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Let's see if we can use our knowledge of Green's theorem

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to solve some actual line integrals.

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And actually, before I show an example, I want to make one

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clarification on Green's theorem.

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All of the examples that I did is I had a region like this,

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and the inside of the region was to the left of

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what we traversed.

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So all my examples I went counterclockwise and so our

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region was to the left of-- if you imagined walking along the

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path in that direction, it was always to our left.

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And that's the situation which Green's theorem would apply.

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So if you were to take a line integral along this path, a

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closed line integral, maybe we could even specify

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it like that.

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You'll see that in some textbooks-- along the curve c

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of f dot dr. This is what equals the double integral over

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this region r of the partial of q with respect to x minus

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the partial of p with respect to y d area.

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And just as reminder, now this q and p are coming from the

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components of f in the situation.

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f would be written as f of xy is equal to P of xy time the i

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component plus Q of xy times the j component.

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So this is a situation where the inside of the region is

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to the left of the direction that we're taking the path.

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If it was in the reverse, then we would put a

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minus sign right here.

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If this arrow went the other way, we'd put a minus sign and

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we can do that because we know that when we're taking line

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integrals through vector fields, if we were reverse

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the direction it becomes the negative of that.

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We showed that, I think, 4 or 5 videos ago.

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With that said, it was convenient to write

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Green's theorem up here.

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Let's actually solve a problem.

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Let's say I have the line integral and let's say

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we're over a curve.

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I'll define the curve in a second.

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But let's say that the integral we're trying to solve is x

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squared minus y squared dx plus 2xy dy.

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And then our curve-- they're giving us the boundary.

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The boundary is the region.

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I'll do it in a different color.

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So the curve is boundary of the region given by all of the

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points x,y such that x is a greater than or equal to 0,

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less than or equal to 1.

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And then y is greater than or equal to 2x squared and

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less than or equal to 2x.

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So let's draw this region that we're dealing with right now.

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So let me draw my x-axis or my y-axis, sorry.

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My y-axis and then my x-axis right there.

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And let's see. x goes from 0 to 1, so if we make--

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that's obviously 0.

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Let's say that that is x is equal to 1, so that's

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all the x values.

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And y varies, it's above 2x squared and below 2x.

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Normally, if you get to large enough numbers, 2x squared is

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larger, but if you're below 1 this is actually going to

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be smaller than that.

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So the upper boundary is 2x, so there's 1 comma 2.

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This is a line y is equal to 2x, so that is the line y is--

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let me draw a straighter line than that.

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The line y is equal to 2x looks something like that.

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That right there is y is equal to 2x.

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Maybe I'll do that in yellow.

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And then the bottom curve right here is y is going to be

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greater than 2x squared.

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It might look something like this.

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And of course, the region that they're talking about is this,

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but we're saying that the curve is the boundary of this region

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and we're going to go in I counterclockwise direction.

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I have to specify that.

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So our curve, we could start at any point really, but we're

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going to go like that.

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Then get to that point and then come back down along that

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top curve just like that.

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And so this met the condition that the inside of the region

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is always going to be our left, so we can just do the straight

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up Green's theorem, we don't have to do the negative of it.

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And let's define our region.

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Let's just do our region.

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This integral right here is going to go-- I'll just do it

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the way-- y varies from y is equal to 2x squared to

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2y is equal to 2x.

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And so maybe we'll integrate with respect to y first.

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And then x, I'll do the outside.

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The boundary of x goes from 0 to 1.

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So they set us up well to do an indefinite integral.

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Now we just have to figure out what goes over

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here-- Green's theorem.

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Our f would look like this in this situation.

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f is f of xy is going to be equal to x squared minus

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y squared i plus 2xy j.

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We've seen this in multiple videos.

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You take the dot product of this with dr, you're going to

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get this thing right here.

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So this expression right here is our P of x y.

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And this expression right here is our Q of xy.

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So inside of here we're just going to apply Green's

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theorem straight up.

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So the partial of Q with respect to x-- so take

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the derivative of this with respect to x.

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We're just going to end up with the 2y.

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And then from that we're going to subtract the partial

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of P with respect to y.

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So if you take the derivative of this with respect to y that

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becomes 0 and then, here you have-- the derivative with

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respect to y here is minus 2y.

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Just like that.

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And so this simplifies to 2y minus minus 2y.

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That's 2y plus plus 2y.

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I'm just subtracting a negative.

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Or this inside, and just to save space, this

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inside-- that's just 4y.

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I don't want to have to rewrite the boundaries.

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That right there is the same thing as 4y.

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The partial of Q with respect to y, 2y minus the partial

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of P with respect to y.

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Which is minus 2y.

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You subtract a negative you get a positive.

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You have 4y.

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Let's take the antiderivative of the inside with respect

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to y and we're going to get 2y squared.

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Let me do it a little bit lower.

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We're going to get 2y squared if you take the partial with

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respect to y you're going to get 4y.

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We're going to evaluate that from y is equal to 2x squared

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to y is equal to 2x.

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And then of course, we still have the outside

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integral still there.

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x goes from 0 to 1 dx.

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This thing is going to be equal to the integral from 0 to 1 and

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then we evaluate it first at 2x.

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So you put a 2x in here, 2x squared is 4x squared.

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2 squared, x squared, so 4x squared times 2 is

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going to be 8x squared.

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Minus-- put this guy in there.

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2x squared squared is 2x to the fourth.

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4x to the fourth times 2 is 8x to the fourth.

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Did I do that right?

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2x squared-- going to put it there for y,

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substitute y with it.

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That squared is 4 X to the fourth.

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Times 2 is 8x x to the fourth.

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Very good.

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All right.

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Now dx.

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Now this is just a straightforward

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one-dimensional integral.

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This is going to be equal to-- I'll just do it here.

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This is going to be equal to the antiderivative of 8x

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squared is 8/3 x to the third.

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And then the antiderivative of 8x to the fourth is

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minus 8/5 x to the fifth.

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Then we're going to have to evaluate that from 0 to 1, or

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we can put a little line there.

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When you put 1 in there you get-- I'll do it

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in a different color.

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We get 8/5 times 1 to the third, which is 8/5

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minus 8/5 minus 8/5.

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And then we're going to have minus-- when you put 0 in

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here you're just going to get a bunch of 0's.

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Oh sorry, I made a mistake.

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It would have been a blunder.

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It's 8/3.

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8/3 times 1 to the third minus 8/5 times 1 to the fifth,

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so that's minus 8/5.

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And then, when you subtract the 0, you evaluate 0 here. you're

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just going to get a bunch of 0's so you don't have

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to do anything else.

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So now we just have to subtract these two fractions.

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So let's get a common denominator of 15.

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8/3 is the same thing if we multiply the numerator

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and denominator by 5.

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That is 40/15.

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And then if we multiply this numerator and denominator by

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3, that's going to be 24/15.

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So minus 24/15 and we get it being equal to 16/15.

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And so using Green's theorem we were able to find the answer

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to this integral up here.

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It's equal to 16/15.

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Hopefully you found that useful.

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I'll do one more example in the next video.

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