Line integral

For some scalar field ${\displaystyle f\colon U\to \mathbb {R} }$ where ${\displaystyle U\subseteq \mathbb {R} ^{n}}$, the line integral along a piecewise smooth curve ${\displaystyle {\mathcal {C}}\subset U}$ is defined as

${\displaystyle \int _{\mathcal {C}}f(\mathbf {r} )\,ds=\int _{a}^{b}f\left(\mathbf {r} (t)\right)\left|\mathbf {r} '(t)\right|\,dt.}$

where ${\displaystyle \mathbf {r} \colon [a,b]\to {\mathcal {C}}}$ is an arbitrary bijective parametrization of the curve ${\displaystyle {\mathcal {C}}}$ such that r(a) and r(b) give the endpoints of ${\displaystyle {\mathcal {C}}}$ 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 ${\displaystyle {\mathcal {C}}}$ is the domain of integration, and the symbol ds may be intuitively interpreted as an elementary arc length of the curve ${\displaystyle {\mathcal {C}}}$ (i.e., a differential length of ${\displaystyle {\mathcal {C}}}$). Line integrals of scalar fields over a curve ${\displaystyle {\mathcal {C}}}$ do not depend on the chosen parametrization r of ${\displaystyle {\mathcal {C}}}$.[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 ${\displaystyle {\mathcal {C}}}$ and the graph of f. See the animation to the right.

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 [t_i−1, t_i] of length Δt = (b − a)/n, then r(t_i) denotes some point, call it a sample point, on the curve C. We can use the set of sample points {r(t_i): 1 ≤ i ≤ n} to approximate the curve C as a polygonal path by introducing the straight line piece between each of the sample points r(t_i−1) and r(t_i). (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 Δs_i. The product of f(r(t_i)) and Δs_i can be associated with the signed area of a rectangle with a height and width of f(r(t_i)) and Δs_i, respectively. Taking the limit of the sum of the terms as the length of the partitions approaches zero gives us

${\displaystyle I=\lim _{\Delta s_{i}\to 0}\sum _{i=1}^{n}f(\mathbf {r} (t_{i}))\,\Delta s_{i}.}$

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

${\displaystyle \Delta s_{i}=\left|\mathbf {r} (t_{i}+\Delta t)-\mathbf {r} (t_{i})\right|\approx {d}\left|\mathbf {r} (t_{i})\right|}$

Substituting this in the above Riemann sum yields

${\displaystyle I=\lim _{\Delta t\to 0}\sum _{i=1}^{n}f(\mathbf {r} (t_{i}))\left|\mathbf {r} '(t_{i})\right|\Delta t}$

which is the Riemann sum for the integral

${\displaystyle I=\int _{a}^{b}f(\mathbf {r} (t))\left|\mathbf {r} '(t)\right|dt.}$

For a vector field F: U ⊆ Rⁿ → Rⁿ, the line integral along a piecewise smooth curve C ⊂ U, in the direction of r, is defined as

${\displaystyle \int _{C}\mathbf {F} (\mathbf {r} )\cdot \,d\mathbf {r} =\int _{a}^{b}\mathbf {F} (\mathbf {r} (t))\cdot \mathbf {r} '(t)\,dt}$

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.

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 = (b − a)/n. Letting t_i be the ith point on [a, b], then r(t_i) 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, Δr_i. 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

${\displaystyle I=\lim _{\Delta t\to 0}\sum _{i=1}^{n}\mathbf {F} (\mathbf {r} (t_{i}))\cdot \Delta \mathbf {r} _{i}}$

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

${\displaystyle \Delta \mathbf {r} _{i}=\mathbf {r} (t_{i}+\Delta t)-\mathbf {r} (t_{i})\approx \mathbf {r} '(t_{i})\,\Delta t.}$

Substituting this in the above Riemann sum yields

${\displaystyle I=\lim _{\Delta t\to 0}\sum _{i=1}^{n}\mathbf {F} (\mathbf {r} (t_{i}))\cdot \mathbf {r} '(t_{i})\,\Delta t,}$

which is the Riemann sum for the integral defined above.

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

${\displaystyle \mathbf {F} =\nabla G,}$

then by the multivariable chain rule the derivative of the composition of G and r(t) is

${\displaystyle {\frac {dG(\mathbf {r} (t))}{dt}}=\nabla G(\mathbf {r} (t))\cdot \mathbf {r} '(t)=\mathbf {F} (\mathbf {r} (t))\cdot \mathbf {r} '(t)}$

which happens to be the integrand for the line integral of F on r(t). It follows, given a path C , that

${\displaystyle \int _{C}\mathbf {F} (\mathbf {r} )\cdot \,d\mathbf {r} =\int _{a}^{b}\mathbf {F} (\mathbf {r} (t))\cdot \mathbf {r} '(t)\,dt=\int _{a}^{b}{\frac {dG(\mathbf {r} (t))}{dt}}\,dt=G(\mathbf {r} (b))-G(\mathbf {r} (a)).}$

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.

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]

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

${\displaystyle {\begin{aligned}\oint _{L}{\frac {1}{z}}\,dz&=\int _{0}^{2\pi }{\frac {1}{e^{it}}}ie^{it}\,dt=i\int _{0}^{2\pi }e^{-it}e^{it}\,dt\\&=i\int _{0}^{2\pi }dt=i(2\pi -0)=2\pi i.\end{aligned}}}$

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

Viewing complex numbers as 2-dimensional vectors, the line integral of a complex-valued function ${\displaystyle f(z)}$ has real and complex parts equal to the line integral and the flux integral of the vector field corresponding to the conjugate function ${\displaystyle {\overline {f(z)}}.}$ Specifically, if ${\displaystyle \mathbf {r} (t)=(x(t),y(t))}$ parametrizes L, and ${\displaystyle f(z)=u(z)+iv(z)}$ corresponds to the vector field ${\displaystyle \mathbf {F} (x,y)={\overline {f(x+iy)}}=(u(x+iy),-v(x+iy)),}$ then:

${\displaystyle {\begin{aligned}\int _{L}f(z)\,dz&=\int _{L}(u+iv)(dx+i\,dy)\\&=\int _{L}(u,-v)\cdot (dx,dy)+i\int _{L}(u,-v)\cdot (dy,-dx)\\&=\int _{L}\mathbf {F} (\mathbf {r} )\cdot d\mathbf {r} +i\int _{L}\mathbf {F} (\mathbf {r} )\cdot d\mathbf {r} ^{\perp }.\end{aligned}}}$

By Cauchy's theorem, the left-hand integral is zero when ${\displaystyle f(z)}$ 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 ${\displaystyle \mathbf {F} ={\overline {f(z)}}}$ is irrotational (curl-free) and incompressible (divergence-free). In fact, the Cauchy-Riemann equations for ${\displaystyle f(z)}$ 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 ${\displaystyle L}$ is given by the integral ${\displaystyle \textstyle {\frac {1}{2i}}\int _{L}{\overline {z}}\,dz.}$ This fact is used, for example, in the proof of the area theorem.