Line integrals are used in many applications. For example, in electricity and magnetism, they provide simple methods for computing the electric and magnetic fields for simple geometries.

By a smooth curve C in the xy-plane will be meant a curve representable in the form under standard Cartesian coordinates:

where x and y are continuous and have continuous derivatives for 𝑎 ≤ t ≤ b. The curve C can be assigned a direction, which will usually be that of increasing t. If A denotes the point [ϕ(𝑎), ψ(𝑎)] and B denotes the point [ϕ(b), ψ(b)],then C can be thought of as the path of a point moving continuously from A to B. This path may cross itself. If the initial point A and terminal point B coincide, C is termed a closed curve; if, in addition, (x, y ) moves from A to B = A without retracing any other point, C is called a simple closed curve.

 

A scalar field has a value associated to each point in space. Examples of scalar fields are height, temperature or pressure maps. In a two-dimensional field, the value at each point can be thought of as a height of a surface embedded in three dimensions. The line integral of a curve along this scalar field is equivalent to the area under a curve traced over the surface defined by the field.

Let us consider a curve C on ℝ² (or ℝ³) parametrized by r =r(t), 𝑎 ≤ t ≤ b, which is a piecewise smooth vector-valued function. We can find its length by a limiting process on piecewise linear curves that approximate C.

Let 𝑎 = t₀ < t₁ < ⋯ < t_n = b be a partition of closed interval [𝑎, b] and Δ_k = t_k − t_k-1. For n = 1, we have

where \( \displaystyle \quad \mathbf{r}(t) = \left( x(t), y(t) \right) \ \) and \( \displaystyle \ \mathbf{i} = \hat{\bf x} , \ \mathbf{j} = \hat{\bf y} \ \) are unit vectors in ℝ².

RJB Figure 7.5 on page 560 AEM by Tyrin In Figure 7.5, the curve C is in black, and connecting points on the curve are line segments.

The distance between two points r(t₁) and r(t₀) on the curve C is

where ∥ ⋅ ∥ is the Euclidean norm of the 2D vector. An approximation for L, the total arclength of the curve C, is given by because Δt_k is positive for every k. Take the limit, as both n → ∞ and each Δt_k → 0, to get the exact total arclength: From engineering point of view, this makes sense because \( \displaystyle \quad \| \dot{\bf r}(t) \| \ \) is the speed of traveled point when t is the time. This formula leads to the following definition.     Example 1: Arclength     Using these notations and definition of the arclength, we introduce the line integral. Note that \( \displaystyle \quad \| \dot{\bf r} (t) \| = + \sqrt{\dot{x}^2 (t) + \dot{y}^2 (t) + \dot{z}^2 (t)} \ \) (with positive branch of square root). We remark that the value of a line integral on C does not depend on the particular parametrization of C, but only on the order in which the points of C are traced.     Example 2: Line integral over helix     Example 3: The average value of a function     Besides the line integrals defined by Eq.\eqref{EqLine.1}, we have the following: This formulas reduce the integrals to ordinary definite integrals in Riemann sense and are thus essential for computation of particular integrals.     Example 4: ?????

In many applications, the path C is not itself smooth but is composed of a finite number of arcs, each of which is smooth. Then C might be a broken line. In this case, C is termed piecewise smooth. The line integral along C is simply, by definition, the sum of the integrals along the pieces.

 

Suppose F = F(x, y, z) is a vector field, that is, a vector-valued function. We define a line integral of a vector field as follows.     Example 5: ????     Eq.\eqref{EqLine.2} can be rewritten in an equivalent form: where F(r) = P(x, y, z) i + Q(x, y, z) j + R(x, y, z) k is the given vector field and dr = dx i + dy j + dz k is the differential.     Example 6: ??????

For simplicity, let us consider a two-dimensional case when the vector field has only two components:     F(r) = P(x, y) i + Q(x, y) j. Then the line integral

has then a simple vector interpretation as follows: where F_T denotes the tangential component of F, that is, the component of F in the direction of the tangent vector T in the direction of increasing arclength s: Therefore, Then     Example 7: Polar coordinates     In the two-dimensional case in which F = P i + Q j is a force field, the integral \( \displaystyle \quad \int \mathbf{F}_T {\text d}s = \int P\,{\text d}x + Q\,{\text d}y \quad \) represents the work done by this force in moving a particle from one end of C to the another end. For the work done is defined in mechanics as "force times distance" or, more precisely, as follows: where Here Δr_i = Δx_ii + Δy_ij is a small increment. Line integral \eqref{EqLine.3} can then be written in the vector form If C is represented in terms of a parameter t, then     Example 8: ??????     If r is the position vector of a particle of mass m moving on C and F is the force applied, then by Newton's Second Law, So if \v| = v, we have Applying the Fundamental Theorem of Calculus, we obtain So the work done equals the gain in kinetic energy. This is a basic law of mechanics.     Example 9: ??????

Given a smooth path C on the plane, we can assign two orthogonal vectors at every point on C---a tangent vector T and a normal vector to the tangent line:

because the vector product v × k of a vector v in the xy-plane with k is a vector u having the same magnitude as v and π/2 behind v. For any vector field F = P i + Q i on the plane. we can construct two projections on these two orthogonal vectors, T and n, using dot product: where n is a vector orthogonal to the tangent vector T, so n ⊥ T. It is convenient to normalize vectors n and T by considering the unit vectors that are denoted by \( \displaystyle \quad \hat{\bf n} = {\bf n}/\|{\bf n}\| \ \) and \( \displaystyle \quad \hat{\bf T} = \dot{\bf r}/ \| \dot{\bf r} \| . \ \) Correspondingly, we can build the line integral:     Example 10: Normal field

 

Subsection: Exercises

 

References