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Well, let's see.
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So before we actually start reviewing for the test,
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I'm going to, I still have to tell you a few small things,
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because I promised to say a few words about what's the difference,
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you know, more precisely, what's the difference between
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curbing zero and a field being a gradient field.
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And why we had this assumption that our vector field
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had to be defined everywhere for a field with curled zero
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to actually be conservative, or test for gradient fields
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to be valid.
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So more about the validity of the gradient sphere
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and things like that.
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So we've seen the statement of the gradient sphere
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in two forms.
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Both of them have to do with comparing a line integral along
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a closed curve to a double integral of a region inside
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and closed by the curve.
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So one of them says the line integral for the work done
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by a vector field along a closed curve counterclockwise
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is equal to the double integral of a curl of a field
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over the enclosed region.
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And the other one says the total flux out of a region.
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So the flux through the curve is equal to the double
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integral of divergence of a field in the region.
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So in both cases, we need the vector field to be defined.
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Not only the left-hand side makes sense if a vector field
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is just defined on the curve, because it's just a line
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integral on C. We don't care what happens inside.
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But for the right-hand side to make sense
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and therefore for the equality to make sense,
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we need the vector field to be defined everywhere
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inside the region.
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So I said if there's a point somewhere in here
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where my vector field is not defined,
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then it doesn't work.
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And actually, we've seen that example.
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So this only works if f and its derivatives
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are defined everywhere in the region R. Otherwise, we're
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in trouble.
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So we've seen, for example, that if I
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gave you the vector field minus yi plus xj
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over x squared plus y squared, so that's
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the same vector field that was on that problem set
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a couple of weeks ago, then f is not defined at the origin,
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but it's defined everywhere else.
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And wherever it's defined, its curl is 0.
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So everywhere else.
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And so if we have a closed curve in the plane,
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well, there's two situations.
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One is if it does not enclose the origin, then yes,
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we can apply the sphere and it will tell us that it's equal
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to the double integral in here of curl f, dA, which
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will be 0, because this is 0.
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However, if I have a curve that encloses the origin,
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let's say, no, like this, for example, then, well,
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I cannot use the same method, because the vector field
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and its curl are not defined at the origin.
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And in fact, you know that, you know,
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ignoring the problem and saying, well, curl is still 0
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everywhere, would give you the wrong answer,
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because we've seen an example.
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We've seen that along the unit circle,
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the total work is 2 pi, not 0.
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So we can't use green.
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Or rather, we can't use it directly.
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So there's an extended version of green sphere info
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that tells you the following thing.
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Well, it tells me that even though I can't do things for just
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this region enclosed by C prime, I can still do things
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for the region in between two different curves.
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So let me show you what I have in mind.
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So let's see if I have my curve C prime,
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where's my yellow choke?
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Oh, yeah.
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So I have this curve C prime.
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I can't apply green sphere into the region inside it.
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But let's cut out a smaller thing.
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So let me call that curve.
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So that one I'm going to make going clockwise.
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You'll see why.
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Then I can say, well, sorry, let me change my mind.
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Structure is not very well prepared.
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That's because my vitalism is drank.
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OK.
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So let's say we have C prime and C double prime,
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both going clockwise.
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Then I claim that green sphere m still applies
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and tells me that the line integral along C prime
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minus the line integral along C double prime
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is equal to the double integral of our region in between.
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So here now it's this region with a hole of the curve.
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And, well, in our case, that will turn out to be 0
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because curve is 0.
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So this doesn't tell us what each of these two line
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integrals is, but actually tells us that they are equal to each
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of them.
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And so by computing one, you can see, actually,
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that for this vector field, if you take any curve that goes
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counterclockwise around the origin,
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you will get 2 pi, no matter what the curve is.
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So how do you get to this?
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And why is this not like conceptually a new theorem?
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Well, just think of the following thing.
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I'm not going to do it on top of that
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because it's going to be messy if I draw too many things.
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But so here I have my C double prime.
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Here I have C prime.
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Let me actually make a slit that will connect them
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to each other like this.
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So now if I take, see, I can form a single closed curve
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that will enclose all of this region.
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Well, with an infinitive thin slit here, counterclockwise.
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And so if I go counterclockwise around this region,
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basically I go counterclockwise along the outer curve,
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then I go along the slit.
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Then I go clockwise along the inside curve,
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then back along the slit, and then I'm done.
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So if I take the line integral along this big curve,
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consisting of all these pieces, now I
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can apply green sphere onto that.
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Because it is a usual counterclockwise curve
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that goes around the region where my field is well defined.
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I've eliminated the origin from the picture.
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And so the total line integral for this thing
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is equal to the integral along C prime, guess the outer one.
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Then I need to also have what I do along the inner side.
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And the inner side is going to be C double prime,
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but going backwards.
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Because now I'm going clockwise on C prime,
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so that I'm going counterclockwise around the shaded
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region.
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Plus, well, of course, there will be contributions
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from the line integral along this white segment.
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But I do it twice once each way, so the cancel out.
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So the white segments cancel out.
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You probably shouldn't in your notes
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find done white segments because probably they're not white
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on your paper.
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But hopefully you get a meaning of what I'm trying to say.
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So basically, that tells you you can still play tricks
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with green sphere and when the region has holes in it.
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Just you have to be careful and somehow subtract some other curve
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so that, together, things will work out.
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There's a similar thing with the divergence theorem,
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of course, with flux and the ball integral of div f.
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You can apply exactly the same argument.
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So basically, you can apply a green sphere
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and for a region that has several boundary curves.
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You just have to be careful that the outer boundary must go
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counterclockwise.
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If you inner boundary, either goes clockwise
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or you put a minus sign.
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And a last cultural note.
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So a definition.
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We say that a region in the plane,
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so I should say a connected region in the plane.
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So that means it consists of a single piece.
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So connected, there's a single piece.
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This two guys together are not connected.
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But if I join them, then this is a connected region.
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We say it's simply connected.
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If any closed curve in it, so I need to get a name to my region,
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let's say r, any closed curve in r
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bounds of any closed curve in r, is also contained in r.
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So concretely, what does that mean?
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That means the region does not have any holes inside it.
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So maybe I should draw two pictures
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to explain what that mean.
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So this guy here is simply connected.
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This guy here is not simply connected.
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Because if I take this curve, that's a curve inside my region.
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But the piece that it bounds is not actually
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entirely contained in my region.
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And so why is that relevant?
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Well, if you know that your vector field
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is defined everywhere in a simply connected region,
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then you don't have to worry about this question of can
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I apply a QN sphere?
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I'm to the inside.
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You know it's automatically OK.
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Because if you have a closed curve, then the vector field
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is defined on the curve.
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It will also be defined inside.
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So if the domain of definition of a vector field,
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it's defined and differentiable.
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So it is simply connected, then we can always apply
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a QN sphere.
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And of course, provided that we do it on a curve where the
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vector field is defined.
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Otherwise, your line integral doesn't make sense.
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So there's nothing to compute.
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But if you have, you know, so again, the argument would be,
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well, if a vector field is defined on the curve, it's also
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defined inside.
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So we're happy.
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And see, so the problem with a vector field here is
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precisely, but its domain of definition is not simply
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connected because there's a hole, namely the origin.
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So for this guy, domain of definition, which is plain minus
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the origin, with the origin removed, is not simply connected.
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And so that's why you have this line integral that makes
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perfect sense, but you can't apply a QN sphere on to it.
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So now, what does that mean in particular?
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Well, we've seen this criterion that if a curve of a
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vector field is 0, and it's defined in the entire plane, then
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the vector field is conservative and it's a gradient field.
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And the argument to prove that is precisely to use
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green sphere.
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So in fact, the actual optimal statement you can make is if
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a vector field is defined in a simply connected region, and
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it's currently 0, then it's a gradient field.
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So let me just write that down for the correct statement.
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The previous one we've seen is also correct, but this one
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is somehow better and closer to what exactly is needed.
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If curl f is 0, and the domain of definition, where f is defined,
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is simply connected, then f is conservative, and that means
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that it's also a gradient field.
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That's the same thing.
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OK?
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Any questions on this?
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No?
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OK, some good news.
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What I've just said here won't come up on the test on
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Thursday.
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OK.
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Still, it's stuff that you should be aware of generally speaking,
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because it will be useful, say, on the next week's problem
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set.
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And maybe on the final, there won't be any really, really
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complicated things, probably.
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But there might be some, you might need to be at least
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vaguely aware of this issue of things being simply connected.
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And by the way, this is also somehow the starting point
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of topology, which is the branch of math that studies
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the shapes of regions.
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So in particular, you can try to distinguish domains
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in the planes by looking at whether they are simply
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connected or not.
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And what kinds of features they have in terms of how you can
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join points with kinds of curves existing them.
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And since that's the branch of math in which I work,
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I thought I should tell you a bit about it.
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OK.
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So now back to reviewing for the exam.
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So we're going to basically list topics.
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And if time permits, I will say a few things about problems
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from practice exam 3D.
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I'm hoping that you have it on your neighbor's
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head or you can somehow get it.
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Anyway, given time, I'm not sure how much I will say
258
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about the problems in the verb themselves.
259
00:19:06.400 --> 00:19:10.560
So the main thing to know about this exam
260
00:19:10.560 --> 00:19:14.040
is how to set up and evaluate double integrals
261
00:19:14.040 --> 00:19:15.600
and line integrals.
262
00:19:15.600 --> 00:19:17.720
If you know how to do these two things,
263
00:19:17.720 --> 00:19:19.720
then you're in much better shape than if you don't.
264
00:19:19.720 --> 00:19:40.640
So the first thing we've seen,
265
00:19:40.640 --> 00:19:50.360
there's two main objects.
266
00:19:50.360 --> 00:19:54.960
And it's kind of important to not confuse them with each other.
267
00:19:54.960 --> 00:19:58.800
There's double integrals of our regions of some quantity,
268
00:19:58.800 --> 00:20:00.600
dA.
269
00:20:00.600 --> 00:20:05.600
The other one is the line integral along a curve
270
00:20:05.600 --> 00:20:12.600
of a vector field, f.dr or f.nDS.
271
00:20:12.600 --> 00:20:17.600
Depending on whether it's work or flux that we're trying to do.
272
00:20:17.600 --> 00:20:21.600
And so we should know how to set up these things
273
00:20:21.600 --> 00:20:23.600
and how to evaluate them.
274
00:20:23.600 --> 00:20:26.600
And roughly speaking, in this one,
275
00:20:26.600 --> 00:20:29.600
you start by drawing a picture of a region
276
00:20:29.600 --> 00:20:32.600
then deciding which way you will integrate it.
277
00:20:32.600 --> 00:20:36.600
Could be the xdydydx or the rd theta.
278
00:20:36.600 --> 00:20:39.600
And then you will set up the bounds carefully
279
00:20:39.600 --> 00:20:43.600
by slicing it and studying how the bounds for the inner variable
280
00:20:43.600 --> 00:20:46.600
depend on the outer variable.
281
00:20:46.600 --> 00:20:54.600
So the first topic will be setting up double integrals.
282
00:20:54.600 --> 00:21:00.600
And so remember, maybe I should make this more explicit.
283
00:21:00.600 --> 00:21:05.600
We want to look for a picture of r
284
00:21:05.600 --> 00:21:10.600
and take slices in the chosen way
285
00:21:10.600 --> 00:21:17.600
so that we get an iterated integral.
286
00:21:17.600 --> 00:21:24.600
Okay, so let's do just a quick example.
287
00:21:24.600 --> 00:21:31.600
So if I look at problem 1 on xm3b,
288
00:21:31.600 --> 00:21:41.600
it says to look at the line integral from 0 to 1,
289
00:21:41.600 --> 00:21:44.600
the line integral from x to 2x,
290
00:21:44.600 --> 00:21:47.600
of possibly something, but dydydx.
291
00:21:47.600 --> 00:21:50.600
And it says, let's look at how we would set this up
292
00:21:50.600 --> 00:21:57.600
the other way around by exchanging x and y.
293
00:21:57.600 --> 00:22:04.600
So we should get to something that will be the same integral
294
00:22:04.600 --> 00:22:08.600
dxdy. I mean, if you have a function of x and y,
295
00:22:08.600 --> 00:22:11.600
then it will be the same function, but of course the bounds change.
296
00:22:11.600 --> 00:22:14.600
So how do we exchange the order of integration?
297
00:22:14.600 --> 00:22:19.600
Well, the only way to do it consistently is to draw a picture.
298
00:22:19.600 --> 00:22:22.600
So let's see, what does this mean?
299
00:22:22.600 --> 00:22:27.600
Here it means we integrate from y equals x to y equals 2x,
300
00:22:27.600 --> 00:22:29.600
x between 0 and 1.
301
00:22:29.600 --> 00:22:31.600
So we should draw a picture.
302
00:22:31.600 --> 00:22:34.600
The lower bound for y is y equals x.
303
00:22:34.600 --> 00:22:39.600
So let's draw y equals x.
304
00:22:39.600 --> 00:22:41.600
That seems to be here.
305
00:22:41.600 --> 00:22:44.600
And we'll go up to y equals 2x,
306
00:22:44.600 --> 00:22:48.600
which is a line also, but with bigger slope.
307
00:22:48.600 --> 00:22:53.600
And then, so for each value of x,
308
00:22:53.600 --> 00:22:58.600
my region will go from x to 2x.
309
00:22:58.600 --> 00:23:03.600
Well, and I do this for all values of x that go from 0 to x equals 1.
310
00:23:03.600 --> 00:23:06.600
So I stop at x equals 1, which is here.
311
00:23:06.600 --> 00:23:13.600
And then my region is something like this.
312
00:23:13.600 --> 00:23:19.600
So this point here, in case you're wondering,
313
00:23:19.600 --> 00:23:22.600
well, when x equals 1, y is 1,
314
00:23:22.600 --> 00:23:26.600
and that point here is 1, 2.
315
00:23:26.600 --> 00:23:29.600
Any questions about that, so far?
316
00:23:29.600 --> 00:23:31.600
So that's somehow that's the first skill.
317
00:23:31.600 --> 00:23:34.600
When you see an integral, how to figure out what it means,
318
00:23:34.600 --> 00:23:36.600
how to draw the region.
319
00:23:36.600 --> 00:23:38.600
And then there's the converse skill,
320
00:23:38.600 --> 00:23:41.600
which is given the region, how to set up the integral for it.
321
00:23:41.600 --> 00:23:45.600
So if we want to set it up instead dx dy,
322
00:23:45.600 --> 00:23:51.600
then that means we are going to actually look at the converse question,
323
00:23:51.600 --> 00:23:54.600
which is for a given value of y,
324
00:23:54.600 --> 00:23:56.600
what is the range of values of x?
325
00:23:56.600 --> 00:23:59.600
So if we fix y,
326
00:23:59.600 --> 00:24:02.600
well, where do we enter the region,
327
00:24:02.600 --> 00:24:04.600
and where do we leave it?
328
00:24:04.600 --> 00:24:06.600
So we seem to enter on this side,
329
00:24:06.600 --> 00:24:09.600
and we seem to leave on that side.
330
00:24:09.600 --> 00:24:12.600
So that seems to be true for the first few values of y that I choose.
331
00:24:12.600 --> 00:24:15.600
But hey, if I take a larger value of y,
332
00:24:15.600 --> 00:24:17.600
then I will enter on this side,
333
00:24:17.600 --> 00:24:20.600
and I will leave on this vertical side,
334
00:24:20.600 --> 00:24:21.600
not on that one.
335
00:24:21.600 --> 00:24:23.600
So I seem to have two different things going on.
336
00:24:23.600 --> 00:24:25.600
The place y, enter my region,
337
00:24:25.600 --> 00:24:28.600
is always y equals 2x,
338
00:24:28.600 --> 00:24:39.600
which is the same as x equals y over 2.
339
00:24:39.600 --> 00:24:46.600
So x seems to always start at y over 2.
340
00:24:46.600 --> 00:24:50.600
But where I leave could be either x equals y,
341
00:24:50.600 --> 00:24:55.600
or here, x equals 1.
342
00:24:55.600 --> 00:24:57.600
And that depends on the value of y.
343
00:24:57.600 --> 00:25:00.600
In fact, I have to break this into two different integrals.
344
00:25:00.600 --> 00:25:04.600
I have to treat separately the case where y is between 0 and 1,
345
00:25:04.600 --> 00:25:07.600
and between 1 and 2.
346
00:25:07.600 --> 00:25:15.600
So what I do in that case is I just make two integrals.
347
00:25:15.600 --> 00:25:18.600
So I say both of them start at y over 2.
348
00:25:18.600 --> 00:25:20.600
But in the first case,
349
00:25:20.600 --> 00:25:23.600
we'll stop at x equals y.
350
00:25:23.600 --> 00:25:27.600
And now what are the values of y for each case?
351
00:25:27.600 --> 00:25:30.600
Where the first case is 1y is between 0 and 1.
352
00:25:30.600 --> 00:25:33.600
The second case is 1y is between 1 and 2,
353
00:25:33.600 --> 00:25:37.600
which I guess this picture now is completely unredible.
354
00:25:37.600 --> 00:25:41.600
But hopefully you've been following what's been going on.
355
00:25:41.600 --> 00:25:44.600
Or as you can see it in the solutions to the problem.
356
00:25:44.600 --> 00:25:47.600
And so that's our final answer.
357
00:25:47.600 --> 00:25:48.600
Okay.
358
00:25:48.600 --> 00:25:53.600
Any questions about how to set up the component integrals
359
00:25:53.600 --> 00:25:57.600
in xy coordinates?
360
00:25:57.600 --> 00:25:59.600
No.
361
00:25:59.600 --> 00:26:02.600
Okay. Who feels comfortable with this kind of problem?
362
00:26:02.600 --> 00:26:04.600
Okay. Good.
363
00:26:04.600 --> 00:26:08.600
I'm happy to see the vast majority.
364
00:26:08.600 --> 00:26:13.600
So the bad news is we have to be able to do it not only in xy coordinates,
365
00:26:13.600 --> 00:26:17.600
but also in the case of xy coordinates.
366
00:26:17.600 --> 00:26:25.600
But also in polar coordinates.
367
00:26:25.600 --> 00:26:27.600
So when you go to polar coordinates,
368
00:26:27.600 --> 00:26:31.600
basically all you have to remember on the side of the integrand
369
00:26:31.600 --> 00:26:36.600
is that x becomes our cosine theta.
370
00:26:36.600 --> 00:26:39.600
y becomes our sine theta.
371
00:26:39.600 --> 00:26:45.600
And the xdy becomes R dr d theta.
372
00:26:45.600 --> 00:26:48.600
In terms of how you slice through your region,
373
00:26:48.600 --> 00:26:52.600
well, you will be integrating first of our R.
374
00:26:52.600 --> 00:26:57.600
So that means what you're doing is you're fixing a value of theta.
375
00:26:57.600 --> 00:27:00.600
And for that value of theta, you ask yourself,
376
00:27:00.600 --> 00:27:02.600
for what range of values of R,
377
00:27:02.600 --> 00:27:05.600
am I going to be inside my region?
378
00:27:05.600 --> 00:27:07.600
So if my region looks like this,
379
00:27:07.600 --> 00:27:09.600
then for this value of theta,
380
00:27:09.600 --> 00:27:12.600
R would go from zero to whatever this distance is.
381
00:27:12.600 --> 00:27:16.600
And of course, I have to find how this distance depends on theta.
382
00:27:16.600 --> 00:27:19.600
And then I will find the extreme values of theta.
383
00:27:19.600 --> 00:27:21.600
Now, of course, if the region is really looking like this,
384
00:27:21.600 --> 00:27:23.600
then you're not going to do it in polar coordinates.
385
00:27:23.600 --> 00:27:27.600
But if it's like a circle or a half circle or things like that,
386
00:27:27.600 --> 00:27:30.600
then even if a problem doesn't tell you to do it in polar coordinates,
387
00:27:30.600 --> 00:27:33.600
you might want to seriously consider it.
388
00:27:33.600 --> 00:27:35.600
So that's, I mean, I'm not going to do it,
389
00:27:35.600 --> 00:27:41.600
but problem two on the practice exam is a good example of doing something in polar coordinates.
390
00:27:41.600 --> 00:27:46.600
Okay?
391
00:27:46.600 --> 00:27:52.600
So in terms of things that we do with polar with double integrals,
392
00:27:52.600 --> 00:27:57.600
there's a few formulas that I'd like you to remember about, you know,
393
00:27:57.600 --> 00:28:01.600
applications that we've seen of double integrals.
394
00:28:01.600 --> 00:28:06.600
So quantities that we can compute with double integrals
395
00:28:06.600 --> 00:28:13.600
include things like the area of the region, its mass, if it has a density,
396
00:28:13.600 --> 00:28:17.600
the average value of some function, for example,
397
00:28:17.600 --> 00:28:22.600
the average value of the x and y coordinates, which we called the center of mass,
398
00:28:22.600 --> 00:28:24.600
are moments of inertia.
399
00:28:24.600 --> 00:28:33.600
So these are just formulas to remember.
400
00:28:33.600 --> 00:28:40.600
So for example, the area of the region is the double integral of just dA.
401
00:28:40.600 --> 00:28:43.600
If it helps you, one dA if you want.
402
00:28:43.600 --> 00:28:46.600
You're integrating the function of one.
403
00:28:46.600 --> 00:28:52.600
You have to remember formulas for mass for the average value of a function,
404
00:28:52.600 --> 00:29:01.600
say f bar, in particular, x bar y bar, which is the center of mass,
405
00:29:01.600 --> 00:29:08.600
and the moment of inertia,
406
00:29:08.600 --> 00:29:17.600
also the polar moment of inertia, which is moment of inertia about the origin.
407
00:29:17.600 --> 00:29:24.600
So that's double integral of x squared plus y squared, then ct dA,
408
00:29:24.600 --> 00:29:29.600
but also moments of inertia about the x and y axis,
409
00:29:29.600 --> 00:29:34.600
which are given by just taking one of these guys.
410
00:29:34.600 --> 00:29:36.600
Don't worry about moment of inertia about an arbitrary line.
411
00:29:36.600 --> 00:29:42.600
I won't ask you for a moment of inertia about some weird line or something like that.
412
00:29:42.600 --> 00:29:45.600
But these, you should know.
413
00:29:45.600 --> 00:29:50.600
Now, what if you somehow, in the spur of a moment, you forget what's the formula for moment of inertia?
414
00:29:50.600 --> 00:29:54.600
I mean, it prefer if you know, but if you have a complete blank in your memory,
415
00:29:54.600 --> 00:29:58.600
there will still be partial credit for setting up the bounds and everything else.
416
00:29:58.600 --> 00:30:03.600
So, you know, the general rule, with the exam, will be, if you're stuck in a calculation
417
00:30:03.600 --> 00:30:07.600
or if you're missing a little piece of a puzzle, try to do as much as you can,
418
00:30:07.600 --> 00:30:10.600
in particular, try to at least set up the bounds of the integral,
419
00:30:10.600 --> 00:30:17.600
there will be partial credit for that always.
420
00:30:17.600 --> 00:30:25.600
So while we're at it about ground rules, how about evaluation,
421
00:30:25.600 --> 00:30:29.600
integrals?
422
00:30:29.600 --> 00:30:32.600
So once you've set it up, you know, you have to sometimes compute it.
423
00:30:32.600 --> 00:30:36.600
Well, first of all, check, you know, just in case the problem says set up but do not evaluate.
424
00:30:36.600 --> 00:30:40.600
Then don't waste your time evaluating it.
425
00:30:40.600 --> 00:30:48.600
If the problem says to compute it, then, you know, you have to compute it.
426
00:30:48.600 --> 00:30:51.600
So what kinds of integration techniques do you need to know?
427
00:30:51.600 --> 00:30:57.600
So you need to know, you must know, well, how to integrate the usual functions,
428
00:30:57.600 --> 00:31:05.600
you know, like one of our x, x to the n, exponential sine, cosine, things like that.
429
00:31:05.600 --> 00:31:15.600
So the usual integrals, you must know what I will call easy trigonometry.
430
00:31:15.600 --> 00:31:17.600
I don't want to give you a complete list.
431
00:31:17.600 --> 00:31:22.600
And the more you ask me about which ones are on the list, the more I will add to the list.
432
00:31:22.600 --> 00:31:26.600
But you know, those that you know that you should know, you should know.
433
00:31:26.600 --> 00:31:29.600
Those that you think you shouldn't know, you don't have to know.
434
00:31:29.600 --> 00:31:33.600
Because I will say, well, of what I will say soon.
435
00:31:33.600 --> 00:31:38.600
You should know also substitution.
436
00:31:38.600 --> 00:31:43.600
You know, how to set, you equal something and then see, oh, this becomes u times du.
437
00:31:43.600 --> 00:31:48.600
And so, you know, substitution method, yes?
438
00:31:48.600 --> 00:31:50.600
Well, do I mean by easy trigonometrics?
439
00:31:50.600 --> 00:31:52.600
Well, certainly you should know how to integrate sine.
440
00:31:52.600 --> 00:31:54.600
You should know how to integrate cosine.
441
00:31:54.600 --> 00:32:00.600
You should be aware that sine square plus cosine square simplifies to 1.
442
00:32:00.600 --> 00:32:02.600
And you should be aware of general things like that.
443
00:32:02.600 --> 00:32:08.600
I would like you to know maybe the double angle sine 2x and cosine 2x, know what these are.
444
00:32:08.600 --> 00:32:11.600
And, you know, the kinds of easy things you can do with that.
445
00:32:11.600 --> 00:32:18.600
Also, things that, you know, involve substitution setting like u equals sine t or u equals cosine t.
446
00:32:18.600 --> 00:32:23.600
I mean, let me instead, you know, give an example of harping that you don't need to know.
447
00:32:23.600 --> 00:32:25.600
And then I will answer.
448
00:32:25.600 --> 00:32:32.600
Okay, so not needed on Thursday.
449
00:32:32.600 --> 00:32:34.600
It doesn't mean that I don't want you to know them.
450
00:32:34.600 --> 00:32:39.600
I would love you to know, you know, every single integral formula.
451
00:32:39.600 --> 00:32:41.600
But shouldn't be your top priority.
452
00:32:41.600 --> 00:32:46.600
So you don't need to know things like hard trigonometric ones.
453
00:32:46.600 --> 00:32:50.600
So let me give you an example.
454
00:32:50.600 --> 00:32:55.600
Okay, so if I ask you to, you know, do this one, then actually I will give you, you know,
455
00:32:55.600 --> 00:32:59.600
maybe like the, I will reprint the formula from the notes or something like that.
456
00:32:59.600 --> 00:33:02.600
Okay, so that one, you don't need to know.
457
00:33:02.600 --> 00:33:07.600
I would love if you happen to know it, but if you need it, it will be given to you.
458
00:33:07.600 --> 00:33:13.600
So these kinds of things, that you cannot compute by any easy method.
459
00:33:13.600 --> 00:33:18.600
And integration by parts.
460
00:33:18.600 --> 00:33:25.600
I believe that I've successfully test solved all the problems without doing any single integration by parts.
461
00:33:25.600 --> 00:33:30.600
Again, in general, it's something that I would like you to know, but, you know,
462
00:33:30.600 --> 00:33:35.600
if it shouldn't be your top priority for this week.
463
00:33:35.600 --> 00:33:40.600
Okay, so you had a question or?
464
00:33:40.600 --> 00:33:42.600
Inverse trigonometric functions.
465
00:33:42.600 --> 00:33:45.600
Let's say the most easy ones.
466
00:33:45.600 --> 00:33:51.600
I would like you to know the easiest inverse trig functions, but not much.
467
00:33:51.600 --> 00:33:52.600
Okay?
468
00:33:52.600 --> 00:34:03.600
So be aware that these functions exist, but it's not a top priority.
469
00:34:03.600 --> 00:34:07.600
I should say, I mean, the more I tell you, I don't need you to know the more, you know,
470
00:34:07.600 --> 00:34:11.600
your physics and other teachers might complain, but, oh, these guys don't know how to integrate.
471
00:34:11.600 --> 00:34:14.600
So try not to forget everything, but, yes?
472
00:34:14.600 --> 00:34:19.600
For substitution, we have to do the Jacobian with that.
473
00:34:19.600 --> 00:34:22.600
No, no, here I just mean like for evaluating just a single variable integral.
474
00:34:22.600 --> 00:34:27.600
I will get to change variables in Jacobian soon, but I'm thinking of this as a different topic.
475
00:34:27.600 --> 00:34:31.600
You know, what I mean by this one is, if I'm asking you to integrate,
476
00:34:31.600 --> 00:34:33.600
I don't know, what's a good example?
477
00:34:33.600 --> 00:34:39.600
0 to 1 T dT of our square root of 1 plus T squared,
478
00:34:39.600 --> 00:34:44.600
then you should think of maybe substituting u equals 1 plus T squared,
479
00:34:44.600 --> 00:34:52.600
and then it becomes easier.
480
00:34:52.600 --> 00:34:59.600
Okay, so this kind of trig, that's what I have in mind here, specifically.
481
00:34:59.600 --> 00:35:04.600
And again, if you're stuck, you know, in particular, if you hit this dreaded guy
482
00:35:04.600 --> 00:35:08.600
and you don't actually have a formula giving you what it is,
483
00:35:08.600 --> 00:35:10.600
it means one of two things.
484
00:35:10.600 --> 00:35:12.600
One is something's wrong with your solution.
485
00:35:12.600 --> 00:35:14.600
The other option is something's wrong with my problem.
486
00:35:14.600 --> 00:35:17.600
So, you know, either way, you know, check quickly what you've done,
487
00:35:17.600 --> 00:35:23.600
and if you can't find a mistake, then, you know, just move ahead to the next problem.
488
00:35:23.600 --> 00:35:26.600
Yes?
489
00:35:26.600 --> 00:35:28.600
Which one? This one?
490
00:35:28.600 --> 00:35:31.600
Oh, yeah, I mean, if you can do it, you know, if you know how to do it,
491
00:35:31.600 --> 00:35:32.600
whichever thing is fair.
492
00:35:32.600 --> 00:35:36.600
I mean, generally speaking, you know, give enough evidence that you found the solution
493
00:35:36.600 --> 00:35:39.600
by yourself, not like, you know.
494
00:35:39.600 --> 00:35:43.600
You know, it didn't somehow come to you by magic.
495
00:35:43.600 --> 00:35:46.600
But, yeah, yeah, I mean, if you know how to integrate this without doing the substitution,
496
00:35:46.600 --> 00:35:49.600
that's absolutely fine by me.
497
00:35:49.600 --> 00:35:53.600
Just, you know, show enough work, general role is show enough work that we see that,
498
00:35:53.600 --> 00:35:57.600
you know, you knew what you were doing.
499
00:35:57.600 --> 00:36:17.600
Okay. Now, another thing we've seen with the bon integrals is how to do more complicated changes of variables.
500
00:36:17.600 --> 00:36:27.600
So, you know, when you want to replace x and y by some variables u and v, given by some formulas in terms of x and y.
501
00:36:27.600 --> 00:36:32.600
So, you need to remember, basically, you know, how to do them.
502
00:36:32.600 --> 00:36:36.600
So, you need to remember that the method consists of three steps.
503
00:36:36.600 --> 00:36:44.600
So, one is you have to find the Jacobian.
504
00:36:44.600 --> 00:36:51.600
And you can choose to do either this Jacobian or the inverse one, depending on what's easiest given what you're given.
505
00:36:51.600 --> 00:36:55.600
You know, you don't have to worry about solving for things that are going around,
506
00:36:55.600 --> 00:36:58.600
just compute one of these Jacobians.
507
00:36:58.600 --> 00:37:08.600
And then the rule is that du dv is absolute value of the Jacobian, dx dy.
508
00:37:08.600 --> 00:37:15.600
So, that takes care of dx dy how to convert that into du dv.
509
00:37:15.600 --> 00:37:30.600
The second thing to know is that, well, you need to, of course, substitute any x and y's in the integrand to convert them to u and v's,
510
00:37:30.600 --> 00:37:36.600
so that you have, you know, a valid integrand involving only u and v.
511
00:37:36.600 --> 00:37:51.600
And then, the last part is setting up the bounds.
512
00:37:51.600 --> 00:37:57.600
And you've seen that probably you've seen on p sets, and an example with an lecture that this can be complicated.
513
00:37:57.600 --> 00:38:01.600
But now, in real life, you know, you do this actually to simplify the integrals.
514
00:38:01.600 --> 00:38:06.600
Probably the one that will be there on Thursday, you know, if there's a problem about that on Thursday,
515
00:38:06.600 --> 00:38:12.600
it will be a situation where the bounds that you get after changing variables are reasonably easy.
516
00:38:12.600 --> 00:38:17.600
Okay, I'm not saying that they will be completely obvious necessarily, but it will be a fairly easy situation.
517
00:38:17.600 --> 00:38:24.600
So, the general method is you look at your region R, and you know, it might have, you know, various sides,
518
00:38:24.600 --> 00:38:32.600
then on each side, you ask yourself, what do I know about x and y, and how to convert that in terms of u and v?
519
00:38:32.600 --> 00:38:39.600
And maybe you will find that, you know, the equation might be just u equals 0, for example, or u equals v, or something like that.
520
00:38:39.600 --> 00:38:52.600
And then, I'd substitute you to decide what you want to do, but maybe the easiest usually is to draw a new picture in terms of u and v coordinates of what your region will look like in the new coordinates.
521
00:38:52.600 --> 00:38:59.600
And it might be that it will be actually much easier. It should be easier looking than what you started with.
522
00:38:59.600 --> 00:39:11.600
Okay, so that's the general idea. There's one change of variable problem on each of the two practice exams to give you a feeling for, you know, what's realistic.
523
00:39:11.600 --> 00:39:20.600
The one, the problem that's on practice exam 3b actually is on the hard side of things, because it's, you know, the question is kind of hidden in a way.
524
00:39:20.600 --> 00:39:27.600
So if you look at problem 6, you might find that it's not telling you very clearly what you have to do.
525
00:39:27.600 --> 00:39:34.600
That's because it's, it was meant to be the hardest problem on that test.
526
00:39:34.600 --> 00:39:41.600
But once you've reduced it to an actual change of variables problem, I expect you to be able to know how to do it.
527
00:39:41.600 --> 00:39:55.600
And on practice exam 3a, there's also, I think it's problem 5 on the other practice exam, and that one is actually pretty stand out and straightforward.
528
00:39:55.600 --> 00:39:58.600
Okay, time to move on. Sorry.
529
00:39:58.600 --> 00:40:21.600
So we've also seen about line integrals.
530
00:40:21.600 --> 00:40:29.600
Okay, so line integrals.
531
00:40:29.600 --> 00:40:33.600
So the main thing to know about them.
532
00:40:33.600 --> 00:40:37.600
So the line integral for work, which is the integral of f the dr.
533
00:40:37.600 --> 00:40:44.600
So let's say that your vector field has components m and n.
534
00:40:44.600 --> 00:40:51.600
So the line integral for work becomes incarnates integral of mdx plus ndy.
535
00:40:51.600 --> 00:41:01.600
While we've also seen line integral for flux.
536
00:41:01.600 --> 00:41:13.600
So the integral of f dot nds becomes the integral along c of just to make sure that I give it to you correctly.
537
00:41:13.600 --> 00:41:22.600
So remember that I don't want to make the mistake in front of you.
538
00:41:22.600 --> 00:41:29.600
So tds is dx, dy, and the normal vector.
539
00:41:29.600 --> 00:41:38.600
So it's, you know, tds goes along the curve and ds goes clockwise perpendicular to the curve.
540
00:41:38.600 --> 00:41:47.600
So it's going to be, well, it's going to be dy and negative dx.
541
00:41:47.600 --> 00:42:00.600
So you will be integrating negative n dx plus m dy.
542
00:42:00.600 --> 00:42:09.600
So if you are blanking and don't remember the signs, then you can just draw the spectrum and make sure that you get it right.
543
00:42:09.600 --> 00:42:17.600
So you should know a little bit about, you know, generating interpretation and how to see easily that it's going to be zero in some cases.
544
00:42:17.600 --> 00:42:22.600
But mostly you should know how to compute, you know, setup and compute these things.
545
00:42:22.600 --> 00:42:26.600
So what do we do when we are here? Well, here we have both x and y's together.
546
00:42:26.600 --> 00:42:30.600
But we want to, because it's a line integral, there should be only one variable.
547
00:42:30.600 --> 00:42:39.600
So the important thing to know is we want to reduce everything to a single parameter.
548
00:42:39.600 --> 00:42:59.600
So the evaluation method is always by reducing to a single parameter.
549
00:42:59.600 --> 00:43:09.600
So you can call, you know, maybe x and y are both functions of some variable t.
550
00:43:09.600 --> 00:43:20.600
And then express everything in terms of some integral of some quantity involving t dt.
551
00:43:20.600 --> 00:43:27.600
Could be that you will just express everything in terms of x or in terms of y or in terms of some angle or something.
552
00:43:27.600 --> 00:43:34.600
So when you are there, it's a usual one variable integral with a single variable in there.
553
00:43:34.600 --> 00:43:37.600
Okay, so that's the general method of calculation.
554
00:43:37.600 --> 00:43:46.600
But we've seen a shortcut for work when we can show that the field is the gradient of the potential.
555
00:43:46.600 --> 00:43:59.600
One thing to know is if the curl of f, which is nx minus my, happens to be zero.
556
00:43:59.600 --> 00:44:05.600
Well, and now I can say, and the domain is simply connected.
557
00:44:05.600 --> 00:44:10.600
Or if you want, if the field is defined everywhere.
558
00:44:10.600 --> 00:44:19.600
Then f is actually a gradient field.
559
00:44:19.600 --> 00:44:29.600
So that means, you know, just to make it more concrete, that means we can find a function little f called the potential, such that it's derivative respect to x is m.
560
00:44:29.600 --> 00:44:32.600
And it's derivative respect to y is n.
561
00:44:32.600 --> 00:44:39.600
We can solve these two conditions for the same function f simultaneously.
562
00:44:39.600 --> 00:44:43.600
And how do we find this function little f?
563
00:44:43.600 --> 00:44:51.600
Okay, so that's the same as saying that the field big f is the gradient of little f.
564
00:44:51.600 --> 00:44:54.600
And how do we find this function little f? Well, we've seen two methods.
565
00:44:54.600 --> 00:45:03.600
One of them involves computing a line integral from the origin to a point in the plane by going first on the x-axis, then vertically.
566
00:45:03.600 --> 00:45:10.600
The other method was to first figure out what this one tells us by integrating it with respect to x.
567
00:45:10.600 --> 00:45:18.600
And then we differentiate our answer with respect to y, and we compare with that to get the complete answer.
568
00:45:18.600 --> 00:45:27.600
Okay, so why is that relevant? Well, first of all, it's relevant in physics, but it's also relevant just to, you know, calculation of line integrals.
569
00:45:27.600 --> 00:45:46.600
We've seen the fundamental theorem of calculus for line integrals, which says if we're integrating a gradient field, and we know what the potential is, then we just have to, well, the line integral is just the change in value of the potential.
570
00:45:46.600 --> 00:45:59.600
So we take the value of the potential at the starting point, sorry, we take value of the potential at the end point minus the value at the starting point, and that will give us the line integral.
571
00:45:59.600 --> 00:46:05.600
So important, this is only for work. There's no statement like that for flux.
572
00:46:05.600 --> 00:46:10.600
Okay, so don't, you know, don't try to apply this in the problem about flux.
573
00:46:10.600 --> 00:46:19.600
And usually you'll see, you know, if you look at the practice exams, you'll see it's pretty clear that there's one problem in which you're supposed to do things this way.
574
00:46:19.600 --> 00:46:25.600
It's kind of that give away, but it's probably not too bad.
575
00:46:25.600 --> 00:46:33.600
Okay, and the other thing we've seen, so I mentioned it at the beginning, but let me mention it again, to conclude things.
576
00:46:33.600 --> 00:46:47.600
So, the graph, a green sphere one, let's us compute line, well, let's us forget, you know, evaluate find, find the value of a line integral along the closed curve by reducing it to a double integral.
577
00:46:47.600 --> 00:47:05.600
So, the graph or work says, says this, and you should remember in there, so C is a closed curve that goes counterclockwise, and R is the region inside.
578
00:47:05.600 --> 00:47:20.600
So, the way you would, you know, if you had to compute both sides separately, you would do them in extremely different ways. Right? This one is a line integral, so you use the method that I explained here, namely you express x and y in terms of a single variable.
579
00:47:20.600 --> 00:47:26.600
You know, say that you're doing a circle, I want to see a theta, I don't want to see an R, R is not a variable, you're on the circle.
580
00:47:26.600 --> 00:47:35.600
This one is a double integral, so if you're doing it say on a disk, you would have both R and theta, if you're using polar coordinates, you would have both x and y.
581
00:47:35.600 --> 00:47:41.600
Here you have two variables of integration, here you should have only one after you parameterize the curve.
582
00:47:41.600 --> 00:47:50.600
And, you know, the fact that it says curl f, I mean curl f is just an x minus m y, it's just any, you know, it's just like any function of x and y.
583
00:47:50.600 --> 00:48:06.600
Okay, the fact that we called it curl f doesn't change how you compute it, you know, you have first to compute the curl of f, say you find I don't know, x, y minus x squared, where it becomes just the usual, the whole of the usual function x, y minus x squared.
584
00:48:06.600 --> 00:48:10.600
There's nothing special to it because it's a curl.
585
00:48:10.600 --> 00:48:24.600
And the other one is the counter part for flux, so it says this, and remember this is mx plus ny.
586
00:48:24.600 --> 00:48:35.600
I mean what's important about these statements is not only remembering, you know, if you just know this formula by heart, you're still in trouble because you need to know what actually the symbol is in here mean.
587
00:48:35.600 --> 00:48:41.600
So remember, you know, what is this line integral and what's the divergence of a field.
588
00:48:41.600 --> 00:48:47.600
So, you know, just something to remember.
589
00:48:47.600 --> 00:48:52.600
And so I guess I'll let you, you know, figure out practice problems because it's time.
590
00:48:52.600 --> 00:49:07.600
But I think that's basically the list of all we've seen and well, that should be it.