## TTU Math2450 Calculus3 Sec 12.3

• 0:00 - 0:02
MAGDALENA TODA: I'm
starting early, am I?
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It's exactly 12:30.
• 0:04 - 0:07
The weather is getting
better, hopefully,
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and not too many people
should miss class today.
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Can you start an attendance
sheet for me [INAUDIBLE]?
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I know I can count on you.
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OK.
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I have good markers today.
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double integrals in
polar coordinates.
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These are all friends of yours.
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You've seen until now
only double integrals that
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involve the rectangles, either
a rectangle, we saw [INAUDIBLE],
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and we saw some type
of double integrals,
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of course that involved
x and y, so-called type
• 1:18 - 1:21
1 and type 2
regions, which were--
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so we saw the rectangular case.
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You have ab plus
cd, a rectangle.
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You have what other kind
of a velocity [INAUDIBLE]
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over the the main of the shape
x between a and be and y.
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You write wild and happy
from bottom to top.
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That's called the wild--
not wild, the vertical strip
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method, where y will be
between the bottom function
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f of x and the top
function f of x.
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And last time I
took examples where
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f and g were both positive, but
remember, you don't have to.
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All you have to have is that
g is always greater than f,
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or equal at some point.
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And then what else do
we have for these cases?
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These are all
continuous functions.
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What else did we have?
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• 2:29 - 2:33
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Where what was going on,
we have played a little bit
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around with y between c
and d limits with points.
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These are horizontal,
so we take the domain
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as being defined by these
horizontal strips between let's
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say a function.
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Again, I need to rotate my head,
but I didn't do my yoga today,
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so it's a little bit sticky.
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I'll try.
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x equals F of y, and x equals
G of y, assuming, of course,
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that f of y is always greater
than or equal to g of y,
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and the rest of the
apparatus is in place.
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Those are not so
hard to understand.
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We played around.
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We switched the integrals.
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We changed the order of
integration from dy dx
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to dx dy, so we have
to change the domain.
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We went from
vertical strip method
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to horizontal strip method
or the other way around.
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And for what kind of
example, something
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like that-- I think it
was a leaf like that,
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we said, let's compute
the area or compute
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another kind of double
integral over this leaf in two
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different ways.
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And we did it with
vertical strips,
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and we did the same
with horizontal strips.
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So we reversed the
order of integration,
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and we said, I'm having the
double integral over domain
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of God knows what, f of
xy, continuous function,
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positive, continuous whenever
you want, and we said da.
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We didn't quite specify
the meaning of da.
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We said that da is
the area element,
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but that sounds a little
bit weird, because it makes
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you think of surfaces,
and an area element
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doesn't have to be a
little square in general.
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It could be something like a
patch on a surface, bounded
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by two curves within your
segments in each direction.
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So you think, well, I
don't know what that is.
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I'll tell you
today what that is.
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It's a mysterious thing,
it's really beautiful,
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Now, what did we do last time?
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We applied the two
theorems that allowed
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us to do this both ways.
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Integral from a to b, what was
my usual [? wrist ?] is down,
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f of x is in g of x, right?
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dy dx.
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So if you do it in
this order, it's
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going to be the same as if
you do it in the other order.
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ab are these guys, and then
this was cd on the y-axis.
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This is the range between
c and d in altitudes.
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So we have integral from
c to d, integral from,
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I don't know what they will be.
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This big guy I'm talking--
which one is the one?
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This one, that's going to
be called x equals f of y,
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or g of y, and let's put the
big one G and the smaller one,
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x equals F of y.
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So you have to [? re-denote ?]
these functions,
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these inverse functions, and
use them as functions of y.
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So it makes sense to
say-- what did we do?
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We first integrated respect to
x between two functions of y.
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That was the so-called
horizontal strip method, dy.
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So I have summarized
the ideas from last time
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that we worked with, generally
with corners x and y.
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We were very happy about them.
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domain, where x was between ab
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and y was between cd.
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Then we went to type 1, not
diabetes, just type 1 region,
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type 2, and those
guys are related.
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So if you understood 1 and
understood the other one,
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and if you have a
nice domain like that,
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you can compute the
area or something.
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The area will correspond
to x equals 1.
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So if f is 1, then
that's the area.
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That will also be a
volume of a cylinder based
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on that region with height 1.
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Imagine a can of Coke
that has height 1,
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and-- maybe better,
chocolate cake,
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that has the shape of
this leaf on the bottom,
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and then its height
is 1 everywhere.
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So if you put 1 here, and
you get the area element,
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and then everything
else can be done
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by reversing the order of
integration if f is continuous.
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But for polar
coordinates, the situation
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has to be reconsidered almost
entirely, because the area
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element, da is called
the area element for us,
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was equal to dx dy for the
cartesian coordinate case.
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And here I'm making a
weird face, I'm weird, no?
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Saying, what am I going
to do, what is this
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going to become for
polar coordinates?
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And now you go, oh my God,
not polar coordinates.
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Those were my
enemies in Calc II.
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Many people told me that.
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And I tried to go
into my time machine
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and go back something
like 25 years ago
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and see how I felt about
them, and I remember that.
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I didn't get them from
the first 48 hours
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after I was exposed to them.
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Therefore, let's
do some preview.
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What were those
polar coordinates?
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Polar coordinates were
a beautiful thing,
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these guys from trig.
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And what did we have
in trigonometry?
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a point on a circle.
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This is not the unit
trigonometric circle,
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it's a circle of--
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I'm a little bit shifted
by a phase of phi 0.
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So you have a radius r.
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And let's call that little r.
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And then, we say, OK,
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That's the second
polar coordinate.
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The angle by measuring
from the, what
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is this called, the x-axis.
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Origin, x-axis, o, x,
going counterclockwise,
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because we are mathemeticians.
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Every normal person, when
they mix into a bowl,
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they mix like that.
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Well, I've seen that
most of my colleagues--
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this is just a
psychological test, OK?
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I wanted to see
how they mix when
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they cook, or mix
up-- most of them
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mix in a trigonometric sense.
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I don't know if this has
anything to do with the brain
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connections, but I think
that's [? kind of weird. ?]
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I don't have a statistical
result, but most of the people
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I've seen that, and do
mathematics, mix like that.
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So trigonometric sense.
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What is the connection with the
actual Cartesian coordinates?
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D you know what Cartesian
comes from as a word?
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Cartesian, that sounds weird.
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STUDENT: From Descartes.
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MAGDALENA TODA: Exactly.
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Who said that?
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Roberto, thank you so much.
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I'm impressed.
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Descartes was--
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STUDENT: French.
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MAGDALENA TODA: --a
French mathematician.
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But actually, I mean,
he was everything.
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He was a crazy lunatic.
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He was a philosopher,
a mathematician,
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a scientist in general.
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He also knew a lot
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But at the time, I don't
know if this is true.
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I should check with wiki,
or whoever can tell me.
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One of my professors in college
told me that at that time,
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there was a fashion
that people would
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change their names like they
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So they and change their
name from Francesca
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to Frenchy, from Roberto
to Robby, from-- so
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if they would have to
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how many names correspond to
the ID, I think less than 20%.
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At that time it was the same.
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All of the scientists loved
to romanize their names.
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And of course he was
of a romance language,
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but he said, what if I
made my name a Latin name,
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I changed my name
into a Latin name.
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So he himself, this is what
my professor told me, he
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himself changed his
name to Cartesius.
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"Car-teh-see-yus" actually, in
Latin, the way it should be.
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OK, very smart guy.
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Now, when we look
a x and y, there
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has to be a connection between
x, y as the couple, and r theta
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as the same-- I mean a
couple, not the couple,
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for the same point.
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Yes, sir?
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STUDENT: Cartesius.
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Like meaning flat?
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The name?
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MAGDALENA TODA: These are
the Cartesian coordinates,
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and it sounds like the word map.
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STUDENT: Because the
meaning of carte--
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STUDENT: But look, look.
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Descartes means from the map.
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From the books, or from the map.
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So he thought what his
name would really mean,
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and so he recalled himself.
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There was no fun, no
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So they had to do something
to enjoy themselves.
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Now, when it comes
to these triangles,
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we have to think of the
relationship between x, y
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and r, theta.
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And could somebody tell me what
the relationship between x, y
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and r, theta is?
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x represents
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STUDENT: R cosine theta.
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STUDENT: r cosine
theta, who says that?
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Trigonometry taught us
that, because that's
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the hypotenuse for cosine.
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In terms of sine, you
know what you have,
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so you're going to have
y equals r sine theta,
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and we have to decide
if x and y are allowed
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to be anywhere in plane.
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For the plane,
I'll also write r2.
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R2, not R2 from the movie,
just r2 is the plane,
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and r3 is the space,
the [? intriguing ?]
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space, three-dimensional one.
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r theta, is a couple where?
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That's a little bit tricky.
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We have to make a restriction.
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We allow r to be anywhere
between 0 and infinity.
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So it has to be a
positive number.
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And theta [INTERPOSING VOICES]
between 0 and 2 pi.
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STUDENT: I've been
sick since Tuesday.
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MAGDALENA TODA: I
believe you, Ryan.
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You sound sick to me.
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Take your viruses away from me.
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Take the germs away.
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I don't even have
the-- I'm kidding,
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Alex, I hope you
don't get offended.
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So, I hope this works this time.
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I'm making a
sarcastic-- it's really,
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I hope you're feeling better.
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So you haven't missed much.
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Only the jokes.
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So x equals r cosine theta,
y equals r sine theta.
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Is that your
favorite change that
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was a differential
mapping from the set x,
• 15:56 - 15:59
y to the set r,
theta back and forth.
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And you are going
to probably say, OK
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how do you denote such a map?
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I mean, going from x,
y to r, theta and back,
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let's suppose that we go
from r, theta to x, y,
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and that's going to be a big if.
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And going backwards is going
to be the inverse mapping.
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So I'm going to
call it f inverse.
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So that's a map from a couple
to another couple of number.
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And you say, OK, but
why is that a map?
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All right, guys,
now let me tell you.
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So x, you can do x as a
function of r, theta, right?
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It is a function of r and theta.
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It's a function
of two variables.
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And y is a function
of r and theta.
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It's another function
of two variables.
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They are both nice
and differentiable.
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We assume not only that
they are differentiable,
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but the partial derivatives
will be continuous.
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So it's really
nice as a mapping.
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And you think, could I
write the chain rule?
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That is the whole idea.
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What is the meaning
of differential?
• 17:18 - 17:20
dx differential dy.
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Since I was chatting with
you, once, [? Yuniel ?],
• 17:23 - 17:29
• 17:29 - 17:31
differential again.
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If you were to define,
like Mr. Leibniz did,
• 17:36 - 17:40
the differential of the
function x with respect
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to both variables, that
was the sum, right?
• 17:44 - 17:45
You've done that
in the homework,
• 17:45 - 17:47
• 17:47 - 17:54
So you get x sub r,
dr, plus f x sub what?
• 17:54 - 17:54
STUDENT: Theta.
• 17:54 - 17:57
MAGDALENA TODA:
Sub theta d-theta.
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what if I see skip the dr?
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No, don't do that.
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First of all, WeBWorK is not
• 18:05 - 18:09
But second of all, the
most important stuff
• 18:09 - 18:13
here to remember is that these
are small, infinitesimally
• 18:13 - 18:15
small, displacements.
• 18:15 - 18:32
Infinitesimally small
displacements in the directions
• 18:32 - 18:34
x and y, respectively.
• 18:34 - 18:37
So you would say, what does
that mean, infinitesimally?
• 18:37 - 18:40
It doesn't mean delta-x small.
• 18:40 - 18:44
Delta-x small would be like
me driving 7 feet, when
• 18:44 - 18:49
I know I have to drive fast to
Amarillo to be there in 1 hour.
• 18:49 - 18:50
Well, OK.
• 18:50 - 18:52
Don't tell anybody.
• 18:52 - 18:55
But, it's about 2 hours, right?
• 18:55 - 18:58
So I cannot be there in an hour.
• 18:58 - 19:02
But driving those seven
feet is like a delta x.
• 19:02 - 19:07
Imagine, however, me
measuring that speed of mine
• 19:07 - 19:10
in a much smaller
fraction of a second.
• 19:10 - 19:16
So shrink that time to
something infinitesimally small,
• 19:16 - 19:17
which is what you have here.
• 19:17 - 19:19
That kind of quantity.
• 19:19 - 19:25
And dy will be y sub r dr
plus y sub theta d-theta.
• 19:25 - 19:29
• 19:29 - 19:33
And now, I'm not going
to go by the book.
• 19:33 - 19:35
I'm going to go
a little bit more
• 19:35 - 19:40
in depth, because in the book--
First of all, let me tell you,
• 19:40 - 19:44
if I went by the book,
what I would come with.
• 19:44 - 19:49
And of course the way
we teach mathematics
• 19:49 - 19:52
all through K-12 and through
college is swallow this theorem
• 19:52 - 19:54
and believe it.
• 19:54 - 19:59
So practically you accept
whatever we give you
• 19:59 - 20:02
without controlling it, without
checking if we're right,
• 20:02 - 20:05
without trying to prove it.
• 20:05 - 20:07
Practically, the
theorem in the book
• 20:07 - 20:09
says that if you
have a bunch of x,
• 20:09 - 20:14
y that is continuous
over a domain, D,
• 20:14 - 20:21
and you do change
the variables over--
• 20:21 - 20:23
STUDENT: I forgot my glasses.
• 20:23 - 20:25
So I'm going to sit very close.
• 20:25 - 20:29
MAGDALENA TODA:
What do you wear?
• 20:29 - 20:31
What [INAUDIBLE]?
• 20:31 - 20:32
STUDENT: I couldn't tell you.
• 20:32 - 20:33
I can see from here.
• 20:33 - 20:34
MAGDALENA TODA: You can?
• 20:34 - 20:34
STUDENT: Yeah.
• 20:34 - 20:35
My vision's not terrible.
• 20:35 - 20:42
MAGDALENA TODA: All
right. f of x, y da.
• 20:42 - 20:47
If I change this da
as dx dy, let's say,
• 20:47 - 20:50
to a perspective
of something else
• 20:50 - 20:52
in terms of polar
coordinates, then
• 20:52 - 20:57
the integral I'm going to get is
over the corresponding domain D
• 20:57 - 21:01
star, whatever that would be.
• 21:01 - 21:06
Then I'm going to have f of
x of r theta, y of r theta,
• 21:06 - 21:10
everything expressed
in terms of r theta.
• 21:10 - 21:14
the a-- so we just
• 21:14 - 21:18
feed you this piece of
cake and say, believe it,
• 21:18 - 21:22
believe it and leave us alone.
• 21:22 - 21:22
OK?
• 21:22 - 21:27
That's what it does in
the book in section 11.3.
• 21:27 - 21:33
So without understanding why
you have to-- instead of the r
• 21:33 - 21:36
d theta and multiply it by an r.
• 21:36 - 21:36
What is that?
• 21:36 - 21:38
You don't know why you do that.
• 21:38 - 21:41
And I thought, that's
the way we thought it
• 21:41 - 21:43
for way too many years.
• 21:43 - 21:46
I'm sick and tired
of not explaining why
• 21:46 - 21:51
you multiply that with an r.
• 21:51 - 21:55
So I will tell you something
that's quite interesting,
• 21:55 - 21:58
and something that I learned
• 21:58 - 22:01
• 22:01 - 22:06
I was in my 20s when I
studied differential forms
• 22:06 - 22:08
for the first time.
• 22:08 - 22:13
And differential
forms have some sort
• 22:13 - 22:25
of special wedge product, which
is very physical in nature.
• 22:25 - 22:30
So if you love physics, you
will understand more or less
• 22:30 - 22:34
• 22:34 - 22:41
Imagine that you have two
vectors, vector a and vector b.
• 22:41 - 22:44
• 22:44 - 22:48
For these vectors,
you go, oh my God.
• 22:48 - 22:54
If these would be vectors in
a tangent plane to a surface,
• 22:54 - 22:56
you think, some
of these would be
• 22:56 - 23:00
tangent vectors to a surface.
• 23:00 - 23:02
This is the tangent
plane and everything.
• 23:02 - 23:07
You go, OK, if these
were infinitesimally
• 23:07 - 23:12
small displacements-- which they
are not, but assume they would
• 23:12 - 23:19
be-- how would you do the area
of the infinitesimally small
• 23:19 - 23:22
parallelogram that
they have between them.
• 23:22 - 23:31
This is actually the area
element right here, ea.
• 23:31 - 23:35
So instead of dx dy, you're
not going to have dx dy,
• 23:35 - 23:40
you're going to have some
sort of, I don't know,
• 23:40 - 23:48
this is like a
d-something, d u, and this
• 23:48 - 23:55
is a d v. And when I compute
the area of the parallelogram,
• 23:55 - 23:58
I consider these to
be vectors, and I
• 23:58 - 24:02
say, how did we get
it from the vectors
• 24:02 - 24:06
to the area of
the parallelogram?
• 24:06 - 24:10
We took the vectors,
we shook them off.
• 24:10 - 24:19
of them, and then we
• 24:19 - 24:23
took the norm, the
magnitude of that.
• 24:23 - 24:27
Does this makes sense,
compared to this parallelogram?
• 24:27 - 24:27
Yeah.
• 24:27 - 24:31
Remember, guys, this
was like, how big
• 24:31 - 24:33
is du, a small
infinitesimal displacement,
• 24:33 - 24:36
but that would be like the
width, one of the dimensions.
• 24:36 - 24:40
There's the other of the
dimension of the area element
• 24:40 - 24:44
times-- this area element
is that tiny pixel that
• 24:44 - 24:49
is sitting on the surface
in the tangent plane, yeah?
• 24:49 - 24:54
Sine of the angle
between the guys.
• 24:54 - 24:55
Oh, OK.
• 24:55 - 25:01
So if the guys are not
perpendicular to one another,
• 25:01 - 25:04
if the two displacements are not
perpendicular to one another,
• 25:04 - 25:07
you still have to multiply
the sine of theta.
• 25:07 - 25:09
Otherwise you don't
get the element
• 25:09 - 25:12
of the area of
this parallelogram.
• 25:12 - 25:18
So why did the Cartesian
coordinates not pose a problem?
• 25:18 - 25:20
For Cartesian
coordinates, it's easy.
• 25:20 - 25:23
• 25:23 - 25:23
It's a piece of cake.
• 25:23 - 25:24
Why?
• 25:24 - 25:32
Because this is the x, this is
the y, as little tiny measures
• 25:32 - 25:33
multiplied.
• 25:33 - 25:37
How much is sine of theta
between Cartesian coordinates?
• 25:37 - 25:38
STUDENT: 1.
• 25:38 - 25:41
MAGDALENA TODA: It's 1,
because its 90 degrees.
• 25:41 - 25:43
When they are
orthogonal coordinates,
• 25:43 - 25:47
it's a piece of cake,
because you have 1 there,
• 25:47 - 25:48
becomes easier.
• 25:48 - 25:51
• 25:51 - 25:57
So in general, what
is the area limit?
• 25:57 - 26:02
The area limit for
arbitrary coordinates--
• 26:02 - 26:17
So area limit for some
arbitrary coordinates
• 26:17 - 26:20
should be defined
as the sined area.
• 26:20 - 26:29
• 26:29 - 26:32
And you say, what do you
mean that's a sined area,
• 26:32 - 26:35
and why would you do that.?
• 26:35 - 26:38
Well, it's not so
hard to understand.
• 26:38 - 26:42
Imagine that you have a
convention, and you say,
• 26:42 - 26:55
OK, dx times dy equals
negative dy times dx.
• 26:55 - 26:57
And you say, what, what?
• 26:57 - 27:01
If you change the
order of dx dy,
• 27:01 - 27:07
this wedge stuff works exactly
like the-- what is that called?
• 27:07 - 27:08
Cross product.
• 27:08 - 27:13
So the wedge works just
like the cross product.
• 27:13 - 27:18
Just like the cross product.
• 27:18 - 27:23
In some other ways, suppose
that I am here, right?
• 27:23 - 27:28
And this is a vector, like an
infinitesimal displacement,
• 27:28 - 27:29
and that's the other one.
• 27:29 - 27:34
If I multiply them
one after the other,
• 27:34 - 27:38
and I use this strange wedge
[INTERPOSING VOICES] the area,
• 27:38 - 27:41
I'm going to have an orientation
for that tangent line,
• 27:41 - 27:46
and it's going to go
up, the orientation.
• 27:46 - 27:48
The orientation is important.
• 27:48 - 27:51
But if dx dy and
I switched them,
• 27:51 - 27:56
I said, dy, swap with dx,
what's going to happen?
• 27:56 - 28:02
I have to change to
change to clockwise.
• 28:02 - 28:04
And then the
orientation goes down.
• 28:04 - 28:07
And that's what they use
in mechanics when it comes
• 28:07 - 28:09
to the normal to the surface.
• 28:09 - 28:13
So again, you guys remember,
• 28:13 - 28:16
and we did the cross product,
and we got the normal.
• 28:16 - 28:19
If it's from this
one to this one,
• 28:19 - 28:21
it's counterclockwise
and goes up,
• 28:21 - 28:24
but if it's from this
vector to this other vector,
• 28:24 - 28:27
it's clockwise and goes down.
• 28:27 - 28:30
This is how a
mechanical engineer
• 28:30 - 28:33
will know how the
surface is oriented
• 28:33 - 28:36
based on the partial
velocities, for example
• 28:36 - 28:39
He has the partial
velocities along a surface,
• 28:39 - 28:43
and somebody says, take the
normal, take the unit normal.
• 28:43 - 28:45
He goes, like, are
you a physicist?
• 28:45 - 28:46
No, I'm an engineer.
• 28:46 - 28:49
You don't know how
to take the normal.
• 28:49 - 28:50
And of course, he knows.
• 28:50 - 28:54
He knows the convention
by this right-hand rule,
• 28:54 - 28:55
whatever you guys call it.
• 28:55 - 28:57
I call it the faucet rule.
• 28:57 - 29:01
It goes like this,
or it goes like that.
• 29:01 - 29:04
It's the same for a faucet,
for any type of screw,
• 29:04 - 29:08
for the right-hand
rule, whatever.
• 29:08 - 29:11
What else do you have
to believe me are true?
• 29:11 - 29:15
dx wedge dx is 0.
• 29:15 - 29:18
Can somebody tell me why
that is natural to introduce
• 29:18 - 29:19
such a wedge product?
• 29:19 - 29:22
STUDENT: Because the sine of
the angle between those is 0.
• 29:22 - 29:23
MAGDALENA TODA: Right.
• 29:23 - 29:29
Once you flatten this, once
you flatten the parallelogram,
• 29:29 - 29:30
there is no area.
• 29:30 - 29:31
So the area is 0.
• 29:31 - 29:35
How about dy dy sined area?
• 29:35 - 29:36
0.
• 29:36 - 29:38
So these are all
the properties you
• 29:38 - 29:41
need to know of the
sine area, sined areas.
• 29:41 - 29:44
• 29:44 - 29:47
OK, so now let's
see what happens
• 29:47 - 29:51
if we take this element,
which is a differential,
• 29:51 - 29:55
and wedge it with this element,
which is also a differential.
• 29:55 - 29:56
OK.
• 29:56 - 30:00
Oh my God, I'm shaking
• 30:00 - 30:02
I'm going to get
something weird.
• 30:02 - 30:04
• 30:04 - 30:07
Let's see what happens.
• 30:07 - 30:14
dx wedge dy equals-- do
you guys have questions?
• 30:14 - 30:18
Let's see what the mechanics are
for this type of computation.
• 30:18 - 30:21
• 30:21 - 30:28
I go-- this is like
a-- displacement wedge
• 30:28 - 30:30
this other displacement.
• 30:30 - 30:33
• 30:33 - 30:36
Think of them as true
vector displacements,
• 30:36 - 30:41
and as if you had a cross
product, or something.
• 30:41 - 30:42
OK.
• 30:42 - 30:44
How does this go?
• 30:44 - 30:45
It's distributed.
• 30:45 - 30:48
It's linear functions,
because we've
• 30:48 - 30:51
studied the
properties of vectors,
• 30:51 - 30:53
this acts by linearity.
• 30:53 - 30:58
So you go and say, first
first, times plus first times
• 30:58 - 31:03
second-- and times is
this guy, this weirdo--
• 31:03 - 31:07
plus second times first,
plus second times second,
• 31:07 - 31:09
where the wedge is
the operator that
• 31:09 - 31:11
has to satisfy these functions.
• 31:11 - 31:14
It's similar to
the cross product.
• 31:14 - 31:15
OK.
• 31:15 - 31:21
Then let's go x
sub r, y sub r, dr
• 31:21 - 31:27
wedge dr. Oh, let's 0 go away.
• 31:27 - 31:30
I say, leave me alone,
you're making my life hard.
• 31:30 - 31:38
Then I go plus x sub r--
this is a small function.
• 31:38 - 31:41
y sub theta, another
small function.
• 31:41 - 31:44
What of this
displacement, dr d theta.
• 31:44 - 31:47
I'm like those d
something, d something,
• 31:47 - 31:49
two small displacements
in the cross product.
• 31:49 - 31:53
OK, plus.
• 31:53 - 31:55
Who is telling me what next?
• 31:55 - 31:56
STUDENT: x theta--
• 31:56 - 32:06
MAGDALENA TODA: x theta
yr, d theta dr. Is it fair?
• 32:06 - 32:10
I did the second guy from the
first one with the first guy
• 32:10 - 32:12
from the second one.
• 32:12 - 32:15
And finally, I'm too
lazy to write it down.
• 32:15 - 32:16
What do I get?
• 32:16 - 32:17
STUDENT: 0.
• 32:17 - 32:17
MAGDALENA TODA: 0.
• 32:17 - 32:18
Why is that?
• 32:18 - 32:20
Because d theta,
always d theta is 0.
• 32:20 - 32:27
It's like you are flattening--
there is no more parallelogram.
• 32:27 - 32:28
OK?
• 32:28 - 32:32
So the two dimensions of
the parallelogram become 0.
• 32:32 - 32:37
The parallelogram would
become [? a secant. ?]
• 32:37 - 32:40
What you get is
something really weak.
• 32:40 - 32:42
And we don't talk
• 32:42 - 32:45
but that's called the Jacobian.
• 32:45 - 32:51
dr d theta and d theta dr, once
you introduce the sine area,
• 32:51 - 32:56
you finally understand
why you get this r here,
• 32:56 - 32:58
what the Jacobian is.
• 32:58 - 32:59
If you don't introduce
the sine area,
• 32:59 - 33:02
you will never understand,
and you cannot explain it
• 33:02 - 33:06
to anybody, any student have.
• 33:06 - 33:12
OK, so this guy, d theta,
which the r is just
• 33:12 - 33:14
swapping the two displacements.
• 33:14 - 33:17
So it's going to be
minus dr d theta.
• 33:17 - 33:19
Why is that, guys?
• 33:19 - 33:23
Because that's how I said, every
time I swap two displacements,
• 33:23 - 33:25
I'm changing the orientation.
• 33:25 - 33:28
It's like the cross
product between a and b,
• 33:28 - 33:30
and the cross product
between b and a.
• 33:30 - 33:35
So I'm going up or I'm going
down, I'm changing orientation.
• 33:35 - 33:36
What's left in the end?
• 33:36 - 33:39
It's really just this
guy that's really weird.
• 33:39 - 33:41
I'm going to collect the terms.
• 33:41 - 33:45
One from here, one
from here, and a minus.
• 33:45 - 33:45
• 33:45 - 33:49
STUDENT: Do the wedges
just cancel out?
• 33:49 - 33:50
MAGDALENA TODA: This was 0.
• 33:50 - 33:52
This was 0.
• 33:52 - 33:58
And this dr d theta is nonzero,
but is the common factor.
• 33:58 - 34:00
So I pull him out from here.
• 34:00 - 34:02
I pull him out from here.
• 34:02 - 34:02
Out.
• 34:02 - 34:08
Factor out, and what's
left is this guy over here
• 34:08 - 34:11
who is this guy over here.
• 34:11 - 34:15
And this guy over
here with a minus
• 34:15 - 34:20
who gives me minus d theta yr.
• 34:20 - 34:21
That's all.
• 34:21 - 34:25
So now you will understand
why I am going to get r.
• 34:25 - 34:30
So the general rule will
be that the area element dx
• 34:30 - 34:36
dy, the wedge sined
area, will be--
• 34:36 - 34:39
you have to help me
with this individual,
• 34:39 - 34:43
because he really looks weird.
• 34:43 - 34:46
Do you know of a name for it?
• 34:46 - 34:50
Do you know what
this is going to be?
• 34:50 - 34:52
Linear algebra people,
only two of you.
• 34:52 - 34:57
Maybe you have an idea.
• 34:57 - 35:00
So it's like, I
take this fellow,
• 35:00 - 35:02
and I multiply by that fellow.
• 35:02 - 35:04
• 35:04 - 35:07
Multiply these two.
• 35:07 - 35:13
And I go minus this
fellow times that fellow.
• 35:13 - 35:15
STUDENT: [INAUDIBLE]
• 35:15 - 35:18
MAGDALENA TODA: It's like
a determinant of something.
• 35:18 - 35:23
So when people write
the differential system,
• 35:23 - 35:26
[INTERPOSING VOICES]
51, you will understand
• 35:26 - 35:28
that this is a system.
• 35:28 - 35:28
OK?
• 35:28 - 35:30
It's a system of two equations.
• 35:30 - 35:32
• 35:32 - 35:34
The other little, like,
vector displacements,
• 35:34 - 35:36
you are going to
write it like that.
• 35:36 - 35:46
dx dy will be matrix
multiplication dr d theta.
• 35:46 - 35:50
And how do you multiply
x sub r x sub theta?
• 35:50 - 35:55
So you go first row times
first column give you that.
• 35:55 - 36:00
And second row times the
column gives you this.
• 36:00 - 36:02
y sub r, y sub theta.
• 36:02 - 36:06
This is a magic guy
called Jacobian.
• 36:06 - 36:10
We keep this a secret, and
most Professors don't even
• 36:10 - 36:13
cover 12.8, because
they don't want to tell
• 36:13 - 36:15
people what a Jacobian is.
• 36:15 - 36:17
This is little r.
• 36:17 - 36:21
I know you don't believe me, but
the determinant of this matrix
• 36:21 - 36:23
must be little r.
• 36:23 - 36:25
You have to help me prove that.
• 36:25 - 36:27
And this is the Jacobian.
• 36:27 - 36:30
Do you guys know why
it's called Jacobian?
• 36:30 - 36:33
It's the determinant
of this matrix.
• 36:33 - 36:43
Let's call this
matrix J. And this
• 36:43 - 36:49
is J, determinant
of [? scripture. ?]
• 36:49 - 36:50
This is called Jacobian.
• 36:50 - 36:54
• 36:54 - 36:55
Why is it r?
• 36:55 - 36:58
Let's-- I don't know.
• 36:58 - 37:00
Let's see how we do it.
• 37:00 - 37:04
• 37:04 - 37:07
This is r cosine theta, right?
• 37:07 - 37:10
This is r sine theta.
• 37:10 - 37:15
So dx must be what x sub r?
• 37:15 - 37:20
X sub r, x sub r, cosine theta.
• 37:20 - 37:22
d plus.
• 37:22 - 37:24
What is x sub t?
• 37:24 - 37:26
• 37:26 - 37:29
x sub theta.
• 37:29 - 37:32
I need to differentiate
this with respect to theta.
• 37:32 - 37:34
STUDENT: It's going to
be negative r sine theta.
• 37:34 - 37:36
MAGDALENA TODA: Minus r
sine theta, very good.
• 37:36 - 37:38
And d theta.
• 37:38 - 37:44
Then I go dy was
sine theta-- dr,
• 37:44 - 37:46
I'm looking at these
equations, and I'm
• 37:46 - 37:49
repeating them for my case.
• 37:49 - 37:53
This is true in general for
any kind of coordinates.
• 37:53 - 37:57
So it's a general equation
for any kind of coordinate,
• 37:57 - 37:59
two coordinates,
two coordinates,
• 37:59 - 38:01
any kind of
coordinates in plane,
• 38:01 - 38:05
you can choose any
functions, f of uv, g of uv,
• 38:05 - 38:07
whatever you want.
• 38:07 - 38:09
But for this particular
case of polar coordinates
• 38:09 - 38:12
is going to look really
pretty in the end.
• 38:12 - 38:16
What do I get when I do y theta?
• 38:16 - 38:17
r cosine theta.
• 38:17 - 38:19
Am I right, guys?
• 38:19 - 38:21
Keen an eye on it.
• 38:21 - 38:27
So this will become-- the
area element will become what?
• 38:27 - 38:31
The determinant of this matrix.
• 38:31 - 38:35
Red, red, red, red.
• 38:35 - 38:36
How do I compute a term?
• 38:36 - 38:39
Not everybody knows,
and it's this times
• 38:39 - 38:45
that minus this times that.
• 38:45 - 38:46
OK, let's do that.
• 38:46 - 38:53
So I get r cosine squared
theta minus minus plus r sine
• 38:53 - 38:56
squared theta.
• 38:56 - 38:59
dr, d theta, and our wedge.
• 38:59 - 39:00
What is this?
• 39:00 - 39:01
STUDENT: 1.
• 39:01 - 39:04
MAGDALENA TODA:
Jacobian is r times 1,
• 39:04 - 39:07
because that's the
Pythagorean theorem, right?
• 39:07 - 39:12
So we have r, and this is
the meaning of r, here.
• 39:12 - 39:17
So when I moved from dx dy,
• 39:17 - 39:19
that I didn't tell you about.
• 39:19 - 39:23
And this wedge
becomes r dr d theta,
• 39:23 - 39:27
and that's the
correct way to explain
• 39:27 - 39:30
why you get the Jacobian there.
• 39:30 - 39:32
We don't do that in the book.
• 39:32 - 39:35
We do it later, and we
sort of smuggle through.
• 39:35 - 39:37
We don't do a very thorough job.
• 39:37 - 39:40
When you go into
• 39:40 - 39:43
you would see that again the
way I explained it to you.
• 39:43 - 39:47
If you ever want to
• 39:47 - 39:52
then you need to take the
• 39:52 - 39:58
and 4351 where you are
• 39:58 - 40:01
If you take those as a math
major or engineering major,
• 40:01 - 40:02
it doesn't matter.
• 40:02 - 40:04
When you go to
• 40:04 - 40:07
don't make you take
• 40:07 - 40:09
• 40:09 - 40:13
So it's somewhere borderline
between senior year
• 40:13 - 40:19
the first course you would take
• 40:19 - 40:22
• 40:22 - 40:23
OK.
• 40:23 - 40:30
So an example of
this transformation
• 40:30 - 40:33
where we know what
• 40:33 - 40:39
Let's say I have
a picture, and I
• 40:39 - 40:43
have a domain D, which
is-- this is x squared
• 40:43 - 40:45
plus y squared equals 1.
• 40:45 - 40:48
I have the domain as being
[INTERPOSING VOICES].
• 40:48 - 40:52
• 40:52 - 40:58
And then I say, I would
like-- what would I like?
• 40:58 - 41:04
I would like the
volume of the-- this
• 41:04 - 41:10
is a paraboloid, z equals
x squared plus y squared.
• 41:10 - 41:13
I would like the
volume of this object.
• 41:13 - 41:14
This is my obsession.
• 41:14 - 41:18
I'm going to create a
vase some day like that.
• 41:18 - 41:23
So you want this
piece to be a solid.
• 41:23 - 41:25
In cross section,
it will just this.
• 41:25 - 41:26
In cross section.
• 41:26 - 41:28
And it's a solid of revolution.
• 41:28 - 41:30
In this cross section,
you have to imagine
• 41:30 - 41:36
revolving it around the z-axis,
then creating a heavy object.
• 41:36 - 41:38
From the outside, don't
see what's inside.
• 41:38 - 41:40
It looks like a cylinder.
• 41:40 - 41:42
But you go inside and
you see the valley.
• 41:42 - 41:46
So it's between a
paraboloid and a disc,
• 41:46 - 41:48
a unit disc on the floor.
• 41:48 - 41:51
How are we going
to try and do that?
• 41:51 - 41:54
And what did I
teach you last time?
• 41:54 - 42:02
Last time, I taught you that--
we have to go over a domain D.
• 42:02 - 42:04
But that domain
D, unfortunately,
• 42:04 - 42:06
is hard to express.
• 42:06 - 42:09
How would you express D
in Cartesian coordinates?
• 42:09 - 42:15
• 42:15 - 42:16
You can do it.
• 42:16 - 42:19
It's going to be a headache.
• 42:19 - 42:22
x is between minus 1 and 1.
• 42:22 - 42:24
Am I right, guys?
• 42:24 - 42:28
And y will be between--
now I have two branches.
• 42:28 - 42:30
One, and the other one.
• 42:30 - 42:33
One branch would be square--
I hate square roots.
• 42:33 - 42:36
I absolutely hate them.
• 42:36 - 42:40
y is between 1 minus
square root x squared,
• 42:40 - 42:43
minus square root
1 minus x squared.
• 42:43 - 42:48
So if I were to ask you to do
the integral like last time,
• 42:48 - 42:51
how would you set
up the integral?
• 42:51 - 42:53
You go, OK, I know what this is.
• 42:53 - 43:01
Integral over D of
f of x, y, dx dy.
• 43:01 - 43:03
This is actually a wedge.
• 43:03 - 43:06
In my case, we avoided that.
• 43:06 - 43:08
We said dh.
• 43:08 - 43:10
And we said, what is f of x, y?
• 43:10 - 43:12
x squared plus y
squared, because I
• 43:12 - 43:16
want everything that's under
the graph, not above the graph.
• 43:16 - 43:19
So everything that's
under the graph.
• 43:19 - 43:27
F of x, y is this guy.
• 43:27 - 43:28
And the I have to
start thinking,
• 43:28 - 43:32
because it's a type 1 or type 2?
• 43:32 - 43:36
It's a type 1 the
way I set it up,
• 43:36 - 43:39
but I can make it
type 2 by reversing
• 43:39 - 43:42
the order of integration
like I did last time.
• 43:42 - 43:44
If I treat it like
that, it's going
• 43:44 - 43:46
to be type 1, though, right?
• 43:46 - 43:51
So I have to put
dy first, and then
• 43:51 - 43:55
change the color of the dx.
• 43:55 - 43:58
And since mister y
is the purple guy,
• 43:58 - 44:03
y would be going between
these ugly square roots that
• 44:03 - 44:04
to go on my nerves.
• 44:04 - 44:10
• 44:10 - 44:17
And then x goes
between minus 1 and 1.
• 44:17 - 44:21
It's a little bit of a headache.
• 44:21 - 44:23
Why is it a headache, guys?
• 44:23 - 44:27
Let's anticipate what we need to
do if we do it like last time.
• 44:27 - 44:32
We need to integrate this
ugly fellow in terms of y,
• 44:32 - 44:36
and when we integrate this in
terms of y, what do we get?
• 44:36 - 44:38
Don't write it, because
it's going to be a mess.
• 44:38 - 44:45
We get x squared times
y plus y cubed over 3.
• 44:45 - 44:47
And then, instead of y, I have
to replace those square roots,
• 44:47 - 44:50
and I'll never get rid
of the square roots.
• 44:50 - 44:53
It's going to be a mess, indeed.
• 44:53 - 44:56
And I may even-- in
general, I may not even
• 44:56 - 44:59
be able to solve the
integral, and that's
• 44:59 - 45:01
because I'll start
• 45:01 - 45:03
crying, I'll get depressed,
I'll take Prozac, whatever
• 45:03 - 45:05
you take for depression.
• 45:05 - 45:08
I don't know, I never took it,
because I'm never depressed.
• 45:08 - 45:11
So what do you do in that case?
• 45:11 - 45:12
STUDENT: Switch to polar.
• 45:12 - 45:14
MAGDALENA TODA: You
switch to polar.
• 45:14 - 45:19
So you use this big polar-switch
theorem, the theorem that
• 45:19 - 45:24
tells you, be smart,
apply this theorem,
• 45:24 - 45:31
and have to understand that
• 45:31 - 45:33
in [INTERPOSING VOICES]
Cartesian coordinates
• 45:33 - 45:37
is D. If you want express
the same thing as D star,
• 45:37 - 45:40
D star will be in
polar coordinates.
• 45:40 - 45:44
You have to be a little bit
smarter, and say r theta,
• 45:44 - 45:49
where now you have to put
the bounds that limit--
• 45:49 - 45:50
STUDENT: r.
• 45:50 - 45:51
MAGDALENA TODA: r from?
• 45:51 - 45:51
STUDENT: 0 to 1.
• 45:51 - 45:53
MAGDALENA TODA: 0
to 1, excellent.
• 45:53 - 45:57
You cannot let r go to
infinity, because the vase is
• 45:57 - 45:57
increasingly.
• 45:57 - 46:01
You only needs the vase that
has the radius 1 on the bottom.
• 46:01 - 46:09
So r is 0 to 1, and
theta is 0 to 1 pi.
• 46:09 - 46:11
And there you have
• 46:11 - 46:16
So I need to be smart
and say integral.
• 46:16 - 46:18
Integral, what do
you want to do first?
• 46:18 - 46:22
Well, it doesn't matter, dr,
d theta, whatever you want.
• 46:22 - 46:26
So mister theta will
be the last of the two.
• 46:26 - 46:32
So theta will be between 0
and 2 pi, a complete rotation.
• 46:32 - 46:36
r between 0 and 1.
• 46:36 - 46:38
And inside here I
have to be smart
• 46:38 - 46:42
and see that life
can be fun when
• 46:42 - 46:44
I work with polar coordinates.
• 46:44 - 46:46
Why?
• 46:46 - 46:47
What is the integral?
• 46:47 - 46:48
x squared plus y squared.
• 46:48 - 46:51
I've seen him
somewhere before when
• 46:51 - 46:55
it came to polar coordinates.
• 46:55 - 46:56
STUDENT: R squared.
• 46:56 - 46:57
STUDENT: That will be r squared.
• 46:57 - 47:00
MAGDALENA TODA: That
will be r squared.
• 47:00 - 47:04
r squared times-- never
forget the Jacobian,
• 47:04 - 47:08
and the Jacobian is mister r.
• 47:08 - 47:13
And now I'm going to
take all this integral.
• 47:13 - 47:16
I'll finally compute
the volume of my vase.
• 47:16 - 47:20
Imagine if this vase
• 47:20 - 47:22
This is my dream.
• 47:22 - 47:25
So imagine that this
vase would have,
• 47:25 - 47:27
I don't know what dimensions.
• 47:27 - 47:29
I need to find the
volume, and multiply it
• 47:29 - 47:32
by the density of gold
and find out-- yes, sir?
• 47:32 - 47:36
STUDENT: Professor, like in this
question, b time is dt by dr,
• 47:36 - 47:38
but you can't switch it--
• 47:38 - 47:39
MAGDALENA TODA: Yes, you can.
• 47:39 - 47:41
That's exactly my point.
• 47:41 - 47:43
I'll tell you in a second.
• 47:43 - 47:48
When can you replace d theta dr?
• 47:48 - 47:52
You can always do that when
you have something under here,
• 47:52 - 47:56
which is a big
function of theta times
• 47:56 - 48:02
a bit function of r, because
you can treat them differently.
• 48:02 - 48:05
• 48:05 - 48:09
Now, this has no theta.
• 48:09 - 48:14
So actually, the
theta is not going
• 48:14 - 48:19
• 48:19 - 48:22
theta for the time being.
• 48:22 - 48:30
What you have inside is Calculus
I. When you have a product,
• 48:30 - 48:31
you can always switch.
• 48:31 - 48:34
And I'll give you
a theorem later.
• 48:34 - 48:39
0 over 1, r cubed,
thank God, this
• 48:39 - 48:42
is Calc I. Integral
from 0 to 1, r
• 48:42 - 48:47
cubed dr. That's Calc
I. How much is that?
• 48:47 - 48:48
I'm lazy.
• 48:48 - 48:50
I don't want to do it.
• 48:50 - 48:51
STUDENT: 1/4.
• 48:51 - 48:52
MAGDALENA TODA: It's 1/4.
• 48:52 - 48:53
Very good.
• 48:53 - 48:54
Thank you.
• 48:54 - 48:58
And if I get further, and I'm a
little bi lazy, what do I get?
• 48:58 - 49:02
1/4 is the constant,
it pulls out.
• 49:02 - 49:03
STUDENT: So, they don't--
• 49:03 - 49:10
MAGDALENA TODA: So I get 2 pi
over 4, which is pi over 2.
• 49:10 - 49:11
Am I right?
• 49:11 - 49:11
STUDENT: Yeah.
• 49:11 - 49:13
MAGDALENA TODA: So
this constant gets out,
• 49:13 - 49:14
integral comes in through 2 pi.
• 49:14 - 49:16
It will be 2 pi, and
• 49:16 - 49:20
So pi over 2 is the volume.
• 49:20 - 49:23
If I have a 1-inch
diameter, and I
• 49:23 - 49:27
have this vase made of gold,
which is a piece of jewelry,
• 49:27 - 49:34
really beautiful, then I'm going
to have pi over 2 the volume.
• 49:34 - 49:36
That will be a little
bit hard to see
• 49:36 - 49:39
what we have in square inches.
• 49:39 - 49:44
We have 1.5-something
square inches, and then--
• 49:44 - 49:45
STUDENT: More.
• 49:45 - 49:46
MAGDALENA TODA:
And then multiply
• 49:46 - 49:50
by the density of
gold, and estimate,
• 49:50 - 49:58
based on the mass, how much
money that's going to be.
• 49:58 - 50:00
What did I want to
tell [? Miteish? ?]
• 50:00 - 50:03
I don't want to forget what
• 50:03 - 50:04
was a smart question.
• 50:04 - 50:09
When can we reverse the
order of integration?
• 50:09 - 50:12
In general, it's
hard to compute.
• 50:12 - 50:15
But in this case, I'm you
are the luckiest person
• 50:15 - 50:17
in the world, because
just take a look at me.
• 50:17 - 50:22
I have, let's see, my
r between 0 and 2 pi,
• 50:22 - 50:29
and my theta between 0 and 2
pi, and my r between 0 and 1.
• 50:29 - 50:32
Whatever, it doesn't matter,
it could be anything.
• 50:32 - 50:36
And here I have a function of r
and a function g of theta only.
• 50:36 - 50:38
And it's a product.
• 50:38 - 50:41
The variables are separate.
• 50:41 - 50:46
When I do-- what do I
do for dr or d theta?
• 50:46 - 50:49
dr. When I do dr--
with respect to dr,
• 50:49 - 50:53
this fellow goes, I
don't belong in here.
• 50:53 - 50:56
I'm mister theta that
doesn't belong in here.
• 50:56 - 50:57
I'm independent.
• 50:57 - 50:59
I want to go out.
• 50:59 - 51:02
And he wants out.
• 51:02 - 51:10
So you have some integrals
that you got out a g of theta,
• 51:10 - 51:16
and another integral, and you
have f of r dr in a bracket,
• 51:16 - 51:21
and then you go d theta.
• 51:21 - 51:23
What is going to happen next?
• 51:23 - 51:27
You solve this integral, and
it's going to be a number.
• 51:27 - 51:30
This number could be 8,
7, 9.2, God knows what.
• 51:30 - 51:33
Why don't you pull that
constant out right now?
• 51:33 - 51:35
So you say, OK, I can do that.
• 51:35 - 51:37
It's just a number.
• 51:37 - 51:38
Whatever.
• 51:38 - 51:42
That's going to be
integral f dr, times
• 51:42 - 51:44
what do you have left
when you pull that out?
• 51:44 - 51:45
A what?
• 51:45 - 51:46
STUDENT: Integral.
• 51:46 - 51:49
MAGDALENA TODA: Integral of
G, the integral of g of theta,
• 51:49 - 51:51
d theta.
• 51:51 - 51:54
So we just proved a theorem
that is really pretty.
• 51:54 - 51:59
If you have to integrate,
and I will try to do it here.
• 51:59 - 52:03
• 52:03 - 52:04
No--
• 52:04 - 52:06
STUDENT: So essentially, when
you're integrating with respect
• 52:06 - 52:11
to r, you can treat any function
of only theta as a constant?
• 52:11 - 52:12
MAGDALENA TODA: Yeah.
• 52:12 - 52:15
I'll tell you in a second
what it means, because--
• 52:15 - 52:16
STUDENT: Sorry.
• 52:16 - 52:17
MAGDALENA TODA: You're fine.
• 52:17 - 52:22
Integrate for domain,
rectangular domains,
• 52:22 - 52:26
let's say u between alpha,
beta, u between gamma,
• 52:26 - 52:30
delta, then what's
going to happen?
• 52:30 - 52:35
As you said very well,
integral from-- what
• 52:35 - 52:38
do you want first, dv or du?
• 52:38 - 52:41
dv, du, it doesn't matter.
• 52:41 - 52:44
v is between gamma, delta.
• 52:44 - 52:47
v is the first guy inside, OK.
• 52:47 - 52:49
Gamma, delta.
• 52:49 - 52:50
I should have cd.
• 52:50 - 52:51
It's all Greek to me.
• 52:51 - 52:55
Why did I pick
that three people?
• 52:55 - 53:00
If this is going to be a product
of two functions, one is in u
• 53:00 - 53:06
and one is in v. Let's
say A of u and B of v,
• 53:06 - 53:11
I can go ahead and say
product of two constants.
• 53:11 - 53:14
And who are those two
constants I was referring to?
• 53:14 - 53:16
You can do that directly.
• 53:16 - 53:19
If the two variables are
separated through a product,
• 53:19 - 53:23
you have a product of
two separate variables.
• 53:23 - 53:26
A is only in u, it
depends only on u.
• 53:26 - 53:31
And B is only on v. They have
nothing to do with one another.
• 53:31 - 53:35
Then you can go ahead and do
the first integral with respect
• 53:35 - 53:43
to u only of a of u, du,
u between alpha, beta.
• 53:43 - 53:46
• 53:46 - 53:49
Times this other constant.
• 53:49 - 53:54
Integral of B of v,
where v is moving,
• 53:54 - 53:59
v is moving between
gamma, delta.
• 53:59 - 54:01
beta, gamma, delta,
• 54:01 - 54:04
put any numbers you want.
• 54:04 - 54:05
OK?
• 54:05 - 54:06
This is the lucky case.
• 54:06 - 54:09
So you're always hoping
that on the final,
• 54:09 - 54:13
you can get something
where you can separate.
• 54:13 - 54:14
Here you have no theta.
• 54:14 - 54:16
This is the luckiest
case in the world.
• 54:16 - 54:19
So it's just r
cubed times theta.
• 54:19 - 54:21
But you can still
have a lucky case
• 54:21 - 54:25
when you have something
like a function of r
• 54:25 - 54:26
times a function of theta.
• 54:26 - 54:29
And then you have
another beautiful polar
• 54:29 - 54:32
coordinate integral
that you're not going
• 54:32 - 54:35
to struggle with for very long.
• 54:35 - 54:37
OK, I'm going to erase here.
• 54:37 - 54:56
• 54:56 - 55:02
For example, let me
give you another one.
• 55:02 - 55:04
Suppose that somebody
was really mean to you,
• 55:04 - 55:08
and wanted to kill
you in the final,
• 55:08 - 55:10
and they gave you the
following problem.
• 55:10 - 55:13
• 55:13 - 55:17
Assume the domain D-- they
don't even tell you what it is.
• 55:17 - 55:19
They just want to
challenge you--
• 55:19 - 55:25
will be x, y with the
property that x squared plus y
• 55:25 - 55:32
squared is between a 1 and a 4.
• 55:32 - 55:36
• 55:36 - 55:53
Compute the integral over D of
r [? pan ?] of y over x and da,
• 55:53 - 55:57
where bi would be ds dy.
• 55:57 - 56:01
So you look at this
cross-eyed and say, gosh,
• 56:01 - 56:04
whoever-- we don't do that.
• 56:04 - 56:05
But I've seen schools.
• 56:05 - 56:09
I've seen this given at a
school, when they covered
• 56:09 - 56:12
this particular
example, they've covered
• 56:12 - 56:15
something like the previous
one that I showed you.
• 56:15 - 56:16
But they never covered this.
• 56:16 - 56:18
And they said,
OK, they're smart,
• 56:18 - 56:20
let them figure this out.
• 56:20 - 56:23
And I think it was Texas A&M.
They gave something like that
• 56:23 - 56:26
without working this in class.
• 56:26 - 56:29
They assumed that
the students should
• 56:29 - 56:31
be good enough to
figure out what
• 56:31 - 56:35
this is in polar coordinates.
• 56:35 - 56:40
So in polar coordinates,
what does the theorem say?
• 56:40 - 56:44
We should switch to a domain
D star that corresponds to D.
• 56:44 - 56:48
Now, D was given like that.
• 56:48 - 56:51
But we have to say
the corresponding D
• 56:51 - 56:55
star, reinterpreted
in polar coordinates,
• 56:55 - 57:00
r theta has to be also
written beautifully out.
• 57:00 - 57:04
Unless you draw the picture,
first of all, you cannot do it.
• 57:04 - 57:08
So the prof at Texas A&M didn't
even say, draw the picture,
• 57:08 - 57:11
and think of the
meaning of that.
• 57:11 - 57:15
What is the meaning of
this set, geometric set,
• 57:15 - 57:17
geometric locus of points.
• 57:17 - 57:19
STUDENT: You've
got a circle sub-
• 57:19 - 57:22
MAGDALENA TODA: You
have concentric circles,
• 57:22 - 57:27
sub-radius 1 and 2, and it's
like a ring, it's an annulus.
• 57:27 - 57:30
And he said, well,
I didn't do it.
• 57:30 - 57:33
I mean they were smart.
• 57:33 - 57:35
I gave it to them to do.
• 57:35 - 57:41
So if the students don't see
at least an example like that,
• 57:41 - 57:45
they have difficulty,
in my experience.
• 57:45 - 57:47
OK, for this kind
of annulus, you
• 57:47 - 57:51
here, but the dotted part
• 57:51 - 57:53
is not included in your domain.
• 57:53 - 57:57
So you have to be smart and
say, wait a minute, my radius
• 57:57 - 57:59
is not starting at 0.
• 57:59 - 58:02
It's starting at 1
and it's ending at 2.
• 58:02 - 58:06
And I put that here.
• 58:06 - 58:11
And theta is the whole
ring, so from 0 to 2 pi.
• 58:11 - 58:14
• 58:14 - 58:18
Whether you do that
over the open set,
• 58:18 - 58:21
that's called annulus
without the boundaries,
• 58:21 - 58:25
or you do it about the
one with the boundaries,
• 58:25 - 58:28
it doesn't matter, the integral
is not going to change.
• 58:28 - 58:33
And you are going to learn
• 58:33 - 58:37
it doesn't matter that if
you remove the boundary,
• 58:37 - 58:39
you put back the boundary.
• 58:39 - 58:43
That is a certain set of a
• 58:43 - 58:46
It's not going to
• 58:46 - 58:49
So no matter how you
express it-- maybe
• 58:49 - 58:52
you want to express
it like an open set.
• 58:52 - 58:55
You still have
the same integral.
• 58:55 - 58:58
Double integral
of D star, this is
• 58:58 - 59:02
going to give me a headache,
unless you help me.
• 59:02 - 59:06
What is this in
polar coordinates?
• 59:06 - 59:06
STUDENT: [INAUDIBLE]
• 59:06 - 59:10
• 59:10 - 59:11
MAGDALENA TODA: I
know when-- once I've
• 59:11 - 59:13
figured out the
integrand, I'm going
• 59:13 - 59:17
to remember to always
multiply by an r,
• 59:17 - 59:19
because if I don't,
I'm in big trouble.
• 59:19 - 59:24
And then I go dr d theta.
• 59:24 - 59:26
But I don't know what this is.
• 59:26 - 59:28
STUDENT: r.
• 59:28 - 59:34
MAGDALENA TODA: Nope, but
you're-- so r cosine theta is
• 59:34 - 59:38
x, r sine theta is y.
• 59:38 - 59:41
When you do y over
x, what do you get?
• 59:41 - 59:44
Always tangent of theta.
• 59:44 - 59:48
And if you do arctangent
of tangent, you get theta.
• 59:48 - 59:51
So that was not hard,
but the students did
• 59:51 - 59:53
not-- in that
class, I was talking
• 59:53 - 59:57
to whoever gave the exam,
70-something percent
• 59:57 - 59:59
of the students did
not know how to do it,
• 59:59 - 60:01
seen something similar,
• 60:01 - 60:07
and they didn't think how
to express this theta in r.
• 60:07 - 60:09
So what do we mean to do?
• 60:09 - 60:12
We mean, is this a product?
• 60:12 - 60:13
It's a beautiful product.
• 60:13 - 60:18
They are separate variables like
[INAUDIBLE] [? shafts. ?] Now,
• 60:18 - 60:20
you see, you can separate them.
• 60:20 - 60:27
The r is between 1 and 2,
so I can do-- eventually I
• 60:27 - 60:28
can do the r first.
• 60:28 - 60:33
And theta is between 0 and
2 pi, and as I taught you
• 60:33 - 60:38
by the previous theorem, you
can separate the two integrals,
• 60:38 - 60:40
because this one gets out.
• 60:40 - 60:41
It's a constant.
• 60:41 - 60:47
So you're left with integral
from 0 to 2 pi theta d
• 60:47 - 61:05
theta, and the integral from 1
to 2 r dr. r dr theta d theta.
• 61:05 - 61:06
This should be a piece of cake.
• 61:06 - 61:14
The only thing we have to
do is some easy Calculus I.
• 61:14 - 61:18
So what is integral
of theta d theta?
• 61:18 - 61:20
I'm not going to rush anywhere.
• 61:20 - 61:27
Theta squared over 2
between theta equals 0 down
• 61:27 - 61:31
and theta equals 2 pi up.
• 61:31 - 61:32
Right?
• 61:32 - 61:34
STUDENT: [INAUDIBLE]
• 61:34 - 61:35
MAGDALENA TODA: Yeah.
• 61:35 - 61:36
I'll do that later.
• 61:36 - 61:37
I don't care.
• 61:37 - 61:41
This is going to be r squared
over 2 between 1 and 2.
• 61:41 - 61:44
So the numerical
• 61:44 - 61:51
how to do any math like
that, is going to be--
• 61:51 - 61:52
STUDENT: 2 pi squared.
• 61:52 - 61:54
MAGDALENA TODA: 2 pi
squared, because I
• 61:54 - 61:58
have 4 pi squared over
2, so the first guy
• 61:58 - 62:08
is 2 pi squared, times-- I
get a 4 and 4 minus 1-- are
• 62:08 - 62:09
you guys with me?
• 62:09 - 62:13
So I get a-- let me
write it like that.
• 62:13 - 62:17
4 over 2 minus 1 over 2.
• 62:17 - 62:19
What's going to
happen to the over 2?
• 62:19 - 62:20
We'll simplify.
• 62:20 - 62:24
So this is going
to be 3 pi squared.
• 62:24 - 62:25
Okey Dokey?
• 62:25 - 62:25
Yes, sir?
• 62:25 - 62:28
STUDENT: How did you split it
into two integrals, right here?
• 62:28 - 62:31
MAGDALENA TODA: That's exactly
what I taught you before.
• 62:31 - 62:34
taught you before,
• 62:34 - 62:37
how did I prove that theorem?
• 62:37 - 62:41
The theorem that was
before was like that.
• 62:41 - 62:44
What was it?
• 62:44 - 62:49
Suppose I have a function of
theta, and a function of r,
• 62:49 - 62:53
and I have d theta
dr. And I think
• 62:53 - 62:56
this weather got to us,
because several people have
• 62:56 - 62:58
the cold and the flu.
• 62:58 - 62:59
• 62:59 - 63:04
It's full of--
mathematicians full of germs.
• 63:04 - 63:09
So theta, you want theta to
be between whatever you want.
• 63:09 - 63:11
Any two numbers.
• 63:11 - 63:12
Alpha and beta.
• 63:12 - 63:15
And r between gamma, delta.
• 63:15 - 63:18
This is what I
explained last time.
• 63:18 - 63:22
So when you integrate with
respect to theta first inside,
• 63:22 - 63:26
g of r says I have nothing
to do with these guys.
• 63:26 - 63:28
They're not my type,
they're not my gang.
• 63:28 - 63:31
I'm going out, have
a beer by myself.
• 63:31 - 63:39
So he goes out and
joins the r group,
• 63:39 - 63:41
because theta and r
have nothing in common.
• 63:41 - 63:45
They are separate variables.
• 63:45 - 63:46
This is a function
of r only, and that's
• 63:46 - 63:48
a function of theta only.
• 63:48 - 63:50
This is what I'm talking about.
• 63:50 - 63:52
OK, so that's a constant.
• 63:52 - 63:56
That constant pulls out.
• 63:56 - 64:00
So in the end, what you have is
that constant that pulled out
• 64:00 - 64:06
is going to be alpha, beta, f of
beta d theta as a number, times
• 64:06 - 64:08
what's left inside?
• 64:08 - 64:11
Integral from gamma
to delta g of r
• 64:11 - 64:18
dr. So when the two functions
F and G are functions of theta,
• 64:18 - 64:22
respectively, r only, they have
nothing to do with one another,
• 64:22 - 64:25
and you can write
the original integral
• 64:25 - 64:29
as the product of integrals,
and it's really a lucky case.
• 64:29 - 64:33
But you are going to encounter
this lucky case many times
• 64:33 - 64:39
in your final, in the midterm,
in-- OK, now thinking of what
• 64:39 - 64:41
I wanted to put on the midterm.
• 64:41 - 64:45
• 64:45 - 64:48
Somebody asked me if I'm going
• 64:48 - 64:52
at the homework and at the
• 64:52 - 64:58
are we going to have something
like the area of the cardioid?
• 64:58 - 65:01
Maybe not necessarily
that-- or area
• 65:01 - 65:05
between a cardioid and a circle
that intersect each other.
• 65:05 - 65:10
Those were doable
even with Calc II.
• 65:10 - 65:13
Something like that, that
was doable with Calc II,
• 65:13 - 65:16
I don't want to do it with a
double integral in Calc III,
• 65:16 - 65:23
and I want to give some problems
that are relevant to you guys.
• 65:23 - 65:27
• 65:27 - 65:29
The question, what's going
to be on the midterm?
• 65:29 - 65:33
is not-- OK, what's going
to be on the midterm?
• 65:33 - 65:36
It's going to be something
very similar to the sample
• 65:36 - 65:38
that I'm going to write.
• 65:38 - 65:41
included in that sample
• 65:41 - 65:45
the volume of a
• 65:45 - 65:50
So how do you compute out
the weight-- exercise 3 or 4,
• 65:50 - 66:07
whatever that is-- we compute
the volume of a sphere using
• 66:07 - 66:08
double integrals.
• 66:08 - 66:17
• 66:17 - 66:20
I don't know if we have time to
do this problem, but if we do,
• 66:20 - 66:25
that will be the last problem--
• 66:25 - 66:29
why is the volume inside the
sphere, volume of a ball,
• 66:29 - 66:30
actually.
• 66:30 - 66:33
Well, the size-- the solid ball.
• 66:33 - 66:36
Why is it 4 pi r cubed over 2?
• 66:36 - 66:38
Your, did she tell
you, or she told
• 66:38 - 66:43
Mr. [? Jaime ?], for example?
• 66:43 - 66:48
They were supposed to tell
you that you can prove that
• 66:48 - 66:49
with Calc II or Calc III.
• 66:49 - 66:51
It's not easy.
• 66:51 - 66:53
It's not an elementary formula.
• 66:53 - 66:54
In the ancient
times, they didn't
• 66:54 - 66:57
know how to do it, because
they didn't know calculus.
• 66:57 - 67:00
So what they tried to is
to approximate it and see
• 67:00 - 67:03
how it goes.
• 67:03 - 67:07
Assume you have the
• 67:07 - 67:09
and r is from here
to here, and I'm
• 67:09 - 67:13
going to go ahead and draw the
equator, the upper hemisphere,
• 67:13 - 67:19
the lower hemisphere, and
you shouldn't help me,
• 67:19 - 67:25
because isn't enough to say
it's twice the upper hemisphere
• 67:25 - 67:29
volume, right?
• 67:29 - 67:34
So if I knew the--
what is this called?
• 67:34 - 67:37
If I knew the
expression z equals
• 67:37 - 67:41
f of x, y of the spherical
cap of the hemisphere,
• 67:41 - 67:45
of the northern hemisphere,
• 67:45 - 67:50
So if somebody even
tries-- one of my students,
• 67:50 - 67:53
I gave him that, he didn't know
polar coordinates very well,
• 67:53 - 67:58
so what he tried to do,
he was trying to do,
• 67:58 - 68:04
let's say z is going
to be square root of r
• 68:04 - 68:10
squared minus z squared minus
y squared over the domain.
• 68:10 - 68:13
So D will be what
domain? x squared
• 68:13 - 68:22
plus y squared between 0 and
r squared, am I right guys?
• 68:22 - 68:26
So the D is on
the floor, means x
• 68:26 - 68:29
squared plus y squared
between 0 and r squared.
• 68:29 - 68:32
This is the D that we have.
• 68:32 - 68:36
This is D So twice what?
• 68:36 - 68:37
f of x, y.
• 68:37 - 68:40
• 68:40 - 68:42
The volume of the
upper hemisphere
• 68:42 - 68:45
is the volume of everything
under this graph, which
• 68:45 - 68:46
is like a half.
• 68:46 - 68:50
It's the northern hemisphere.
• 68:50 - 68:53
dx dy, whatever is dx.
• 68:53 - 68:55
So he tried to do
it, and he came up
• 68:55 - 68:58
with something very ugly.
• 68:58 - 69:02
Of course you can imagine
what he came up with.
• 69:02 - 69:03
What would it be?
• 69:03 - 69:04
I don't know.
• 69:04 - 69:06
Oh, God.
• 69:06 - 69:10
x between minus r to r.
• 69:10 - 69:31
y would be between 0
and-- you have to draw it.
• 69:31 - 69:32
STUDENT: It's
going to be 0 or r.
• 69:32 - 69:32
STUDENT: Yeah.
• 69:32 - 69:33
STUDENT: Oh, no.
• 69:33 - 69:35
MAGDALENA TODA: So x
is between minus r--
• 69:35 - 69:36
STUDENT: It's going to
be as a function of x.
• 69:36 - 69:38
MAGDALENA TODA: And this is x.
• 69:38 - 69:39
And it's a function of x.
• 69:39 - 69:45
And then you go square root
r squared minus x squared.
• 69:45 - 69:47
It looks awful in
Cartesian coordinates.
• 69:47 - 69:54
And then for f, he just
plugged in that thingy,
• 69:54 - 69:56
and he said dy dx.
• 69:56 - 69:58
And he would be
right, except that I
• 69:58 - 70:00
just looking
• 70:00 - 70:04
at it, because it's a mess.
• 70:04 - 70:06
It's a horrible, horrible mess.
• 70:06 - 70:09
I don't like it.
• 70:09 - 70:14
So how am I going to solve
this in polar coordinates?
• 70:14 - 70:16
I still have the 2.
• 70:16 - 70:17
I cannot get rid of the 2.
• 70:17 - 70:21
How do I express--
in polar coordinates,
• 70:21 - 70:26
the 2 would be one for the upper
part, one for the lower part--
• 70:26 - 70:29
How do I express in polar
coordinates the disc?
• 70:29 - 70:31
Rho or r.
• 70:31 - 70:38
r between 0 to R, and theta,
all the way from 0 to 2 pi.
• 70:38 - 70:41
So I'm still sort of lucky
• 70:41 - 70:47
I go 0 to 2 pi
integral from 0 to r,
• 70:47 - 70:51
and for that guy, that
is in the integrand,
• 70:51 - 70:54
I'm going to say squared of z.
• 70:54 - 71:04
z is r squared minus-- who
is z squared plus y squared
• 71:04 - 71:07
in polar coordinates?
• 71:07 - 71:10
r squared. very good. r squared.
• 71:10 - 71:14
Don't forget that
• 71:14 - 71:20
you have to say times r,
the Jacobian, dr d theta.
• 71:20 - 71:24
Can we solve this, and
find the correct formula?
• 71:24 - 71:26
• 71:26 - 71:27
We need the u substitution.
• 71:27 - 71:31
Without the u substitution,
• 71:31 - 71:33
But I don't know how
to do u substitution,
• 71:33 - 71:35
• 71:35 - 71:38
Of course you can help me.
• 71:38 - 71:39
Who is the constant?
• 71:39 - 71:41
R is the constant.
• 71:41 - 71:43
It's a number.
• 71:43 - 71:46
Little r is a variable.
• 71:46 - 71:48
Little r is a variable.
• 71:48 - 71:54
• 71:54 - 71:56
STUDENT: r squared
is going to be the u.
• 71:56 - 71:57
MAGDALENA TODA: u, very good.
• 71:57 - 71:59
r squared minus r squared.
• 71:59 - 72:02
How come this is
working so well?
• 72:02 - 72:07
Look why du will be
constant prime 0 minus 2rdr.
• 72:07 - 72:10
• 72:10 - 72:18
So I take this couple
called rdr, this block,
• 72:18 - 72:22
and I identify the
block over here.
• 72:22 - 72:31
And rdr represents du
over minus 2, right?
• 72:31 - 72:33
So I have to be
smart and attentive,
• 72:33 - 72:37
because if I make a mistake
at the end, it's all over.
• 72:37 - 72:41
So 2 tiomes integral
from 0 to 2 pi.
• 72:41 - 72:45
I could get rid of
that and say just 2 pi.
• 72:45 - 72:46
Are you guys with me?
• 72:46 - 72:53
I could say 1 is theta-- as
the product, go out-- times--
• 72:53 - 72:57
and this is my integral that
I'm worried about, the one only
• 72:57 - 73:00
in r.
• 73:00 - 73:02
Let me review it.
• 73:02 - 73:07
• 73:07 - 73:09
This is the only one
• 73:09 - 73:11
This is a piece of cake.
• 73:11 - 73:13
This is 2, this is 2 pi.
• 73:13 - 73:14
This whole thing is 4 pi a.
• 73:14 - 73:18
At least I got some 4 pi out.
• 73:18 - 73:20
What have I done in here?
• 73:20 - 73:23
I've applied the u
substitution, and I
• 73:23 - 73:25
have to be doing a better job.
• 73:25 - 73:31
I get 4 pi times what is
it after u substitution.
• 73:31 - 73:37
This guy was minus
1/2 du, right?
• 73:37 - 73:40
This fellow is squared
u, [? squared ?]
• 73:40 - 73:42
squared u as a power.
• 73:42 - 73:43
STUDENT: u to the 1/2.
• 73:43 - 73:45
MAGDALENA TODA: u
to the one half.
• 73:45 - 73:52
And for the integral, what
in the world do I write?
• 73:52 - 73:53
STUDENT: r squared--
• 73:53 - 73:54
MAGDALENA TODA: OK.
• 73:54 - 74:03
So when little r is 0, u
is going to be r squared.
• 74:03 - 74:09
When little r is
big R, you get 0.
• 74:09 - 74:11
Now you have to
help me finish this.
• 74:11 - 74:13
It should be a piece of cake.
• 74:13 - 74:16
I cannot believe it's hard.
• 74:16 - 74:19
What is the integral of 4 pi?
• 74:19 - 74:21
Copy and paste.
• 74:21 - 74:25
Minus 1/2, integrate
y to the 1/2.
• 74:25 - 74:27
STUDENT: 2/3u to the 3/2.
• 74:27 - 74:34
MAGDALENA TODA: 2/3 u to the
3/2, between u equals 0 up,
• 74:34 - 74:38
and u equals r squared down.
• 74:38 - 74:39
• 74:39 - 74:40
STUDENT: You've got
a negative sign.
• 74:40 - 74:42
MAGDALENA TODA: I've
got a negative sign.
• 74:42 - 74:43
STUDENT: Where is it--
• 74:43 - 74:46
MAGDALENA TODA: So when
I go this minus that,
• 74:46 - 74:48
it's going to be very nice.
• 74:48 - 74:48
Why?
• 74:48 - 74:56
I'm going to say minus 4
pi over 2 times 2 over 3.
• 74:56 - 74:59
I should have simplified
them from the beginning.
• 74:59 - 75:05
I have minus 5 pi over
3 times at 0 I have 0.
• 75:05 - 75:09
At r squared, I have r
squared, and the square root
• 75:09 - 75:12
is r, r cubed.
• 75:12 - 75:13
r cubed.
• 75:13 - 75:20
• 75:20 - 75:22
Oh my God, look how
beautiful it is.
• 75:22 - 75:24
Two minuses in a row.
• 75:24 - 75:27
Multiply, give me a plus.
• 75:27 - 75:28
• 75:28 - 75:30
MAGDALENA TODA: Plus.
• 75:30 - 75:37
4 pi up over 3 down, r cubed.
• 75:37 - 75:41
So we proved something
that is essential,
• 75:41 - 75:43
and we knew it from
when we were in school,
• 75:43 - 75:46
but they told us that
we cannot prove it,
• 75:46 - 75:51
because we couldn't prove that
the volume of a ball was 4 pi r
• 75:51 - 75:52
cubed over 3.
• 75:52 - 75:53
Yes, sir?
• 75:53 - 75:56
STUDENT: Why are the limits
of integration reversed?
• 75:56 - 75:57
Why is r squared on the bottom?
• 75:57 - 76:02
MAGDALENA TODA: Because
first comes little r, 0,
• 76:02 - 76:06
and then comes little r to
be big R. When I plug them
• 76:06 - 76:10
in in this order-- so
let's plug them in first,
• 76:10 - 76:11
little r equals 0.
• 76:11 - 76:16
I get, for the bottom part,
I get u equals r squared,
• 76:16 - 76:19
and when little
r equals big R, I
• 76:19 - 76:22
get big R squared minus
big R squared equals 0.
• 76:22 - 76:24
And that's the good
thing, because when
• 76:24 - 76:29
I do that, I get a minus, and
• 76:29 - 76:30
I get a plus.
• 76:30 - 76:33
And the volume is a positive
volume, like every volume.
• 76:33 - 76:36
4 pi [INAUDIBLE].
• 76:36 - 76:39
So that's it for today.
• 76:39 - 76:42
We finished 12-- what is that?
• 76:42 - 76:44
12.3, polar coordinates.
• 76:44 - 76:50
And we will next time
do some homework.
• 76:50 - 76:52
Ah, I opened the
homework for you.
• 76:52 - 76:55
So go ahead and do at least
the first 10 problems.
• 76:55 - 76:58
If you have difficulties,
let me know on Tuesday,
• 76:58 - 77:02
so we can work some in class.
• 77:02 - 77:05
STUDENT: [? You do ?] so much.
• 77:05 - 77:10
STUDENT: So, I went to the
• 77:10 - 77:10
[INTERPOSING VOICES]
• 77:10 - 77:14
• 77:14 - 77:16
[SIDE CONVERSATION]
• 77:16 - 78:34
• 78:34 - 78:36
STUDENT: Can you
imagine two years
• 78:36 - 78:38
of a calculus that's the
equivalent to [? American ?]
• 78:38 - 78:40
and only two credits?
• 78:40 - 78:41
MAGDALENA TODA:
• 78:41 - 78:44
everything was pretty
much accelerated.
• 78:44 - 78:47
STUDENT: Yeah, and
they say, no, no, no--
• 78:47 - 78:48
• 78:48 - 78:53
[INAUDIBLE] calculus,
in two years,
• 78:53 - 78:56
that is only equivalent
to two credits.
• 78:56 - 78:58
I was like--
• 78:58 - 79:00
MAGDALENA TODA:
Anyway, what happens
• 79:00 - 79:03
is that we used to have
very good evaluators
• 79:03 - 79:06
in the registrar's office, and
most of those people retired
• 79:06 - 79:09
or they got promoted in other
• 79:09 - 79:12
So they have three new hires.
• 79:12 - 79:15
Those guys, they don't
know what they are doing.
• 79:15 - 79:17
Imagine, you would
• 79:17 - 79:20
next week, you go
for the registrar.
• 79:20 - 79:22
You don't know
what you're doing.
• 79:22 - 79:22
You need time.
• 79:22 - 79:23
Yes?
• 79:23 - 79:25
• 79:25 - 79:27
I'll wait for [INAUDIBLE].
• 79:27 - 79:28
MAGDALENA TODA: It's OK.
• 79:28 - 79:30
Do you have secrets?
• 79:30 - 79:32
STUDENT: No, I don't.
• 79:32 - 79:34
MAGDALENA TODA: Homework
is due the 32st.
• 79:34 - 79:35
question from the homework.
• 79:35 - 79:35
Like I had a problem that I
was working on, and I was like
• 79:35 - 79:37
MAGDALENA TODA:
From the homework.
• 79:37 - 79:39
OK You can wait.
• 79:39 - 79:42
You guys have other,
more basic questions?
• 79:42 - 79:43
[INTERPOSING VOICES]
• 79:43 - 79:49
• 79:49 - 79:51
MAGDALENA TODA: There
is only one meeting.
• 79:51 - 79:54
Oh, you mean-- Ah.
• 79:54 - 79:55
Yes, I do.
• 79:55 - 80:00
I have the following
three-- Tuesday,
• 80:00 - 80:05
Wednesday, and Friday-
no, Tuesday, Wednesday,
• 80:05 - 80:06
and Thursday.
• 80:06 - 80:10
On Friday we can have something,
some special arrangement.
• 80:10 - 80:12
This Friday?
• 80:12 - 80:17
• 80:17 - 80:23
Today, I have--
I have right now.
• 80:23 - 80:24
2:00.
• 80:24 - 80:27
students will come later.
• 80:27 - 80:29
So you can just right now.
• 80:29 - 80:32
And tomorrow around 11:15,
because I have meetings
• 80:32 - 80:35
before 11 at the college.
• 80:35 - 80:37
STUDENT: Do you mind if I go
get something to eat first?
• 80:37 - 80:39
Or how long do you think
• 80:39 - 80:40
MAGDALENA TODA:
Even if they come,
• 80:40 - 80:42
I'm going to stop
them and talk to you,
• 80:42 - 80:44
• 80:44 - 80:44
STUDENT: Thank you very much.
• 80:44 - 80:45
I'll see you later.
• 80:45 - 80:46
STUDENT: I just wanted to say
I'm sorry for coming in late.
• 80:46 - 80:47
I slept in a little
bit this morning--
• 80:47 - 80:49
MAGDALENA TODA: Did you
get the chance to sign?
• 80:49 - 80:50
STUDENT: Yes.
• 80:50 - 80:51
MAGDALENA TODA:
There is no problem.
• 80:51 - 80:51
I'm--
• 80:51 - 80:56
STUDENT: I woke up at like
12:30-- I woke up at like 11:30
• 80:56 - 81:00
and I just fell right back
asleep, and then I got up
• 81:00 - 81:01
and I looked at my
phone and it was 12:30,
• 81:01 - 81:03
and I was like, I
have class right now.
• 81:03 - 81:05
And so what happened was like--
• 81:05 - 81:06
MAGDALENA TODA: You were tired.
• 81:06 - 81:07
You were doing
homework until late.
• 81:07 - 81:09
STUDENT: --homework
and like, I usually
• 81:09 - 81:11
am on for an
earlier class, and I
• 81:11 - 81:13
didn't go to bed earlier
than I did last night,
• 81:13 - 81:15
and so I just overslept.
• 81:15 - 81:17
MAGDALENA TODA: I
did the same, anyway.
• 81:17 - 81:19
I have similar experience.
• 81:19 - 81:20
STUDENT: You have
a very nice day.
• 81:20 - 81:21
MAGDALENA TODA: Thank you.
• 81:21 - 81:22
You too.
• 81:22 - 81:24
So, show me what
• 81:24 - 81:26
STUDENT: There it was.
• 81:26 - 81:27
I looked at that
problem, and I thought,
• 81:27 - 81:30
that's extremely
simple, acceleration--
• 81:30 - 81:32
MAGDALENA TODA: Are they
independent, really?
• 81:32 - 81:32
STUDENT: Huh?
• 81:32 - 81:35
MAGDALENA TODA: Are they--
b and t are independent?
• 81:35 - 81:37
I need to stop.
• 81:37 - 81:39
STUDENT: But I
didn't even bother.
Title:
TTU Math2450 Calculus3 Sec 12.3
Description:

Double Integral in polar coordinates

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Video Language:
English
 jackie.luft edited English subtitles for TTU Math2450 Calculus3 Sec 12.3