# Green's theorem exampleÂ 2

## Video transcript

Let's say I have a path in the
xy plane that's essentially the unit circle. So this is my y-axis, this is
my x-axis, and our path is going to be the unit circle. And we're going to traverse
it just like that. We're going to traverse
it clockwise. I think you get the idea. And so its equation
is the units circle. So the equation of this is x
squared plus y squared is equal to 1; has a radius
of 1 unit circle. And what we're concerned
with is the line integral over this curve c. It's a closed curve c. It's actually going in that
direction of 2y dx minus 3x dy. So, we are probably
tempted to use Green's theorem and why not? So let's try. So this is our path. So Green's theorem tells us
that the integral of some curve f dot dr over some path where f
is equal to-- let me write it a little nit neater. Where f of x,y is equal to P
of x, y i plus Q of x, y j. That this integral is equal to
the double integral over the region-- this would be the
region under question in this example. Over the region of the partial
of Q with respect to x minus the partial of
P with respect to y. All of that dA, the
differential of area. And of course, the region
is what I just showed you. Now, you may or may not
remember-- well, there's a slight, subtle thing in
this, which would give you the wrong answer. In the last video we said that
Green's theorem applies when we're going counterclockwise. Notice, even on this little
thing on the integral I made it go counterclockwise. In our example, the
curve goes clockwise. The region is to our right. Green's theorem-- this applies
when the region is to our left. So in this situation when the
region is to our right and we're going-- so this
is counterclockwise. So in our example, where we're
going clockwise, the region is to our right, Green's theorem
is going to be the negative of this. So in our example, we're going
to have the integral of c and we're going to go in the
clockwise direction. So maybe I'll draw it like that
of f dot dr. This is going to be equal to the double
integral over the region. You could just swap these two--
the partial of P with respect to y minus the partial of
Q with respect to x da. So let's do that. So this is going to be equal
to, in this example, the integral over the region--
let's just keep it abstract for now. We could start setting the
boundaries, but let's just keep the region abstract. And what is the partial of P
with respect-- let's remember, this right here is our-- I
think we could recognize right now that if we take f dot dr
we're going to get this. The dr contributes
those components. The f contributes
these two components. So this is P of x,y. And then this is Q of x,y. And we've seen it. I don't want to go into the
whole dot dr and take the dot product over and over again. I think you can see that
this is the dot product of two vectors. This is the x component
of f, y component of f. This is the x component of dr,
y component of dr. So let's take the partial of P
with respect to y. You take the derivative of this
with respect to y, you get 2. Derivative of 2y is just 2. So you get 2, and then,
minus the derivative of Q with respect to x. Derivative of this with
respect to x is minus 3. So we're going to get minus
3, and then all of that da. And this is equal to the
integral over the region. What's this, it's
2 minus minus 3? That's the same
thing as 2 plus 3. So it's the integral over
the region of 5 dA. 5 is just a constant, so we can
take it out of the integral. So this is going to turn out
to be quite a simple problem. So this is going to be equal to
5 times the double integral over the region R dA. Now what is this thing? What is this thing right here? It looks very abstract,
but we can solve this. This is just the
area of the region. That's what that double
integral represents. You just sum up all
the little dA's. That's a dA, that's a dA. You sum up the infinite
sums of those little dA's over the region. Well, what's the area
of this unit circle? Here we just break out a little
bit of ninth grade-- actually, even earlier than that--
pre-algebra or middle school geometry. Area is equal to pi r squared. What's our radius? So unit circle,
our radius is 1. Length is 1. So the area here is pi. So this thing over here,
that whole thing is just equal to pi. So the answer to our line
integral is just 5 pi, which is pretty straightforward. I mean, we could have taken the
trouble of setting up a double integral where we take the
antiderivative with respect to y first and write y is equal to
the negative square root of 1 minus x squared y is equal to
the positive square root. x goes from minus 1 to 1. But that would have been
super hairy and a huge pain. And we just have to realize,
no, this is just the area. And the other interesting thing
is I challenge you to solve the same integral without
using Green's theorem. You know, after generating a
parameterization for this curve, going in that direction,
taking the derivatives of x of t and y of t. Multiplying by the appropriate
thing and then taking the antiderivative-- way hairier
than what we just did using Green's theorem to get 5 pi. And remember, the reason why
it wasn't minus 5 pi here is because we're going in
a clockwise direction. If we were going in a
counterclockwise direction we could have applied the straight
up Green's theorem, and we would have gotten minus 5 pi. Anyway, hopefully you
found that useful.