Line Integral Example 2 (part 2) Part 2 of an example of taking a line integral over a closed path
Line Integral Example 2 (part 2)
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- In the last video, we set out to figure out the surface area
- of the walls of this weird-looking building, where
- the ceiling of the walls was defined by the function f of xy
- is equal to x plus y squared, and then the base of this
- building, or the contour of its walls, was defined by the path
- where we have a circle of radius 2 along here, then we go
- down along the y-axis, and then we take another left, and we go
- along the x-axis, and that was our building.
- And in the last video, we figured out this first
- wall's surface area.
- In fact, you can think of it, our original problem is, we
- wanted to figure out the line integral along the closed path,
- so it was a closed line integral, along the closed path
- c of f of xy, and we're always multiplying f of xy times a
- little bit, a little, small distance of our path, ds.
- We're writing this in the most abstract way possible.
- And what we saw in the last video is, the easiest way to do
- this is to break this up into multiple paths, or into
- multiple problems.
- So you can imagine, this whole contour, this whole path we
- call c, but we could call this part, we figured out in
- the last video, c1.
- This part we can call, let me make a point, c2, and this
- point right here is c3.
- So we could redefine, or we can break up, this line integral,
- this closed-line integral, into 3 non-closed line integrals.
- This will be equal to the line integral along the path c1 of f
- of xy ds, plus the line integral along c2 of f of x y
- ds plus the line integral, you might have guessed it, along c3
- of f of xy ds, and in the last video, we got as far as
- figuring out this first part, this first curvy wall
- all right here.
- Its surface area, we figured out, was 4 plus 2 pi.
- Now we've got to figure out the other 2 parts.
- So let's do C2, let's do this line integral next.
- And in order to do it, we need to do another
- parameterization of x and y.
- It's going to be different than what we did for this part.
- We're no longer along this circle, we're
- just along the y-axis.
- So as long as we're there, x is definitely going
- to be equal to 0.
- So that's my parameterization, x is equal to 0.
- If we're along the y-axis, x is definitely equal to 0.
- And then y, we could say it starts off at y is equal to 2.
- Maybe we'll say y is equal to 2 minus t, for t is between 0, t
- is greater than or equal to 0, less than or equal to 2.
- And that should work.
- When t is equal to 0, we're at this point right there, and
- then as t increases towards 2, we move down the y-axis, until
- eventually when t is equal to 2, we're at that
- point right there.
- So that's our parameterization.
- And so let's evaluate this line, and we could do our
- derivatives, too, if we like.
- What's the derivative, I'll write it over here.
- What's dx dt?
- Pretty straightforward.
- Derivative of 0 is 0, and dy dt is equal to the
- derivative of this.
- It's just minus 1, right?
- 2 minus t, derivative of minus t, is just minus 1.
- And so let's just break it up.
- So we have this thing right here, so we have the
- integral along c2.
- But let's, instead of writing c2, I'll leave c2 there, but
- we'll say were going from t is equal to 0 to 2 of f of xy.
- f of xy is this thing right here, is x plus y squared,
- and then times ds.
- And we know from the last several videos, ds can be
- rewritten as the square root of dx dt squared, so 0 squared,
- plus dy dt squared, so minus 1 squared is 1, all
- of that times dt.
- And obviously, this is pretty nice and clean.
- This is 0 plus 1, square root, this just becomes 1.
- And then what is x?
- x, if we write it in terms of our parameterization, is always
- going to be equal to 0, and then y squared is going to
- be 2 minus is t squared.
- So this is going to be 2 minus t squared.
- So this whole crazy thing simplified to, we're going to
- go from t is equal to 0 to t is equal to 2, the x disappears in
- our parameterization, x just stays 0, regardless of what
- t is, and then you have y squared, but y is the same
- thing as 2 minus t, so 2 minus t squared, and then you have
- your dt sitting out there.
- This is pretty straightforward.
- I always find it easier when you're finding an
- antiderivative of this, although you can do this in
- your head, I like to just actually multiply
- out this binomial.
- So this is going to be equal to the antiderivative from t is
- equal to 0 to t is equal to 2 of 4 minus 2 minus 4t plus t
- squared, plus t squared, just like that dt.
- And this is pretty straightforward.
- This is going to be, the antiderivative of this is 4
- t minus 2 t squared, right?
- When you take the derivative, there's 2 times minus 2 is
- minus 4 t, and then you have plus 1/3 t to the third, right?
- These are just simple antiderivatives, and we need
- to evaluate it from 0 to 2.
- And so let's evaluate it at 2.
- 4 times 2 is 8, let me pick a new color.
- 4 times 2 is 8, minus 2 times 2 squared, so 2 times 4, so minus
- 8, plus 1/3 times 2 to the third power.
- So 1/3 times 8.
- So these cancel out.
- We have 8 minus 8, and we just have 8/3.
- So this just becomes 8/3.
- And then we have to put a 0 in, minus 0 evaluate here, but
- it's just going to be 0.
- We have 4 times 0, two times 0, all of these are going to be 0.
- So minus 0.
- So just like that, we found our surface area
- of our second wall.
- This turned out being, this right here is 8/3.
- And now we have our last wall, and then we can
- just add them up.
- So we have our last wall.
- I'll do another parameterization.
- I want to have the graph there.
- Well, maybe I can paste it again.
- So there's the graph again.
- And now we're going to do our last wall.
- So our last wall is this one right here, which is, we
- could write it, you know, this was c3.
- Let me switch colors here.
- So this is c, we're going to go along contour c3 of f of xy ds,
- which is the same thing as, let's do a parameterization.
- Along this curve, if we just say, x is equal to t, very
- straight forward, for t is greater than or equal to 0,
- less than or equal to 2, and this whole time that we're
- along the x-axis, y is going to be equal to 0.
- That's pretty straightforward parameterization.
- So this is going to be equal to, we're going to go from t is
- equal to 0 to t is equal to 2 of f of xy, which is, I'll
- write in terms of x right now, x and y, x plus y
- squared times ds.
- Now, what is dx-- well, let me write ds right here.
- Times ds.
- That's what we're dealing with.
- Now we know what ds is.
- ds is equal to the square root of dx dt squared plus
- dy dt squared times dt.
- We proved that in the first video.
- Or we didn't rigorously prove it, but we got the sense
- of why this is true.
- And what's the derivative of x with respect to t?
- Well, that's just 1, so this is just going to be a 1,
- 1 squared, same thing.
- And the derivative of y with respect to z is 0.
- So this is is 0, 1 plus 0 is 1, square root of 1 is 1.
- So this thing just becomes dt.
- ds is going to be equal to dt, in this case.
- So this just becomes a dt.
- And then our x is going to be equal to a t, that's part
- of our definition of our parameterization, and y is
- zero, so we can ignore it.
- So this was a super-simple integral.
- So this simplified down to, we're going to go from 0 to 2
- of t dt, which is equal to the antiderivative of t is just 1/2
- t squared, and we're going to go 0 to 2, which is equal
- to 1/2 times 2 squared.
- 2 squared is 4, times 1/2 is 2, and then minus 1/2
- times 0 squared, minus 0.
- So this third wall's area right there is just 2.
- Pretty straightforward.
- So that right there, the area there, is just 2.
- And so to answer our question, what was this line integral
- evaluated over this closed path of f of xy?
- Well, we just add up these numbers.
- We have 4 plus 2 pi plus 8/3 plus 2, well, what is this.
- 8/3 is same thing as 2 and 2/3, so we have 4 plus 2 and 2/3 is
- 6 and 2/3, plus another 2 is 8 and 2/3, so this whole thing
- becomes 8 and 2/3, if we write it as a mixed
- number, plus 2 pi.
- And we're done!
- And we're done.
- Now we can start trying to do line integrals with
- vector-valued functions.
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