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CHRISTINE BREINER: Welcome
back to recitation.

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In this video, I'd like us to
work on the following problem.

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We're going to let F be the
vector field that's defined by

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r to the n, times the
quantity xi plus yj.

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And r in this case is x squared
plus y squared to the

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1/2 as it usually is.

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The square root of x squared
plus y squared.

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And then I'd like us to
do the following.

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Use extended Green's theorem to
show that F is conservative

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for all integers n.

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And then find a potential
function.

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So there are two parts.

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The first part is that you
want to show that F is

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conservative.

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And then once you know it's
conservative, you can find a

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potential function.

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So why don't you take a little
while to work on that.

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And then when you're feeling
good about your answer, bring

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the video back up, and I'll
show you what I did.

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OK, welcome back.

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So again, what was the
point of this video?

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We want to do two things.

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We want to work on
two problems.

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The first is to show that this
vector field F I've given you

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is conservative.

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And then we want to find
a potential function.

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And we want to be able to show
it's conservative for all

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integers n.

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And what I want to point out
is that for certain integer

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values of n, we're going to run
into some problems with

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differentiability
at the origin.

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OK?

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So we're going to try and deal
with all of it at once, and

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simultaneously deal with all
of the integers by allowing

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ourselves to show that F is
conservative even if we don't

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include the origin
in our region.

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OK.

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So I want to point out a few
things first. And the first

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thing I want to point out is if
we denote F as we usually

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do in two dimensions as (M,N),
then the curl of F is going to

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be N sub x minus M sub y.

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OK.

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I actually calculated these
earlier, but I want to point

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out that M sub y is actually
equal to n times r to the n

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minus 2, times xy.

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Let me make sure I wrote
that correctly.

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Yes.

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But that is also exactly
equal to N sub x.

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And so what does that give us?

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Since N sub x minus M sub y is
the curl of F when we have

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this vector (M,N), we know that
the curl of F is equal to

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0 by this work.

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OK, now if our vector field
was defined on a

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simply-connected region, then
that's enough to show that F

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is conservative.

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OK?

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We just use Green's theorem
right away.

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Right?

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But the problem is that we
are not necessarily on a

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simply-connected region because
we could have problems

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at the origin.

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And so I'm going to deal
with this in a

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slightly different way.

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To show that F is conservative,
what

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do I want to show?

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I want to show that when I take
the line integral F dot

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dr over any closed loop
that I get 0.

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That's ultimately what
I'm trying to show.

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So there are fundamentally two
types of curves that I'm

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concerned with.

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Two closed curves in r2 that
I'm concerned with, and I'm

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going to draw a picture of those
two types of curves.

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So in r2 I'm going to have
curves that miss the origin.

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Some curve like this,
which I'll call C1.

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And then I'm going to have
curves that go around the

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origin, and I'll call this C2.

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OK?

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Fundamentally, there's a
difference between this curve

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and this curve, because this
curve contains the region

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where F is defined and
differentiable, right?

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Every point on the interior of
this curve, F is defined and

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differentiable and therefore,
I can apply regular old

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Green's theorem here.

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OK?

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So I know by Green's theorem,
the integral over the closed

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curve C1 of F dot dr is equal
to 0, and that's simply

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because the curl of
F is equal to 0.

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Right?

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We can immediately use Green's
theorem because we know that

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the integral over this loop C1
is equal to the integral over

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this region of the curl of F.
That's just Green's theorem.

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So I can apply Green's
theorem here.

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Now the problem here is I can't
apply Green's theorem

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because this origin
is a trouble spot.

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Right?

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I'm not necessarily
differentiable there, so I

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have to be a little
more careful.

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OK, and so what I do is I'm
going to explain why

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immediately I can get the
integral over C2 is

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actually also 0.

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And what I'm going to do is I'm
actually going to draw,

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hopefully, a circle
that contains C2.

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So I'm going to draw a circle.

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It's a lot of curves,
but this is supposed

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to look like a circle.

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Sorry about that.

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It's a little big on the low
side, but it's a circle.

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OK.

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This is a circle.

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And I'm going to call this C3.

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Now, I can tell you right away
that the integral over the

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curve C3 of F dot dr is equal to
0, and let me explain why.

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OK?

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F is a normal vector field
relative to a circle.

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Let's look at this again.

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It's radial, and that's
why we know this.

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F is a radial vector field.

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It's really the vector field
(x,y) times a scalar,

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depending on the radius.

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So if I look at this
picture right here,

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then F is going to--

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let me draw it in color--

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F is going to, at any
given point, be

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in the radial direction.

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But that is exactly normal
to the tangent

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direction of this curve.

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So this is the direction F
points, and this is the

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direction the tangent vector
points to the curve.

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But remember, F dot dr is the
same as F dotted with the

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tangent vector ds.

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OK?

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And so that is why for this
circle, it's immediately

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obvious that F dot
dr is equal to 0.

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Because at any given point on
this circle, I'm taking a

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vector field, I'm dotting it
with a vector field that's

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orthogonal to it, so
I get 0, and when I

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integrate 0 I get 0.

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OK?

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So that's why this is 0.

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And now where the extended
version of Green's theorem

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comes in, is the fact that, if
I look in this region, F is

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defined and differentiable.

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Right?

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F is defined and differentiable
in this entire

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region that I've just shaded.

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Which is the region between
my circle and my curve C2.

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And what that tells me is that
because this one is 0--

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when I integrate along this
curve it's 0-- the integral

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along this curve also
has to be 0, right?

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That's what you actually have
seen already when you talked

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about the extended version
of Green's theorem.

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You can compare the integral
along this curve to the

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integral along this curve
because in the region between

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them, F is everywhere defined
and differentiable.

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So you can apply Green's
theorem there.

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It just now has two boundary
components, instead of in this

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case where it just has one
boundary component.

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And so since the integral on
this curve is 0, and the curl

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of F is 0, and F is defined and
differentiable everywhere

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in this region, that tells you
that the integral on the curve

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C2 is also 0.

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Let me say that one
more time, OK?

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I'm going to label it in
blue so you can see it.

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I'm going to call this region
that's shaded R. So Green's

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theorem says that the double
integral in R of the curl of F

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is equal to the integral
around this curve.

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And then I come in and I go
around this direction and I

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come back out, and that gives me
the entire integral of the

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curl of F on this region.

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Right?

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The curl of F is 0 everywhere
in this region, so that

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integral is 0.

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And so the sum of the integral
on C3 minus the integral on C2

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has to be 0.

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Since this one is 0,
that one is 0.

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So you've seen this before.

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I just want to remind
you about where

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that's coming from.

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All right.

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So now we have to do one other
thing, and that's we have to

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find a potential function.

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OK, so let's talk about how to
find a potential function.

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I'm going to do this by one of
the methods we saw in lecture.

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I'm in R2, and I'm going to
start at a certain point and

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I'm going to integrate
up to (x1,y1) from

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this certain point.

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And then I'm going to
figure out what the

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function is that way.

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So what I'm going
to do again--

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I'll write it this way--

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I'm going to figure out f of
(x1,y1) by integrating along a

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certain curve, F dot dr.

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Now I can't do exactly what I
did previously, because for

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certain values of n, I run into
trouble with integrating

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F from the origin.

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So what I'm going to do is
instead of integrating from

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the origin, I'm going
to integrate from

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the point (1, 1).

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OK?

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So I'm going to start at the
point (1,1), and I'm going to

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integrate in the y-direction,
and then I'm going to

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integrate in the x-direction.

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So this will be my first
curve and this will

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be my second curve.

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And I will land at (x1,y1).

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So again, this is one of the
strategies we've seen

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previously.

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This is the idea that I'm going
to integrate in the

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y-direction, from y equals
1 to y equals y1.

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So this will be the point
(1,y1), so x is fixed there.

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And I'm going to integrate in
the x-direction, from x equals

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1 to x equals x1, when
y is equal to y1.

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So let's break this down.

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And let me remind you, also, the
integral along this curve

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C of F dot dr should
be P dx plus Q dy.

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Right?

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And so I'm going to look at what
P dx is and what Q dy is

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on C1 and on C2.

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All right.

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So let's do that.

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00:10:06,020 --> 00:10:09,350

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OK, so I have to remind myself
what P and Q actually are in

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order to do this.

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So let me write that down,
because this will be

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helpful: (P, Q).

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P is r to the n, x, and
Q is r to the n, y.

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All right?

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So that's what we're
dealing with here.

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I'm going to come back to this
picture, and then I'm going to

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come back and forth a little
bit at this point.

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So if I want to integrate P dx
plus Q dy on the curve C1,

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what I need to observe
first is that x is

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fixed, so dx is 0.

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So I'm actually just going
to integrate Qdy.

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All right.

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So the first integral
along C1 is just a

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parameterization in y.

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So it's the integral from 0 to
y1 of Q evaluated at x equal

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1, and y going from 1 to y1.

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00:11:04,860 --> 00:11:06,090
SPEAKER 1: 1 to y1.

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00:11:06,090 --> 00:11:08,060
CHRISTINE BREINER: y
going from 1 to y1.

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OK?

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Sorry.

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Yes. y going from 1 to y1.

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Sorry about that.

251
00:11:11,760 --> 00:11:11,970
Right?

252
00:11:11,970 --> 00:11:14,340
I was avoiding the origin, so
I'd better not put a 0 down

253
00:11:14,340 --> 00:11:17,670
there, because that's where I
was running into problems. OK.

254
00:11:17,670 --> 00:11:20,780
So Q is r to the n, y.

255
00:11:20,780 --> 00:11:27,680
So I have to remember what r
is. r is x squared plus y

256
00:11:27,680 --> 00:11:28,960
squared to the 1/2.

257
00:11:28,960 --> 00:11:35,250
So in this case, Q is: x is 1,
and then I square it and I get

258
00:11:35,250 --> 00:11:39,490
1, and then I have y squared,
and then to the n over 2--

259
00:11:39,490 --> 00:11:44,370
so this is my r to the n part
along the curve C1--

260
00:11:44,370 --> 00:11:47,740
and then I multiply by y,
and then I take dy.

261
00:11:47,740 --> 00:11:49,760
So there are a lot of pieces
here, so let me just make sure

262
00:11:49,760 --> 00:11:51,070
we understand what's
happening.

263
00:11:51,070 --> 00:11:55,980
I am interested in this entire
thing, P dx plus Q dy along

264
00:11:55,980 --> 00:11:58,110
the curve C1.

265
00:11:58,110 --> 00:12:02,140
dx is 0 along that
curve. x is 1.

266
00:12:02,140 --> 00:12:04,690
And y is going from 1 to y1.

267
00:12:04,690 --> 00:12:06,540
So if I come back over
here, I see I'm only

268
00:12:06,540 --> 00:12:08,340
interested in the Qdy part.

269
00:12:08,340 --> 00:12:10,740
y is going from 1 to y1.

270
00:12:10,740 --> 00:12:17,230
And then this is r to the n,
when x is 1 and y is y.

271
00:12:17,230 --> 00:12:18,340
And this is the y part.

272
00:12:18,340 --> 00:12:22,000
So this is exactly Qdy
on the curve C1.

273
00:12:22,000 --> 00:12:24,150
Now let's look at what happens
on the curve C2.

274
00:12:24,150 --> 00:12:26,780
So if I come back over here
again, I want to have P dx

275
00:12:26,780 --> 00:12:29,870
plus Q dy on the curve C2.

276
00:12:29,870 --> 00:12:34,160
Notice y is fixed at y1
there, so dy is 0.

277
00:12:34,160 --> 00:12:36,760
And so I'm only interested
in the P dx part.

278
00:12:36,760 --> 00:12:38,650
Everything is going to
be in terms of x.

279
00:12:38,650 --> 00:12:40,920
And let's see if we can do
the same kind of thing.

280
00:12:40,920 --> 00:12:44,370
I'm going to be integrating
from 1 to x1.

281
00:12:44,370 --> 00:12:48,590
Now r is going to be of the
form x plus y1 squared,

282
00:12:48,590 --> 00:12:50,360
to the n over 2.

283
00:12:50,360 --> 00:12:51,280
And then--

284
00:12:51,280 --> 00:12:56,450
P has an x here and not
a y-- times x dx.

285
00:12:56,450 --> 00:13:00,820
So again, P is r to the n times
x, so this is r to the n

286
00:13:00,820 --> 00:13:03,790
times x exactly on
the curve C2.

287
00:13:03,790 --> 00:13:07,060
Because on C2, y is fixed at y1,
so that's why I actually

288
00:13:07,060 --> 00:13:08,750
substituted in a y1 here.

289
00:13:08,750 --> 00:13:13,270
It's the same reason I
substituted in a 1 here for x,

290
00:13:13,270 --> 00:13:16,130
because x was fixed at
1 on the curve C1.

291
00:13:16,130 --> 00:13:20,080
So now I have to integrate
these two things.

292
00:13:20,080 --> 00:13:22,360
I'm going to just write down
what you get in both cases,

293
00:13:22,360 --> 00:13:24,460
because it's really
single-variable calculus at

294
00:13:24,460 --> 00:13:27,120
this point in both cases.

295
00:13:27,120 --> 00:13:30,070
The easiest way to do this,
probably, in my mind, is to do

296
00:13:30,070 --> 00:13:31,320
a u-substitution.

297
00:13:31,320 --> 00:13:33,340

298
00:13:33,340 --> 00:13:35,070
Oops, I made a mistake.

299
00:13:35,070 --> 00:13:36,320
This should be an x squared.

300
00:13:36,320 --> 00:13:36,990
I apologize.

301
00:13:36,990 --> 00:13:38,605
This should be an x squared,
because this is supposed to be

302
00:13:38,605 --> 00:13:39,720
a radius, right?

303
00:13:39,720 --> 00:13:42,990
It's x squared plus whatever y
is squared, to the n over 2.

304
00:13:42,990 --> 00:13:45,320
So if you didn't see the squared
here, and you got

305
00:13:45,320 --> 00:13:46,390
nervous, you were correct.

306
00:13:46,390 --> 00:13:48,210
There should be a
squared here.

307
00:13:48,210 --> 00:13:49,700
So anyway, I'm going to go
back to what I was saying

308
00:13:49,700 --> 00:13:50,130
previously.

309
00:13:50,130 --> 00:13:56,430
To integrate these things, the
easiest thing to do is to take

310
00:13:56,430 --> 00:13:59,080
what is inside the parentheses
and set it equal to u, and

311
00:13:59,080 --> 00:14:00,880
then do a u-substitution
from there.

312
00:14:00,880 --> 00:14:03,290
So again, I'm not going to
actually do that for you, but

313
00:14:03,290 --> 00:14:04,700
I'm going to tell you
what you get.

314
00:14:04,700 --> 00:14:06,980
Now, there are two different
situations.

315
00:14:06,980 --> 00:14:10,920
And the situations follow when
n is any integer except

316
00:14:10,920 --> 00:14:14,290
negative 2, and then when
n is negative 2.

317
00:14:14,290 --> 00:14:17,110
And the reason is because when n
is negative 2, this exponent

318
00:14:17,110 --> 00:14:18,550
is a minus 1.

319
00:14:18,550 --> 00:14:21,560
So when you integrate, you end
up with a natural log.

320
00:14:21,560 --> 00:14:24,820
So let me just point out the
two things that you get in

321
00:14:24,820 --> 00:14:27,210
each case, and then we'll
evaluate and see what the

322
00:14:27,210 --> 00:14:28,710
solutions are in each case.

323
00:14:28,710 --> 00:14:31,760
So I'm just going to at this
point write down what I got,

324
00:14:31,760 --> 00:14:33,920
because this is your
single-variable calculus.

325
00:14:33,920 --> 00:14:41,260
OK, so what I got when n was not
equal to minus 2, you get

326
00:14:41,260 --> 00:14:42,490
the following thing.

327
00:14:42,490 --> 00:14:51,600
You get 1 plus y squared,
evaluated at n plus 2, over 2,

328
00:14:51,600 --> 00:14:53,420
over n plus 2.

329
00:14:53,420 --> 00:14:56,480
And this is evaluated
from 1 to y1.

330
00:14:56,480 --> 00:14:59,110
And then this one you get a
similar thing there, but now

331
00:14:59,110 --> 00:15:00,250
the y1 is fixed here.

332
00:15:00,250 --> 00:15:05,860
So you get an x squared plus y1
squared, to the n plus 2,

333
00:15:05,860 --> 00:15:11,840
over 2, over n plus 2, evaluated
from 1 to x1.

334
00:15:11,840 --> 00:15:14,180
So here, the y1 is fixed and
it's the x-values that are

335
00:15:14,180 --> 00:15:16,730
changing, and here the y-values
are changing.

336
00:15:16,730 --> 00:15:19,440
So when n is not equal to 2, I
get exactly this quantity when

337
00:15:19,440 --> 00:15:21,570
I integrate these two
terms. And so now,

338
00:15:21,570 --> 00:15:23,280
let's see what happens.

339
00:15:23,280 --> 00:15:23,960
OK?

340
00:15:23,960 --> 00:15:26,170
Exactly what happens
is the following.

341
00:15:26,170 --> 00:15:29,760
Notice that when I put in y1
here, I get a 1 plus y1

342
00:15:29,760 --> 00:15:32,590
squared, to the n plus 2
over 2, over n plus 2.

343
00:15:32,590 --> 00:15:33,770
Right?

344
00:15:33,770 --> 00:15:35,496
I'm not going to write it down,
because I'm going to

345
00:15:35,496 --> 00:15:37,270
show you it gets killed
off immediately.

346
00:15:37,270 --> 00:15:39,000
Where does it get killed off?

347
00:15:39,000 --> 00:15:43,140
It gets killed off when I
evaluate this one at 1.

348
00:15:43,140 --> 00:15:43,460
OK?

349
00:15:43,460 --> 00:15:47,090
So the upper bound here is the
same as the lower bound here.

350
00:15:47,090 --> 00:15:50,000
When I put in a 1 here, I get
1 plus y1 squared to the n

351
00:15:50,000 --> 00:15:52,940
plus 2 over 2 over n plus 2.

352
00:15:52,940 --> 00:15:54,450
It's a lot of n's and 2's.

353
00:15:54,450 --> 00:15:57,950
But the point is that when I
evaluate this one at y1 and I

354
00:15:57,950 --> 00:16:00,570
evaluate this one at 1, I get
exactly the same thing, but

355
00:16:00,570 --> 00:16:05,120
the signs are opposite and
so they subtract off.

356
00:16:05,120 --> 00:16:07,420
In the final answer, I'm not
going to see this upper bound

357
00:16:07,420 --> 00:16:09,270
and I'm not going to see this
lower bound, because they're

358
00:16:09,270 --> 00:16:10,460
going to subtract off.

359
00:16:10,460 --> 00:16:13,040
And what I'm actually left with
is just two terms. And

360
00:16:13,040 --> 00:16:15,120
those two terms I'm going
to write up here.

361
00:16:15,120 --> 00:16:20,730
Those two terms are going to be
x1 squared plus y1 squared

362
00:16:20,730 --> 00:16:25,650
to the n plus 2, over
2, over n plus 2.

363
00:16:25,650 --> 00:16:28,130
Minus, 1 plus 1--
which is just--

364
00:16:28,130 --> 00:16:32,450
2 to the n plus 2, over
2, over n plus 2.

365
00:16:32,450 --> 00:16:33,570
What it this really?

366
00:16:33,570 --> 00:16:37,410
This is just r to the
n plus 2, over n

367
00:16:37,410 --> 00:16:39,880
plus 2, plus a constant.

368
00:16:39,880 --> 00:16:42,425
Because this is just a
constant for any n.

369
00:16:42,425 --> 00:16:45,610
And notice n is not
equal to minus 2--

370
00:16:45,610 --> 00:16:46,480
negative 2.

371
00:16:46,480 --> 00:16:47,610
That was the place we
were going to run

372
00:16:47,610 --> 00:16:49,080
into trouble otherwise.

373
00:16:49,080 --> 00:16:51,550
And so when n is not equal to
negative 2-- when you do all

374
00:16:51,550 --> 00:16:52,480
the integration--

375
00:16:52,480 --> 00:16:55,890
you should arrive at this as
your potential function.

376
00:16:55,890 --> 00:16:57,000
OK?

377
00:16:57,000 --> 00:17:00,100
And again, what I did was I
evaluated to make it simpler

378
00:17:00,100 --> 00:17:02,140
on ourselves so we didn't have
to write everything out.

379
00:17:02,140 --> 00:17:05,620
I noticed that if I evaluate
this at the two bounds, and

380
00:17:05,620 --> 00:17:07,370
evaluate this at the two
bounds, and I add them

381
00:17:07,370 --> 00:17:11,510
together, that the evaluation
here plus the evaluation here

382
00:17:11,510 --> 00:17:14,640
are the same numerically but
opposite in sign, and so they

383
00:17:14,640 --> 00:17:16,240
subtract off.

384
00:17:16,240 --> 00:17:17,570
And then I just have
to evaluate at this

385
00:17:17,570 --> 00:17:19,300
one and this one.

386
00:17:19,300 --> 00:17:26,000
So that's n not equal
to negative 2.

387
00:17:26,000 --> 00:17:28,790
Now let's do the n equal
to negative 2 case.

388
00:17:28,790 --> 00:17:32,320
OK, so now I'm integrating this
exact same thing in the n

389
00:17:32,320 --> 00:17:34,090
equal to negative 2 case.

390
00:17:34,090 --> 00:17:35,980
And I'll just write down
again what I get by the

391
00:17:35,980 --> 00:17:37,080
substitution.

392
00:17:37,080 --> 00:17:43,270
And what I get is natural log of
one plus y squared, over 2,

393
00:17:43,270 --> 00:17:45,740
evaluated from 1 to y1.

394
00:17:45,740 --> 00:17:52,420
Plus, natural log of x squared
plus y1 squared, over 2,

395
00:17:52,420 --> 00:17:54,140
evaluated from 1 to x1.

396
00:17:54,140 --> 00:17:55,640
Let me make sure I
have that right.

397
00:17:55,640 --> 00:17:56,550
Yes.

398
00:17:56,550 --> 00:17:59,070
And the same kind of thing is
going to happen that happened

399
00:17:59,070 --> 00:18:03,690
before, in terms of when I put
y1 in here, and I put 1 in

400
00:18:03,690 --> 00:18:08,370
here, I get the same thing but
with an opposite sign.

401
00:18:08,370 --> 00:18:09,980
Here it's a positive.

402
00:18:09,980 --> 00:18:12,540
It's natural log 1 plus
y1 squared over 2.

403
00:18:12,540 --> 00:18:15,850
And here it's natural log 1 plus
y1 squared over 2, but

404
00:18:15,850 --> 00:18:18,240
because it's the lower bound,
it's a negative sign.

405
00:18:18,240 --> 00:18:22,080
So whatever I get here and what
I get here subtract off.

406
00:18:22,080 --> 00:18:24,680
And then in the end, I wind up
getting just the following two

407
00:18:24,680 --> 00:18:33,680
terms. I get x1 squared plus
y1 squared over 2, minus

408
00:18:33,680 --> 00:18:36,500
natural log of 2 over 2.

409
00:18:36,500 --> 00:18:39,580
So this term comes from
evaluating this at x1.

410
00:18:39,580 --> 00:18:43,120
And this term comes from
evaluating this one at y

411
00:18:43,120 --> 00:18:44,260
equaling 1.

412
00:18:44,260 --> 00:18:45,900
And if you notice,
what is this?

413
00:18:45,900 --> 00:18:50,110
This is exactly natural log
of r plus a constant.

414
00:18:50,110 --> 00:18:53,830
So let me step to the other side
so we can see it clearly.

415
00:18:53,830 --> 00:18:58,710
So this is natural log of r
squared, but by log rules,

416
00:18:58,710 --> 00:19:02,510
that's really 2 times natural
log of r, so it divides by 2

417
00:19:02,510 --> 00:19:04,660
and I'm just left with
natural log of r, and

418
00:19:04,660 --> 00:19:06,220
this is just a constant.

419
00:19:06,220 --> 00:19:10,900
And so my potential function
in that case is exactly

420
00:19:10,900 --> 00:19:12,812
natural log of r plus
a constant.

421
00:19:12,812 --> 00:19:14,870
All right, this was a long
problem, so I'm just going to

422
00:19:14,870 --> 00:19:17,100
remind us where we came from
and what we were doing.

423
00:19:17,100 --> 00:19:18,350
So let's go back to
the beginning.

424
00:19:18,350 --> 00:19:21,400

425
00:19:21,400 --> 00:19:24,060
So what we did initially, was we
had this vector field F. It

426
00:19:24,060 --> 00:19:29,010
was a radial vector field. r
to the n times xi plus yj.

427
00:19:29,010 --> 00:19:31,930
And we wanted to first show that
it was conservative for

428
00:19:31,930 --> 00:19:33,880
any integer value of
n, and then to find

429
00:19:33,880 --> 00:19:35,350
its potential function.

430
00:19:35,350 --> 00:19:37,570
And obviously we do it in that
order, because if it's not

431
00:19:37,570 --> 00:19:41,880
conservative, we're not going to
find a potential function.

432
00:19:41,880 --> 00:19:45,100
In this case, what I observed
first was that the

433
00:19:45,100 --> 00:19:47,860
curl of F was 0.

434
00:19:47,860 --> 00:19:53,110
And so in places where I had
a closed curve that didn't

435
00:19:53,110 --> 00:19:55,930
contain the origin, I knew that
the integral all around

436
00:19:55,930 --> 00:19:59,230
that closed curve was 0 just
by Green's theorem.

437
00:19:59,230 --> 00:20:01,350
But if I had a closed curve
that contained the origin,

438
00:20:01,350 --> 00:20:04,000
because F is not differentiable
for all the n

439
00:20:04,000 --> 00:20:06,790
values there, I have to
be a little careful.

440
00:20:06,790 --> 00:20:08,550
It's actually even 0, right?

441
00:20:08,550 --> 00:20:10,040
When x is 0 and y is
0, I'm going to

442
00:20:10,040 --> 00:20:12,020
get something 0 there.

443
00:20:12,020 --> 00:20:16,120
So I need to figure out
a way to determine the

444
00:20:16,120 --> 00:20:18,090
line integral on C2.

445
00:20:18,090 --> 00:20:18,310
Right?

446
00:20:18,310 --> 00:20:19,070
And that was my goal.

447
00:20:19,070 --> 00:20:22,220
For any C2 that contains the
origin, how do I figure out F

448
00:20:22,220 --> 00:20:23,100
dot dr.

449
00:20:23,100 --> 00:20:27,650
And so I just compared it to
what I get when I take F dot

450
00:20:27,650 --> 00:20:29,260
dr around a circle.

451
00:20:29,260 --> 00:20:32,050
Because I know that I can always
find a circle bigger,

452
00:20:32,050 --> 00:20:34,380
and then I can say I've
got this region here--

453
00:20:34,380 --> 00:20:36,800
in between-- on which F is
defined everywhere, so I can

454
00:20:36,800 --> 00:20:39,950
apply Green's theorem to
that inside region.

455
00:20:39,950 --> 00:20:44,090
And I know that the curl of F on
the inside region is 0, and

456
00:20:44,090 --> 00:20:48,550
so the integral on C2 and C3
is going to agree, right?

457
00:20:48,550 --> 00:20:50,720
Because the integral on
C3 I showed was 0 just

458
00:20:50,720 --> 00:20:52,240
geometrically.

459
00:20:52,240 --> 00:20:56,140
And then the integral on
C2 then has to be 0.

460
00:20:56,140 --> 00:20:56,595
All right?

461
00:20:56,595 --> 00:20:59,420
And so that was just when you
were using the extended

462
00:20:59,420 --> 00:21:01,460
version of Green's theorem.

463
00:21:01,460 --> 00:21:05,740
And then to find a potential
function, we came over here.

464
00:21:05,740 --> 00:21:08,130
And we had to avoid the
origin because of the

465
00:21:08,130 --> 00:21:09,910
differentiability problem
at the origin.

466
00:21:09,910 --> 00:21:10,600
So we started--

467
00:21:10,600 --> 00:21:13,250
instead of where we usually
start, which is from (0, 0)--

468
00:21:13,250 --> 00:21:15,630
we started from the
point (1, 1).

469
00:21:15,630 --> 00:21:18,470
And we just determined the
potential function going from

470
00:21:18,470 --> 00:21:23,360
the point (1, 1) to the point
(x1, y1) along a curve that

471
00:21:23,360 --> 00:21:26,040
went straight up so x was fixed,
and then along the

472
00:21:26,040 --> 00:21:28,640
curve that went straight
over so y was fixed.

473
00:21:28,640 --> 00:21:31,040
And so then we were able to
break up this thing where I'm

474
00:21:31,040 --> 00:21:31,940
integrating over C--

475
00:21:31,940 --> 00:21:33,800
P dx plus Q dy--

476
00:21:33,800 --> 00:21:37,170
into two separate pieces, and
each of them was fairly simple

477
00:21:37,170 --> 00:21:38,300
to write down.

478
00:21:38,300 --> 00:21:41,930
So let's look at
what they were.

479
00:21:41,930 --> 00:21:45,180
This first one was where
we were moving up.

480
00:21:45,180 --> 00:21:48,440
And there was no dx. x
was just fixed at 1.

481
00:21:48,440 --> 00:21:51,180
And y was going from 1 to y1.

482
00:21:51,180 --> 00:21:51,980
Right?

483
00:21:51,980 --> 00:21:54,260
And so x is fixed at 1,
so I put a 1 there.

484
00:21:54,260 --> 00:21:55,370
And y is going from 1 to y1.

485
00:21:55,370 --> 00:21:59,240
So I evaluate Q dy
on that curve.

486
00:21:59,240 --> 00:22:01,740
And then the next one was P dx
on the curve where I'm moving

487
00:22:01,740 --> 00:22:03,620
straight across.

488
00:22:03,620 --> 00:22:06,970
Right? dy is 0 there, so I
just pick up the P dx.

489
00:22:06,970 --> 00:22:10,220
And my y value was fixed
at y1, and x was

490
00:22:10,220 --> 00:22:12,320
varying from 1 to x1.

491
00:22:12,320 --> 00:22:15,260
And so then I just had to
be a little bit careful.

492
00:22:15,260 --> 00:22:17,670
I didn't show you exactly how
you integrate, but using a

493
00:22:17,670 --> 00:22:18,950
substitution trick--

494
00:22:18,950 --> 00:22:20,630
single-variable calculus--

495
00:22:20,630 --> 00:22:22,970
shouldn't be too bad for
you at this point.

496
00:22:22,970 --> 00:22:25,770
We distinguished between when n
was not equal to negative 2

497
00:22:25,770 --> 00:22:27,460
and when n was equal
to negative 2.

498
00:22:27,460 --> 00:22:32,110
In the case n not equal to
negative 2, we determined the

499
00:22:32,110 --> 00:22:35,470
integral, we simplified, and
we got to a place where the

500
00:22:35,470 --> 00:22:38,550
potential function was exactly
equal to r to the n plus 2

501
00:22:38,550 --> 00:22:42,160
over n plus 2, plus
some constant.

502
00:22:42,160 --> 00:22:46,550
Then in the case where n was
equal to negative 2, when you

503
00:22:46,550 --> 00:22:48,410
do the substitution, you get
a different integral.

504
00:22:48,410 --> 00:22:50,430
And in that case, you get
into natural log.

505
00:22:50,430 --> 00:22:52,630
And so again, we just
had the natural log.

506
00:22:52,630 --> 00:22:54,450
We have these different
functions.

507
00:22:54,450 --> 00:22:55,470
We're evaluating the
natural log of

508
00:22:55,470 --> 00:22:56,820
these different functions.

509
00:22:56,820 --> 00:22:58,080
We have the bounds.

510
00:22:58,080 --> 00:23:00,940
We simplify everything, and we
get exactly to the place where

511
00:23:00,940 --> 00:23:03,650
you have natural log of
r plus a constant.

512
00:23:03,650 --> 00:23:06,500
And so we found our potential
function in the case n is

513
00:23:06,500 --> 00:23:10,190
equal to negative 2, and
then any other n value.

514
00:23:10,190 --> 00:23:12,160
So, a very long problem.

515
00:23:12,160 --> 00:23:13,700
I hope you got something
out of it.

516
00:23:13,700 --> 00:23:14,950
And this is where I will stop.

517
00:23:14,950 --> 00:23:18,265