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JOEL LEWIS: Hi.

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Welcome back to recitation.

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In lecture, you've been learning
about line integrals

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of vector fields.

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And I have a couple of nice
questions on that subject for

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you right here.

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So I want F to be the vector
field whose first coordinate

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is xy and whose second
coordinate is x

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squared plus y squared.

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And so what I'd like you do is
compute the line integral of F

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around two different curves C.
So both curves start at the

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point 1, 1 and they end
at the point 2, 4.

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So in part a, the curve is just
the straight line that

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connects the point
1, 1 to 2, 4.

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And in part b, the curve is
this sort of piecewise.

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It's two sides of a
rectangle, right?

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It goes straight up until it
gets to the point 1, 4, and

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then it goes across
to the point 2, 4.

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So it's a piecewise, smooth,
curved path that connects

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those two points.

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So I'd like you to compute the
integral over each of these

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curves of F dot dr. So why don't
you pause the video,

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have a go at that,
come back, and we

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can work it out together.

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So, when you're computing a line
integral over a curve,

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really the thing that you want
to do is you want to

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parametrize the curve, and then
that gives you stuff that

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you can plug in.

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You'll have expressions
for x and y in

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terms of your parameter.

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So you can plug it in and you
just turn this integral right

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into a nice single variable
integral, and then you can

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compute it.

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So that's our basic strategy
for computing integrals of

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this form line integrals
of vector fields.

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So let's have a go.

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Let's start with part a.

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So in part a, what we need to
do to apply this method is

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that we need to parametrize
the curve in question.

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So this is a straight line.

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And if you look at it, it's the
line through the points 1,

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1 and 2, 4.

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So this line has equation
y equals 3x minus 2.

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That's our line, and OK.

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So we need to choose some
parameter that will give us

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this segment of this line.

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So a natural thing to
do on this case

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is --it's easy enough--

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y is already written in terms
of x, so it's natural enough

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just to take a parameter
that's equal to x.

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So it's up to you whether you
introduce the letter t in

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order to do this, or not.

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I'm going to do it with the
letter t here in part a, but

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you could do this problem
exactly the same way just

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using the letter x.

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So what I'm going to do is I'm
going to let x equals t so

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that y is equal to 3t minus 2.

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OK, so that is the parametric
equation for the entire line,

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but we only want the
part between the

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points 1, 1 and 2, 4.

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So the part between the lines
1, 1 and 2, 4 is the part

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where t is between 1 and 2.

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Where x is between 1 and 2.

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OK.

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So this is our parametrization.

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So now we need to figure out
what is the field F in this

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parametrization, and what is
dr. And then after we have

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those, we can just put them into
our integral and compute.

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So F. In this parametrization,
well, we take the equation for

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F, which is xy comma x squared
plus y squared, and

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we just plug in.

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So in this case, xy is going to
be 3t minus 2 times t, so

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that's 3t squared minus 2t.

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And x squared plus y squared,
well that's t squared, plus 3t

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minus 2 quantity squared.

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So that's what F is.

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And also we have that dr--

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well, we just take the
differentials of x and y-- so

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this is going to be
dt comma 3dt.

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Or if you like, 1 comma 3 times
dt if you like to factor

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out your dt.

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So that's what F and dr is.

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So now we need to compute
our integral.

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So the integral over C
of F dot dr, well,

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you just plug in.

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So this is the integral
over C now.

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So OK.

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So now we need to look
at our bounds.

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So the integral over C means
the integral as t varies in

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the range that we
need to cover.

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That whole curve.

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So in this case, we said that
was from t equals 1 to 2.

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So it's the integral as t goes
from 1 to 2 of F dot dr.

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So in the first coordinates,
let me factor out

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the dt at the end.

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So that's going to be 3t squared
minus 2t, times 1,

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plus --OK well, let's expand
this out now--.

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3t minus 2 quantity squared
that's going to give me a 9t

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squared minus 12t plus 4
--so this is 9t squared

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minus 12t plus 4--

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and then we have to add
t squared to it.

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So this is plus 10t squared
minus 12t plus 4, times 3, and

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then this whole thing is dt.

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dt is the whole integrand,
there.

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I could even put in another pair
of parentheses just to

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emphasize that, perhaps.

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OK.

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Now this is straightforward.

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I mean, it's a little
complicated looking, but it's

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just an integral of
a polynomial.

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Easy enough to do.

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Let's first just combine terms.
OK, so let's look at

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the t squareds.

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We have a 10t squared times 3.

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So 30t squared, and then
another 3 times 1.

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So 33t squared minus 2t minus
36t is minus 38t plus 12--

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4 times 3--

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dt.

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OK, and now we integrate.

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So this is equal
to 11t cubed--

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that's a 3--

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minus 19t squared plus 12t as
t varies between 1 and 2.

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And all right.

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OK, so now we've got to plug
in and evaluate and so on.

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So at 2, this is 88 minus
76 plus 24, minus 11

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minus 19 plus 12.

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So you do some arithmetic
and this is going

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to work out to 32.

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OK, so there's part a.

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It's a nice, simple curve,
so we had a nice, simple

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parametrization.

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We computed F and dr, then we
dotted them, and integrated.

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OK, so now we're going to do the
same exact thing for part

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b, but in part b, the curve is
a little more complicated.

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Let's come over here where we've
got some empty space.

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So in part b, our curve
looks like this.

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So it starts at the point 1, 1,
and then it goes up to the

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point 1, 4, and then it goes
over to the point 2, 4.

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All right?

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So it's hard to parametrize in
one fell swoop something that

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makes a sharp right
angle like that.

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So a natural thing to do is to
split the integral over this

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whole curve into the integrals
over the two different pieces.

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So let's call this vertical
part C1 and this

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horizontal part C2.

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And so we know that the integral
over C of F dot dr is

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equal to the integral over C1 of
F dot dr plus the integral

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over C2 of F dot dr. And so
now, it's easy enough to

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parametrize these two separate
curves separately.

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C1, for example, is the straight
line segment that

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goes from 1, 1 to 1, 4.

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So C1.

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So that means we have x equal to
1, and 1 less than or equal

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to y less than or equal to 4.

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So a natural parametrization
here is just the

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parametrization that uses
the parameter y.

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Right?

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So in this one, I'm not going
to bother introducing

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a new letter t.

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I'm just going to stick
with x and y.

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So we have x equals 1, and y is
our parameter and it goes

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from 1 to 4.

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So now let's look at
what F and dr are.

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So in this case,
F is equal to--

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its first coordinate is xy,
and x is just 1 here--

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so this is y.

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And its second coordinate was x
squared plus y squared, and

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so that's going to be
1 plus y squared.

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And dr--

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well, r here is 1 comma y--

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so dr is equal to 0 comma dy.

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Or 0, 1 times dy, if you wanted
to factor that dy out

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to the end.

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OK.

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Good.

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So we're all set to do
that first integral.

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So let's do that.

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So we have the integral over
C1 of F dot dr is equal to

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what we dot these two
things together.

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And the first term gives me y
times 0, and that's just 0.

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So that's going to die, and
all we're left with is the

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second term.

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So it's the integral
of 1 plus y squared

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dy, but we need bounds.

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Right?

191
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OK, so y was going from 1
to 4 in this integral.

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So it's the integral from 1 to
4 of 1 plus y squared dy.

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OK.

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So we can either continue and
evaluate this now, or we could

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go and do the second one.

196
00:10:39,920 --> 00:10:43,210

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Let's finish evaluating it since
we've already got it

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written up here.

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So this is equal to y, plus y
cubed over 3, between 1 and 4.

200
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So what is this?

201
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This is 4 plus 64/3,
minus 1 plus 1/3.

202
00:11:08,080 --> 00:11:11,880
So that looks like
it's 24 to me.

203
00:11:11,880 --> 00:11:14,540
OK, so we get 24 for
the first part.

204
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Now, let's do the second part.

205
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So it's C2 here.

206
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I'll draw a little line there
to separate them.

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Now on curve C2 --let's go
back and look at it--

208
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OK, so curve C2 is the segment
connecting the points 1, 4 and

209
00:11:28,960 --> 00:11:31,230
the point 2, 4.

210
00:11:31,230 --> 00:11:36,690
OK, so y is always 4 on this
curve, and x goes from 1 to 2.

211
00:11:36,690 --> 00:11:41,305
So 1 is less than or equal to
x less than or equal to 2, y

212
00:11:41,305 --> 00:11:43,820
is equal to 4.

213
00:11:43,820 --> 00:11:47,060
So a natural parametrization
here again, is just to take x

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to be our parameter.

215
00:11:48,230 --> 00:11:50,260
And again, I'm not going to
introduce a letter t.

216
00:11:50,260 --> 00:11:52,330
We're just using x
as our parameter.

217
00:11:52,330 --> 00:11:56,340
So in this case, F--

218
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well, it's xy, so x is
just x and y is 4--

219
00:12:01,366 --> 00:12:04,160
so that's 4x comma--

220
00:12:04,160 --> 00:12:07,050
and the second coordinate is x
squared plus y squared-- so

221
00:12:07,050 --> 00:12:10,980
that's x squared plus 16.

222
00:12:10,980 --> 00:12:22,210
And dr is equal to dx comma 0.

223
00:12:22,210 --> 00:12:24,440
OK, so that's F and dr.

224
00:12:24,440 --> 00:12:30,320
So the integral that I want now
is the integral over C2 of

225
00:12:30,320 --> 00:12:36,900
F dot dr. OK, so we just plug
in here what we've got.

226
00:12:36,900 --> 00:12:41,330
So this is equal to
the integral of--

227
00:12:41,330 --> 00:12:44,530
well, the first coordinates
are 4x dx and the second

228
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coordinates just give me
0-- so it's 4x dx.

229
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And again, I need my bounds.

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Well, I had-- over here--

231
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I had 1 less than or equal to x
is less than or equal to 2.

232
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So that's the integral
between 1 and 2.

233
00:12:55,620 --> 00:12:58,350

234
00:12:58,350 --> 00:12:59,200
4x--

235
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integrate that-- and I get 2x
squared between 1 and 2, which

236
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is equal to 8 minus 2, or 6.

237
00:13:09,295 --> 00:13:11,870

238
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All right.

239
00:13:12,650 --> 00:13:16,080
So let's see what we've got.

240
00:13:16,080 --> 00:13:17,820
So we had--

241
00:13:17,820 --> 00:13:18,990
back here--

242
00:13:18,990 --> 00:13:22,050
we had our curve C, which
we split into the two

243
00:13:22,050 --> 00:13:24,040
parts, C1 and C2.

244
00:13:24,040 --> 00:13:27,360
And we wanted to know what the
integral over C was, and we've

245
00:13:27,360 --> 00:13:32,965
separately computed the
integral over C1.

246
00:13:32,965 --> 00:13:35,790
And we computed that to be 24.

247
00:13:35,790 --> 00:13:39,075
And we computed the integral
over C2, and that was 6.

248
00:13:39,075 --> 00:13:42,380

249
00:13:42,380 --> 00:13:49,190
So the integral over the whole
curve of F dot dr is equal to

250
00:13:49,190 --> 00:13:55,170
24 plus 6, which is 30.

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00:13:55,170 --> 00:13:56,350
OK.

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00:13:56,350 --> 00:13:58,250
So there's your answer
for the second part.

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00:13:58,250 --> 00:14:01,120
Now one thing I'd like you to
notice is that over this curve

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00:14:01,120 --> 00:14:03,150
C in part b--

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00:14:03,150 --> 00:14:05,060
over the whole curve
in part b--

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00:14:05,060 --> 00:14:09,645
we got that the integral
of this field F was 30.

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00:14:09,645 --> 00:14:14,400
And now if you remember, right
here, in the first part, in

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00:14:14,400 --> 00:14:17,640
part a, we computed the integral
over a different

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00:14:17,640 --> 00:14:21,340
curve that connected the
two same endpoints.

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00:14:21,340 --> 00:14:24,300
And we found that the integral
came out to 32.

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00:14:24,300 --> 00:14:26,770
So one thing you should take
away from this is that the

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00:14:26,770 --> 00:14:30,310
integral over a curve joining
two points can depend on which

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00:14:30,310 --> 00:14:32,200
curve you choose, right?

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00:14:32,200 --> 00:14:34,440
So we had two different curves
and we got two different

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00:14:34,440 --> 00:14:36,150
answers, even though
the two curves

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00:14:36,150 --> 00:14:37,920
connected the same points.

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00:14:37,920 --> 00:14:39,190
So that's interesting.

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00:14:39,190 --> 00:14:40,890
And the other thing to take
away from this is just the

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00:14:40,890 --> 00:14:42,190
general approach.

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00:14:42,190 --> 00:14:48,080
Which is that whenever you have
a problem like this, what

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00:14:48,080 --> 00:14:50,430
you want to do is you want
to take your curve--

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00:14:50,430 --> 00:14:56,360
so whether it be well, in part
a we had this straight line,

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00:14:56,360 --> 00:14:57,640
slanted line.

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00:14:57,640 --> 00:15:00,820
In part b where we had this
nice piecewise linear with

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00:15:00,820 --> 00:15:03,060
these vertical and horizontal
parts--

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00:15:03,060 --> 00:15:07,280
you want to break it into nice
pieces and parametrize them.

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00:15:07,280 --> 00:15:09,870
You know, sometimes you only
need one piece when it's an

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00:15:09,870 --> 00:15:12,770
easy-to-parametrize
curve like that.

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00:15:12,770 --> 00:15:15,060
Sometimes, if it has corners
or so on, you

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00:15:15,060 --> 00:15:16,700
might want more pieces.

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00:15:16,700 --> 00:15:19,380
Break it into pieces, choose a
nice parametrization, and that

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00:15:19,380 --> 00:15:21,910
reduces your problem just to
computing integrals, just like

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00:15:21,910 --> 00:15:26,370
we've done in Calculus
I-- in 18.01--

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00:15:26,370 --> 00:15:28,620
and then you just integrate.

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00:15:28,620 --> 00:15:29,400
All right.

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00:15:29,400 --> 00:15:31,020
I'll end there.

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00:15:31,020 --> 00:15:31,531