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So last time we've seen the curl of vectoring
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vector fields with components m and n,
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we defined that to be n sub x minus m sub y.
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And we said this measures how far that vector field is
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from being conservative.
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If the curl is 0 and if the field is defined everywhere,
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then it's going to be conservative.
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And so when I take a line integral along a closed curve,
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I don't have to compute it.
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I know it's going to be 0.
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But now let's say that I have a general vector field.
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So the curl will not be 0.
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And I still want to compute a line integral along a closed
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curve.
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Well, I could compute it directly.
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Or there's another way.
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And that's what we are going to see today.
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So say that I have a closed curve c,
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and I want to find the work.
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So there's two options.
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One is direct calculation.
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And the other one is grain theorem.
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So grain theorem is another way to avoid calculating
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line integral if we don't want to.
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So what does it say?
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It says if c is a closed curve and closing origin R
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in the plane.
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And I have to insist c should go counterclockwise.
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And if I have a vector field, that's defined and
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defraniable everywhere, not only on the curve c, which is
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what I need to define the line integral, but also on the
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origin inside.
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Then the line integral for the work done along c is actually
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equal to a double integral of our origin inside of curl f
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v a.
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So that's the conclusion.
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And if you want me to write it in coordinates, maybe I
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should do that.
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So the line integral in terms of a components that's the
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integral of mdx plus ndy.
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And the curl is nx minus mi dA.
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So that's the other way to state it.
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So that's a really strange statement if you think about it.
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Because the left hand side is a line integral.
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So the way we compute it is we take this expression mdx plus
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ndy.
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And we parameterize the curve.
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We express x and y in terms of some variable t, maybe
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or whatever you want to call it.
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And then you will do a one variable integral over t.
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This right hand side here, it's a double integral dA.
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So we do it the way that we learned how to a couple of weeks
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ago.
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You take your origin, you slice it in the x direction or in the
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y direction, and you integrate dx dy after setting up the
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bounds carefully.
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Or maybe in polar coordinates are the RD theta.
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Let's see, the way in which you compute these things is
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completely different.
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This one on the left hand side lives only on the curve.
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While the right hand side lives everywhere in this region
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inside.
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So here x and y are related, believe it the curve.
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Here x and y are independent.
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There are just some bounds between them.
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And of course, what you're integrating is different.
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Here it's a line integral for work.
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Here it's a double integral of some function of x and y.
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So it's a very publicizing statement at first.
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It's a very powerful tool.
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So we're going to try to see how it works
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concretely, what it says, what are the consequences, how we
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could convince ourselves that yes, this works, and so on.
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That's going to be the topic for today.
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Any questions about the statement first?
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OK.
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Oh, yeah, one remark.
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Sorry.
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So here it says counterclockwise.
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What if I have a curve that goes clockwise?
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Well, you could just take the negative and integrate counter
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clockwise.
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I mean, somehow, why does the theorem
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choose counterclockwise over clockwise?
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How does it know that it's counterclockwise,
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rather than clockwise?
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Well, the answer is basically in our convention for curl.
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We've said curl is nx minus my and not the other way around.
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And that's a convention as well.
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So somehow, the two conventions match with each other.
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That's the best answer I can give you.
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So if you met somebody from a different planet,
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they might have a green sphere with the opposite conventions,
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with curves going clockwise and the curl
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is going to be different if you have a way around.
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Probably if you met an alien, then I'm
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not sure if you would be discussing green sphere
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I'm first, but just in case, you know.
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OK.
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So that being said, so there's a warning here, which
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is that this is only for closed curves.
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So if I give you a curve that's not closed,
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and I tell you, well, compute the 9 integral,
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then you have to do it by hand.
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You have to parameterize the curve.
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Or if you really don't like that 9 integral,
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you could close the path by adding some other 9 integral
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to it, and then compute using green sphere.
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But you can't use green sphere on directly if the curve is not
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closed.
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So let's do a quick example.
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So let's say that I give you c, the circle of radius 1,
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centered at the point 2, 0.
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So you know, it's out here.
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That's make of c.
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And let's say that I do it counterclockwise so that it will
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match with the statement of a theorem.
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And let's say that I want you to compute the 9 integral
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along c of y e to the minus x dx plus 1 half of x squared
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minus e to the minus x dy.
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And you know, that's a kind of a statistic example,
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but maybe I will ask you to do that.
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So how would you do it directly?
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Well, to do it directly, you would have to parameterize
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this curve so that we'd probably involve setting x equals 2
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plus cosine theta y equals sine theta.
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But I'm using a parameter of the angle around the circle.
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It's like the unit circle, the usual one, but shifted
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by 2 in the x direction.
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And then I would set dx equals minus sine theta d theta.
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I would set dy equals cos theta d theta.
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And I would substitute, and I would integrate from 0 to 2 pi.
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And I would probably run into a bit of trouble
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because I would have this e to the minus x, which
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would give me something that I really don't want to integrate.
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So instead of doing that, which looks pretty much no doomed,
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I'm going to instead of going to use grain sphere.
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So using grain sphere, for where I will do it,
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is I will instead compute a double integral.
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OK?
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So I will compute for double integral of our origin
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inside of curl f da.
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So let's call this m.
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Let's call this n.
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And then I will actually just use the form coordinates
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nx minus mi da.
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And what is r here?
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Well, r is the disk in here.
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So of course, it might not be that pleasant,
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because we'll also have to set up this double integral.
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And for that, we'll have to figure out a way
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to slice this region nicely.
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We could do it dx dy.
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We could do it dy dx.
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Or maybe we will want to actually make a change of variables
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to first shift this to the origin, change x to x minus 2.
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And then switch to polar coordinates.
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Well, let's see what happens later.
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So what is, so this is r.
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So what is n sub x?
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Well, n sub x is x plus e to the minus x minus what
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is m sub y e to the minus x.
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This is nx.
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This is mi da.
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Well, it seems to simplify a bit.
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I will just get the ball integral over r of x da, which
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looks suddenly a lot more pleasant.
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And of course, I made up the example in that way,
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so that it simplifies when you use green sphere.
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But you know, that gives you an example
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where you can turn a really hard line integral
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into an easier double integral.
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Now, how do we compute that double integral?
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Well, so one way would be to set it up
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or let's actually be a bit smarter and observe
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that this is actually the area of the region r times the x
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coordinate of its center of mass.
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If I look at the definition of the center of mass,
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it's the average value of x.
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So it's one of the area times the double integral of x da.
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Well, possibly with the density.
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But here I'm taking uniform density 1.
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And now I think I know just by looking at the picture
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on where the center of mass of this circle will be.
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I mean, it will be right in the middle.
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So that is 2, if you want, by symmetry.
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And the area of this guy is just pi
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because it's a disk of radius 1.
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So I will just get 2 pi.
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And of course, if you didn't see that,
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then you could also compute that double integral directly.
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It's a nice exercise.
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But see, here using geometry helps you
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to actually streamline the calculation.
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Any questions?
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Yes?
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I'll actually be the mass part.
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OK, yes, let me just repeat for last part.
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So I say we have to compute the double integral of x da
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over this region here, which is a disk of radius 1,
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centered at this point is 2, 0.
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So instead of setting up the integral with bounds
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and integrating the x dy or the y dx or in polar coordinates,
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I'm just going to say, well, let's remember
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the definition of a center of mass.
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It's the average value of a function x in the region.
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So it's one over the area of the region
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times the double integral of x da.
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If you look again at the definition of x bar,
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it's one over the area double integral of x da.
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Well, maybe if there's a density,
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then it's one over mass times double integral of x density da.
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But if density is one, then it just becomes this.
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So switching the area moving the area to the other side,
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I'll get double integral of x da is the area of the region
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times the x coordinate of the center of mass.
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The area of the region is pi because it's a unit disk.
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And the center of mass is the center of the disk,
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so its x bar is 2.
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And I get 2 pi.
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OK.
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Now, I didn't actually have to do this in my example today,
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but I thought that would be a good review.
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It would remind you of center of mass and all that.
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OK.
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Any other questions?
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OK.
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So let's see.
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Now that we've seen how to use it in practice,
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how to avoid calculating a line integral if we don't want to,
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let's try to convince ourselves that this theorem makes sense.
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So well, let's start with an easy case
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where we should be able to know the answer to both sides.
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So let's look at a special case.
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Let's look at the case where curl f is 0.
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Then, well, we'd like to conclude that f is conservative.
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That's what we've said.
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Well, let's see what happens.
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So grain theorem says that if I have a closed curve,
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then the line integral of f is equal to the double integral
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of curl on the region inside.
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And if the curl is 0, then I will be integrating 0.
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So this is actually how you prove that if your vector field has curl 0,
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then it's conservative.
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OK, so in particular, if you have a vector field
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that's defined everywhere in the plane,
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then you take any closed curve, well,
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you will get that the line integral will be 0.
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Strictly speaking, that will only work here if the curve goes counterclockwise.
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But otherwise, just look at the various loops that it makes
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and orient each of them counterclockwise and some things together.
259
00:18:19.080 --> 00:18:26.080
So let me state that again.
260
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So a consequence of grain theorem is that
261
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if f is defined everywhere in the plane
262
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and the curl of f is 0 everywhere, then f is conservative.
263
00:19:14.920 --> 00:19:21.920
And so this actually is the input we needed to justify our criterion.
264
00:19:21.920 --> 00:19:26.920
The test that we saw last time saying, well, to check if something is a gradient field,
265
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if it's conservative, we just have to compute the curl and check whether it's 0.
266
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So how do we prove that now carefully?
267
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Well, you just take a closed curve in the plane,
268
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you switch the orientation if needed so that it becomes counterclockwise.
269
00:19:54.920 --> 00:20:01.920
And then you look at the region inside and then you know that the line integral
270
00:20:01.920 --> 00:20:08.920
will be equal to the double integral of curl,
271
00:20:08.920 --> 00:20:15.920
which is the double integral of 0, therefore that's 0.
272
00:20:15.920 --> 00:20:20.920
But see, so now let's say that we try to do that for the vector field that was on your problem set,
273
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that was not defined at the origin.
274
00:20:24.920 --> 00:20:28.920
So if you've done the problem set and found the same answers that I did,
275
00:20:28.920 --> 00:20:32.920
then you will have found that the vector field had curl 0 everywhere,
276
00:20:32.920 --> 00:20:37.920
but still it wasn't conservative because if you went around the unit circle,
277
00:20:37.920 --> 00:20:40.920
then you got an integral that was 2 pi.
278
00:20:40.920 --> 00:20:46.920
Or if you compared the two halves, you got different answers for two paths that go from the same point to the same point.
279
00:20:46.920 --> 00:20:51.920
So you know, it fails this property, but that's because it's not defined everywhere.
280
00:20:51.920 --> 00:20:53.920
So what goes wrong with this argument?
281
00:20:53.920 --> 00:20:58.920
Well, if I take the vector field that was on the problem set,
282
00:20:58.920 --> 00:21:05.920
and if I do things, you know, say that I look at the unit circle,
283
00:21:05.920 --> 00:21:09.920
that's a closed curve, so I would like to use grain sphere.
284
00:21:09.920 --> 00:21:16.920
And grain sphere would tell me the line integral along this loop is equal to the double integral of curl
285
00:21:16.920 --> 00:21:20.920
over this region here, the unit disk.
286
00:21:20.920 --> 00:21:25.920
And of course, the curl is 0. Well, except at the origin.
287
00:21:25.920 --> 00:21:28.920
At the origin, the vector field is not defined.
288
00:21:28.920 --> 00:21:31.920
You cannot take the derivatives, and the curl is not defined.
289
00:21:31.920 --> 00:21:34.920
And somehow that messes things up.
290
00:21:34.920 --> 00:21:41.920
You cannot apply grain sphere to the vector field.
291
00:21:41.920 --> 00:21:59.920
So you cannot apply grain sphere to the vector field on problems at a problem 2,
292
00:21:59.920 --> 00:22:08.920
when c and closes the origin.
293
00:22:08.920 --> 00:22:15.920
And so that's why this guy, even though it has curl 0, is not conservative.
294
00:22:15.920 --> 00:22:19.920
You know, there's no contradiction.
295
00:22:19.920 --> 00:22:23.920
And somehow you have to imagine that, well, the curl here is really not defined,
296
00:22:23.920 --> 00:22:27.920
but somehow it becomes infinite so that when you do the double integral,
297
00:22:27.920 --> 00:22:30.920
you actually get 2 pi instead of 0.
298
00:22:30.920 --> 00:22:37.920
I mean, that doesn't make any sense, of course, but that's one way to think about it.
299
00:22:37.920 --> 00:22:44.920
Okay, any questions? Yes?
300
00:22:44.920 --> 00:22:53.920
But we just know the word, the redo.
301
00:22:53.920 --> 00:22:58.920
So we find the curl at that point.
302
00:22:58.920 --> 00:23:05.920
Well, so actually it's not defined, see, because the curl is 0 every well.
303
00:23:05.920 --> 00:23:10.920
So if the curl was well defined at the origin, you know, you try to then take the double integral,
304
00:23:10.920 --> 00:23:12.920
well, no matter what value you put for a function.
305
00:23:12.920 --> 00:23:15.920
If you have a function that's 0 everywhere except at the origin,
306
00:23:15.920 --> 00:23:19.920
and some other value at the origin, the integral is still 0.
307
00:23:19.920 --> 00:23:22.920
So it's worse than that. It doesn't have, it's not only that you can't compute it,
308
00:23:22.920 --> 00:23:29.920
it's really that it's not defined.
309
00:23:29.920 --> 00:23:36.920
Anyway, that's, you know, like a slightly pathological example.
310
00:23:36.920 --> 00:23:41.920
Yes?
311
00:23:41.920 --> 00:23:45.920
Well, we wouldn't be able to because the curl is not defined at the origin,
312
00:23:45.920 --> 00:23:48.920
so you can't actually integrate it.
313
00:23:48.920 --> 00:23:50.920
Okay, so that's the problem.
314
00:23:50.920 --> 00:23:54.920
I mean, if you try to integrate, you know, we've said everywhere where it's defined,
315
00:23:54.920 --> 00:23:57.920
the curl is 0. So what you would be integrating would be 0.
316
00:23:57.920 --> 00:24:02.920
But that doesn't work because at the origin, it's not defined.
317
00:24:02.920 --> 00:24:08.920
Yes?
318
00:24:08.920 --> 00:24:13.920
Ah, so if you take a curl that makes a figure 8, then indeed my proof of a various faults.
319
00:24:13.920 --> 00:24:18.920
So I kind of tricked you, you know, it's not actually correct.
320
00:24:18.920 --> 00:24:26.920
So if the curl does a figure 8, then what you would do is you would actually cut it into its two halves,
321
00:24:26.920 --> 00:24:29.920
and for each of them you would apply a range theorem.
322
00:24:29.920 --> 00:24:33.920
And then you would still get that you know, if the curl is 0, then this line integral is 0,
323
00:24:33.920 --> 00:24:36.920
that one is also 0, so that's from a 0.
324
00:24:36.920 --> 00:24:41.920
Okay, small details that you don't really need too much about,
325
00:24:41.920 --> 00:24:49.920
but indeed, if you want to be careful with details, then my proof is not quite complete.
326
00:24:49.920 --> 00:24:54.920
But the conclusion is still true.
327
00:24:54.920 --> 00:24:57.920
Let's move on.
328
00:24:57.920 --> 00:25:07.920
So I want to tell you how to prove grain's theorem.
329
00:25:07.920 --> 00:25:17.920
Because you know, it's such a strange formula that, you know, where can it come from possibly?
330
00:25:17.920 --> 00:25:35.920
So let me remind you, first of all, the statement we want to prove is that the line integral along the closed curve of mdx plus ndy is equal to the double integral of the region inside of nx minus my dA.
331
00:25:35.920 --> 00:26:04.920
And you know, let's simplify our lives a bit by proving easier statements. So actually, the first observation will actually prove something easier, namely that the line integral, let's see, of mdx along the closed curve is equal to the double integral of the region inside of minus mdx.
332
00:26:04.920 --> 00:26:18.920
So that's the special case where n is 0, where you have only an x component for your vector field.
333
00:26:18.920 --> 00:26:32.920
Now, why is that good enough? Well, the claim is if I can prove this, I claim you will be able to do the same thing to prove the other case where there's only a y component.
334
00:26:32.920 --> 00:26:38.920
And then, if you add them together, you'll get the general case.
335
00:26:38.920 --> 00:26:41.920
So let me explain.
336
00:26:41.920 --> 00:27:10.920
Okay, so, a similar argument, which I will not do to save time,
337
00:27:10.920 --> 00:27:28.920
will show, so it will be just the same thing but switching the rules of x and y, that if I integrate along a closed curve and the y, then I'll get the double integral of, sorry, and sub x dA.
338
00:27:28.920 --> 00:27:43.920
And so now, if I have proved these two formulas separately, then if you sum them together, we'll get the correct statement.
339
00:27:43.920 --> 00:27:51.920
Well, let me not write it. We get green sphere.
340
00:27:51.920 --> 00:28:00.920
So we've simplified outtask a little bit, we'll just be trying to prove the case where there's only an x component.
341
00:28:00.920 --> 00:28:03.920
So let's do it.
342
00:28:03.920 --> 00:28:10.920
Well, we have another problem, which is the region that we're looking at. The curve that we're looking at might be very complicated.
343
00:28:10.920 --> 00:28:26.920
If I give you, let's say I give you, I don't know, a curve that does something like this, well, it will be kind of tricky to set up a double integral of other region inside.
344
00:28:26.920 --> 00:28:35.920
So maybe we first want to look at curves that are simpler, that will actually allow us to set up the double integral easily.
345
00:28:35.920 --> 00:28:42.920
So the second observation, so that was the first observation, sorry.
346
00:28:42.920 --> 00:29:00.920
The second observation is that we can decompose R into simpler regions.
347
00:29:00.920 --> 00:29:07.920
So what do I mean by that?
348
00:29:07.920 --> 00:29:16.920
Well, let's say that I have, you know, a region, and I'm going to cut it into two. So I will have R1 and R2.
349
00:29:16.920 --> 00:29:21.920
And then, of course, I need to have the curves that go around them.
350
00:29:21.920 --> 00:29:35.920
So I had my initial curve C was going around everybody. And they have curves C1 that goes around R1.
351
00:29:35.920 --> 00:29:41.920
And C2 goes around R2.
352
00:29:41.920 --> 00:29:57.920
So what I would like to say is if we can prove that the statement is true, so let's see.
353
00:29:57.920 --> 00:30:21.920
So for C1 and also for C2, then I claim we can prove the statement for C.
354
00:30:21.920 --> 00:30:28.920
So how do we do that? Well, we just add these two equalities together. Why does that work?
355
00:30:28.920 --> 00:30:35.920
There's something fishy going on because C1 and C2 have this piece here in the middle, that's not very in C.
356
00:30:35.920 --> 00:30:41.920
So if you add the line integral along C1 and C2, you get, you know, these unwanted pieces.
357
00:30:41.920 --> 00:30:46.920
But the good news is actually you go twice for that edge in the middle.
358
00:30:46.920 --> 00:31:00.920
So this appears once in C1 going up and once in C2 going down. So in fact, when you will do the work, you know, when you will sum the work, you will add these two guys together, they will cancel.
359
00:31:00.920 --> 00:31:18.920
So the line integral along C will be, then, it will be the sum of the line integral on C1 and C2.
360
00:31:18.920 --> 00:31:36.920
And that will equal, therefore, the double integral of R1 plus the double integral of R2, which is the double integral of R of negative MY.
361
00:31:36.920 --> 00:31:54.920
The reason for this equality here is because we go twice for the inner part, what do we want to say?
362
00:31:54.920 --> 00:32:19.920
For the boundary, well, we go twice, sorry, along the boundary between R1 and R2, with opposite orientations.
363
00:32:19.920 --> 00:32:31.920
So the extra things cancel out. Okay, so that means I just need to look at, you know, smaller pieces if that makes my life easier.
364
00:32:31.920 --> 00:32:40.920
So now what will make my life easy? Well, let's say that I have, you know, a curve like that.
365
00:32:40.920 --> 00:32:47.920
Well, I guess I should have really draw a pumpkin or something like that, because it would be more seasonal, but, okay.
366
00:32:47.920 --> 00:33:06.920
Well, I don't really know how to draw pumpkins. So, okay, so I'm going to, you know, what I will do is I will cut this into smaller regions for which I have well-defined lower and upper boundaries so that I will be able to set up a double integral, the Y, the X, easily.
367
00:33:06.920 --> 00:33:23.920
So, originally like this, I will actually cut it here, here, and here, into five smaller pieces so that each small piece will let me set up the double integral, the Y, the X.
368
00:33:23.920 --> 00:33:40.920
Okay, so we'll cut R into what I will call vertically simple regions.
369
00:33:40.920 --> 00:34:00.920
So what's a vertically simple region? That's a region that's given by looking at X between A and B for some values of A and B, and for each value of X, Y is between some function of X and some other function of X.
370
00:34:00.920 --> 00:34:14.920
So, for example, this guy is vertically simple, C X runs from this value of X to that value of X, and for each X, Y goes between this value to that value.
371
00:34:14.920 --> 00:34:38.920
And same with each of these.
372
00:34:38.920 --> 00:35:02.920
Okay, so now we are down to the main step that we have to do, which is to prove this identity.
373
00:35:02.920 --> 00:35:27.920
If C is, sorry, if R is vertically simple, and C is the boundary of R going counterclockwise.
374
00:35:27.920 --> 00:35:39.920
Okay, so let's look at how we would do it.
375
00:35:39.920 --> 00:35:53.920
So we say vertically simple region looks like X goes between A and B, and Y goes between two values that are given by functions of X.
376
00:35:53.920 --> 00:36:10.920
Okay, so this is Y equals F2FX. This is Y equals F1FX. This is A. This is B. Our region is this thinking here.
377
00:36:10.920 --> 00:36:25.920
So let's compute both sides. And when I say compute, of course, we will not get numbers, because we don't know what M is, we don't know what F1 and F2 are.
378
00:36:25.920 --> 00:36:40.920
So let's start with the line integral. How do I compute the line integral along this curve that goes all around here? Well, it looks like there will be four pieces.
379
00:36:40.920 --> 00:37:05.920
So we actually have four things to compute, C1, C2, C3, and C4. Okay. Well, let's start with C1. So if we integrate on C1 MDX, how do we do that?
380
00:37:05.920 --> 00:37:16.920
Well, we know that on C1, Y is given by a function of X. So we can just get rid of Y and express everything in terms of X. Okay.
381
00:37:16.920 --> 00:37:31.920
So we know Y is F1FX, and X goes from A to B. So that will be the integral from A to B of, well, I have to take the function M.
382
00:37:31.920 --> 00:37:50.920
So M depends normally on X and Y. Maybe I should put X and Y here. And then I will plug Y equals F1FX, DX. And then I have a single variable integral, and that's what I have to compute.
383
00:37:50.920 --> 00:38:09.920
Of course, I cannot compute it here because I don't know what this is. So it has to stay this way. Okay. Next one. The integral along C2. Well, let's think for a second.
384
00:38:09.920 --> 00:38:24.920
So if we can see two X equals B, it's constant. So DX is zero, and you know, you would integrate actually over a variable Y. But well, we don't have a Y component.
385
00:38:24.920 --> 00:38:33.920
See, this is the reason why we made the first observation. We got rid of the other term because it simplifies our life here.
386
00:38:33.920 --> 00:38:51.920
We just get zero. Okay. Just looking quickly ahead. There's another one that will be zero as well. Right. Which one? Yeah. C4. So gives me zero.
387
00:38:51.920 --> 00:39:20.920
What about C3? Well, C3 will look a lot like C1. Okay. So we're going to use the same kind of thing that we did with C1.
388
00:39:20.920 --> 00:39:40.920
Okay. So, you know, C3. Well, let's see. So on C3, Y is a function of X again. And so we're using as our variable X. But now X goes down from B to A.
389
00:39:40.920 --> 00:40:00.920
So it will be the integral from B to A of M of X and F2 of X DX. Now, if you prefer, that's negative integral from A to B of M of X and F2 of X DX.
390
00:40:00.920 --> 00:40:29.920
Okay. So now, if I sum all these pieces together, I get that the length integral along the closed curve is the integral from A to B of M of X, F1 of X DX minus the integral from A to B of M of X, F2 of X DX.
391
00:40:29.920 --> 00:40:46.920
So that's the left hand side. Next, I should try to look at my double integral and see if I can make it equal to that.
392
00:40:46.920 --> 00:41:05.920
So let's look at the other guy. Double integral over R of negative MYD. Well, first I'll take the minus sign out. It will make my life a little bit easier.
393
00:41:05.920 --> 00:41:21.920
And second, so I say that I will try to set this up in the way that's the most efficient. And my choice of this kind of region means that it's easier to set up the Y DX.
394
00:41:21.920 --> 00:41:44.920
So if I set it up the Y DX, then I know for a given value of X, Y goes from F1 of X to F2 of X. And X goes from A to B.
395
00:41:44.920 --> 00:42:01.920
Is that okay with everyone? Okay, so now if I compute the inner integral, well, what do I get if I integrate partial and partial Y with respect to Y? I'll get M back.
396
00:42:01.920 --> 00:42:22.920
Okay, so I will get M at the point X F2 of X minus M at the point X F1 of X.
397
00:42:22.920 --> 00:42:43.920
And so this becomes the integral from A to B. I guess that was a minus sign of M of X F2 of X minus M of X F1 of X DX.
398
00:42:43.920 --> 00:43:01.920
And so that's the same as up there. So that's the end of the proof because we've checked that for this special case when we have only an X component and a vertically simple region, things work.
399
00:43:01.920 --> 00:43:16.920
Then we can remove the assumption that things are vertically simple using this second observation. We can just glue various pieces together and prove it for any region. Then we do the same thing with the Y component.
400
00:43:16.920 --> 00:43:28.920
That's the first observation when we add the things together we get grain sphere remnants full generality.
401
00:43:28.920 --> 00:43:38.920
Okay, so let me finish with a cool example.
402
00:43:38.920 --> 00:43:49.920
So there's one place in real life where grain sphere are used to be extremely useful.
403
00:43:49.920 --> 00:43:58.920
I say used to because computers have actually made that obsolete.
404
00:43:58.920 --> 00:44:09.920
So let me show you the picture of this device. This is called a plan meter. And what it does.
405
00:44:09.920 --> 00:44:26.920
And what it does is it measures areas. So it used to be that when you, you know, when you were an experimental scientist, you would run your chemical or biological experiment or whatever.
406
00:44:26.920 --> 00:44:36.920
And you would have all of these recording devices. And the data would go well not on to a floppy disk or a hard disk or whatever because you didn't have those at the time. You didn't have a computer on your lab.
407
00:44:36.920 --> 00:44:45.920
They would go on to a piece of graph paper. You know, so you would have your graph paper and you would have, you know, some curve on it.
408
00:44:45.920 --> 00:44:53.920
And very often you wanted to know what's the total amount of product that you have synthesized or whatever question might be.
409
00:44:53.920 --> 00:45:01.920
It might relate with, you know, the area under your curve. So you'd say, oh, that's easy. That's just integrate. Well, except you don't have a function.
410
00:45:01.920 --> 00:45:10.920
You can put that into your calculator. The next thing you could do is, well, let's count the little squares. But you know, if you've seen a piece of graph paper, that's kind of time consuming.
411
00:45:10.920 --> 00:45:19.920
So people invented these things called plan meters. Well, so it's something where there's a really heavy thing based at one corner.
412
00:45:19.920 --> 00:45:30.920
And there's a lot of dials and gauges and everything. And there's one arm that you move. And so what you do is you take the moving arm and you just slide it all around your curve.
413
00:45:30.920 --> 00:45:46.920
And you look at one of the dials and suddenly what comes, you know, as you go around, it gives you complete garbage. But when you come back here, that dials, suddenly gives you the value of the area of this region.
414
00:45:46.920 --> 00:45:55.920
So how does it work? You know, this gadget never knows about the region inside because you don't take it all over here. You only take it along the curve.
415
00:45:55.920 --> 00:46:05.920
So what it does actually is it computes a line integral. Okay. So, you know, it has the system of wheels and everything that compute for you.
416
00:46:05.920 --> 00:46:14.920
The line integral along C of well, depends on the model, but some of them compute the line integral of X, D, Y, some of them compute different line integrals.
417
00:46:14.920 --> 00:46:29.920
Okay. And now if you apply green sphere, you see that when you have a counterclockwise curve, this will be just the area of the region inside.
418
00:46:29.920 --> 00:46:36.920
And so that's, you know, that's how it works. I mean, of course, now you use a computer and it does the sums. Yes?
419
00:46:36.920 --> 00:46:44.920
That costs several thousand dollars, possibly more. So that's why I didn't bring one.