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From Wikipedia, the free encyclopedia

In mathematics, a line integral is an integral where the function to be integrated is evaluated along a curve.[1] The terms path integral, curve integral, and curvilinear integral are also used; contour integral is used as well, although that is typically reserved for line integrals in the complex plane.

The function to be integrated may be a scalar field or a vector field. The value of the line integral is the sum of values of the field at all points on the curve, weighted by some scalar function on the curve (commonly arc length or, for a vector field, the scalar product of the vector field with a differential vector in the curve). This weighting distinguishes the line integral from simpler integrals defined on intervals. Many simple formulae in physics, such as the definition of work as , have natural continuous analogues in terms of line integrals, in this case , which computes the work done on an object moving through an electric or gravitational field F along a path .

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  • Introduction to the line integral | Multivariable Calculus | Khan Academy
  • Line Integrals - Evaluating a Line Integral
  • Vector Calculus - Line Integrals of Vector Field | Example & Solution
  • What is a LINE INTEGRAL? // Big Idea, Derivation & Formula
  • Evaluating Line Integrals

Transcription

If we're just dealing with two dimensions, and we want to find the area under a curve, we have good tools in our toolkit already to do it, and I'll just remind us of our tools. so let's say, that's the x-axis, that's the y-axis, let me draw some arbitrary function right here, and that's my function f of x. And let's say we want to find the area between x is equal to a, so that's x equal to a, and x is equal to b. We saw this many, many, many videos ago. The way you can think about it, is you take super small widths of x, or super small changes in x. We could call them delta x's, but because they're so small, we're going to call them a dx. Super, infinitesimally small changes in x. And then you multiply them times the value of f of x at that point. So you multiply it times the height at that point, which is the value of f of x. So you get f of x times each of these infinitesimally small bases, that'll give you the area of this infinitesimally narrow rectangle right there. And since each of these guys are infinitely small, you're going to have an infinite number of these rectangles in order to fill the space. You're going to have an infinite number of these, right? And so the tool we use was the definite integral. The definite integral is a sum, is an infinite sum of these infinitely small areas, or these infinitely small rectangles. And the notations that we use, they would go from a b. And we've done many videos on how do you evaluate these things. I just want to remind you, conceptually, what this is saying. This is conceptually saying, let's take a small change in x, multiply it times the height at that point, and you're going to have an infinite number of these, because these x's are super small, they're infinitely small, so you're going to have an infinite number of those. So take an infinite sum of all of those, from x is equal to a to x is equal to b. And that's just our standard definite integral. Now what I want to do in this video is extend this, broaden this a little bit, to solve, I guess it maybe could say a harder or a broader class of problems. Let's say that we are, let's go to three dimensions now. And I'll just draw the x-y plane first. Maybe I'll keep this, just to kind of make the analogy clear. I'm going to kind of flatten this, so we have some perspective. So let's say that this right here is the y-axis, kind of going behind the screen. You can imagine if I just pushed on this and knocked it down. So that's the y-axis, and that is my x-axis right there. And let's say I some path in the x-y plane. And in order to really define a path in the x-y plane, I'll have to parameterize both the x and y variables. So let's say that x is equal to, let me switch colors. I'm using that orange too much. Let's say that x is is equal to some function of some parameter t, and let's say y is equal to some other function of that same parameter t, and let's say we're going to start, we're going to have t go from, t is going to be greater than or equal to a, and then less than or equal to b. Now this will define a path in the x-y plane, and if this seems confusing, you might want to review the videos on parametric equations. But essentially, when t is equal to a, you're going to have x is equal to, so t is equal to a, you're going to have x is equal to g of a, and you're going to have y is equal to h of a. So you're going to have this point right here, so maybe it might be, I don't know, I'll just draw a random point here. When t is equal to a, you're going to plot the coordinate point g of a. That's going to be our x-coordinate. This is g of a, right here. And then our y-coordinate is going to be h of a. Right? You just put t is equal to a in each of these equations, and then you get a value for x and y. So this coordinate right here would be h of a. And then, you would keep incrementing t larger and larger, until you get to b, but you're going to get a series of points that are going to look something like that. That right there is a curve, or it's a path, in the x-y plane. And you know, you're saying, how does that relate to that right now? What are we doing? Well, let me just write a c here, for saying, that's our curve, our that's our path. Now, let's say I have another function that associates every point in the x-y plane with some value. So let's say I have some function, f of x y. What it does is associate every point on the x-y plane with some value. So let me plot f of x y. Let me make a vertical axis here. We could do a different color. Call it the f of x y axis, maybe we could even call it the z-axis, if you want to. But some vertical axis right there. And for every point, so if you give me an x and a y, and put into my f of x y function, it's going to give you some point. So I can just draw some type of a surface that f of x y represents. And this'll all become a lot more concrete when I do some concrete examples. So let's say that f of x y looks something like this. I'm going to try my best to draw it. I'll do a different color. Let's say f of x y. Some surface. I'll draw part of it. It's some surface that looks, let's say it looks something like that. That is f of x y. And remember, all this is, is you give me an x, you give me a y, you pop it into f of x y, it's going to give me some third value that we're going to plot in this vertical axis right here. I mean, example, f of x y? It could be, I'm not saying this is a particular case, it could be x plus y. It could be f of x y. These are just examples. It could be x times y. If x is 1, y is 2, f of x y will be 1 times 2. But let's say when you plot, for every point on the x-y plane, when you plot f of x y you get this surface up here, and we want to do something interesting. We want to figure out, not the area under this curve, this was very simple when we did it the first time. I want to find the area if you imagine a curtain, or a fence, that goes along this curve. You can imagine this being a very straight linear path, going just along the x-axis from a to b. Now we have this kind of crazy, curvy path that's going along the x-y plane. And you can imagine if you drew a wall, or curtain, or a fence that went straight up from this to my f of x y, let me do my best effort to draw that. Let me draw it. So it's going to go up to there, and maybe this point corresponds to there. And when you draw that curtain up, it's going to intersect it something like that. Let's say it looks something like that. So this point right here corresponds to that point right there. So if you imagine, you have a curtain, f of x y is the roof, and this is a, what I've drawn here, this curve, this kind of shows you the bottom of a wall. This is some kind of crazy wall. And let me say, this point it corresponds to, well, actually, let me draw it little bit different. This point will correspond to some point up here, so when you trace where it intersects, it will look something maybe like that, I don't know. Something like that. And I'm trying my best to help you visualize this. So maybe I'll shade this in to make it a little solid, let's say f of x y is little transparent. You can see. But you have this curvy-looking wall here. And the whole point of this video is, how can we figure out the area of this curvy-looking wall, that's essentially the wall or the fence that happens if you go from this curve and jump up, and hit the ceiling at this f of x y? So let's think a little bit about how we can do it. Well, if we just use the analogy of what we did previously, we could say, well look. Let's make a little change in distance of our curve. Let's call that ds. That's a little change in distance of my curve, right there. And if I multiply that change in distance of the curve times f of x y at that point, I'm going to get the area of that little rectangle right there. Right? So if I take the ds, my change in my, you can imagine the arc length of this curve at that point, so let me write, you know, ds is equal to super small change in arc length of our path, or of our curve. That's our ds. So you can imagine, the area of that little rectangle right there, along my curvy wall, is going to be ds, I'll make it a capital S, ds times the height at that point. Well, that's f of x y. And then if I take the sum, because these are infinitely narrow, these ds's have infinitely small width, if I were take the infinite sum of all of those guys, from t is equal to a to t is equal to b, right, from t is equal to a, I keep taking the sum of those rectangles, to t is equal to b, right there, that will give me my area. I'm just using the exact same logic as I did up there. I'm not being very mathematically rigorous, but I want to give you the intuition of what we're doing. We're really just bending the base of this thing to get a curvy wall instead of a straight, direct wall like we had up here. But you're saying, Sal, this is all abstract, and how can I even calculate something like this, this makes no sense to me, I have an s here, I have an x and a y, I have a t, what can I do with this? And let's see if we can make some headway. And I promise you, when we do it with a tangible problem, the end product of this video is going to be a little bit hairy to look at. But when we do it with an actual problem, it'll actually, I think, be very concrete, and you'll see it's not too hard to deal with. But let's see if we can get all of this in terms of t. So first of all, let's focus just on this ds. So let me re-pick up the x-y axis. So if I were to reflip the x-y, let me switch colors, this is just getting a little monotonous. So if I were to reflip the x-y axis like that, actually, let me do that with that same green, so you know we're dealing with the same x-y axis. So that's my y-axis, that is my x-axis. And so this path right here, if I were to just draw it straight up like this, it would look something like this. Right? That's my path, my arc. You know, this is when t is equal to a, so this is t is equal to a, this is t is equal to b. Same thing, I just kind of picked it back up so you can visualize it. And we say that we have some change in arc length, let's say, let me switch colors. Let's say that this one right here. Let's say that's some small change in arc length, and we're calling that ds. Now, is there some way to relate ds to infinitely small changes in x or y? Well, if we think about it, if we really-- and this is all a little bit hand-wavy, I'm not being mathematically rigorous, but I think it'll give you the correct intuition-- if you imagine this is, you can figure out the length of ds if you know the length of these super small changes in x and super small changes in y. So if this distance right here is ds, infinitesimally small change in x, this distance right here is dy, infinitesimally small change in y, right? Then we could figure out ds from the Pythagorean Theorem. You can say that ds is going to be, it's the hypotenuse of this triang.e It's equal to the square root of dx squared plus dy squared. So that seems to make things a little bit, you know, we can get rid of the ds all of a sudden. So let's rewrite this little expression here, using this sense of what ds, is really the square root of dx squared plus dy squared. And I'm not being very rigorous, and actually it's very hard to be rigorous with differentials, but intuitively I think it makes a lot of sense. So we can say that this integral, the area of this curvy curtain, is going to be the integral from t is equal to a to t is equal to b of f of x y, instead of writing ds, we can write this, times the square root of dx squared plus dy squared. Now we at least got rid of this big capital S, but we still haven't solved the problem of, how do you solve something, you know, an integral, a definite integral that looks like this? We have it in terms of t here, but we only have it in terms of x's and y's here. So we need to get everything in terms of t. Well, we know x and y are both functions of t, so we can actually rewrite it like this. We can rewrite it as from t is equal to a, to t is equal to b. And f of x y, we can write it, f is a function of x, which is a function of t, and f is also a function of y, which is also a function of t. So you give me a t, I'll be able to give you an x or y, and once you give me an x or y, I can figure out what f is. So we have that, and then we have this part right here. I'll do it in orange. Square root of dx squared plus dy squared. But we still don't have things in terms of t. We need a dt someplace here in order be able to evaluate this integral. And we'll see that in the next video, when I do a concrete problem. But I really want to give you a sense for the end product, the formula we're going to get at the end product of this video, where it comes from. So one thing we can do, is if we allow ourselves to algebraically manipulate differentials, what we can do is let us multiply and divide by dt. So one way to think about it, you could rewrite, so let me just do this orange part right here. Let's do a little side right here. So if you take this orange part, and write it in pink, and you have dx squared, and then you have plus dy squared, and let's say you just multiply it times dt over dt, right? That's a small change in t, divided by a small change in t. That's 1, so of course you can multiply it by that. If we're to bring in this part inside of the square root sign, right, so let me rewrite this. This is the same thing as 1 over dt times the square root of dx squared plus dy squared, and then times that dt. Right? I just wanted to write it this way to show you I'm just multiplying by 1. And here, I'm just taking this dt, writing it there, and leaving this over here. And now if I wanted to bring this into the square root sign, this is the same thing, this is equal to, and I'll do it very slowly, just to make sure, I'll allow you to believe that I'm not doing anything shady with the algebra. This is the same thing as the square root of 1 over dt squared, let me make the radical a little bit bigger, times dx squared plus dy squared, and all of that times dt, right? I didn't do anything, you could just take the square root of this and you'd get 1 over dt. And if I just distribute this, this is equal to the square root, and we have our dt at the end, of dx squared, or we could even write, dx over dt squared, plus dy over dt squared. Right? dx squared over dt squared is just dx over dt squared, same thing with the y's. And now all of a sudden, this starts to look pretty interesting. Let's substitute this expression with this one. We said that these are equivalent. And I'll switch colors, just for the sake of it. So we have the integral. From t is equal to a. Let me get our drawing back, if I-- from t is equal to a to t is equal to b of f of x of t times, or f of x of t and f of, or and y of t, they're both functions of t, and now instead of this expression, we can write the square root of, well, what's dx, what's the change in x with respect to, whatever this parameter is? What is dx dt? dx dt is the same thing as g prime of t. Right? x is a function of t. The function I wrote is g prime of t. And then dy dt is same thing as h prime of t. We could say that, you know, this function of t. So I just wanted to make that clear. We know these two functions, so we can just take their derivatives with respect to t. But I'm just going to leave it in that form. So the square root, and we take the derivative of x with respect to t squared, plus the derivative of y with respect to t squared, and all of that times dt. And this might look like some strange and convoluted formula, but this is actually something that we know how to deal with. We've now simplified this strange, you know, this arc-length problem, or this line integral, right? That's essentially what we're doing. We're taking an integral over a curve, or over a line, as opposed to just an interval on the x-axis. We've taken the strange line integral, that's in terms of the arc length of the line, and x's and y's, and we've put everything in terms of t. And I'm going to show you that in the next video, right? Everything is going to be expressed in terms of t, so this just turns into a simple, definite integral. So hopefully that didn't confuse you too much. I think you're going to see in the next video that this, right here, is actually a very straightforward thing to implement. And just to remind you where it all came from, I think I got the parentheses right. This right here was just a change in our arc length. That whole thing right there was just a change in arc length. And this is just the height of our function at that point. And we're just summing it, doing an infinite sum of infinitely small lengths. So this was a change in our arc length times the height. This is going to have an infinitely narrow width, and they're going to take an infinite number of these rectangles to get the area of this entire fence, or this entire curtain. And that's what this definite integral will give us, and we'll actually apply it in the next video.

Vector calculus

In qualitative terms, a line integral in vector calculus can be thought of as a measure of the total effect of a given tensor field along a given curve. For example, the line integral over a scalar field (rank 0 tensor) can be interpreted as the area under the field carved out by a particular curve. This can be visualized as the surface created by z = f(x,y) and a curve C in the xy plane. The line integral of f would be the area of the "curtain" created—when the points of the surface that are directly over C are carved out.

Line integral of a scalar field

The line integral over a scalar field f can be thought of as the area under the curve C along a surface z = f(x,y), described by the field.

Definition

For some scalar field where , the line integral along a piecewise smooth curve is defined as

where is an arbitrary bijective parametrization of the curve such that r(a) and r(b) give the endpoints of and a < b. Here, and in the rest of the article, the absolute value bars denote the standard (Euclidean) norm of a vector.

The function f is called the integrand, the curve is the domain of integration, and the symbol ds may be intuitively interpreted as an elementary arc length of the curve (i.e., a differential length of ). Line integrals of scalar fields over a curve do not depend on the chosen parametrization r of .[2]

Geometrically, when the scalar field f is defined over a plane (n = 2), its graph is a surface z = f(x, y) in space, and the line integral gives the (signed) cross-sectional area bounded by the curve and the graph of f. See the animation to the right.

Derivation

For a line integral over a scalar field, the integral can be constructed from a Riemann sum using the above definitions of f, C and a parametrization r of C. This can be done by partitioning the interval [a, b] into n sub-intervals [ti−1, ti] of length Δt = (ba)/n, then r(ti) denotes some point, call it a sample point, on the curve C. We can use the set of sample points {r(ti): 1 ≤ in} to approximate the curve C as a polygonal path by introducing the straight line piece between each of the sample points r(ti−1) and r(ti). (The approximation of a curve to a polygonal path is called rectification of a curve, see here for more details.) We then label the distance of the line segment between adjacent sample points on the curve as Δsi. The product of f(r(ti)) and Δsi can be associated with the signed area of a rectangle with a height and width of f(r(ti)) and Δsi, respectively. Taking the limit of the sum of the terms as the length of the partitions approaches zero gives us

By the mean value theorem, the distance between subsequent points on the curve, is

Substituting this in the above Riemann sum yields

which is the Riemann sum for the integral

Line integral of a vector field

Definition

For a vector field F: URnRn, the line integral along a piecewise smooth curve CU, in the direction of r, is defined as

where · is the dot product, and r: [a, b] → C is a bijective parametrization of the curve C such that r(a) and r(b) give the endpoints of C.

A line integral of a scalar field is thus a line integral of a vector field, where the vectors are always tangential to the line of the integration.

Line integrals of vector fields are independent of the parametrization r in absolute value, but they do depend on its orientation. Specifically, a reversal in the orientation of the parametrization changes the sign of the line integral.[2]

From the viewpoint of differential geometry, the line integral of a vector field along a curve is the integral of the corresponding 1-form under the musical isomorphism (which takes the vector field to the corresponding covector field), over the curve considered as an immersed 1-manifold.

Derivation

The trajectory of a particle (in red) along a curve inside a vector field. Starting from a, the particle traces the path C along the vector field F. The dot product (green line) of its tangent vector (red arrow) and the field vector (blue arrow) defines an area under a curve, which is equivalent to the path's line integral. (Click on image for a detailed description.)

The line integral of a vector field can be derived in a manner very similar to the case of a scalar field, but this time with the inclusion of a dot product. Again using the above definitions of F, C and its parametrization r(t), we construct the integral from a Riemann sum. We partition the interval [a, b] (which is the range of the values of the parameter t) into n intervals of length Δt = (ba)/n. Letting ti be the ith point on [a, b], then r(ti) gives us the position of the ith point on the curve. However, instead of calculating up the distances between subsequent points, we need to calculate their displacement vectors, Δri. As before, evaluating F at all the points on the curve and taking the dot product with each displacement vector gives us the infinitesimal contribution of each partition of F on C. Letting the size of the partitions go to zero gives us a sum

By the mean value theorem, we see that the displacement vector between adjacent points on the curve is

Substituting this in the above Riemann sum yields

which is the Riemann sum for the integral defined above.

Path independence

If a vector field F is the gradient of a scalar field G (i.e. if F is conservative), that is,

then by the multivariable chain rule the derivative of the composition of G and r(t) is
which happens to be the integrand for the line integral of F on r(t). It follows, given a path C, that

In other words, the integral of F over C depends solely on the values of G at the points r(b) and r(a), and is thus independent of the path between them. For this reason, a line integral of a conservative vector field is called path independent.

Applications

The line integral has many uses in physics. For example, the work done on a particle traveling on a curve C inside a force field represented as a vector field F is the line integral of F on C.[3]

Flow across a curve

For a vector field , F(x, y) = (P(x, y), Q(x, y)), the line integral across a curve CU, also called the flux integral, is defined in terms of a piecewise smooth parametrization r: [a,b] → C, r(t) = (x(t), y(t)), as:

Here is the dot product, and is the clockwise perpendicular of the velocity vector .

The flow is computed in an oriented sense: the curve C has a specified forward direction from r(a) to r(b), and the flow is counted as positive when F(r(t)) is on the clockwise side of the forward velocity vector r'(t).

Complex line integral

In complex analysis, the line integral is defined in terms of multiplication and addition of complex numbers. Suppose U is an open subset of the complex plane C, f : UC is a function, and is a curve of finite length, parametrized by γ: [a,b] → L, where γ(t) = x(t) + iy(t). The line integral

may be defined by subdividing the interval [a, b] into a = t0 < t1 < ... < tn = b and considering the expression

The integral is then the limit of this Riemann sum as the lengths of the subdivision intervals approach zero.

If the parametrization γ is continuously differentiable, the line integral can be evaluated as an integral of a function of a real variable:

When L is a closed curve (initial and final points coincide), the line integral is often denoted sometimes referred to in engineering as a cyclic integral.

The line integral with respect to the conjugate complex differential is defined[4] to be

The line integrals of complex functions can be evaluated using a number of techniques. The most direct is to split into real and imaginary parts, reducing the problem to evaluating two real-valued line integrals. The Cauchy integral theorem may be used to equate the line integral of an analytic function to the same integral over a more convenient curve. It also implies that over a closed curve enclosing a region where f(z) is analytic without singularities, the value of the integral is simply zero, or in case the region includes singularities, the residue theorem computes the integral in terms of the singularities. This also implies the path independence of complex line integral for analytic functions.

Example

Consider the function f(z) = 1/z, and let the contour L be the counterclockwise unit circle about 0, parametrized by z(t) = eit with t in [0, 2π] using the complex exponential. Substituting, we find:

This is a typical result of Cauchy's integral formula and the residue theorem.

Relation of complex line integral and line integral of vector field

Viewing complex numbers as 2-dimensional vectors, the line integral of a complex-valued function has real and complex parts equal to the line integral and the flux integral of the vector field corresponding to the conjugate function Specifically, if parametrizes L, and corresponds to the vector field then:

By Cauchy's theorem, the left-hand integral is zero when is analytic (satisfying the Cauchy–Riemann equations) for any smooth closed curve L. Correspondingly, by Green's theorem, the right-hand integrals are zero when is irrotational (curl-free) and incompressible (divergence-free). In fact, the Cauchy-Riemann equations for are identical to the vanishing of curl and divergence for F.

By Green's theorem, the area of a region enclosed by a smooth, closed, positively oriented curve is given by the integral This fact is used, for example, in the proof of the area theorem.

Quantum mechanics

The path integral formulation of quantum mechanics actually refers not to path integrals in this sense but to functional integrals, that is, integrals over a space of paths, of a function of a possible path. However, path integrals in the sense of this article are important in quantum mechanics; for example, complex contour integration is often used in evaluating probability amplitudes in quantum scattering theory.

See also

References

  1. ^ Kwong-Tin Tang (30 November 2006). Mathematical Methods for Engineers and Scientists 2: Vector Analysis, Ordinary Differential Equations and Laplace Transforms. Springer Science & Business Media. ISBN 978-3-540-30268-1.
  2. ^ a b Nykamp, Duane. "Line integrals are independent of parametrization". Math Insight. Retrieved September 18, 2020.
  3. ^ "16.2 Line Integrals". www.whitman.edu. Retrieved 2020-09-18.
  4. ^ Ahlfors, Lars (1966). Complex Analysis (2nd ed.). New York: McGraw-Hill. p. 103.

External links

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