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In the last video, we set out to figure out the surface area

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of the walls of this weird-looking building, where

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the ceiling of the walls was defined by the function f of xy

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is equal to x plus y squared, and then the base of this

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building, or the contour of its walls, was defined by the path

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where we have a circle of radius 2 along here, then we go

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down along the y-axis, and then we take another left, and we go

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along the x-axis, and that was our building.

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And in the last video, we figured out this first

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wall's surface area.

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In fact, you can think of it, our original problem is, we

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wanted to figure out the line integral along the closed path,

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so it was a closed line integral, along the closed path

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c of f of xy, and we're always multiplying f of xy times a

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little bit, a little, small distance of our path, ds.

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We're writing this in the most abstract way possible.

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And what we saw in the last video is, the easiest way to do

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this is to break this up into multiple paths, or into

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multiple problems.

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So you can imagine, this whole contour, this whole path we

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call c, but we could call this part, we figured out in

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the last video, c1.

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This part we can call, let me make a point, c2, and this

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point right here is c3.

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So we could redefine, or we can break up, this line integral,

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this closed-line integral, into 3 non-closed line integrals.

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This will be equal to the line integral along the path c1 of f

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of xy ds, plus the line integral along c2 of f of x y

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ds plus the line integral, you might have guessed it, along c3

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of f of xy ds, and in the last video, we got as far as

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figuring out this first part, this first curvy wall

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all right here.

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Its surface area, we figured out, was 4 plus 2 pi.

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Now we've got to figure out the other 2 parts.

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So let's do C2, let's do this line integral next.

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And in order to do it, we need to do another

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parameterization of x and y.

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It's going to be different than what we did for this part.

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We're no longer along this circle, we're

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just along the y-axis.

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So as long as we're there, x is definitely going

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to be equal to 0.

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So that's my parameterization, x is equal to 0.

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If we're along the y-axis, x is definitely equal to 0.

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And then y, we could say it starts off at y is equal to 2.

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Maybe we'll say y is equal to 2 minus t, for t is between 0, t

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is greater than or equal to 0, less than or equal to 2.

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And that should work.

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When t is equal to 0, we're at this point right there, and

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then as t increases towards 2, we move down the y-axis, until

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eventually when t is equal to 2, we're at that

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point right there.

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So that's our parameterization.

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And so let's evaluate this line, and we could do our

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derivatives, too, if we like.

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What's the derivative, I'll write it over here.

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What's dx dt?

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Pretty straightforward.

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Derivative of 0 is 0, and dy dt is equal to the

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derivative of this.

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It's just minus 1, right?

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2 minus t, derivative of minus t, is just minus 1.

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And so let's just break it up.

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So we have this thing right here, so we have the

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integral along c2.

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But let's, instead of writing c2, I'll leave c2 there, but

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we'll say were going from t is equal to 0 to 2 of f of xy.

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f of xy is this thing right here, is x plus y squared,

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and then times ds.

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And we know from the last several videos, ds can be

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rewritten as the square root of dx dt squared, so 0 squared,

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plus dy dt squared, so minus 1 squared is 1, all

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of that times dt.

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And obviously, this is pretty nice and clean.

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This is 0 plus 1, square root, this just becomes 1.

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And then what is x?

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x, if we write it in terms of our parameterization, is always

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going to be equal to 0, and then y squared is going to

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be 2 minus is t squared.

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So this is going to be 2 minus t squared.

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So this whole crazy thing simplified to, we're going to

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go from t is equal to 0 to t is equal to 2, the x disappears in

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our parameterization, x just stays 0, regardless of what

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t is, and then you have y squared, but y is the same

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thing as 2 minus t, so 2 minus t squared, and then you have

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your dt sitting out there.

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This is pretty straightforward.

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I always find it easier when you're finding an

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antiderivative of this, although you can do this in

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your head, I like to just actually multiply

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out this binomial.

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So this is going to be equal to the antiderivative from t is

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equal to 0 to t is equal to 2 of 4 minus 2 minus 4t plus t

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squared, plus t squared, just like that dt.

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And this is pretty straightforward.

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This is going to be, the antiderivative of this is 4

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t minus 2 t squared, right?

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When you take the derivative, there's 2 times minus 2 is

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minus 4 t, and then you have plus 1/3 t to the third, right?

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These are just simple antiderivatives, and we need

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to evaluate it from 0 to 2.

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And so let's evaluate it at 2.

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4 times 2 is 8, let me pick a new color.

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4 times 2 is 8, minus 2 times 2 squared, so 2 times 4, so minus

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8, plus 1/3 times 2 to the third power.

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So 1/3 times 8.

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So these cancel out.

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We have 8 minus 8, and we just have 8/3.

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So this just becomes 8/3.

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And then we have to put a 0 in, minus 0 evaluate here, but

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it's just going to be 0.

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We have 4 times 0, two times 0, all of these are going to be 0.

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So minus 0.

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So just like that, we found our surface area

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of our second wall.

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This turned out being, this right here is 8/3.

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And now we have our last wall, and then we can

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just add them up.

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So we have our last wall.

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I'll do another parameterization.

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I want to have the graph there.

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Well, maybe I can paste it again.

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Edit.

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So there's the graph again.

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And now we're going to do our last wall.

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So our last wall is this one right here, which is, we

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could write it, you know, this was c3.

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Let me switch colors here.

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So this is c, we're going to go along contour c3 of f of xy ds,

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which is the same thing as, let's do a parameterization.

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Along this curve, if we just say, x is equal to t, very

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straight forward, for t is greater than or equal to 0,

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less than or equal to 2, and this whole time that we're

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along the x-axis, y is going to be equal to 0.

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That's pretty straightforward parameterization.

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So this is going to be equal to, we're going to go from t is

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equal to 0 to t is equal to 2 of f of xy, which is, I'll

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write in terms of x right now, x and y, x plus y

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squared times ds.

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Now, what is dx-- well, let me write ds right here.

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Times ds.

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That's what we're dealing with.

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Now we know what ds is.

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ds is equal to the square root of dx dt squared plus

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dy dt squared times dt.

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We proved that in the first video.

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Or we didn't rigorously prove it, but we got the sense

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of why this is true.

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And what's the derivative of x with respect to t?

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Well, that's just 1, so this is just going to be a 1,

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1 squared, same thing.

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And the derivative of y with respect to z is 0.

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So this is is 0, 1 plus 0 is 1, square root of 1 is 1.

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So this thing just becomes dt.

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ds is going to be equal to dt, in this case.

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So this just becomes a dt.

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And then our x is going to be equal to a t, that's part

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of our definition of our parameterization, and y is

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zero, so we can ignore it.

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So this was a super-simple integral.

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So this simplified down to, we're going to go from 0 to 2

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of t dt, which is equal to the antiderivative of t is just 1/2

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t squared, and we're going to go 0 to 2, which is equal

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to 1/2 times 2 squared.

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2 squared is 4, times 1/2 is 2, and then minus 1/2

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times 0 squared, minus 0.

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So this third wall's area right there is just 2.

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Pretty straightforward.

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So that right there, the area there, is just 2.

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And so to answer our question, what was this line integral

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evaluated over this closed path of f of xy?

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Well, we just add up these numbers.

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We have 4 plus 2 pi plus 8/3 plus 2, well, what is this.

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8/3 is same thing as 2 and 2/3, so we have 4 plus 2 and 2/3 is

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6 and 2/3, plus another 2 is 8 and 2/3, so this whole thing

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becomes 8 and 2/3, if we write it as a mixed

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number, plus 2 pi.

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And we're done!

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And we're done.

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Now we can start trying to do line integrals with

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vector-valued functions.

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