Line Integrals


The process of evaluating a function along a path/curve is line integral as the term named.

  • Line integral over a continuous scalar field $f$ along a smooth(at-least piecewise smooth) curve $C$  parameterized as $\vec{r}(t)$ in an interval $a\leq t\leq b$ is defined as -
    $$\int_C f\,ds=\int_a^b f(\vec r(t))\,||\vec r^{\prime}(t)||\, dt =\lim_{n\to \infty}\sum_{i=1}^{n} f_i\, \Delta s_i$$
    Where $||\vec r^{\prime}(t)||$ is the norm/magnitude of $\vec r^{\prime}(t)$ and if we divide the curve $C$ into n small arcs then $\Delta s_i$ is the length of $i$-th arc and $f_i$ is value of the function at any point  in that small arc.
  •   The line integral over continuous vector field $\vec F$ along a curve $C$ parameterized as $\vec{r}(t)$ ($a\leq t\leq b$) is defined as -
     $$\int_C \vec F \cdot d\vec r=\int_a^b \vec F(\vec r(t))\cdot\vec r^{\prime}(t)\, dt= \int_C \vec F \cdot \vec T ds$$
     Where, $\vec{r}^{\prime}(t)=\dfrac{d\vec r}{dt}$ and $\vec T\left( t \right) = \dfrac{{\vec r'\left( t \right)}}{{\left\| {\vec r'\left( t \right)} \right\|}}$ is unit tangent vector and $ds$ is elementary curve length .




Much like double integrals, line integration is a process by which we can integrate multivariable functions. However, with line integral, we integrate the function along a curve.

In other words, through line integration, we can find the length of a curve in space. Line integrals are used to calculate work done along a path in a force field, flux thorough a two-dimensional curve . Close line integrals in a physical field gives us insight about the nature (i.e. conservative) of the field.



Bird's Eye View       

Compared to single variable definite integral, in the line integral, we have a multivariable function integrated along a curvy path.

For doing the integral along a line usually, we first have to parameterize the equation of the line/path with a single variable, WHY? From the definition of the line integral ($\int_C f\, ds$) we see that we have a single integral so, we need a single variable which ties up all the information (different coordinate variables) of the path into one, for representing the path of integration!
Then we use this parametric relation to represent the integrand in terms of that parameter. Then what remains is more or less a single integral.

If we have a scalar field $f(x,y,z)$ to be integrated along a curve $C$ parameterized as ,

\begin{align} x & =g_x(t), y=g_y(t), z=g_z(t) \\\text{or, }\, \vec r & =g_x(t)\hat i+g_y(t)\hat j+g_z(t) \hat k\qquad (a \leq t \leq  b)\end{align}

then the line integral is-
$$\int_C f(x,y,z) ds =\int_a^b f(g_x,g_y,g_z)\,||V(t)||\,dt \qquad ,
V(t)=\dfrac{d\vec r}{dt}$$

In some cases parametrization may not be needed i.e. if the integration path is along one of the coordinate axes. As an example where the integration path is along $X$ axis, the number of variables reduces to one eliminating the need of further parametrization  -
$$\int_C f(x,y) ds = \int_{x_1}^{x_2} f(x,0) dx$$

In case of vector fields each coordinate is associated with a scalar function. Here we have to put the parametrization information of the curve $C$ into each of that scalar function. And then take dot product of the vector function with the tangent of the curve $\dfrac{d\vec r}{dt}.$
If we have a vector field $\vec F(x,y,z)$ to be evaluated along the same curve $C$ the line integral becomes-
$$\int_C \vec F(x,y,z).d\vec r= \int_a^b\left[ \vec F(g_x,g_y,g_z) \cdot \dfrac{d\vec r}{dt}\right] dt$$

Line integral on a Scalar Field or a Vector field is somewhat similar.

enlightened Remarks:

  • The curve has to be in the domain of the integrand-function i.e. if we have a function of two variables($x,y$) we can't integrate it along a curve which is defined by more than two variables(i.e. $x,y,z$).
  • Different parametrization of a single curve $C$ gives the same answer always, as long as the parametrization is smooth.
  • Some problems can be done without parametrization but it can also be inefficient.



Context of the Definition

 As we see there are mainly two types of line integrals --

  1.  Scalar line integrals
  2.  Vector line integrals

♦  Scalar line integrals

Here we have scalar function (let $f(x,y,z)$) and a curve $C$. We will first see the line integral in terms of the sum of values(for insight) then how to evaluate it as an integral.

In case of sum method the curve $C$ is chopped up into $n$ small pieces, then the length of that small pieces $\Delta s_i$ multiplied with the value of the function $f(x_i,y_i,z_i)$ at some point($x_i,y_i,z_i$) in that small piece. Then such values of all $n$ small pieces summed up to get the value of the line integral. As we make the pieces very small we get close to the absolute value of line integral of $f(x,y,z)$ along $C$.
$$\int_C f(x,y,z) ds=\lim_{n\to \infty}\sum_{i=1}^{n} f(x_i,y_i,z_i)\, \Delta s_i$$
This is somewhat a generalization of Riemann sums to any curve in space[8]. We can do any line integral by this elementary definition but it is a numerical approach. For good illustration in 3D, we step down to $\mathbb R^2$ domain with $z=f(x,y)$ but this formula is applicable to any dimension.




Animation 1: Line in integral in scalar field $f(x,y)$ as sum.

When the intervals of the arc become infinitesimally  small we get the integral form--
$$\int_C f(x,y,z) ds=\int_a^b f(\vec r(t))\,||\vec r^{\prime}(t)||\, dt $$
Where $\vec r (t)$ is the parametric equation of $C$.

Better to see an example --

Evaluate $\int_C (2+x^2y)\, ds$, where $C$ is the upper half ($y\geq 0$) of the unit circle $x^2+y^2=1$.[4]

The circle can be parameterized as,
$$x=\cos(t)\quad y=\sin(t)\quad
\text{or,}\quad \vec r =\cos(t) \hat i+ \sin(t) \hat j$$
$0\leq t\leq \pi$ describes the upper part of the circle.
In Cartesian coordinates ,
\begin{align} ds=\sqrt {dx^2+dy^2} &=\sqrt {\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\, dt\\
&= \sqrt {\sin^2t+cos^2t}\, dt = dt \end{align}
So the integral becomes,
\begin{align} \int_C (2+x^2y)\, ds & =\int_0^{\pi}(2+\cos^2t\, \sin t)dt\\
&= \left[2t-\dfrac{\cos^3t}{3}\right]_0^{\pi}\\
&= 2\pi+\frac 2 3

One can also do it by directly substituting $y=\sqrt {1-x^2}$ and calculating $ds$ as $\left(\sqrt{1+\dfrac{dy}{dx}}\right)dx$.



Animation 2: Line Integration in scalar field $f(x,y)$ as area

Vector line integrals

Let $\vec F(x,y,z)=P(x,y,z)\hat i+Q(x,y,z)\hat j+R(x,y,z)\hat k$ is a continuous vector field in $\mathbb{R}^3$ to be evaluated along a curve $C$.

In the spirit of the previous section, $C$ is chopped up into $n$ small subarcs each having a length $\Delta s_i$. In case of vector fields we have to multiply $\Delta s_i$ with the dot product of $\vec F(x_i,y_i,z_i)$ and the unit tangent vector $\vec T(x_i,y_i,z_i)$ at any point $(x_i,y_i,z_i)$ on that subarc. Then all such values for $n$ small subarcs summed up to get the value of the line integral approximately.

$$\int_C \vec F\cdot d\vec r \approx \sum_{i=1}^{n} \vec F(x_i,y_i,z_i)\cdot \vec T(x_i,y_i,z_i)\, \Delta s_i$$



Animation 3: Line in integral in vector field $\vec F(x,y)$ as sum.

When the subarcs become infinitely small the real value of the integral is obtained.
$$\int_C \vec F\cdot d\vec r=\lim_{n\to \infty}\sum_{i=1}^{n} \vec F(x_i,y_i,z_i)\cdot \vec T(x_i,y_i,z_i)\, \Delta s_i=\int_C \vec F\cdot \vec T \, ds$$

if $\vec r(t)$ is the parametric equation of $C$ then $\vec T =\dfrac{\vec r'(t)}{|\vec r'(t)|},\quad \vec r'(t)=\dfrac{d\vec r}{dt}$


Example : Find the line integral in the vector field $\vec F(x,y)=x^2\hat i-xy\hat j$ along a quarter circular path parameterized as $\vec r(t)=\cos t\hat i+\sin t\hat j,\,0\leq t\leq \pi/2$. [4]

Writing $\vec F$ in terms of $t$ :$\vec F(\vec r(t)) \cos^2t\, \hat i-\cos t\sin t\, \hat j$\
$\vec r'(t)=-\sin t\hat i+\cos t\hat j$

So the line integral becomes-
\begin{align} & \int_0^{\pi/2} \vec F(\vec r(t))\cdot \vec r'(t)\, dt\\
&= \int_0^{\pi/2} (-2\cos^2t\,\sin t)\,dt\\
&=\left[\dfrac{2\cos^3t}{3}\right]_0^{\pi/2}=-\frac 2 3



Animation 4 : Line integration in vector field of the previous example.(vectors are not to scale)

Properties of Line Integrals

  •  Scalar fields

  1. $\int_{-C} f\, ds =\int_{C} f\, ds$
  2. $\int_{C} \alpha f\, ds =\alpha\int_{C} f\, ds$,   $\alpha$ is a scalar factor
  3. $\int_{C} (f+g)\, ds =\int_{C} f\, ds+\int_{C} g\, ds$
  •  Vector fields

  1. $\int_{-C} f\, ds =-\int_{C} f\, ds$
  2. $\int_C \alpha \vec F\cdot d\vec r=\alpha\int_C \vec F\cdot d\vec r$,  $\alpha$ is a scalar factor
  3. $\int_C (\vec F+\vec G)\cdot d\vec r=\int_C \vec F\cdot d\vec r+\int_C \vec G\cdot d\vec r$

There is also one type of line integral defined in complex plane named as contour integral. It is has relation with vector line integrals.



1. Finding length or mass of a wire 

A wire in space is described by the parametric equation-
  $$\vec r(t)=(\cos t)\hat i+(\sin t)\hat j+t \hat k,\quad 0\leq t\leq 2\pi$$



Animation 5 : Parametric equation of helix

      If the mass density in space is $\rho(x,y,z)=1$ then the length and mass values will be same --
      Length= $\int_C \mathrm{d}s$,      Mass= $\int_C \rho \, \mathrm{d}s$
      \vec r'(t)&=(-\sin t)\hat i+(\cos t)\hat j+\hat k\\
      ||\vec r'(t)||&=\sqrt{(-\sin t)^2+(\cos t)^2+1}=\sqrt 2\\
      \int_{\text{wire}} \mathrm{d}s=\int_0^{2\pi}& ||\vec r'(t)||\,\mathrm dt=\sqrt{2}\int_0^{2\pi} \mathrm dt= 2\sqrt{2}\cdot \large\pi
     The length and mass of the wire has the value $2\sqrt{2}\large\pi$

Similarly, you can find the moment of inertia and center of mass of the wire.

2. Calculating work done in a force field

     Find the work done in a force field $\vec F=(y-x^2)\hat i+(z-y^2)\hat j+(x-z^2)\hat k$ along a path $\vec r(t)=t \hat i+t^2 \hat j+t^3 \hat k,\quad 0\leq t\leq 1$.
  First evaluate the $\vec F$ on the curve.
  \begin{align}\vec F &=(y-x^2)\hat i+(z-y^2)\hat j+(x-z^2)\hat k \qquad &\\
  \implies\vec F &=(t^2-t^2)\hat i+(t^3-t^4)\hat j+(t-t^6)\hat k
  Take the dot product with the tangent(${\mathrm d\vec r}/{\mathrm dt}$) of the curve ..
  $$\dfrac{\mathrm d\vec r}{\mathrm dt}=\hat i+2t\hat j+3t^2\hat k,\qquad
  \vec F \cdot \dfrac{\mathrm d\vec r}{\mathrm dt}=2t^4-2t^5+3t^3-3t^8\qquad$$
  Now the work done is-
  $$\int_{\text{path}}\vec F \cdot \dfrac{\mathrm d\vec r}{\mathrm dt}\,\mathrm dt=\int_0^1(2t^4-2t^5+3t^3-3t^8) dt=\frac{29}{60}$$

3. Detecting conservative fields

 If close line integral along any path in a Vector field is zero then the field is a conservative one. In those fields, line integrals depend on the final and initial point, not on the evaluation path of the line integral.

Line integrals have applications in finding flux across a curve, current inside a loop (Ampere's law), emf induced in a wire (Faraday's law), and many more.



Line integrals were invented in the early 19th century to solve problems involving fluid flow, forces, electricity, and magnetism[4].

Pause and Ponder

One can see the vector field is composed of scalar functions associated with different directions. More explicitly, in the scalar field, each point in space is associated with some scalar value, but in a vector field, each point in space is associated with some vector.

In the case of vector line integral after taking the dot product of the vectors with the tangent of the curve, we get some values, scalar values (similar to the scalar field but along the curve only). Although the initial definitions and context of the application of line integral in Scalar and Vector fields are little different they have similarities in the calculations.

With Line Integral you can calculate the work it takes to put a satellite in orbit, the amount of fuel a vehicle ( rovers !) need to go through some specific path over a hilly surface.
It also has applications in economics!



  1.  B.R. Kusse and E.A. Westwig. Mathematical Physics: Applied Mathematics for Scientists and Engineers. Wiley, 2nd edition, 1998
  2.  Tom M. Apostol. Calculus, volume 2. Wiley, 2nd edition, 1969.
  3.  Erwin Kreyszig. Advanced Engineering Mathematics, chapter Vector Integral Calculus. Integral Theorems, pages 413–426. Wiley, 10th edition.
  4.  James Stewart. Calculus : Concepts and Contexts, chapter Vector Calculus. Line Integrals, pages 913–924. Brooks/Cole, Cengage Learning, 4th edition.

Further Reading

  1.  For step by step solutions and more application based examples: George B.Thomas and Ross L. Finney. Calculus and Analytic Geometry,chapter Integration in Vector Fields. 9th edition, 1998.
  2. For good problem sets: James Stewart. Calculus : Concepts and Contexts, chapter Vector Calculus. Line Integrals, Brooks/Cole, Cengage Learning, 4th edition
  3. Howard Anton, Irl bivens, Stephen Davis."Line Integrals." Calculus: Early Transcendentals,10th Edition, Wiley Publishers,ISBN 978-0-470-64769-1, page-1094


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