
• 44:04 - 44:08
Imagine that the
density is-- we have
• 44:08 - 44:11
a piece of cake with layers.
• 44:11 - 44:13
And again, you see
how hungry I am.
• 44:13 - 44:19
So you have a layer, and
then cream, or whipped cream,
• 44:19 - 44:22
or mousse, and another
layer, and another mousse.
• 44:22 - 44:25
The density will vary.
• 44:25 - 44:29
But then there are bodies in
physics where the density is
• 44:29 - 44:31
even a smooth function.
• 44:31 - 44:38
It doesn't matter that you have
such a discontinuous function.
• 44:38 - 44:39
What would you do?
• 44:39 - 44:40
You just split.
• 44:40 - 44:47
You have triple row 1 for the
first layer, then triple row 2
• 44:47 - 44:50
for the second later,
the layer of mousse,
• 44:50 - 44:53
and then let's
say it's tiramisu,
• 44:53 - 44:58
you have another layer, row
three, dV3 for the top layer
• 44:58 - 44:59
of the tiramisu.
• 44:59 - 45:00
STUDENT: Can any row
be kept constant?
• 45:00 - 45:03
MAGDALENA TODA: So
these are discontinuous.
• 45:03 - 45:05
They are all constant, though.
• 45:05 - 45:07
That would be the
• 45:07 - 45:11
because presumably mousse would
have the constant density,
• 45:11 - 45:15
the dough has a constant,
homogeneous density, and so on.
• 45:15 - 45:19
But what if the density
varies in that body from point
• 45:19 - 45:20
to point?
• 45:20 - 45:23
Then nobody can do
it by approximation.
• 45:23 - 45:27
You'd say volume, mass 1 plus
mass 2 plus mass 3 plus mass 1.
• 45:27 - 45:31
You have to have a triple
integral where this row varies,
• 45:31 - 45:33
constantly varies.
• 45:33 - 45:35
And for an engineer,
that would be a puzzle.
• 45:35 - 45:39
Poor engineers says,
oh my God, the density
• 45:39 - 45:40
is different from
one point to another.
• 45:40 - 45:43
I have to find an
approximated function
• 45:43 - 45:48
for that density moving from one
point to another on that body.
• 45:48 - 45:55
And then the only way to do it
would be to solve an integral.
• 45:55 - 45:57
Imagine that somebody--
now it just occurred,
• 45:57 - 46:02
I never thought
about it-- we would
• 46:02 - 46:05
be measured in terms of
this type of integral.
• 46:05 - 46:11
Of course, people would be able
to measure mass right away.
• 46:11 - 46:13
But then, if you were
to know the density--
• 46:13 - 46:18
you cannot even know the density
at every point of the body.
• 46:18 - 46:21
It varies a lot, so
every point of our bodies
• 46:21 - 46:26
has a different
material and a density.
• 46:26 - 46:27
OK.
• 46:27 - 46:28
STUDENT: Tiramisu. [INAUDIBLE]
• 46:28 - 46:31
• 46:31 - 46:32
MAGDALENA TODA: Huh?
• 46:32 - 46:33
STUDENT: So you
use the tiramasu,
• 46:33 - 46:34
you're making me hungry.
• 46:34 - 46:35
MAGDALENA TODA:
Yeah, because now,
• 46:35 - 46:38
OK take your mind
off the tiramisu.
• 46:38 - 46:40
Think about an exam.
• 46:40 - 46:42
Then you don't--
• 46:42 - 46:45
STUDENT: Now I'm sick.
• 46:45 - 46:46
MAGDALENA TODA: Exactly.
• 46:46 - 46:50
Now you need something
against nausea.
• 46:50 - 46:54
Let's see what else
is interesting to do.
• 46:54 - 46:58
• 46:58 - 47:00
I'll give you ten minutes.
• 47:00 - 47:02
How much did I steal from you?
• 47:02 - 47:07
I stole constantly about
five minutes of your breaks
• 47:07 - 47:10
for the last few Tuesdays.
• 47:10 - 47:12
STUDENT: So the integral--
• 47:12 - 47:14
MAGDALENA TODA: The
integral of that.
• 47:14 - 47:19
I think I would be fair
to give you 10 minutes
• 47:19 - 47:22
as a gift today to compensate.
• 47:22 - 47:28
OK, so remind me to let
you go 10 minutes early.
• 47:28 - 47:32
Especially since
spring break is coming.
• 47:32 - 47:37
We have a 3D application.
• 47:37 - 47:40
We have several 3D applications.
• 47:40 - 47:44
Let me see which one
I want to mimic first.
• 47:44 - 47:49
• 47:49 - 47:51
Yeah.
• 47:51 - 47:56
I'm going to pick my favorite,
because I just want to.
• 47:56 - 48:02
• 48:02 - 48:16
So imagine you
have a disc that is
• 48:16 - 48:19
x squared plus y squared
equals 1 would be the circle.
• 48:19 - 48:22
That's the unit
disc on the floor.
• 48:22 - 48:27
• 48:27 - 48:38
And then I have the plane
x plus y plus z equals 8.
• 48:38 - 48:40
Then I'm going to
draw that plane.
• 48:40 - 48:41
I'll try my best.
• 48:41 - 48:49
• 48:49 - 48:51
It's similar to two
examples from the book,
• 48:51 - 48:54
but I did not want to
repeat the ones in the book
• 48:54 - 48:57
because I want you to
• 48:57 - 48:59
That's kind of the idea.
• 48:59 - 49:06
So you have this
picture, and you
• 49:06 - 49:11
realize that we had that
in the first octant before.
• 49:11 - 49:16
So I say, I don't
want the volume
• 49:16 - 49:21
of the body over the whole disc,
only over the part of the disc
• 49:21 - 49:24
which is in the first octant.
• 49:24 - 49:33
So I say, I want this domain
D, which is going to be what?
• 49:33 - 49:36
x squared plus y squared
less than or equal to 1
• 49:36 - 49:40
in plane, with x
positive, y positive.
• 49:40 - 49:43
Do you know what we call
that in trigonometry?
• 49:43 - 49:47
• 49:47 - 49:50
Does anybody know what we
call this in trigonometry?
• 49:50 - 49:58
• 49:58 - 50:00
Let me put the points
while you think.
• 50:00 - 50:03
Hopefully, you are
• 50:03 - 50:07
This is 1 in x-axis.
• 50:07 - 50:14
1, 0, 0, and this is
0, 1, 0, and this is y.
• 50:14 - 50:18
If I were to go up
until I meet the plane,
• 50:18 - 50:20
what point would this--
• 50:20 - 50:22
STUDENT: [INAUDIBLE]
• 50:22 - 50:27
MAGDALENA TODA: What point
would this-- on the thing.
• 50:27 - 50:30
• 50:30 - 50:35
STUDENT: 1, 0, 7
and then 0, 1, 7.
• 50:35 - 50:36
MAGDALENA TODA: 1, 0, 7.
• 50:36 - 50:41
• 50:41 - 50:47
This would be you said 0, 1, 6.
• 50:47 - 50:49
And this would be 1, 0, 7.
• 50:49 - 50:51
• 50:51 - 50:52
How do you know?
• 50:52 - 50:53
STUDENT: Y plus z--
• 50:53 - 50:57
MAGDALENA TODA: Because z,
because it's on the y-axis,
• 50:57 - 51:01
and since you are on the
x-axis here, y has to be 0.
• 51:01 - 51:02
So you're right.
• 51:02 - 51:02
Very good.
• 51:02 - 51:03
Excellent.
• 51:03 - 51:11
Now I'm going to
say, I'd like to know
• 51:11 - 51:34
the-- compute the volume of
the body that is bounded above
• 51:34 - 51:52
from above by x plus y plus
z equals 8, who's projection
• 51:52 - 52:01
on the floor is the
domain D. And I'll say
• 52:01 - 52:02
volume of the cylindrical body.
• 52:02 - 52:10
• 52:10 - 52:15
So how could you obtain
such a, again-- No,
• 52:15 - 52:17
this is Murphy's Law.
• 52:17 - 52:25
OK, how could you obtain such
an object, such a cylinder?
• 52:25 - 52:27
STUDENT: Take a pencil,
and cut it into fourths.
• 52:27 - 52:28
MAGDALENA TODA: Huh?
• 52:28 - 52:31
STUDENT: Take like a cylindrical
pencil and cut it into fourths.
• 52:31 - 52:34
MAGDALENA TODA: Take a
salami, a piece of salami.
• 52:34 - 52:40
Cut that piece of salami into
four, into four quarters.
• 52:40 - 52:44
• 52:44 - 52:50
And then we take, we slice,
and we slice like that.
• 52:50 - 52:52
So we have something like--
• 52:52 - 52:54
STUDENT: I tried to think
of a non- food example.
• 52:54 - 52:56
MAGDALENA TODA: --a quarter.
• 52:56 - 52:59
How can I draw this?
• 52:59 - 53:01
OK, this is what it means.
• 53:01 - 53:03
You don't see this one.
• 53:03 - 53:05
You don't see this part.
• 53:05 - 53:06
You don't see this part.
• 53:06 - 53:07
This is curved.
• 53:07 - 53:11
And here, instead of cutting
with another perpendicular
• 53:11 - 53:13
plane, along the
salami-- so this
• 53:13 - 53:18
is the axis of the salami--
instead of taking the knife
• 53:18 - 53:21
and cutting like that, I'm
cutting an oblique plane,
• 53:21 - 53:26
and this is what this
oblique plane will do.
• 53:26 - 53:28
STUDENT: If you cut
that way, then you
• 53:28 - 53:31
would have only squares.
• 53:31 - 53:32
MAGDALENA TODA: Hmm?
• 53:32 - 53:43
So I'm going to have some
oblique-- I cannot draw better.
• 53:43 - 53:45
I don't know how to draw better.
• 53:45 - 53:47
So it's going to be an
oblique cut in the salami.
• 53:47 - 53:50
• 53:50 - 53:53
Let's think how we
do this problem.
• 53:53 - 53:55
Elementary, it will
be a piece of cake--
• 53:55 - 53:58
it would be a piece of--
• 53:58 - 53:59
STUDENT: A piece of salami.
• 53:59 - 54:00
MAGDALENA TODA: No.
• 54:00 - 54:01
It wouldn't be apiece of salami.
• 54:01 - 54:03
STUDENT: It could be done.
• 54:03 - 54:06
MAGDALENA TODA: How could we do
that quickly with the Calculus
• 54:06 - 54:07
III we know?
• 54:07 - 54:09
STUDENT: Find the
triple integral.
• 54:09 - 54:12
Oh, you want us to do
the double integral?
• 54:12 - 54:14
MAGDALENA TODA: Double, triple,
I don't know what to do.
• 54:14 - 54:16
What do you think is best?
• 54:16 - 54:18
Let's do that triple
integral first,
• 54:18 - 54:21
and you'll see that it's the
same thing as double integral.
• 54:21 - 54:31
Triple integral over B, the
body of the salami, 1 dV.
• 54:31 - 54:34
How can we set it up?
• 54:34 - 54:37
Well, this is a
little bit tricky.
• 54:37 - 54:39
It's going to be like that.
• 54:39 - 54:42
• 54:42 - 54:45
We can say, I have a double
integral over my domain,
• 54:45 - 54:52
D. When it comes to the z,
mister z has to be first.
• 54:52 - 54:54
So mister z says, I'm first.
• 54:54 - 54:56
I know where I'm going.
• 54:56 - 55:00
You guys, x and y
are bound together,
• 55:00 - 55:03
mired in the element
of area of the circles.
• 55:03 - 55:06
This is like dx dy.
• 55:06 - 55:08
But I am independent from you.
• 55:08 - 55:09
I am z.
• 55:09 - 55:13
So I'm going all the way
from the floor to what?
• 55:13 - 55:15
You taught me that.
• 55:15 - 55:20
8 minus x minus y, and 1.
• 55:20 - 55:22
This is the way to do
it as a triple integral,
• 55:22 - 55:25
but then Alex will
say, I could have
• 55:25 - 55:26
done this as a double integral.
• 55:26 - 55:29
Let me show you how.
• 55:29 - 55:33
I could have done it over
the domain D in plane.
• 55:33 - 55:36
Put the function,
8 minus x minus y
• 55:36 - 55:38
is [? B and ?] z from
the very beginning,
• 55:38 - 55:43
because that's my altitude
function, f of x and y.
• 55:43 - 55:47
So then I say dx dy, dx
dy, it doesn't matter.
• 55:47 - 55:48
That's the only theory element.
• 55:48 - 55:49
Fine.
• 55:49 - 55:51
It's the same thing.
• 55:51 - 55:54
This is what I wanted
you to observe.
• 55:54 - 55:56
Whether you view it like the
triple integral like that,
• 55:56 - 55:58
or you view it as the
double integral like that,
• 55:58 - 56:02
it's the same thing.
• 56:02 - 56:03
This is not a headache.
• 56:03 - 56:06
The headache is coming next.
• 56:06 - 56:08
This is not a headache.
• 56:08 - 56:11
So you can do it in two ways.
• 56:11 - 56:14
And I'd like to
look at the-- check
• 56:14 - 56:20
the two methods of doing this.
• 56:20 - 56:24
• 56:24 - 56:27
And set up the integrals
without solving them.
• 56:27 - 56:38
• 56:38 - 56:39
Can you read my mind?
• 56:39 - 56:42
Do you realize what I'm asking?
• 56:42 - 56:44
Imagine that would
be on the midterm.
• 56:44 - 56:47
What do you think I'm
asking, the two methods?
• 56:47 - 56:49
This can be interpreted
in many ways.
• 56:49 - 56:50
There are two methods.
• 56:50 - 56:54
I mean, one method by
doing it with Cartesian
• 56:54 - 56:56
coordinates x and y.
• 56:56 - 56:58
The other method is switching
to polar coordinates
• 56:58 - 57:01
and set up the integral
without solving.
• 57:01 - 57:04
And you say, why not solving?
• 57:04 - 57:05
Because I'm going to cheat.
• 57:05 - 57:08
I'm going to use a
TI-92 to solve it,
• 57:08 - 57:11
or I'm going to use
a Matlab or Maple.
• 57:11 - 57:13
If it looks a little
bit complicated,
• 57:13 - 57:15
then I don't want
to spend my time.
• 57:15 - 57:19
Actually, engineers,
after taking Calc III,
• 57:19 - 57:20
they know a lot.
• 57:20 - 57:23
They understand a lot
• 57:23 - 57:27
But do you think if you work on
a real-life problem like that,
• 57:27 - 57:29
that your boss will
let you waste your time
• 57:29 - 57:31
and do the integral by hand?
• 57:31 - 57:31
STUDENT: No.
• 57:31 - 57:34
MAGDALENA TODA: Most integrals
are really complicated
• 57:34 - 57:35
in everyday life.
• 57:35 - 57:36
So what you're
going to do is going
• 57:36 - 57:40
to be a scientific software,
like Matlab, which is primarily
• 57:40 - 57:44
for engineers, Mathematica,
which is similar to Matlab,
• 57:44 - 57:46
but is mainly for
mathematicians.
• 57:46 - 57:48
It was invented
at the University
• 57:48 - 57:51
of Illinois Urbana-Champaign,
and they're still
• 57:51 - 57:52
very proud of it.
• 57:52 - 57:54
I prefer Matlab
because I feel Matlab
• 57:54 - 57:58
is stronger, has higher
capabilities than Mathematica.
• 57:58 - 57:59
You can use Maple.
• 57:59 - 58:05
Maple lets you set up the
endpoints even as functions.
• 58:05 - 58:08
And then it's user
friendly, you type in this,
• 58:08 - 58:10
you type in the endpoints.
• 58:10 - 58:12
It has little windows, here.
• 58:12 - 58:14
You don't need to
know any programming.
• 58:14 - 58:18
It's made for people who
have no programming skills.
• 58:18 - 58:20
So it's going to show
a little window on top,
• 58:20 - 58:22
here, here, here and here.
• 58:22 - 58:24
You [? have ?] those,
and you press Enter,
• 58:24 - 58:27
and it's going to spit
the answer back at you.
• 58:27 - 58:29
So this is how
engineers actually
• 58:29 - 58:31
solve the everyday integrals.
• 58:31 - 58:33
Not by hand.
• 58:33 - 58:36
I want to be able to
set it up in both ways
• 58:36 - 58:39
before I go home or
eat something, right?
• 58:39 - 58:44
So we don't have to spend
a lot of time on it.
• 58:44 - 58:48
But if you want to tell me
how I am going to set it up,
• 58:48 - 58:50
I would be very grateful.
• 58:50 - 58:54
So this is Cartesian,
and this is polar.
• 58:54 - 59:06
• 59:06 - 59:07
All right.
• 59:07 - 59:08
Who helps me?
• 59:08 - 59:10
In Cartesian-- which
one do you prefer?
• 59:10 - 59:11
I mean, it doesn't matter.
• 59:11 - 59:14
You guys are good
and smart, and you'll
• 59:14 - 59:16
figure out what I need to do.
• 59:16 - 59:19
If I want to do it in terms
of vertical strip-- so
• 59:19 - 59:21
for vertical strip
method-- first
• 59:21 - 59:25
I integrate with respect to
y, and then with respect to x.
• 59:25 - 59:27
And maybe, to test
• 59:27 - 59:30
let me change the
order of integrals
• 59:30 - 59:33
and see how much you
understood from that last time.
• 59:33 - 59:35
STUDENT: [INAUDIBLE]
• 59:35 - 59:39
MAGDALENA TODA: So x is
between what and what?
• 59:39 - 59:40
STUDENT: 0 and 1.
• 59:40 - 59:41
MAGDALENA TODA: Look
at this picture.
• 59:41 - 59:43
I have to reproduce
this picture like that.
• 59:43 - 59:47
0 to 1, says Alex,
and he's right.
• 59:47 - 59:49
And why will he decide against--
• 59:49 - 59:50
STUDENT: 1 minus x squared.
• 59:50 - 59:53
MAGDALENA TODA: --square
root 1 minus x squared.
• 59:53 - 59:55
So we know very well
what we are going to do,
• 59:55 - 59:57
what Maple is
going to do for us.
• 59:57 - 60:00
1 square root 1 minus x squared.
• 60:00 - 60:02
And then what do I put here?
• 60:02 - 60:03
8 minus x minus y.
• 60:03 - 60:06
Can I do it by hand?
• 60:06 - 60:08
Yes, I guarantee to you
I can do it by hand.
• 60:08 - 60:10
Let me tell you why.
• 60:10 - 60:14
Because when we integrate
with respect to y, I get xy.
• 60:14 - 60:18
So I get xy, and y will be
plugged in 1 minus x squared.
• 60:18 - 60:24
How am I going to solve
an integral like this?
• 60:24 - 60:28
I can the first one with a
table, the second one with a u
• 60:28 - 60:30
substitution.
• 60:30 - 60:33
On the last one is a
little bit painful.
• 60:33 - 60:35
I'm going to have
y squared over 2--
• 60:35 - 60:36
STUDENT: That's
the easiest part.
• 60:36 - 60:39
MAGDALENA TODA: According
to Alex, yes, you're right.
• 60:39 - 60:41
Maybe that is the easiest.
• 60:41 - 60:43
STUDENT: That's the [INAUDIBLE]
part you can integrate--
• 60:43 - 60:45
MAGDALENA TODA: And I can
integrate one at a time,
• 60:45 - 60:47
and I'm going to
waste all my time.
• 60:47 - 60:49
So if I want to be an
efficient engineer,
• 60:49 - 60:54
and my boss is waiting for
the end-of-the-day project,
• 60:54 - 60:56
of course I'm not going
to do this by hand.
• 60:56 - 60:59
How about the other integral?
• 60:59 - 61:02
Same integral.
• 61:02 - 61:05
Same idea, y between 0 and 1.
• 61:05 - 61:07
And x between 0 and
• 61:07 - 61:08
STUDENT: Square root
of 1 minus y squared.
• 61:08 - 61:11
MAGDALENA TODA: Square
root of 1 minus y squared.
• 61:11 - 61:16
Because I'll do this guy
with horizontal strips,
• 61:16 - 61:19
the vertical strips.
• 61:19 - 61:23
And here's the y-- I rotate
my head and it cracks,
• 61:23 - 61:28
so that means that
I need some yoga.
• 61:28 - 61:31
y is between 0 and 1.
• 61:31 - 61:31
Or gymnastics.
• 61:31 - 61:36
So x is between
0 and square root
• 61:36 - 61:40
1 minus y squared. [INAUDIBLE].
• 61:40 - 61:42
And I'll leave it
here on the meter.
• 61:42 - 61:46
And I'm going to make a
sample like I promised.
• 61:46 - 61:49
OK, good.
• 61:49 - 61:53
How would you do this to set up
the polar coordinate integral?
• 61:53 - 61:58
And that is why Alex said
maybe that's a pain because
• 61:58 - 62:00
of a reason.
• 62:00 - 62:04
And he's right, it's a little
bit painful to solve by hand.
• 62:04 - 62:07
But again, once you
switch to polar,
• 62:07 - 62:11
you can solve it with a
calculator or a computer
• 62:11 - 62:14
software, scientific
software in no time.
• 62:14 - 62:19
In Maple, you just have
to plug in the numbers.
• 62:19 - 62:22
You cannot plug in theta,
I think, as a symbol.
• 62:22 - 62:23
I'm not sure.
• 62:23 - 62:27
But you can put theta
as t and r will be r,
• 62:27 - 62:29
or you can use
whatever letters you
• 62:29 - 62:31
want that are roman letters.
• 62:31 - 62:35
So you have to
integrate smartly, here,
• 62:35 - 62:38
switching to r and
theta, and think
• 62:38 - 62:41
about the meaning of that.
• 62:41 - 62:45
So first of all, if
I put dr d theta,
• 62:45 - 62:49
I'm not worried that you won't
be able to get r and theta,
• 62:49 - 62:51
because I know you can do it.
• 62:51 - 62:56
You can prove it to me
right now. r between 0 and
• 62:56 - 62:57
STUDENT: 1.
• 62:57 - 62:58
MAGDALENA TODA: Excellent.
• 62:58 - 63:01
And theta, pay
attention, between 0 and
• 63:01 - 63:02
STUDENT: pi over 2
• 63:02 - 63:03
MAGDALENA TODA: Excellent.
• 63:03 - 63:04
I'm proud.
• 63:04 - 63:04
Yes, sir?
• 63:04 - 63:07
STUDENT: Is it supposed
to be r dr 2 theta,
• 63:07 - 63:08
or are you going
to add that later?
• 63:08 - 63:10
MAGDALENA TODA: I
will add it here.
• 63:10 - 63:13
So the integrand
will contain the r.
• 63:13 - 63:17
Now what do I put
in terms of this?
• 63:17 - 63:19
I left enough room.
• 63:19 - 63:23
STUDENT: Is it pi over 2,
or is it negative pi over 2?
• 63:23 - 63:25
MAGDALENA TODA:
It doesn't matter,
• 63:25 - 63:31
because I'll have to take that--
we assume always theta to go
• 63:31 - 63:35
counterclockwise, and go
between 0 and pi over 2,
• 63:35 - 63:39
so that when you start--
let me make this motion.
• 63:39 - 63:41
You are here at theta equals 0.
• 63:41 - 63:42
STUDENT: Oh, OK.
• 63:42 - 63:42
Sorry.
• 63:42 - 63:44
I got my coordinates
mixed around--
• 63:44 - 63:46
MAGDALENA TODA: --and
counterclockwise to pi over 2.
• 63:46 - 63:47
[INTERPOSING VOICES]
• 63:47 - 63:49
• 63:49 - 63:50
MAGDALENA TODA: Yeah.
• 63:50 - 63:55
So you go in the trigonometric--
Here, you have 8 minus,
• 63:55 - 63:58
and who tells me what
I'm supposed to type?
• 63:58 - 63:59
STUDENT: r over x.
• 63:59 - 64:08
MAGDALENA TODA: r
cosine theta minus
• 64:08 - 64:09
STUDENT: Sine.
• 64:09 - 64:11
MAGDALENA TODA: r sine theta.
• 64:11 - 64:14
And let mister
whatever his name is,
• 64:14 - 64:17
the computer, find the answer.
• 64:17 - 64:19
Can I do it by hand?
• 64:19 - 64:21
Actually, I can.
• 64:21 - 64:27
I can, but again, it's not worth
it, because it drives me crazy.
• 64:27 - 64:30
How would I do it by hand?
• 64:30 - 64:32
I would split the
integral into three,
• 64:32 - 64:36
and I would easily
compute 8 times r,
• 64:36 - 64:37
integrand is going to be easy.
• 64:37 - 64:38
Right?
• 64:38 - 64:39
Agree with me?
• 64:39 - 64:41
Then what am I going to do?
• 64:41 - 64:47
I'm going to say, an r out
times an r, out comes r squared.
• 64:47 - 64:51
And I have integral of r
squared times a function
• 64:51 - 64:54
of theta only,
which is going to be
• 64:54 - 64:56
sine theta plus cosine theta.
• 64:56 - 64:59
We are going to say, yes,
with a minus, with a minus.
• 64:59 - 65:02
• 65:02 - 65:07
Now, when I compute
r and theta thingy,
• 65:07 - 65:10
theta will be between
0 and pi over 2.
• 65:10 - 65:12
r will be between 0 and 1.
• 65:12 - 65:16
But I don't care, because
Matthew reminded me,
• 65:16 - 65:19
if you have a product
of separate variables,
• 65:19 - 65:22
life becomes all of the
sudden easier for you.
• 65:22 - 65:25
STUDENT: You've also got to
add your integral of [? 8r ?]
• 65:25 - 65:25
[? dr. ?]
• 65:25 - 65:26
MAGDALENA TODA: Yeah.
• 65:26 - 65:28
At the end, I'm going to
add the integral of 8r.
• 65:28 - 65:30
So I take them separately.
• 65:30 - 65:33
I just look at one chunk.
• 65:33 - 65:35
And this chunk will be what?
• 65:35 - 65:39
Can you even see how easy it's
going to be with the naked eye?
• 65:39 - 65:41
Firs of all,
integral from 0 to 1,
• 65:41 - 65:44
r squared dr is a piece of cake.
• 65:44 - 65:46
How much is that--
piece of salami.
• 65:46 - 65:47
STUDENT: 1/3.
• 65:47 - 65:49
MAGDALENA TODA: 1/3.
• 65:49 - 65:50
Right?
• 65:50 - 65:52
Because it's r cubed over 3.
• 65:52 - 65:53
Then you have 1/3.
• 65:53 - 65:55
That's easy.
• 65:55 - 65:58
With a minus in front, but I
don't care about it in the end.
• 65:58 - 66:04
What is the integral of
sine theta cosine theta?
• 66:04 - 66:06
STUDENT: Negative [INAUDIBLE].
• 66:06 - 66:11
MAGDALENA TODA: Minus
cosine theta plus sine theta
• 66:11 - 66:14
taken between 0 and pi over 2.
• 66:14 - 66:16
Will this be hard?
• 66:16 - 66:20
Who's going to tell me what,
or how I'm going to get what--
• 66:20 - 66:23
we don't compute it now,
but I just give you.
• 66:23 - 66:24
Cosine of pi over 3 is?
• 66:24 - 66:25
STUDENT: 0.
• 66:25 - 66:25
MAGDALENA TODA: 0.
• 66:25 - 66:27
Sine of pi over 2 is?
• 66:27 - 66:28
STUDENT: Oh yeah.
• 66:28 - 66:28
1.
• 66:28 - 66:28
MAGDALENA TODA: 1.
• 66:28 - 66:31
So this is going to
be 1 minus, what's
• 66:31 - 66:35
the whole thingy computed at 0?
• 66:35 - 66:35
STUDENT: [INAUDIBLE].
• 66:35 - 66:39
MAGDALENA TODA: It's going to
be minus 1, but minus minus 1
• 66:39 - 66:40
is plus 1.
• 66:40 - 66:43
So I have 2.
• 66:43 - 66:46
So only this chunk of the
integral would be easy.
• 66:46 - 66:47
Minus 2/3.
• 66:47 - 66:48
OK?
• 66:48 - 66:51
So it can be done by hand,
but why waste the time when
• 66:51 - 66:53
you can do it with Maple?
• 66:53 - 66:53
Yes, sir?
• 66:53 - 66:56
STUDENT: Where did
you get rid of 8?
• 66:56 - 66:58
On the second, after the 8--
• 66:58 - 67:00
MAGDALENA TODA: No, I didn't.
• 67:00 - 67:02
That's exactly what
we were talking.
• 67:02 - 67:06
Alex says, but you just
talked about integral of 8r,
• 67:06 - 67:07
but you didn't want to do it.
• 67:07 - 67:09
I said, I didn't want to do it.
• 67:09 - 67:13
This is just the second
chunk of this integral.
• 67:13 - 67:17
So I know that I can do integral
of integral of 8r in no time.
• 67:17 - 67:20
Then I would need to
take this and add that,
• 67:20 - 67:21
and get the number.
• 67:21 - 67:23
I don't care about the number.
• 67:23 - 67:25
I just care about the method.
• 67:25 - 67:25
Yes, sir?
• 67:25 - 67:28
STUDENT: Why are the limits
from 0 to 1 instead of like 0
• 67:28 - 67:30
to r squared?
• 67:30 - 67:33
Because didn't we say
earlier the domain
• 67:33 - 67:35
is x squared plus y squared?
• 67:35 - 67:38
Wouldn't that be r squared?
• 67:38 - 67:38
MAGDALENA TODA: No.
• 67:38 - 67:39
No, wait.
• 67:39 - 67:40
This is r squared.
• 67:40 - 67:41
STUDENT: Right.
• 67:41 - 67:44
Why didn't we plug r
squared into the 1 again.
• 67:44 - 67:47
MAGDALENA TODA: And that means
r is between 0 and 1, right?
• 67:47 - 67:48
STUDENT: Oh, OK.
• 67:48 - 67:50
MAGDALENA TODA: r squared
being less than 1.
• 67:50 - 67:52
That means r is between 0 and 1.
• 67:52 - 67:53
OK?
• 67:53 - 67:57
And one last problem-- no.
• 67:57 - 67:59
No last problem.
• 67:59 - 68:01
We have barely 10 minutes.
• 68:01 - 68:04
So you read from the book some.
• 68:04 - 68:08
I will come back to this
section, and I'll do review.
• 68:08 - 68:11
Have a wonderful
spring break, and I'm
• 68:11 - 68:14
going to see you after
spring break on Tuesday.
• 68:14 - 68:16
[INTERPOSING VOICES]
• 68:16 - 68:18
Title:
TTU Math2450 Calculus3 Sec 12.5
Description:

Triple Integrals

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Video Language:
English
 jackie.luft edited English subtitles for TTU Math2450 Calculus3 Sec 12.5