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In the last video, we began[br]to explore Stokes' theorem.

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And what I want to[br]do in this video

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is to see whether it's[br]consistent with some

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of what we have already seen.

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And to do that, let's imagine--[br]so let me draw my axes.

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So that's my z-axis.

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That is my x-axis.

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And then that is my y-axis.

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And now let's imagine a[br]region in the xy plane.

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So let me draw it like this.

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So let's say this is my[br]region in the xy plane.

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I will call that[br]region R. And I also

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have a boundary of that region.

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And let's say we care[br]about the direction

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that we traverse the boundary.

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And let's say we're[br]going to traverse it

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in a counterclockwise direction.

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So we have this path that[br]goes around this region.

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We can call that c.

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So we'll call that[br]c, and we're going

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to traverse it in the[br]counterclockwise direction.

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And let's say that we also[br]have a vector field f.

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That essentially its[br]i component is just

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going to be a[br]function of x and y.

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And its j component[br]is only going

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to be a function of x and y.

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And let's say it[br]has no k component.

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So the vector field[br]on this region,

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it might look[br]something like this.

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I'm just drawing random things.

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And then if you go[br]off that region,

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if you go in the[br]z direction, it's

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just going to look the same[br]as you go higher and higher.

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So that vector,[br]it wouldn't change

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as you change your z component.

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And all of the vectors[br]would essentially

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be parallel to, or[br]if z is 0, actually

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sitting on the xy plane.

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Now given this, let's[br]think about what

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Stokes' theorem would tell us[br]about the value of the line

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integral over the[br]contour-- let me

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draw that a little[br]bit neater-- the line

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integral over the[br]contour c of f dot

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dr, f dot lowercase dr,[br]Where dr is obviously

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going along the contour.

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So if we take Stokes'[br]theorem, then this quantity

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right over here should[br]be equal to this quantity

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right over here.

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It should be equal to the double[br]integral over the surface.

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Well this region is[br]really just a surface

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that's sitting in the xy plane.

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So it should really just[br]be the double integral--

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let me write that[br]in that same-- it'll

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be the double integral over[br]our region, which is really

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just the same thing[br]as our surface,

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of the curl of f dot n.

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So let's just think about[br]what the curl of f dot n is.

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And then d of s would[br]just be a little chunk

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of our region, a little chunk[br]of our flattened surface

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right over there.

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So instead of ds,[br]I'll just write da.

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But let's think of what curl[br]of f dot n would actually be.

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So let's work on[br]curl of f first.

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So the curl of f--[br]and the way I always

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remember it is[br]we're going to take

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the determinant of[br]this ijk partial

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with respect to x,[br]partial with respect to y,

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partial with respect to z.

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This is just the definition[br]of taking the curl.

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We're figuring out how[br]much this vector field

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would cause something to spin.

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And then we want the[br]i component, which

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is our function p, which is[br]just a function of x and y,

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j component, which is[br]just the function q.

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And there was no z[br]component over here, so 0.

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And so this is going[br]to be equal to-- well

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if we look at the[br]i component, it's

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going to be the[br]partial of y of 0.

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That's just going to be 0, minus[br]the partial of q with respect

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to z.

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Well what's the partial[br]of q with respect to z?

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Well q isn't a[br]function of z at all.

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So that's also going to be[br]0-- let me write this out

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just so it's not too confusing.

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So our i component, it's[br]going to be partial of 0

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with respect to y.

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Well that's just going to[br]be 0 minus the partial of q

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with respect to z.

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Well the partial of[br]q with respect to z

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is just going to be 0.

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So we have a 0 i component.

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And then we want to[br]subtract the j component.

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And then the j component partial[br]of 0 with respect to x is 0.

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And then from that you're going[br]to subtract the partial of p

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with respect to z.

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Well once again, p is not[br]a function of z at all.

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So that's going to be 0 again.

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And then you have plus k times[br]the partial of q with respect

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to x.

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Remember this is just the[br]partial derivative operator.

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So the partial of q[br]with respect to x.

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And from that we're going[br]to subtract the partial of p

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with respect to y.

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So the curl of f just simplifies[br]to this right over here.

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Now what is n?

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What is the unit normal vector.

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Well we're in the xy plane.

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So the unit normal[br]vector is just

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going to be straight[br]up in the z direction.

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It's going to have[br]a magnitude of 1.

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So in this case, our[br]unit normal vector

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is just going to[br]be the k vector.

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So we're essentially just going[br]to take-- so curl of f is this.

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And our unit normal[br]vector is just

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going to be equal to the k.

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It's just going to[br]be the k unit vector.

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It's going to go straight up.

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So what happens if we[br]take the curl of f dot k?

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If we just dot this with k.

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We're just dotting[br]this with this.

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Well, we're just going to end up[br]with this part right over here.

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So curl of f dot the unit[br]normal vector is just

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going to be equal[br]to this business.

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It's just going to be equal to[br]the partial of q with respect

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to x minus the partial[br]of p with respect to y.

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And this is neat because[br]using Stokes' theorem

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in the special case,[br]where we're dealing

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with a flattened-out[br]surface in the xy plane,

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in this situation, this just[br]boiled down to Green's theorem.

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This thing right over here just[br]boiled down to Green's theorem.

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So we see that Green's theorem[br]is really just a special case--

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let me write theorem[br]a little bit neater.

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We see that Green's[br]theorem is really

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just a special case[br]of Stokes' theorem,

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where our surface is flattened[br]out, and it's in the xy plane.

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So that should make us feel[br]pretty good, although we still

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have not proven Stokes' theorem.

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But the one thing that I do[br]like about this is seeing

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that Green's theorem and[br]Stokes' theorem is consistent

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is now it starts to make[br]sense of this right over here.

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When we first learned Green's[br]theorem, we were like,

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what is this?

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what's going on over here?

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But now this is[br]telling us this is just

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taking the curl in this[br]region along this surface.

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And now starts to make a lot[br]of sense based on the intuition

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that we saw in the last video.