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Let's say we have the same path that we had in the last video.

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Draw my y-axis, that is my x-axis.

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Let's say the path looks like this, it looks

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something like this.

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It's the same one we had in the last video.

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Might not look exactly like it, let me see what I

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did in the last video.

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It looked like that in the last video, but close enough.

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Let's say we're dealing with the exact same

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curve as the last video.

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We could call that curve c.

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Now last video, we dealt with a vector field that only had

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vectors in the i direction.

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Let's build with another vector field that only has vectors in

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j direction, or the vertical direction.

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So let's say that q, the vector field q of xy, say it's equal

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to capital Q of xy times j, and we are going to concern

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ourselves we're going to concern ourselves with though

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closed line integral around the path c of q dot dr. And

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we've seen it already.

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dr can be rewritten as dx times i, plus dy times j.

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So if we were to take the dot product of these two, this

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line integral is going to be the exact same thing.

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This is going to be the same thing as the closed line

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integral over c of q dot dr. Well, Q only has a j-component.

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So if you take [? its ?]

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0i, so 0 times dxs is 0, and then you're going

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to have Q xy times dy.

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They had no i-component, so this is just going to be Q,

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I'll switch to that same color again, Q of x and y times dy.

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That's the dot product.

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There was no i-component, that's why we lose the dx.

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Now let's see if there's any way that we can solve this line

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integral without having to resort to a third parameter, t.

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Just like we did in the last video.

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Actually, it will be almost identical, we're just dealing

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with y's now instead of x's.

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So what we can do is, we could say, well, what's our

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minimum y and our maximum y?

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So our minimum y, let's say it's right here.

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The minimum y, let's call that a.

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Let's say our maximum y that we attain is right over there.

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Let's call that b.

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Oh, and just like in the last, I forgot to tell you the

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direction of the curve.

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But this is the same path as last time, so we're going in a

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counterclockwise direction.

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The exact same curve, exact same path.

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So we're going in that direction.

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Now in the last video, we broke it up into two functions of

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x's, two y's as a function of x.

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Now we want to deal with y's.

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Let's break it up into two functions of y.

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So if we break this path into two paths, those are kind of

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our extreme points, let's call this past right here, let's

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00:02:59,719 --> 00:03:06,109
call that path right there, let's call that y, let's call

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that x of-- so here, along this path, x is equal to-- or I

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could just write path 2, or a call it c2.

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We could say it's,

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x is equal to x2 of y.

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That's that path.

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And then the first path, or it doesn't have to be the

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first path, depending where you start.

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You can start anywhere.

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Let's say this one in magenta.

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We'll call that path 1, and we could say, that that's defined

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as x is equal to x1 of y.

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It's a little confusing when you have x as a function of y,

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but it's really completely analogous to what we

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did in the last video.

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We're literally just swapping x's and y.

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We're now expressing x as a function of y, instead

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of y as a function of x.

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So we have these two curves.

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You can imagine just flipping it, and we're doing the exact

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same thing that we did in the last videos, just

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now in terms of y.

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00:03:55,819 --> 00:03:59,609
But if you look at it this way, this line integral can be

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rewritten as being equal to the integral, let's

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00:04:07,590 --> 00:04:11,650
just do c2 first.

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This is the integral from b to a.

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00:04:15,379 --> 00:04:17,860
We start at b, and we go to a.

84
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This is, we're coming back from a high y to a low y.

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The integral from b to a of Q of-- in that gray color.

86
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Q of, instead of having an x there, we know along this curve

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right here, x is equal to, we want everything in terms of y.

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So here, x is equal to x2 of y.

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00:04:40,209 --> 00:04:47,799
So Q of x2 of y, x2 of y comma y, maybe I'm using too many

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colors here, but I think you get the idea.

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dy.

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93
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So this is the part of the line integral, just over

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this left hand curve.

95
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And then we're going to add to that the line integral, or

96
00:05:03,939 --> 00:05:07,980
really just a regular integral now, from y is equal to a to y

97
00:05:07,980 --> 00:05:18,325
is equal to b of Q of, instead of x being equal to x2, now x

98
00:05:18,324 --> 00:05:21,170
is equal to x1 of y, it's equal to this curve, this

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00:05:21,170 --> 00:05:22,780
other function.

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So x1 of y, x1 of y comma y, dy, and we can do exactly what

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00:05:33,060 --> 00:05:34,899
we did in the previous video.

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Instead of, we don't like the larger number on the bottom, so

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let's swap these two around.

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00:05:39,759 --> 00:05:42,870
So if you swap these two, if you make this into an a, and

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this into a b, that makes it the negative of the integral,

106
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when you swap the two, change the direction.

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This is exactly what we did in the last video, so hopefully

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it's nothing too fancy.

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But now that we have the same boundaries of integration,

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these two definite integrals, we can just write them as

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one definite integral.

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00:06:00,279 --> 00:06:05,469
So this is going to be equal to the integral from a to b.

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And I'll write this one first, since it's positive.

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Of, I'll write in this one.

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Q of x1 of y comma y, minus this one.

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Right?

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We have the minus sign here.

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Minus q of x2 of y and y dy.

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Let me do that in that neutral color.

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dy, that's multiplied by all of these things.

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I distributed out the dy, I think you get the idea.

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This is identical to what we did in the last video.

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And this could be rewritten as, this is equal to the integral

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from a to b of, and inside of the integral, we're evaluating

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the function of Q of xy from the boundaries, the upper

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boundary, where, the upper boundary is going to be from x

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is equal to x1 of y, and the lower boundary is

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x is equal to x2 of y.

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Right?

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All the x's we substituted with that, and then we get some

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expression, and then from that, we subtract this with x

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substituted as x2 of y.

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That's exactly what we did, and just like I said in last video,

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we're going kind of the reverse direction that we normally

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go in definite integrals.

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We normally get to this, and then the next

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step is, we get this.

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But now we're going in the reverse direction, but it's

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all the same difference.

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And all of that times dy.

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And just like we saw in the last video, this, let me do it

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in orange, this expression right here, actually let me

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draw that dy a little further out so it doesn't

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get all congested.

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Let me do the dy out here.

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This expression, this entire expression, is the same exact

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thing as the integral from x is equal to, I can just write it

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00:08:07,410 --> 00:08:11,280
here, let me write it in the same colors.

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x2 of y to x1 of y to x1 of y the partial of Q

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with respect to x dx.

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I want to make it very clear.

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This is, at least in my mind, the first part,

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a little confusing.

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But if you just saw an integral like this, this is the

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inside of a double integral.

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And it is.

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The outside is what we saw there, the integral

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from a to b, dy.

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But if you just saw this in a double integral, what you would

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do is you would take the antiderivative of this, the

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00:08:55,620 --> 00:08:58,639
antiderivative of this with respect to x, the

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00:08:58,639 --> 00:09:01,429
antiderivative of the partial of Q with respect to x with

165
00:09:01,429 --> 00:09:04,509
respect to x, is going to be just Q of xy.

166
00:09:04,509 --> 00:09:07,120
And since it's a definite integral, you would evaluate it

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00:09:07,120 --> 00:09:11,850
at x1 of y, and then subtract from that, this function

168
00:09:11,850 --> 00:09:14,370
evaluated x2 of y, which is exactly what we did.

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So hopefully you appreciate that.

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And then we got our result, which is very similar

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to the last result.

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What does this double integral represent?

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00:09:24,220 --> 00:09:28,389
It represents, well, anything, if you have any double integral

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that goes from-- if you imagine, this is some function,

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let me draw it in three dimensions.

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This is really almost a review of what we did

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in the last video.

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If that's the y-axis, that's our x-axis, that's our z-axis.

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This is some function of x and y, so some surface you

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00:09:45,720 --> 00:09:48,595
can imagine on xy plane.

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It's some surface.

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So we could call that the partial of q with respect to x.

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00:09:56,159 --> 00:09:59,259
And what this double integral is, this is essentially

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defining a region, and you can kind of view this dx times

185
00:10:02,490 --> 00:10:05,549
dy as kind of a small differential of area.

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00:10:05,549 --> 00:10:09,689
So the region under question, the boundary points, are from y

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00:10:09,690 --> 00:10:14,460
going from, y goes from, at the bottom, it goes from x2 of y,

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which we saw was a curve that looks something like this.

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That's the lower y, and over here, if we draw it in two

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00:10:19,620 --> 00:10:22,600
dimensions, this was the lower y-curve.

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The upper y-curve is x1 of y, so the upper y-curve looks

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something like that.

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00:10:29,210 --> 00:10:32,570
The upper y-curve goes something like that.

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So x varies from the lower y-curve to the upper

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00:10:35,509 --> 00:10:36,649
y-curve, right?

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00:10:36,649 --> 00:10:38,539
That's what we're doing right here.

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And then y varies from a to b.

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And so this is essentially saying, let's take the double

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00:10:44,340 --> 00:10:51,019
integral over this region right here of this function.

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00:10:51,019 --> 00:10:53,949
So it's essentially the volume, if this is the ceiling,

201
00:10:53,950 --> 00:10:56,300
and this boundary is essentially the wall.

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00:10:56,299 --> 00:10:57,679
It's the volume of that room.

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00:10:57,679 --> 00:10:58,750
And I don't know what it would look like when

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00:10:58,750 --> 00:11:00,139
it comes up here.

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00:11:00,139 --> 00:11:01,840
But you can kind of imagine something like that.

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00:11:01,840 --> 00:11:02,990
It would be the volume of that.

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00:11:02,990 --> 00:11:04,690
So that's what we're taking.

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00:11:04,690 --> 00:11:07,490
This is the identical result we got in the last video.

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00:11:07,490 --> 00:11:09,060
And this is a pretty neat thing.

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00:11:09,059 --> 00:11:13,599
So all of a sudden, this vector, that-- and Q of xy, I

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00:11:13,600 --> 00:11:16,060
didn't draw it out like I did the last time, Q of xy

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00:11:16,059 --> 00:11:16,629
only has [? things ?]

213
00:11:16,629 --> 00:11:19,475
in the j-direction, so it only has, if I were to draw its

214
00:11:19,475 --> 00:11:21,920
vector field, the vectors only go up and down.

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00:11:21,919 --> 00:11:25,959


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00:11:25,960 --> 00:11:28,650
They have no horizontal component to them.

217
00:11:28,649 --> 00:11:32,269
But we saw, when you start with a vector field like this, you

218
00:11:32,269 --> 00:11:35,809
take the line integral around this closed loop, and I'll

219
00:11:35,809 --> 00:11:39,000
rewrite it right here, you take this line integral around this

220
00:11:39,000 --> 00:11:44,200
closed loop of q dot dr, which is equal to the integral around

221
00:11:44,200 --> 00:11:49,560
the closed loop of Q of xy dy.

222
00:11:49,559 --> 00:11:53,239
We just figured out that that's equivalent to the double

223
00:11:53,240 --> 00:11:58,490
integral over the region.

224
00:11:58,490 --> 00:12:00,659
This is the region.

225
00:12:00,659 --> 00:12:00,919
Right?

226
00:12:00,919 --> 00:12:02,379
That's exactly what we're doing over here.

227
00:12:02,379 --> 00:12:03,950
If I just gave you the region, you'd have to define it, you'd

228
00:12:03,950 --> 00:12:07,116
say, well, x is going from, this is going from this

229
00:12:07,115 --> 00:12:09,779
function to that function, and y is going from a to b, and you

230
00:12:09,779 --> 00:12:11,309
might want to review the double integral videos, if

231
00:12:11,309 --> 00:12:11,969
that confuses you.

232
00:12:11,970 --> 00:12:15,269
So we're taking the double integral over the region of the

233
00:12:15,269 --> 00:12:20,370
partial of Q with respect to x, d-- well, you could write dx

234
00:12:20,370 --> 00:12:23,490
dy, or you can even right a little da, right?

235
00:12:23,490 --> 00:12:26,110
The differential of area, right, that we can imagine

236
00:12:26,110 --> 00:12:28,669
as a da, which is the same thing as a dx dy.

237
00:12:28,669 --> 00:12:31,389
And if we combine that with the last video, and this is kind of

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00:12:31,389 --> 00:12:35,029
the neat bringing it all together part, the result of

239
00:12:35,029 --> 00:12:36,759
the last video was this.

240
00:12:36,759 --> 00:12:39,939
That if I had a function that's defined completely in terms of

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00:12:39,940 --> 00:12:42,270
x, we had this, right here.

242
00:12:42,269 --> 00:12:43,090
We had that result.

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00:12:43,090 --> 00:12:46,280
Actually, let me copy and paste both of these to a nice clean

244
00:12:46,279 --> 00:12:51,100
part of my whiteboard, and then we can do the

245
00:12:51,100 --> 00:12:53,029
exciting conclusion.

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00:12:53,029 --> 00:12:57,169
Let me copy and paste that.

247
00:12:57,169 --> 00:12:59,299
So that's what we got in the last video.

248
00:12:59,299 --> 00:13:03,199
And this video, we got this result.

249
00:13:03,200 --> 00:13:07,080
I'll just copy and paste that part right there.

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You might already predict where this is going.

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And then let me paste it over here.

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This is the result from this video.

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Now let's think about an arbitrary vector field but is

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defined as, I'll do that in pink, let's say F is a vector

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field defined over the xy plane, and F is equal to P

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of xy i plus Q of xy j.

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You can almost imagine F being the addition of our vector

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fields, P and Q, that we did in the last two videos.

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Q was this video, and we did P in the video before that.

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But this is really any arbitrary vector field.

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And let's say we want to take the vector field, or sorry, the

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line integral of this vector field, along some path.

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It could be the same one we've done, which has been

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a very arbitrary one.

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It's really any arbitrary path.

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So let me draw some arbitrary path over here.

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Let's say, that is my arbitrary path, my arbitrary curve.

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Let's say it goes in that counterclockwise direction,

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just like that.

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And I'm interested in what the line integral, the closed line

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integral, around that path of F dot dr is.

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And we've seen it multiple times.

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dr is equal to dx i plus dy times j.

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So this line integral can be rewritten as, this is equal

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to the line integral around the path c.

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F dot dr, that's going to be this term times dx, so that's

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p of xy times dx plus this term, Q of xy times dy.

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And this whole thing, essentially this is the same

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thing as the line integral of p of xy dx, plus the line

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integral of Q of xy dy.

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Now what are these things?

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This is what we figured out in the first video, this is what

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we figured out just now in this video.

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This thing right here is the exact same thing

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as that over there.

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So this is going to be equal to the double integral over this

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region right here, of the minus partial of P with respect to y.

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Instead of a dy dx, we could say just over the

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differential of area.

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And then plus this one, this result.

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Q.

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This thing right here is exactly what we just proved,

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is exactly what we just showed in this video.

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So that's plus, I'll leave it up there, maybe I'll do it in

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the yellow, plus the double integral over the same region

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of the partial of Q with respect to x.

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da, where that's just dy dx, or dx dy, you can

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switch the order, it's

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the differential of area.

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And now, we can add these two integrals.

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What do we get?

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So this is equal to, and this is kind of our

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big, grand conclusion.

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Maybe magenta is called for here.

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The double integral over the region of, I'll write this one

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first because it's positive, that one's negative, of the

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partial of Q with respect to x, minus the partial of p with

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respect to y, d, the differential of area.

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This is our big takeaway.

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This is our big takeaway.

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Let me write it here.

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The line integral of the closed loop of F dot dr is equal to

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the double integral of this expression.

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And it's something, just remember.

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We're taking the function that was associated with the

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x-component, or the i-component, we're taking the

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partial with respect to y.

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And the function that was associated with the

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y-component, we're taking the partial with respect to x.

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And that first one, we're taking the negative of.

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That's a good way to remember it.

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00:17:19,910 --> 00:17:23,140
But this result right here, this is, maybe I should

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write it in green, this is Green's Theorem.

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And it's a neat way to relate a line integral of a vector field

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that has these partial derivatives, assuming it has

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these partial derivatives, to the region, to a double

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integral of the region.

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Now, and this is a little bit of a side note, we've seen in

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several videos before, we've learned that if F is

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conservative, which means it's the gradient of some function,

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that it's path-independent, that the closed integral around

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any path is equal to 0.

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And that's still true.

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So that tells us that if F is conservative, this thing right

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00:18:14,720 --> 00:18:17,559
here must be equal to 0.

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That's the only way that you're always going to enforce that

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00:18:20,069 --> 00:18:23,349
this whole integral is going to be equal, is going to be equal

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00:18:23,349 --> 00:18:25,969
to 0 over any, any, any region.

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00:18:25,970 --> 00:18:27,269
I'm sure you could think of situations where they

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cancel each other out, but really over any region.

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That's the only way that this is going to be true.

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That these two things are going to be equal to 0.

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So then you could say, partial of Q with respect to x, minus

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00:18:38,220 --> 00:18:42,370
the partial of P with respect to y, has to be equal to 0, or

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00:18:42,369 --> 00:18:44,969
these two things have to equal each other.

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Or.

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This is kind of a corollary to Green's Theorem.

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Kind of a low-hanging fruit you could have figured out.

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00:18:49,900 --> 00:18:53,009
The partial of Q with respect to x is equal to the partial

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of P with respect to y.

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And when you study exact equations in differential

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00:18:59,069 --> 00:19:01,460
equations, you'll see this a lot more.

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00:19:01,460 --> 00:19:04,140
And actually, well, I won't go into too much, but conservative

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00:19:04,140 --> 00:19:07,330
fields, you're actually, the differential form of what you

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00:19:07,329 --> 00:19:10,500
see in the line integrals, if it's conservative, it would

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00:19:10,500 --> 00:19:11,450
be an exact equation.

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00:19:11,450 --> 00:19:12,610
But we're not going to go into that too much.

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But hopefully you might see the parallels if you've already run

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00:19:15,329 --> 00:19:17,789
into exact equations in differential equations.

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00:19:17,789 --> 00:19:20,980
But this is the big takeaway, and let's maybe do some

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00:19:20,980 --> 00:19:25,779
examples using this takeaway in the next video.

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