CHRISTINE BREINER: Welcome back to recitation.
In this video I would like us to use Green's theorem to
compute the following integral where it's the integral over
the curve c, where c is the circle drawn here.
So the circle is oriented so the interior is on the left
and it's centered at the point where x equals a, y equals 0.
So the integral--
you can read it but I will read it also for you-- it's
the integral of 3x squared y squared dx plus 2x squared
times the quantity 1 plus xy dy.
So you are supposed to use Green's
theorem to compute that.
And why don't you pause the video, give it a shot, and
then when you're ready to see my solution you can bring the
video back up.
OK, welcome back.
Well, what we're going to do, obviously to use Green's
theorem to compute this-- because I gave you a big hint
that you're supposed to use Green's theorem because we
have an integral over a closed curve.
And what we're going to do is now take the integral of
another certain integrand--
obviously related in some way to this--
over the region that I will shade here.
So we're interested in this shaded region now.
So let me write down what Green's theorem is and then
we'll put in the important parts.
OK, so just to remind you, I'm not going to write down all
the hypotheses of Green's theorem that we need, but the
point that I want to make is that if we start the integral
over this closed curve c of mdx plus ndy we can also
integrate over the region that C encloses of N sub x minus M
sub y dy dx.
So that is ultimately what we're going to do.
And so in this case again, as always, m is going to be the
function associated with the dx portion.
It's the ice component of the vector field.
And n is going to be the function associated
here with the dy.
That's standard obviously.
So now what we want to do is transform what we have there
into something that looks like this.
So the region--
I'm just going to keep calling it R for the moment--
but the region R you'll notice is the thing that I shaded in
the drawing.
And so now let's compute what this is--
well you have the capacity to do that.
That's just some straight taking derivative.
So I'm just going to write down what it is and not show
you all the individual pieces.
So I'll just write down what you get and then the
simplified version.
So you get 6x squared y plus 4x is N sub x.
And then M sub y is negative 6x squared y.
And I'm going to call dy dx, I'm just going to refer to it
now as dA, because that makes it a lot easier to write.
Oh, I guess I should call this--
well, dA, this is the area.
The volume form there.
So this simplifies and I just get the integral of 4x over
this region dA.
Now this-- you saw an example of this in lecture of how to
deal with these types of problems. At this point, if I
really wanted to I could figure out what the bounds are
in that region in terms of x and y, and I could do a lot of
work and integrate it all, or I could
remember one simple fact.
Which is that if I have--
I think you see it in class as x bar-- the center of mass
should be equal to 1 over the volume of the region times the
integral of x dA over the region.
That's what we know.
This should be-- this maybe doesn't look like a v. But so
in this case volume is area, isn't it?
Maybe I should write area, that might make you nervous.
So if I take the area--
I could just say A of R-- if I take the area, 1 over the
area, and then I multiply by the integral of x over R with
dy dx with respect to dA, then I get the center of mass.
Well, let's look at what in this picture-- what is the
center of mass here?
If I want to balance this thing on a pencil tip, if I
want to balance this disc-- assuming the density is
everywhere the same on a pencil tip-- where am I going
to put the pencil?
I'm going to put it right at the center.
The x value there is a, the y value there is 0.
So if I had the y center of mass, I would want 0.
But I want the x center of mass, so I want a.
This x bar is actually equal to, from the
picture, is equal to a.
And now let's notice what I've done.
I have taken--
I had this quantity here--
I've taken this quantity except for the 4 and I have a
way of writing explicitly what this quantity is without
actually doing any of the integration.
I haven't done any sophisticated things at this
point except know what the center of mass is.
So also what is the area of the region?
I'm going to need that.
The area of the region-- it's a circle of radius a.
So the area is pi a squared.
So this quantity is pi a squared.
And all this together tells me that the integral of x over R
dA, if I solve for this part, I get a times pi a squared.
So I get pi a cubed.
And that only differs from our answer by one thing.
There's a scalar--
you multiply by 4 and that gives us what we want.
I said differs from our answer.
Differs from what we want by one thing.
We just multiplied by 4 there, so we multiply by 4 here.
So in fact--
maybe there looks like there was a little magic here, so
let me point out some of the key points at the end.
I started off knowing I was trying to integrate 4x over
the region R. And R in this case was a circle centered at
(a, 0) and of radius a.
And then I said, well, I don't want to do a
lot of work for this.
So I'm going to not cheat, but I'm going to use my knowledge
of the center of mass to make this easier.
So the center of mass is equal to 1 divided by the area of
the region times the integral of x over the region.
So I want to find the integral of x over the region, I just
solve for the integral of x over the region.
I just solve for that part.
The center of mass-- from just looking at the picture and
understanding what the center of mass means-- the center of
mass in the x component is a.
The area is pi a squared.
So I end up with a pi a cubed when I solve for
the integral of x.
And then because I wanted the integral of 4x, I just
multiply by 4.
And so the final answer is actually 4 pi a cubed.
So what I was trying to find, if you remember, was I was
trying to find the value when I integrated over this curve
of a certain vector field.
And that one was going to be a little messy
to do it that way.
But Green's theorem, actually there's a lot of cancellation
which makes it much easier, and then the center of mass is
a nice little trick to use at the end.
And then the calculation is quite simple.
So I think that's where I'll stop.