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Green's theorem example 2
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.