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Let's say I have a path in the xy plane that's essentially

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the unit circle.

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So this is my y-axis, this is my x-axis, and our path is

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going to be the unit circle.

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And we're going to traverse it just like that.

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We're going to traverse it clockwise.

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I think you get the idea.

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And so its equation is the units circle.

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So the equation of this is x squared plus y squared is

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equal to 1; has a radius of 1 unit circle.

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And what we're concerned with is the line integral

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over this curve c.

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It's a closed curve c.

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It's actually going in that direction of 2y dx minus 3x dy.

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So, we are probably tempted to use Green's

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theorem and why not?

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So let's try.

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So this is our path.

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So Green's theorem tells us that the integral of some curve

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f dot dr over some path where f is equal to-- let me write

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it a little nit neater.

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Where f of x,y is equal to P of x, y i plus Q of x, y j.

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That this integral is equal to the double integral over the

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region-- this would be the region under question

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in this example.

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Over the region of the partial of Q with respect to x

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minus the partial of P with respect to y.

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All of that dA, the differential of area.

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And of course, the region is what I just showed you.

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Now, you may or may not remember-- well, there's a

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slight, subtle thing in this, which would give

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you the wrong answer.

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In the last video we said that Green's theorem applies when

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we're going counterclockwise.

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Notice, even on this little thing on the integral I made

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it go counterclockwise.

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In our example, the curve goes clockwise.

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The region is to our right.

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Green's theorem-- this applies when the region is to our left.

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So in this situation when the region is to our right and

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we're going-- so this is counterclockwise.

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So in our example, where we're going clockwise, the region is

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to our right, Green's theorem is going to be the

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negative of this.

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So in our example, we're going to have the integral of c and

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we're going to go in the clockwise direction.

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So maybe I'll draw it like that of f dot dr. This is going

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to be equal to the double integral over the region.

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You could just swap these two-- the partial of P with respect

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to y minus the partial of Q with respect to x da.

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So let's do that.

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So this is going to be equal to, in this example, the

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integral over the region-- let's just keep it

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abstract for now.

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We could start setting the boundaries, but let's just

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keep the region abstract.

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And what is the partial of P with respect-- let's remember,

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this right here is our-- I think we could recognize right

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now that if we take f dot dr we're going to get this.

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The dr contributes those components.

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The f contributes these two components.

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So this is P of x,y.

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And then this is Q of x,y.

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And we've seen it.

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I don't want to go into the whole dot dr and take the dot

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product over and over again.

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I think you can see that this is the dot product

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of two vectors.

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This is the x component of f, y component of f.

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This is the x component of dr, y component of dr. So let's

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take the partial of P with respect to y.

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You take the derivative of this with respect to y, you get 2.

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Derivative of 2y is just 2.

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So you get 2, and then, minus the derivative of

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Q with respect to x.

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Derivative of this with respect to x is minus 3.

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So we're going to get minus 3, and then all of that da.

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And this is equal to the integral over the region.

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What's this, it's 2 minus minus 3?

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That's the same thing as 2 plus 3.

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So it's the integral over the region of 5 dA.

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5 is just a constant, so we can take it out of the integral.

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So this is going to turn out to be quite a simple problem.

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So this is going to be equal to 5 times the double integral

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over the region R dA.

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Now what is this thing?

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What is this thing right here?

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It looks very abstract, but we can solve this.

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This is just the area of the region.

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That's what that double integral represents.

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You just sum up all the little dA's.

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That's a dA, that's a dA.

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You sum up the infinite sums of those little

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dA's over the region.

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Well, what's the area of this unit circle?

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Here we just break out a little bit of ninth grade-- actually,

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even earlier than that-- pre-algebra or middle

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school geometry.

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Area is equal to pi r squared.

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What's our radius?

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So unit circle, our radius is 1.

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Length is 1.

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So the area here is pi.

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So this thing over here, that whole thing is

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just equal to pi.

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So the answer to our line integral is just 5 pi, which

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is pretty straightforward.

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I mean, we could have taken the trouble of setting up a double

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integral where we take the antiderivative with respect to

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y first and write y is equal to the negative square root of 1

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minus x squared y is equal to the positive square root.

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x goes from minus 1 to 1.

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But that would have been super hairy and a huge pain.

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And we just have to realize, no, this is just the area.

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And the other interesting thing is I challenge you to solve

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the same integral without using Green's theorem.

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You know, after generating a parameterization for this

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curve, going in that direction, taking the derivatives

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of x of t and y of t.

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Multiplying by the appropriate thing and then taking the

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antiderivative-- way hairier than what we just did using

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Green's theorem to get 5 pi.

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And remember, the reason why it wasn't minus 5 pi here

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is because we're going in a clockwise direction.

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If we were going in a counterclockwise direction we

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could have applied the straight up Green's theorem, and we

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would have gotten minus 5 pi.

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Anyway, hopefully you found that useful.

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