So let’s say I have a function of x and y; f of x and y is equal to x plus y squared. If I try to draw that, let’s see if I can have a good attempt at it.
The following video will tell you about Line integral example 2 (part 1) Multivariable Calculus. You can use youtube to mp3 tool to convert the video to mp3.Notice: download videos and mp3 must be licensed by the version owner and used only for study and research, and not for commercial use and dissemination.
Line integral example 2 (part 1) Multivariable Calculus
That is my y axis– I’m going to do a little perspective here –this is my x axis– I make do the negative x and y axis, could do it in that direction –this is my x axis here. And if I were to graph this when y is 0, it’s going to be just a– let me draw it in yellow –is going to be just a straight line that looks something like that. And then for any given actually, we’re going to have a parabola in y. y is going to look something like that.
I’m just going to it in the positive quadrant. It’s going to look something like that. It’ll actually, when you go into the negative y, you’re going to see the other half of the parabola, but I’m not going to worry about it too much. So you’re going to have this surface.
it looks something like that. Maybe I’ll do to another attempt at drawing it. But this is our ceiling we’re going to deal with again.
And then I’m going to have a path in the xy plane. I’m going to start at the point 2 comma 0. x is equal to 2, y is 0.
And I’m going to travel, just like we did in the last video, I’m going to travel along a circle, but this time the circle’s going to have of radius 2. Move counter clockwise in that circle. This is on the xy plane, just to be able to visualize it properly.
So this right here’s a point 0, 2. And I’m going to come back along the y axis. This is my path; I’m going to come back along the y access and then so I look a left here, and then I’m going to take another left here in and come back along the x axis. I drew it in these two shades of green.
That is my contour. And what I want to do is I want to evaluate the surface area of essentially this little building that has the roof of f of xy is equal to x plus y squared, and I want to find the surface area of its walls. So you’ll have this wall right here, whose base is the x axis. Then you’re going to have this wall, which is along the curve; it’s going to look something like kind of funky wall on that curved side right there.
I’ll try my best effort to try to– it’s going to be curving way up like that and then along the y axis. It’s going to have like a half a parabolic wall right there. I’ll do that back wall along the y axis. I’ll do that in orange, I’ll use magenta.
That is the back wall along the y axis. Then you have this front wall along the x axis. And then you have this weird curvy curtain or wall– do that maybe in blue –that goes along this curve right here, this part of a circle of radius 2.
So hopefully you get that visualization. It’s a little harder; I’m not using any graphic program at this time. But I want to figure out the surface area, the combined surface area of these three walls.
And in very simple notation we could say, well, the surface area of those walls– of this wall plus that wall plus that wall –is going to be equal to the line integral along this curve, or along this contour– however you want to call it –of f of xy,– so that’s x plus y squared –ds, where ds is just a little length along our contour. And since this is a closed loop, we’ll call this a closed line interval. And we’ll sometimes see this notation right here.
Often you’ll see that in physics books. And we’ll be dealing with a lot more. And we’ll put a circle on the interval sign.
And all that means is that the contour we’re dealing with is a closed contour; we get back to where we started from. But how do we solve this thing? A good place to start is to just to find the contour itself. And just to simply it, we’re going to divide it into three pieces and it essentially just do three separate line integrals. Because you know, this isn’t a very continuous contour.
so the first part. Let’s do this first part of the curve where we’re going along a circle of radius 2. And that’s pretty easy to construct if we have x– let me do each part of the contour in a different color, so if I do orange this part of the contour –if we say that x is equal 2 cosine of t and y is equal to 2 sine of t and if we say that t– and this is really just building off what we saw on the last video –if we say that t– and that this is from t is a greater than or equal to 0 and is less than or equal to pi over 2 –t is essentially going to be the angle that we’re going along the circle right here. This will actually describe this path.
And if you know, how I constructed this is little confusing, you might want to review the video on parametric equations. So this is the first part of our path. So if we just wanted to find the surface area of that wall right there, we know we’re going to have to find dx, dt and dy, dt.
So let’s get that out of the way right now. So if we say dx, dt is going to be equal to minus 2, sine of t, dy, dy is going to be equal to 2 cosine of t; just the derivatives of these. We’ve seen that many times before. So it we want this orange wall’s surface area, we can take the integral– and if any of this is confusing, there are two videos before this where we kind of derive this formula –but we could take the integral from t is equal to 0 to pi over 2 our function of x plus y squared and then times the ds.
So x plus y squared will give the height of each little block. And then we want to get the width of each little block, which is ds, but we know that we can rewrite the ds as the square root– give myself some room right here –of dx of the derivative of x with respect to t squared– so that is minus 2 sine of t squared –plus the derivative of y with respect to t squared, dt. This will give us the orange section, and then we can worry about the other two walls.
And so how can we simplify this? Well, this is going to be equal to the integral from 0 to pi over 2 of x plus y squared. And actually, let me write everything in terms of t. So x is equal to 2 cosine of t. So let me write that down.
So it’s 2 cosine of t plus y, which is 2 sine of t, and we’re going to square everything. And then all of that times this crazy radical. Right now it looks like a hard antiderivative or integral to solve, but I we’ll find out it’s not too bad. This is going to be equal to 4 sine squared of t plus 4 cosine squared of t.
We can factor a 4 out. I don’t want to forget the dt. This over here– let me just simplify this expression so I don’t have to keep rewriting it. That is the same thing is the square root of 4 times sine squared of t plus cosine squared of t.
We know what that is: that’s just 1. So this whole thing just simplifies to the square root of 4, which is just 2. So this whole thing simplifies to 2, which is nice for solving our antiderivative. That means simplifying things a lot.
So this whole thing simplifies down to– I’ll do it over here. I don’t want to waste too much space; I have two more walls to figure out –the integral from t is equal to 0 to pi over 2. I want to make it very clear.
I just chose the simplest parametrization I could for x and y. But I could have picked other parametrizations, but then I would have had to change t accordingly. So as long as you’re consistent with how you do it, it should all work out.
There isn’t just one parametrization for this curve; it’s kind of depending on how fast you want to go along the curve. Watch the parametric functions videos if you want a little bit more depth on that. Anyway, this thing simplifies. We have a 2 here; 2 times cosine of t, that’s 4 cosine of t.
And then here we have 2 sine squared sine of t squared. So that’s 4 sine squared of t. And then we have to multiply times this 2 again, so that gives us an 8. 8 time sine squared of t, dt.
And then you know, sine squared of t; that looks like a tough thing to find the antiderivative for, but we can remember that sine squared of, really anything– we could say sine squared of u is equal to 1/2 half times 1 minus cosine of 2u. So we can reuse this identity. I can try the t here; sine squared of t is equal to 1/2 times 1 minus cosine of 2t. Let me rewrite it that way because that’ll make the integral a lot easier to solve.
So we get integral from 0 the pi over 2– and actually I could break up, well I won’t break it up –of 4 cosine of t plus 8 times this thing. 8 times this thing; this is the same thing as sine squared of t. So 8 times this– 8 times 1/2 is 4 –4 times 1 minus cosine of 2t– just use a little trig identity there –and all of that dt. Now this should be reasonably straight forward to get the antiderivative of.
Let’s just take it. The antiderivative of this is antiderivative of cosine of t; that’s a sine of t. The derivative of sine is cosine.
So this is going to be 4 sine of t– the scalars don’t affect anything –and then, well let me just distribute this 4. So this is 4 times 1 which is 4 minus 4 cosine of 2t. So the antiderivative of 4 is 4t– plus 4t –and then the antiderivative of minus 4 cosine of u00b5 t? Let’s see it’s going to be sine of 2t. The derivative of sine of 2t is 2 cosine of 2t.
We’re going to have to have a minus sign there, and put a 2 there, and now it should work out. What’s the derivative of minus 2 sine of t? Take the derivative of the inside 2 times minus 2 is minus 4. And the derivative of sine of 2t with respect to 2t is cosine of 2t. So there we go; we’ve figured out our antiderivative.
Now we evaluate it from 0 the pi over 2. And what do we get? We get 4 sine– let me write this down, for I don’t want to skip too many –sine of pi over 2 plus 4 times pi over 2– that’s just 2 pi minus 2 sine of 2 times pi over 2 sine of pie, and then all of that minus all this evaluated at 0. That’s actually pretty straightforward because sine of 0 is 0. 4 times 0 is 0, and sine of 2 times 0, that’s also 0.
So everything with the 0’s work out nicely. And then what do we have here? Sine of pi over 2– in my head, I think sine of 90 degrees; same thing –that is 1. And then sine of pi is 0, that’s 180 degrees.
So this whole thing cancels out. So we’re left with 4 plus 2 pi. So just like that we were able to figure out the area of this first curvy wall here, and frankly, that’s the hardest part. Now let’s figure out the area of this curve.
And actually you’re going to find out that these other curves as they go along the axes are much, much, much easier, but we’re going to have to find different parametrizations for this. So if we take this curve right here, let’s do a parametrization for that. Actually, you know what? Let me continue this in the next video because I realize I’ve been running a little longer.
I’ll do the next two walls and then we’ll sum them all up.
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