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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

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on that subject
for you right here.

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

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x*y and whose second coordinate
is x 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

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start at the 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 connects

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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),

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and then it goes across
to the point (2, 4).

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So it's a piecewise
smooth curve, path,

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that connects those two points.

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

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of these 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 can work it out

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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

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is you want to
parametrize the curve,

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and then that gives you
stuff that you can plug in.

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You'll have
expressions for x and y

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in 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,

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and then you can compute it.

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So that's our basic strategy
for computing integrals

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of 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

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is 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, 1)

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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 in
this case is-- it's easy enough,

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y is already written
in terms of x,

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so it's natural
enough just to take

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a parameter that's equal to x.

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

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to do this, or not.

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

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but you could do this problem
exactly the same way just using

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the letter x.

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

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so 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 points (1, 1) and

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(2, 4).

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So the part between the
lines (1, 1) and (2, 4)

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is the part 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

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is the field F in
this parametrization,

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

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we can just put them into
our integral and compute.

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So F, in this
parametrization, well,

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we take the equation for F,
which is x*y comma x squared

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plus y squared, and
we just plug in.

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So in this case, x*y is going
to be 3t minus 2 times t,

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so that's 3 t 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-- well,
we just take the differentials

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of x and y-- so this is
going to be dt comma 3*dt.

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Or if you like, 1
comma 3 times dt

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if you like to
factor 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, 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

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as t varies in 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 3 t
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

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going to give me a 9 t
squared minus 12t plus 4--

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so this is 9 t squared
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 10 t squared
minus 12t plus 4, times 3,

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and 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

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to emphasize that, perhaps.

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OK.

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Now this is a
straightforward-- I mean,

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it's a little
complicated looking,

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but it's just an
integral of a polynomial.

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Easy enough to do.

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Let's first just combine terms.

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OK, so let's look
at the t squareds.

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We have a 10 t squared times 3.

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

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So 33 t squared.

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And I've got minus 2t
minus 36t is minus 38t,

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plus 12-- 4 times 3-- dt.

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OK, and now we integrate.

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So this is equal to
11 t cubed-- that's

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a 3-- minus 19 t
squared plus 12t as t

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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 minus 19

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plus 12.

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So you do some
arithmetic and this

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is going 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

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a nice, simple 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 b,

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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),

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and then it goes up
to the point (1, 4),

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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

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something that makes a
sharp right angle like that.

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So a natural thing to do
is to split the integral

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over this whole curve into
the integrals over the two

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different pieces.

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So let's call this vertical part
C_1 and this horizontal part

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C_2.

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And so we know that the
integral over C of F

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dot dr is equal to
the integral over C_1

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of F dot dr plus the integral
over C_2 of F dot dr.

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And so now, it's easy enough
to parametrize these two

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separate curves separately.

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C_1, for example, is the
straight line segment that

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goes from (1, 1) to (1, 4).

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So C_1.

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

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or equal to y less
than or equal to 4.

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So a natural
parametrization here

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is just the 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 a new letter

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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

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and it goes 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--
its first coordinate is x*y,

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and x is just 1
here, so this is y.

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And its second coordinate
was x squared plus y squared,

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

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And dr-- 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

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dy out 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 C_1
of F dot dr is equal to-- well,

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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

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left with is the second term.

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So it's the integral of 1 plus y
squared dy, but we need bounds.

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Right?

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00:10:24,560 --> 00:10:27,975
OK, so y was going from
1 to 4 in this integral.

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00:10:27,975 --> 00:10:34,440
So it's the integral from 1
to 4 of 1 plus y squared dy.

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00:10:34,440 --> 00:10:34,940
OK.

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So we can either continue
and evaluate this now,

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or we could go and
do the second one.

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Let's finish evaluating
it since we've already

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got it written up here.

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

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00:10:56,084 --> 00:10:56,750
So what is this?

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

194
00:11:08,080 --> 00:11:11,880
So that looks like
it's 24 to me.

195
00:11:11,880 --> 00:11:14,540
OK, so we get 24
for the first part.

196
00:11:14,540 --> 00:11:16,465
Now, let's do the second part.

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00:11:16,465 --> 00:11:18,415
So C_2 here.

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00:11:18,415 --> 00:11:21,270
I'll draw a little line
there to separate them.

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00:11:21,270 --> 00:11:24,770
Now, curve C_2-- let's go
back and look at it-- OK,

200
00:11:24,770 --> 00:11:28,820
so curve C_2 is the segment
connecting the points (1, 4)

201
00:11:28,820 --> 00:11:31,230
and the point (2, 4).

202
00:11:31,230 --> 00:11:36,690
OK, so y is always 4 on this
curve, and x goes from 1 to 2.

203
00:11:36,690 --> 00:11:41,070
So 1 is less than or equal
to x less than or equal to 2,

204
00:11:41,070 --> 00:11:43,820
y is equal to 4.

205
00:11:43,820 --> 00:11:45,550
So a natural
parametrization here,

206
00:11:45,550 --> 00:11:48,219
again, is just to take
x to be our parameter.

207
00:11:48,219 --> 00:11:50,260
And again, I'm not going
to introduce a letter t.

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

209
00:11:52,330 --> 00:11:58,280
So in this case,
F-- well, it's x*y,

210
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so x is just x and y is
4-- so that's 4x comma--

211
00:12:04,160 --> 00:12:06,645
and the second coordinate is
x squared plus y squared--

212
00:12:06,645 --> 00:12:10,980
so that's x squared plus 16.

213
00:12:10,980 --> 00:12:22,210
And dr is equal to dx comma 0.

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

215
00:12:24,440 --> 00:12:30,000
So the integral that I want
now is the integral over C_2

216
00:12:30,000 --> 00:12:36,900
of F dot dr. OK, so we just
plug in here what we've got.

217
00:12:36,900 --> 00:12:41,920
So this is equal to
the integral of-- well,

218
00:12:41,920 --> 00:12:45,120
the first coordinates are 4x dx
and the second coordinates just

219
00:12:45,120 --> 00:12:47,875
give me 0-- so it's 4x dx.

220
00:12:47,875 --> 00:12:49,230
And again, I need my bounds.

221
00:12:49,230 --> 00:12:51,507
Well, I had-- over
here, I had 1 less than

222
00:12:51,507 --> 00:12:53,215
or equal to x is less
than or equal to 2.

223
00:12:53,215 --> 00:12:55,620
So that's the integral
between 1 and 2.

224
00:12:58,350 --> 00:13:04,300
4x-- integrate that-- and I get
2 x squared between 1 and 2,

225
00:13:04,300 --> 00:13:09,295
which is equal to
8 minus 2, or 6.

226
00:13:11,870 --> 00:13:12,650
All right.

227
00:13:12,650 --> 00:13:16,080
So let's see what we've got.

228
00:13:16,080 --> 00:13:21,610
So we had-- back here, we had
our curve C, which we split

229
00:13:21,610 --> 00:13:24,040
into the two parts C_1 and C_2.

230
00:13:24,040 --> 00:13:26,830
And we wanted to know what
the integral over C was,

231
00:13:26,830 --> 00:13:32,965
and we've separately computed
the integral over C_1.

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

233
00:13:35,790 --> 00:13:39,075
And we computed the integral
over C_2, and that was 6.

234
00:13:42,380 --> 00:13:45,910
So the integral over
the whole curve of F

235
00:13:45,910 --> 00:13:55,170
dot dr is equal to 24
plus 6, which is 30.

236
00:13:55,170 --> 00:13:56,350
OK.

237
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 --> 00:13:59,930
Now one thing I'd
like you to notice

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is that over this
curve C in part b--

240
00:14:03,150 --> 00:14:05,210
over the whole
curve in part b-- we

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00:14:05,210 --> 00:14:09,645
got that the integral
of this field F was 30.

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00:14:09,645 --> 00:14:13,550
And now if you
remember, right here,

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00:14:13,550 --> 00:14:16,950
in the first part, in part
a, we computed the integral

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00:14:16,950 --> 00:14:20,130
over a different curve
that connected the two

245
00:14:20,130 --> 00:14:21,340
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 --> 00:14:26,160
So one thing you should
take away from this

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00:14:26,160 --> 00:14:29,190
is that the integral over
a curve joining two points

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00:14:29,190 --> 00:14:32,200
can depend on which
curve you choose, right?

250
00:14:32,200 --> 00:14:33,770
So we had two
different curves and we

251
00:14:33,770 --> 00:14:36,620
got two different answers, even
though the two curves connected

252
00:14:36,620 --> 00:14:37,920
the same points.

253
00:14:37,920 --> 00:14:39,190
So that's interesting.

254
00:14:39,190 --> 00:14:40,939
And the other thing
to take away from this

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00:14:40,939 --> 00:14:42,190
is just the general approach.

256
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--

258
00:14:50,430 --> 00:14:56,360
so whether it be-- well, in part
a we had this straight line,

259
00:14:56,360 --> 00:14:57,640
slanted line.

260
00:14:57,640 --> 00:15:00,460
In part b where we had
this nice piecewise

261
00:15:00,460 --> 00:15:03,060
linear with these vertical
and horizontal parts--

262
00:15:03,060 --> 00:15:06,925
you want to break it into
nice pieces, parametrize them.

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00:15:06,925 --> 00:15:08,300
You know, sometimes
you only need

264
00:15:08,300 --> 00:15:11,620
one piece when it's an
easy-to-parametrize curve

265
00:15:11,620 --> 00:15:12,770
like that.

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00:15:12,770 --> 00:15:14,990
Sometimes, if it has
corners or so on,

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00:15:14,990 --> 00:15:16,700
you might want more pieces.

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00:15:16,700 --> 00:15:19,000
Break it into pieces, choose
a nice parametrization,

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00:15:19,000 --> 00:15:21,450
and that reduces your problem
just to computing integrals,

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00:15:21,450 --> 00:15:26,370
just like 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.