JOEL LEWIS: Hi.
Welcome back to recitation.
In lecture, you've been learning about line integrals
of vector fields.
And I have a couple of nice questions on that subject for
you right here.
So I want F to be the vector field whose first coordinate
is xy and whose second coordinate is x
squared plus y squared.
And so what I'd like you do is compute the line integral of F
around two different curves C. So both curves start at the
point 1, 1 and they end at the point 2, 4.
So in part a, the curve is just the straight line that
connects the point 1, 1 to 2, 4.
And in part b, the curve is this sort of piecewise.
It's two sides of a rectangle, right?
It goes straight up until it gets to the point 1, 4, and
then it goes across to the point 2, 4.
So it's a piecewise, smooth, curved path that connects
those two points.
So I'd like you to compute the integral over each of these
curves of F dot dr. So why don't you pause the video,
have a go at that, come back, and we
can work it out together.
So, when you're computing a line integral over a curve,
really the thing that you want to do is you want to
parametrize the curve, and then that gives you stuff that
you can plug in.
You'll have expressions for x and y in
terms of your parameter.
So you can plug it in and you just turn this integral right
into a nice single variable integral, and then you can
compute it.
So that's our basic strategy for computing integrals of
this form line integrals of vector fields.
So let's have a go.
Let's start with part a.
So in part a, what we need to do to apply this method is
that we need to parametrize the curve in question.
So this is a straight line.
And if you look at it, it's the line through the points 1,
1 and 2, 4.
So this line has equation y equals 3x minus 2.
That's our line, and OK.
So we need to choose some parameter that will give us
this segment of this line.
So a natural thing to do on this case
is --it's easy enough--
y is already written in terms of x, so it's natural enough
just to take a parameter that's equal to x.
So it's up to you whether you introduce the letter t in
order to do this, or not.
I'm going to do it with the letter t here in part a, but
you could do this problem exactly the same way just
using the letter x.
So what I'm going to do is I'm going to let x equals t so
that y is equal to 3t minus 2.
OK, so that is the parametric equation for the entire line,
but we only want the part between the
points 1, 1 and 2, 4.
So the part between the lines 1, 1 and 2, 4 is the part
where t is between 1 and 2.
Where x is between 1 and 2.
OK.
So this is our parametrization.
So now we need to figure out what is the field F in this
parametrization, and what is dr. And then after we have
those, we can just put them into our integral and compute.
So F. In this parametrization, well, we take the equation for
F, which is xy comma x squared plus y squared, and
we just plug in.
So in this case, xy is going to be 3t minus 2 times t, so
that's 3t squared minus 2t.
And x squared plus y squared, well that's t squared, plus 3t
minus 2 quantity squared.
So that's what F is.
And also we have that dr--
well, we just take the differentials of x and y-- so
this is going to be dt comma 3dt.
Or if you like, 1 comma 3 times dt if you like to factor
out your dt.
So that's what F and dr is.
So now we need to compute our integral.
So the integral over C of F dot dr, well,
you just plug in.
So this is the integral over C now.
So OK.
So now we need to look at our bounds.
So the integral over C means the integral as t varies in
the range that we need to cover.
That whole curve.
So in this case, we said that was from t equals 1 to 2.
So it's the integral as t goes from 1 to 2 of F dot dr.
So in the first coordinates, let me factor out
the dt at the end.
So that's going to be 3t squared minus 2t, times 1,
plus --OK well, let's expand this out now--.
3t minus 2 quantity squared that's going to give me a 9t
squared minus 12t plus 4 --so this is 9t squared
minus 12t plus 4--
and then we have to add t squared to it.
So this is plus 10t squared minus 12t plus 4, times 3, and
then this whole thing is dt.
dt is the whole integrand, there.
I could even put in another pair of parentheses just to
emphasize that, perhaps.
OK.
Now this is straightforward.
I mean, it's a little complicated looking, but it's
just an integral of a polynomial.
Easy enough to do.
Let's first just combine terms. OK, so let's look at
the t squareds.
We have a 10t squared times 3.
So 30t squared, and then another 3 times 1.
So 33t squared minus 2t minus 36t is minus 38t plus 12--
4 times 3--
dt.
OK, and now we integrate.
So this is equal to 11t cubed--
that's a 3--
minus 19t squared plus 12t as t varies between 1 and 2.
And all right.
OK, so now we've got to plug in and evaluate and so on.
So at 2, this is 88 minus 76 plus 24, minus 11
minus 19 plus 12.
So you do some arithmetic and this is going
to work out to 32.
OK, so there's part a.
It's a nice, simple curve, so we had a nice, simple
parametrization.
We computed F and dr, then we dotted them, and integrated.
OK, so now we're going to do the same exact thing for part
b, but in part b, the curve is a little more complicated.
Let's come over here where we've got some empty space.
So in part b, our curve looks like this.
So it starts at the point 1, 1, and then it goes up to the
point 1, 4, and then it goes over to the point 2, 4.
All right?
So it's hard to parametrize in one fell swoop something that
makes a sharp right angle like that.
So a natural thing to do is to split the integral over this
whole curve into the integrals over the two different pieces.
So let's call this vertical part C1 and this
horizontal part C2.
And so we know that the integral over C of F dot dr is
equal to the integral over C1 of F dot dr plus the integral
over C2 of F dot dr. And so now, it's easy enough to
parametrize these two separate curves separately.
C1, for example, is the straight line segment that
goes from 1, 1 to 1, 4.
So C1.
So that means we have x equal to 1, and 1 less than or equal
to y less than or equal to 4.
So a natural parametrization here is just the
parametrization that uses the parameter y.
Right?
So in this one, I'm not going to bother introducing
a new letter t.
I'm just going to stick with x and y.
So we have x equals 1, and y is our parameter and it goes
from 1 to 4.
So now let's look at what F and dr are.
So in this case, F is equal to--
its first coordinate is xy, and x is just 1 here--
so this is y.
And its second coordinate was x squared plus y squared, and
so that's going to be 1 plus y squared.
And dr--
well, r here is 1 comma y--
so dr is equal to 0 comma dy.
Or 0, 1 times dy, if you wanted to factor that dy out
to the end.
OK.
Good.
So we're all set to do that first integral.
So let's do that.
So we have the integral over C1 of F dot dr is equal to
what we dot these two things together.
And the first term gives me y times 0, and that's just 0.
So that's going to die, and all we're left with is the
second term.
So it's the integral of 1 plus y squared
dy, but we need bounds.
Right?
OK, so y was going from 1 to 4 in this integral.
So it's the integral from 1 to 4 of 1 plus y squared dy.
OK.
So we can either continue and evaluate this now, or we could
go and do the second one.
Let's finish evaluating it since we've already got it
written up here.
So this is equal to y, plus y cubed over 3, between 1 and 4.
So what is this?
This is 4 plus 64/3, minus 1 plus 1/3.
So that looks like it's 24 to me.
OK, so we get 24 for the first part.
Now, let's do the second part.
So it's C2 here.
I'll draw a little line there to separate them.
Now on curve C2 --let's go back and look at it--
OK, so curve C2 is the segment connecting the points 1, 4 and
the point 2, 4.
OK, so y is always 4 on this curve, and x goes from 1 to 2.
So 1 is less than or equal to x less than or equal to 2, y
is equal to 4.
So a natural parametrization here again, is just to take x
to be our parameter.
And again, I'm not going to introduce a letter t.
We're just using x as our parameter.
So in this case, F--
well, it's xy, so x is just x and y is 4--
so that's 4x comma--
and the second coordinate is x squared plus y squared-- so
that's x squared plus 16.
And dr is equal to dx comma 0.
OK, so that's F and dr.
So the integral that I want now is the integral over C2 of
F dot dr. OK, so we just plug in here what we've got.
So this is equal to the integral of--
well, the first coordinates are 4x dx and the second
coordinates just give me 0-- so it's 4x dx.
And again, I need my bounds.
Well, I had-- over here--
I had 1 less than or equal to x is less than or equal to 2.
So that's the integral between 1 and 2.
4x--
integrate that-- and I get 2x squared between 1 and 2, which
is equal to 8 minus 2, or 6.
All right.
So let's see what we've got.
So we had--
back here--
we had our curve C, which we split into the two
parts, C1 and C2.
And we wanted to know what the integral over C was, and we've
separately computed the integral over C1.
And we computed that to be 24.
And we computed the integral over C2, and that was 6.
So the integral over the whole curve of F dot dr is equal to
24 plus 6, which is 30.
OK.
So there's your answer for the second part.
Now one thing I'd like you to notice is that over this curve
C in part b--
over the whole curve in part b--
we got that the integral of this field F was 30.
And now if you remember, right here, in the first part, in
part a, we computed the integral over a different
curve that connected the two same endpoints.
And we found that the integral came out to 32.
So one thing you should take away from this is that the
integral over a curve joining two points can depend on which
curve you choose, right?
So we had two different curves and we got two different
answers, even though the two curves
connected the same points.
So that's interesting.
And the other thing to take away from this is just the
general approach.
Which is that whenever you have a problem like this, what
you want to do is you want to take your curve--
so whether it be well, in part a we had this straight line,
slanted line.
In part b where we had this nice piecewise linear with
these vertical and horizontal parts--
you want to break it into nice pieces and parametrize them.
You know, sometimes you only need one piece when it's an
easy-to-parametrize curve like that.
Sometimes, if it has corners or so on, you
might want more pieces.
Break it into pieces, choose a nice parametrization, and that
reduces your problem just to computing integrals, just like
we've done in Calculus I-- in 18.01--
and then you just integrate.
All right.
I'll end there.