Hi everyone, and welcome to our lecture on line integrals. So let's start off with a little recall of what we have seen so far with integration. We've seen the definite integral of a single variable function f of x dx and the geometric A picture that goes along with this is we have the xy plane, we have some real numbers A to B, or over some particular inner interval. And then we have the function doing this thing, and this area under the curve is represented by the definite integral from a to b. In this example, I want you to think about the function itself. We have the domain of f in this particular case, the interval that we care about. The domain is the interval a to b, this little baby line segment on the x axis, and our function takes this interval from a to b, this little segment on the x axis, and it's a scalar value function and you give me some number, I gave you one back and it does its range, some subset of the reals. Now we have seen, we're trying to generalize all these notions to three dimensions we're trying to take this idea and put it into space. So how do we generalize this what's the next step will now was look at this thing in 3d. Now, consider. Not just the line segment, but consider a curve C in our three inside a space, the xy plane. So let's draw not just the x and y axis, but also a Z axis on here. So we have our 3d picture. I still want some curve floating around, so I'll draw curve maybe it does this thing. Maybe it loops around I don't know. Whatever does so some cursing and I'd like to know, what about a function now that's defined on C, so you give me some point on C. And I'll give you back a number so some multivariable function perhaps or even some vector value function in time. But I'll give you back a function I wanted to find this function on c, where c is the domain. So I think about this as sort of I take the line segment that I started with the one that was on the xy plane, and I kind of imagined my gravity turning off. When I was if these things started to float like a string in space what would I do then? Well, I still like to be able to integrate along this thing I'd like to know what is the integral, I don't have from A to B at the moment, I'm really just integrating over C of this function. And now that I'm in space, we're going to change up our notation because I'm no longer on the x axis. We're going to write this as DS. This integral it turns out will be defined, these has a name and this integral is the line integral. Now I have to talk to you a little bit about the naming of this thing. So if I say the line integral, I have what do I have a curve in space floating around doing this thing and I have a scalar value function defined on that line. So I can just kind of turn gravity off and let my line segment float around. This line integral it's a bad name in my opinion, because well, it assumes that I'm working with a line. However, nothing is forcing me to make my curve a straight line, it can totally bend around via circle, do whatever I want. So I don't love the fact that it's a line in some places you'll see this called a path integral. For that very reason so when you see path integral or line integral, they're really the same thing. But assuming we're if I call it a line integral, the curve most certainly can be a line or but it doesn't have to be. In addition, these lines line integrals, the key here so two parts curve, and scalar valued function. Everything starts with the parameterize curve so you'll always be given a curve see that you'll the parameterization will be given to you or. You will have to know it. We will talk about some common parameterizations in a second, but just remember c will be defined as X of t, y of t if I'm in the plane, or if I am in space, X of t, y of t, or Z of t just depends on context. If I am working inside of the xy plane as a parameterize curve or as a curve in space, the formula. .Here's another one that I want you to keep track of and keep handy for this line integral of a curve and space so little fds, this is for scalar valued functions. This will be defined as I should put a parameterization on this thing. So I'm going to go a to b for my parameter t, so I'm always looking at like a finite segment of my curve or something like that okay. This will be then the definite integral from a to b so very similar to my single variable case when I was a line segment on the x axis of a pay attention here F of. X of t comma y of t, there's the Z party throw that in as well. Okay, and then times the magnitude of the derivative of C prime, where C prime is your parameterization. Sometimes we call it R, sometimes we call it C I have no preference here, but this is just my parameterization my position function. Remember, this is the speed derivative of the position is velocity and if I take the magnitude of this thing, then it is the speed. So I'm sort of like involving this new term, which is my sort of cost my extra piece that I have to make this all work the variable once you do all this, there's only T's floating around. So this is all done dt and this is our formula for a line integral of a scalar value function for any curve In the plane or in space, now, what is the point of all this? What's the geometric meaning so if I have a nice function and multivariable functions small f scalar valued, and I'm going to make it positive for a second, so it's graphed, lives above the xy plane, then the line integral or path integral of FDS as we're calling it in general, this represents the area of one side of a fence with currency. So if you think about the picture, I have some curve and space, I'm going to do my best to draw this here. I apologize in advance I have some curve C floating around in space and if I want this Integral just like before it was the area under a curve. Well now I want you to imagine, it is a sort of fence and I can draw a fence and it's going to wiggle and wave and follow the curve around I'm going from A to B some finite part of the curve, so this will be then the area, the surface area of one side of the fence. This is what the bay see and height, of course the function. So this is what you can what you can think about, if you want to imagine that picture in two dimensions now, again waving and moving out into space. These line integrals, they're defined for a single curve, but if you start gluing curves together, so for example, if you have a piecewise curve, if so if c is a piecewise curve, so it's made up of other pieces, c1, c2, c3, maybe you have, I don't know, a triangle or something like that each piece is a line segment. So c1, c2, and c3, to integrate, over the entire curve, FDS you would take the line integral over each piece. So we take, the line and we'll over c1, we take the line integral over c2, and we take the line integral over c3. So, if you ever have a bunch of line segments that are put together, you have to do this for each one, triangle square any, you know, pick a favorite shape. So you can this this process generalizes quite nicely to more complicated shapes. The work obviously gets longer, then you have more integrals to do, but they're very doable. You just have to do it three times, or four times or whatever times. When you get these kind of problems there's two types of ways that you see them. The first one is that they hand you the parameterization. Here's some crazy helix in space, here's some other shape that you don't know. I'm going to hand you the curve and the parameterization will be given. From that you go off into your line integral and work out the formula however, you are expected to sort of know at this point at this level of your math career, there's some parameterizations that you are expected to know. These are usually not given in problems so I'm just going to list them here for you to have but if you're a circle, and we've done this before, if you're a circle C. And usually you're centered at the origin. So your center is (0, 0), and your radius is r, well, then your parametrization which may or may not be given if I use RFT or I use CFT either one, but the parameterization of course is given by r cos t and then r sint. One lap around the circle, which is usually good enough for what we want to pay attention to the problem. Maybe they want two lapse or whatever is 0 to 2 pi. So we've seen that one before. So this is one that you're supposed to know. The other one is like what if I give you just a good old function like a good old simple function, the parabola, the cubic y = f(x). Once in a while, you get a function and is remembered to parameterize the curve of a function. So the graph, the graph is defined by the parameterization. We let x be the parameter, so we do t and then f(t), and then we just run from a to b, these will be given the bounds whatever we want it to be. But the key is you turn x into t and then you let the y-coordinate just be f(t). So for example if we had the parabola, so I wanted y = x squared and then my parameterization of the curve, which they may or may not give me would just be t, t squared. And that the t run from whatever bounds they wanted. The other one is going to come up a bunch in your saw with the triangle is what's the line segment. So what if I have a line segment somewhere in space. So usually, we'll start at the point P will end that Q. So you can imagine somewhere in space here's P to Q direction will matter. So just pay attention to where you're starting and where you're finish. Well, then the parameterization is always defined by, I'm going to write this sort of in point form, (1- t) P + tQ. And in this particular form, there's more than one way to do this but this particular form is kind of nice, because t runs from 0 to 1. And remember the bounds on the parameter land at the bounds of the integral. So if you can get an integral from 0 to 1, those are the nicest numbers you can work with. There's more than one way to do this, but there's a nice one. And you can check if I plug in t as 0, then I just get back P, right, because this becomes 1 that becomes 0. So I start a P, and if I let t become 1, then I get 0 + 1Q, which is just Q. So this parameterization, the third one and the first one probably come up the most. I'll remind you of them as we go through, but I just want you have these sort of handy that you can reference as needed. This new concept of a path integral or line integral, same thing, it comes with a couple of facts that also mimic the behavior of a regular definite integral of a single variable function. So fact number 1 is that if the curve C has two different parameterizations, let's call them maybe C1 and C2. So maybe perhaps you've come up with another way to parameterize your curve, then the path integral doesn't change. The path integral is independent, so we say independent of the line integral. You can use either one, you'll get the same answer either way. If you think about this geometric intuition, if I describe that fence using a different parameter, my parameter goes from 0 to 1, but you go a little slower and you go from 0 to 10, it's still the area under a fence that's not going to change. So we say it's independent of the parametrization, you will get the same one, usually you pick the one with that's easiest to solve. Fact number 2 is if you change the directions, if you reverse the direction of the parameterization, so basically you have first parameterization and is the opposite. So you maybe you go counterclockwise and clockwise around a circle or you start a P and go to Q, or perhaps the other one is you go to Q and go to P, whatever it is. If you reverse the direction, that is like switching the bounds, so instead of going from a to b, you go from b to a on a definite integral. What is switching the bounce? Do you want a definite integral? Well, it pops out a negative sign. So this negative now corresponds to direction. And if you switch directions, a negative sign pops out. There'll be times when they want you to switch directions, but just know it behaves very much like the definite integral in a single variable case. I just want to write out the formula for a line integral in space, and space means we're in 3D. So the curve C is defined by some parametrization, let's say r(t), I'm in space, so I have three coordinates. So some function x, there's t's floating around in the x position and the y position. And of course, now that we're in a space, the z position, let's bound this parameter from a to b. Then the line integral in space becomes the definite integral from a to b, of f(x(t), y(t), y(t)). So you basically plug in the parameterization, you can think of this as f(r(t)). So you plug in the parameterization, what does that mean? Everywhere I see an x, I put in the expression with t. Everywhere I see a y, I plug in the expression. Everywhere I see a z, I just substitutes, this is a substitution. And then times the magnitude of the derivative vector or the speed of this thing, all dt. Remember, the derivative vector of any position is just the derivative of each of the components. So good old x'(t), y'(t), so we take three derivatives. This is a very involved process although nothing is individually difficult, just lots of steps. I'm integrating and differentiating all in the same process. And then of course, the magnitude of this vector is the square root of each of the components squared, so you have to go through all of this. Some problems get worse than others, sometimes it cleans up nicely, but just work it out. If the intervals terrible at the end of the day, remember, this is a definite integral. And you could always run to a computer and solve this thing. And one thing I want to point out, you've seen a special case before, so a special case that we've actually already seen, if my scalar valued function is the very boring yet important constant function one, so you give me any points and I just kicked back 1. Then when I plug in to the function I just get back 1. So the line integral here of f of sds, becomes the line integral of 1ds. And then by its formula, by its definition, this is like a little formula up here. This just becomes the definite integral from a to b assume over the same curve, of the absolute value or the magnitude of the speed, where have you seen this before? Look at that, hopefully keeping a big formula sheet at this point. Where have you seen an integral from a to b of the square root of each of the components squared? Well, that just turns out to be arc length, this is the length of the curve or the arc length of the curve C. So in this special case where the function is 1, this formula simplifies to give the size of the curve. And you've seen that before. All right, let's do an example. On the screen here we evaluate the line integral or the path integral of (2 + x y squared) ds, where C is the upper half of the unit circle. Whenever I get a question like this, I always like to take my time and sort of draw the picture. This will pay dividends later. The unit circle, remember that centered at the origin of radius 1, so on the upper half of the circle, and going from -1 to 1, it's going to be in the positive direction. So just friendly reminder that's counter clockwise. We assume positive by default unless told otherwise. The scalar valued function, they don't call it f, but again, we have to recognize this, this is little f(x, y), this is (2 + x squared y). If you think about like, just remind yourself this a scalar valued. If it give me an x and y, this will kick back a number. I'm going to need the parameterization of curve, I need to find out what this parameterization is. So this is one that I'm supposed to know, unit circle is given by just cos t, Comma sine of t. I'm only going around the upper half. So now I just got to be careful t is going from zero to pi. We're not doing the full lap, just around two pi. And then I have basically all the information I need to go forth and prosper. So I have f ds this turns out to be from we're going from zero to pi of f of the parameterization and then times the magnitude of the derivative all dt. Let's go off on the side and get what I need. Let's do the derivative vector first, I guess. So let's do our prime of t. The derivative of cosine is negative sine of t derivative sine of t is cosine of t, the magnitude of this vector for particular values of t. This is going to work out nicely. This is sine squared of t, plus cosine squared of t. Of course, that's an identity that we can't forget. That is just a square to 1, which of course is 1. So this magical price we have to pay for being able to do math in three dimensions. This works out nicely to just be one that won't always happen. However it does for this case. And now we have the definite integral from 0 to pi, now we had the function and wherever I see an x, I'm going to plug in anything. This is my x of t and this is my y, I plug in the parameterization. So this becomes 2 Plus cosine squared of t times sine of t, all dt. From here I'm in a basically old territory things that I know how to do more. I just have to go through this and solve this using techniques of normal definite integration. In my mind when I get to this point, most of the work is done. And I just go ahead and integrate the integral of 2 of course, is 2t, the integral of cosine squared of t, well that's a nice substitution with u equals cosine of t. And then d u is negative sine of t. When you make that substitution, and you can check, you get u squared, so the integral is going to become u cubed. I'm doing a little bit of this in my head here, but you can check this if you want. Cosine cubed t over 3. And I evaluate this from zero to pi, and when you plug it all back in, you get 2 pi plus two thirds. It's a weird shape on the graph, but it's a weird number but 2 pi plus two thirds is perfectly fine. Let's do one more example. So let's find the path integral of the scalar valued function, y times sine of z ds. And this curve c now will be our helix. We saw this before on another slide, where it's defined by cosine t, sine t, to t. And we're just going to go one lap around t 0 to 2 pi. So this is just some part of the helix in space. I want to evaluate this I need to find a few things. So this is given to us this time, we're not supposed to know the parameterization of a helix, so they're giving us the position. We're going to need a derivative, so let's go ahead and do that. That will turn out to be the vector and minus sine of t, Cosine of t,1. And then the magnitude of this will be the square root of each of the component squared. This gives me a sine squared plus a cosine squared of plus 1 squared sine squared plus cosine. There's our Pythagorean identity coming up again and I get 1 plus 1 all under the square root gives square root of 2. I'm going to use my formula. Remember my little baby f of this is like x, y and z. There's no x, which is great. So I'm not really going to use the x component of my position function, x, y and z. But I'm going to substitute everywhere I see a y, I put in sign everywhere I see a z I put in t. So, this becomes the definite integral from zero to two pi of, so y becomes sine of t, sine of z becomes sine of t. And then don't forget I have to put in this constant times the derivative the magnitude of the derivative so there's a root two floating around dt. Choose a constant of course, we could bring that right out in front. We'll just keep it there won't get rid of it. And then I have the definite integral from 0 to 2 pi of sine squared of T DT and this is one that we've seen before. This is one we know how to do we could grab a computer at this point and go solve it if you want though if you want to see it and it's closed form. You can rewrite this using your identity. So sine squared becomes one minus cosine of two t. That's a trig identity that's not calculus that substitutes that. And from here, another constant comes out front. So you have root two over two. And then you can go ahead and just integrate this thing with respect to t to get 1 minus one half sine of 2 to plug this in from 0 to 2 pi, and you get, you can check this root 2 times pi. So here's another example just to go through for these line integrals. We saw though that scalar value functions are not the only functions we're going to study in this course. And of course, we can have vector fields, so I can take f of x y and z. And of course map it over to some vector with different components. So if I have a vector field or a vector value function, how do I integrate that over a curve, we call this also the line integral of a vector fields along the curve. Again, we really try to distinguish our vector field by either using capital letters and putting a vector over it. And the notation for this line integral is F.dir, and this is truly a dot because I have two vectors here. So I say that like the dot product, I say dot and the parameterization of the curve will be given where you have to have like one of those common ones. And we're going to go from a to b of f, and I plug in the parameterization, just like I did before. So I make my substitutions putting everything in front of a t. And now the only difference, and this is important is because I have a vector in here, I want to dot this with the derivative of the position vector. So this truly is a dot product. So a little different if you notice I have a vector valued function so I dot it with a vector before when I had a scalar valued function, I found the magnitude, which is a scalar. And this will give me a good old definite integral all in terms of t that I know how to solve. So this is the other formula for line integrals. That is tells you how to do this for vector fields. Let us do one example. Again, we will try to keep it simple, just for these beginning ones. I am going to give you a vector of field that's defined by x squared i minus x y j. Maybe they give it to an ij notation. Maybe they don't, you can always convert it. Friendly reminder this is the same as saying x squared, minus x y, write it in whatever notation you're comfortable with. Let's go around the curve defined by RFT, which is given by cosine of t, sine of t, and we're going to go zero to two pi over 2. This is like a quarter on the unit circle. Okay, let's find what this is. The line integral of f.ds all vectors now is given by the definite integral on the bounds of the parameter. So zero to pi over two. I'm going to plug in my parameterization. So we'll do that in a second dot with the derivative of the position vector. Let's go go well we need the derivative done this a few times now so we'll go quickly is derivative of cosine is negative sine of t the derivative of sine is cosine. And when I plug in the parameterization remember everywhere I see an x, I'm going to put a cosine. Everywhere I put a y I'm going to put a sign. So this becomes the definite integral from 0 to pi over 2 of the vector So x squared becomes cosine squared of t and then minus xy becomes minus cosine t times sine of t. This is dotted with the derivative, the velocity vector here, minus sine of t, cosine of tdt. One more, let's take a dot product of vectors. We know how to do that we multiply the components, and I get minus sine of t, cosine squared of t, and then minus cosine squared of t, sine of t, all dt. That cleans up a little bit, you get zero to pi over two of minus two, sine of t, cosine squared of t. That's a nice, relatively simple, definite integral, we can make a u substitution with u equals cosine of t. And very similar to before we get to times cosine cubed t, all over 3, evaluated from 0 to pi over 2, and you can check, plug all that in you get minus 2 thirds. It's important to realize that all this work at the end of the day gives you back a scalar. This is still the integral of some function. It's some definite integral with some number. The interpretation of this if you're interested in the physics behind this, this is called the work sometimes, this is the work in physics this number. So a line integral of a vector valued function. The line integral of a vector field is sometimes called the work. All right, so you've seen how to do these examples. There's lots of little pieces, there's derivatives, integrals, we're putting it all together at this point. Go ahead and do lots of problems, find as many as you can just constant practice, and really go slow, be careful. Check your work as you go have fun with these. I always compare these to kind of hitting it like a golf shot like it can be a very frustrating game. But when you get it right and you have this beautiful math in front of you, using all your knowledge of calculus, man feels good sometimes. So, go through these, keep your formulas handy, and keep working hard great job on this video. We'll see you next time.
