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Line integral

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Definite integral of a scalar or vector field along a path

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\int _{a}^{b}f'(t)\,dt=f(b)-f(a)

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Inmathematics, aline integral is anintegral where thefunction to be integrated is evaluated along acurve.^[1] The termspath integral,curve integral, andcurvilinear integral are also used;contour integral is used as well, although that is typically reserved forline integrals in the complex plane.

The function to be integrated may be ascalar field or avector 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 (commonlyarc length or, for a vector field, thescalar product of the vector field with adifferential vector in the curve). This weighting distinguishes the line integral from simpler integrals defined onintervals. Many simple formulae in physics, such as the definition ofwork as $W=\mathbf {F} \cdot \mathbf {s}$ , have natural continuous analogues in terms of line integrals, in this case ${\textstyle W=\int _{L}\mathbf {F} (\mathbf {s} )\cdot d\mathbf {s} }$ , which computes thework done on an object moving through an electric or gravitational fieldF along a path $L {\displaystyle L}$ .

Vector calculus

In qualitative terms, a line integral in vector calculus can be thought of as a measure of the total effect of a giventensor 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 byz =f(x,y) and a curveC in thexy plane. The line integral off would be the area of the "curtain" created—when the points of the surface that are directly overC are carved out.

Line integral of a scalar field

The line integral over a scalar fieldf can be thought of as the area under the curveC along a surfacez =f(x,y), described by the field.

Definition

For somescalar field $f\colon U\to \mathbb {R}$ where $U\subseteq \mathbb {R} ^{n}$ , the line integral along apiecewise smooth curve ${\mathcal {C}}\subset U$ is defined as $\int _{\mathcal {C}}f\,ds=\int _{a}^{b}f\left(\mathbf {r} (t)\right)\left|\mathbf {r} '(t)\right|\,dt,$ where $\mathbf {r} \colon [a,b]\to {\mathcal {C}}$ is an arbitrarybijective parametrization of the curve ${\mathcal {C}}$ such thatr(a) andr(b) give the endpoints of ${\mathcal {C}}$ anda <b. Here, and in the rest of the article, the absolute value bars denote thestandard (Euclidean) norm of a vector.

The functionf is called the integrand, the curve ${\mathcal {C}}$ is the domain of integration, and the symbolds may be intuitively interpreted as an elementaryarc length of the curve ${\mathcal {C}}$ (i.e., a differential length of ${\mathcal {C}}$ ). Line integrals of scalar fields over a curve ${\mathcal {C}}$ do not depend on the chosen parametrizationr of ${\mathcal {C}}$ .^[2]

Geometrically, when the scalar fieldf is defined over a plane(n = 2), its graph is a surfacez =f(x,y) in space, and the line integral gives the (signed)cross-sectional area bounded by the curve ${\mathcal {C}}$ and the graph off. See the animation to the right.

Derivation

For a line integral over a scalar field, the integral can be constructed from aRiemann sum using the above definitions off,C and a parametrizationr ofC. This can be done by partitioning theinterval[a,b] inton sub-intervals[t_i−1,t_i] of lengthΔt = (b −a)/n, thenr(t_i) denotes some point, call it a sample point, on the curveC. We can use theset of sample points{r(t_i): 1 ≤i ≤n} to approximate the curveC as apolygonal path by introducing the straight line piece between each of the sample pointsr(t_i−1) andr(t_i). (The approximation of a curve to a polygonal path is calledrectification of a curve, seehere for more details.) We then label the distance of the line segment between adjacent sample points on the curve asΔs_i. The product off(r(t_i)) andΔs_i can be associated with the signed area of a rectangle with a height and width off(r(t_i)) andΔs_i, respectively. Taking thelimit of thesum of the terms as the length of the partitions approaches zero gives us $I=\lim _{\Delta s_{i}\to 0}\sum _{i=1}^{n}f(\mathbf {r} (t_{i}))\,\Delta s_{i}.$

By themean value theorem, the distance between subsequent points on the curve, is $\Delta s_{i}=\left|\mathbf {r} (t_{i}+\Delta t)-\mathbf {r} (t_{i})\right|\approx \left|\mathbf {r} '(t_{i})\Delta t\right|$

Substituting this in the above Riemann sum yields $I=\lim _{\Delta t\to 0}\sum _{i=1}^{n}f(\mathbf {r} (t_{i}))\left|\mathbf {r} '(t_{i})\right|\Delta t$ which is the Riemann sum for the integral $I=\int _{a}^{b}f(\mathbf {r} (t))\left|\mathbf {r} '(t)\right|dt.$

Line integral of a vector field

Definition

For avector fieldF:U ⊆Rⁿ →Rⁿ, the line integral along apiecewise smooth curveC ⊂U, in the direction ofr, is defined as $\int _{C}\mathbf {F} (\mathbf {r} )\cdot d\mathbf {r} =\int _{a}^{b}\mathbf {F} (\mathbf {r} (t))\cdot \mathbf {r} '(t)\,dt$ where· is thedot product, andr: [a,b] →C is a regularparametrization (i.e: $||\mathbf {r} '(t)||\neq 0\;\;\forall t\in [a,b]$ ) of the curveC such thatr(a) andr(b) give the endpoints ofC.

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

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

From the viewpoint ofdifferential geometry, the line integral of a vector field along a curve is the integral of the corresponding 1-form under themusical isomorphism (which takes the vector field to the correspondingcovector field), over the curve considered as animmersed 1-manifold.

Derivation

The trajectory of a particle (in red) along a curve inside a vector field. Starting froma, the particle traces the pathC along the vector fieldF. 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 ofF,C and its parametrizationr(t), we construct the integral from aRiemann sum. We partition theinterval[a,b] (which is the range of the values of theparametert) inton intervals of lengthΔt = (b −a)/n. Lettingt_i be theith point on[a,b], thenr(t_i) gives us the position of theith point on the curve. However, instead of calculating up the distances between subsequent points, we need to calculate theirdisplacement vectors,Δr_i. As before, evaluatingF at all the points on the curve and taking the dot product with each displacement vector gives us theinfinitesimal contribution of each partition ofF onC. Letting the size of the partitions go to zero gives us a sum $I=\lim _{\Delta t\to 0}\sum _{i=1}^{n}\mathbf {F} (\mathbf {r} (t_{i}))\cdot \Delta \mathbf {r} _{i}$

By themean value theorem, we see that the displacement vector between adjacent points on the curve is $\Delta \mathbf {r} _{i}=\mathbf {r} (t_{i}+\Delta t)-\mathbf {r} (t_{i})\approx \mathbf {r} '(t_{i})\,\Delta t.$

Substituting this in the above Riemann sum yields $I=\lim _{\Delta t\to 0}\sum _{i=1}^{n}\mathbf {F} (\mathbf {r} (t_{i}))\cdot \mathbf {r} '(t_{i})\,\Delta t,$

which is the Riemann sum for the integral defined above.

Path independence

Main article:Gradient theorem

If a vector fieldF is thegradient of ascalar fieldG (i.e. ifF isconservative), that is, $\mathbf {F} =\nabla G,$ then by themultivariable chain rule thederivative of thecomposition ofG andr(t) is ${\frac {dG(\mathbf {r} (t))}{dt}}=\nabla G(\mathbf {r} )\cdot \mathbf {r} '(t)=\mathbf {F} (\mathbf {r} (t))\cdot \mathbf {r} '(t)$ which happens to be the integrand for the line integral ofF onr(t). It follows, given a pathC, that $\int _{C}\mathbf {F} (\mathbf {r} )\cdot d\mathbf {r} =\int _{a}^{b}\mathbf {F} (\mathbf {r} (t))\cdot \mathbf {r} '(t)\,dt=\int _{a}^{b}{\frac {dG(\mathbf {r} (t))}{dt}}\,dt=G(\mathbf {r} (b))-G(\mathbf {r} (a)).$

In other words, the integral ofF overC depends solely on the values ofG at the pointsr(b) andr(a), and is thus independent of the path between them. For this reason, a line integral of a conservative vector field is calledpath independent.

Applications

The line integral has many uses in physics. For example, thework done on a particle traveling on a curveC inside a force field represented as a vector fieldF is the line integral ofF onC.^[3]

For another example, seeAmpère's circuital law.

Flow across a curve

For avector field $\mathbf {F} \colon U\subseteq \mathbb {R} ^{2}\to \mathbb {R} ^{2}$ ,F(x,y) = (P(x,y),Q(x,y)), theline integral across a curveC ⊂U, also called theflux integral, is defined in terms of apiecewise smooth parametrizationr: [a,b] →C,r(t) = (x(t),y(t)), as: $\int _{C}\mathbf {F} (\mathbf {r} )\cdot d\mathbf {r} ^{\perp }=\int _{a}^{b}{\begin{bmatrix}P{\big (}x(t),y(t){\big )}\\Q{\big (}x(t),y(t){\big )}\end{bmatrix}}\cdot {\begin{bmatrix}y'(t)\\-x'(t)\end{bmatrix}}~dt=\int _{a}^{b}\left(-Q~dx+P~dy\right).$

Here⋅ is the dot product, and $\mathbf {r} '(t)^{\perp }=(y'(t),-x'(t))$ is the clockwise perpendicular of the velocity vector $\mathbf {r} '(t)=(x'(t),y'(t))$ .

The flow is computed in an oriented sense: the curveC has a specified forward direction fromr(a) tor(b), and the flow is counted as positive whenF(r(t)) is on the clockwise side of the forward velocity vectorr'(t).

Complex line integral

Incomplex analysis, the line integral is defined in terms ofmultiplication andaddition of complex numbers. SupposeU is anopen subset of thecomplex planeC,f :U →C is a function, and $L\subset U$ is a curve of finite length, parametrized byγ: [a,b] →L, whereγ(t) =x(t) +iy(t). The line integral $\int _{L}f(z)\,dz$ may be defined by subdividing theinterval [a,b] intoa =t₀ <t₁ < ... <t_n =b and considering the expression $\sum _{k=1}^{n}f(\gamma (t_{k}))\,[\gamma (t_{k})-\gamma (t_{k-1})]=\sum _{k=1}^{n}f(\gamma _{k})\,\Delta \gamma _{k}.$

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

If the parametrizationγ iscontinuously differentiable, the line integral can be evaluated as an integral of a function of a real variable: $\int _{L}f(z)\,dz=\int _{a}^{b}f(\gamma (t))\gamma '(t)\,dt.$

WhenL is a closed curve (initial and final points coincide), the line integral is often denoted ${\textstyle \oint _{L}f(z)\,dz,}$ sometimes referred to in engineering as acyclic integral.

To establish a complete analogy with the line integral of a vector field, one must go back to the definition of differentiability in multivariable calculus. The gradient is defined fromRiesz representation theorem, and inner products in complex analysis involve conjugacy (the gradient of a function $\gamma$ at some $z\in \mathbb {C}$ would be ${\overline {\gamma '(z)}}$ , and the complex inner product would attribute twice a conjugate to $\gamma '$ in the vector field definition of a line integral).

The line integral with respect to the conjugate complex differential ${\overline {dz}}$ is defined^[4] to be $\int _{L}f(z){\overline {dz}}:={\overline {\int _{L}{\overline {f(z)}}\,dz}}=\int _{a}^{b}f(\gamma (t)){\overline {\gamma '(t)}}\,dt.$

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. TheCauchy integral theorem may be used to equate the line integral of ananalytic function to the same integral over a more convenient curve. It also implies that over a closed curve enclosing a region wheref(z) is analytic withoutsingularities, the value of the integral is simply zero, or in case the region includes singularities, theresidue 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 functionf(z) = 1/z, and let the contourL be the counterclockwiseunit circle about 0, parametrized byz(t) =e^it witht in[0, 2π] using thecomplex exponential. Substituting, we find: ${\begin{aligned}\oint _{L}{\frac {1}{z}}\,dz&=\int _{0}^{2\pi }{\frac {1}{e^{it}}}ie^{it}\,dt=i\int _{0}^{2\pi }e^{-it}e^{it}\,dt\\&=i\int _{0}^{2\pi }dt=i(2\pi -0)=2\pi i.\end{aligned}}$

This is a typical result ofCauchy's integral formula and theresidue theorem.

Relation of complex line integral and line integral of vector field

Viewing complex numbers as 2-dimensionalvectors, the line integral of a complex-valued function $f(z)$ has real and complex parts equal to the line integral and the flux integral of the vector field corresponding to theconjugate function ${\overline {f(z)}}.$ Specifically, if $\mathbf {r} (t)=(x(t),y(t))$ parametrizesL, and $f(z)=u(z)+iv(z)$ corresponds to the vector field $\mathbf {F} (x,y)={\overline {f(x+iy)}}=(u(x+iy),-v(x+iy)),$ then: ${\begin{aligned}\int _{L}f(z)\,dz&=\int _{L}(u+iv)(dx+i\,dy)\\&=\int _{L}(u,-v)\cdot (dx,dy)+i\int _{L}(u,-v)\cdot (dy,-dx)\\&=\int _{L}\mathbf {F} (\mathbf {r} )\cdot d\mathbf {r} +i\int _{L}\mathbf {F} (\mathbf {r} )\cdot d\mathbf {r} ^{\perp }.\end{aligned}}$

ByCauchy's theorem, the left-hand integral is zero when $f(z)$ is analytic (satisfying theCauchy–Riemann equations) for any smooth closed curve L. Correspondingly, byGreen's theorem, the right-hand integrals are zero when $\mathbf {F} ={\overline {f(z)}}$ isirrotational (curl-free) andincompressible (divergence-free). In fact, the Cauchy-Riemann equations for $f(z)$ are identical to the vanishing of curl and divergence forF.

ByGreen's theorem, the area of a region enclosed by a smooth, closed, positively oriented curve $L {\displaystyle L}$ is given by the integral ${\textstyle {\frac {1}{2i}}\int _{L}{\overline {z}}\,dz.}$ This fact is used, for example, in the proof of thearea theorem.

Quantum mechanics

Thepath integral formulation ofquantum mechanics actually refers not to path integrals in this sense but tofunctional integrals, that is, integrals over a space of paths, of a functionof 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 evaluatingprobability amplitudes in quantumscattering theory.

See also

References

^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.
^^a ^bNykamp, Duane."Line integrals are independent of parametrization".Math Insight. RetrievedSeptember 18, 2020.
^"16.2 Line Integrals".www.whitman.edu. Retrieved2020-09-18.
^Ahlfors, Lars (1966).Complex Analysis (2nd ed.). New York: McGraw-Hill. p. 103.

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