Line Integrals

In mathematics, a line integral is an integral where the function to be integrated is evaluated along a curve. 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 $W=\mathbf {F} \cdot \mathbf {s}$, 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 the work done on an object moving through an electric or gravitational field F along a path $L$.

Volume, path (line), and surface integrals

Volume Integrals

Given a scalar field $\rho :\mathbb {R} ^{3}\to \mathbb {R}$ that denotes a density at each specific point, and an arbitrary volume $\Omega \subseteq \mathbb {R} ^{3}$, the total "mass" $M$ inside of $\Omega$ can be determined by partitioning $\Omega$ into infinitesimal volumes. At each position $\mathbf {q} \in \Omega$, the volume of the infinitesimal volume is denoted by the infinitesimal $dV$. This gives rise to the following integral:

$M=\iiint _{\mathbf {q} \in \Omega }\rho (\mathbf {q} )dV$

Path Integrals

Given any oriented path $C$ (oriented means that there is a preferred direction), the differential $d\mathbf {q} =dx\mathbf {i} +dy\mathbf {j} +dz\mathbf {k}$ denotes an infinitesimal displacement along $C$ in the preferred direction. This differential can be used in various path integrals. Letting $f:\mathbb {R} ^{3}\to \mathbb {R}$ denote an arbitrary scalar field, and $\mathbf {F} :\mathbb {R} ^{3}\to \mathbb {R} ^{3}$ denote an arbitrary vector field, various path integrals include:

$\int _{\mathbf {q} \in C}f(\mathbf {q} )d\mathbf {q}$, $\int _{\mathbf {q} \in C}f(\mathbf {q} )|d\mathbf {q} |$, $\int _{\mathbf {q} \in C}\mathbf {F} (\mathbf {q} )\cdot d\mathbf {q}$, $\int _{\mathbf {q} \in C}\mathbf {F} (\mathbf {q} )|d\mathbf {q} |$, and many more.

$\int _{\mathbf {q} \in C}d\mathbf {q}$ denotes the total displacement along $C$, and $\int _{\mathbf {q} \in C}|d\mathbf {q} |$ denotes the total length of $C$.

Calculating Path Integrals

To compute a path integral, the continuous oriented curve $C$ must be parameterized. $\mathbf {q} _{C}(t)$ will denote the point along $C$ indexed by $t$ from the range $[t_{0},t_{1}]$. $\mathbf {q} _{C}(t_{0})=\mathbf {q} _{0}$ must be the starting point of $C$ and $\mathbf {q} _{C}(t_{1})=\mathbf {q} _{1}$ must be the ending point of $C$. As $t$ increases, $\mathbf {q} _{C}(t)$ must proceed along $C$ in the preferred direction. An infinitesimal change in $t$, $dt$, results in the infinitesimal displacement $d\mathbf {q} ={\frac {d\mathbf {q} _{C}}{dt}}dt$ along $C$. In the path integral $\int _{\mathbf {q} \in C}\mathbf {F} (\mathbf {q} )\cdot d\mathbf {q}$, the differential $d\mathbf {q}$ can be replaced with ${\frac {d\mathbf {q} _{C}}{dt}}dt$ to get $\int _{t=t_{0}}^{t_{1}}\mathbf {F} (\mathbf {q} _{C}(t))\cdot {\frac {d\mathbf {q} _{C}}{dt}}dt$

Example 1: As an example, consider the vector field $\mathbf {F} (x,y,z)=3\mathbf {i} -x\mathbf {j} +5y\mathbf {k}$ and the straight line curve $C$ that starts at $(1,1,1)$ and ends at $(7,-1,-2)$. $C$ can be parameterized by $\mathbf {q} _{C}(t)=(1+6t,1-2t,1-3t)$ where $t\in [0,1]$. ${\frac {d\mathbf {q} _{C}}{dt}}=6\mathbf {i} -2\mathbf {j} -3\mathbf {k}$. We can then evaluate the path integral $\int _{\mathbf {q} \in C}\mathbf {F} (\mathbf {q} )\cdot d\mathbf {q}$ as follows:

$\int _{\mathbf {q} \in C}\mathbf {F} (\mathbf {q} )\cdot d\mathbf {q} =\int _{\mathbf {q} \in C}(3\mathbf {i} -x\mathbf {j} +5y\mathbf {k} )\cdot d\mathbf {q} =\int _{t=0}^{1}(3\mathbf {i} -(1+6t)\mathbf {j} +5(1-2t)\mathbf {k} )\cdot (6\mathbf {i} -2\mathbf {j} -3\mathbf {k} )dt$

$=\int _{t=0}^{1}(18+(2+12t)+(-15+30t))dt=\int _{t=0}^{1}(5+42t)dt=(5t+21t^{2}){\bigg |}_{t=0}^{1}=26$

If a vector field $\mathbf {F}$ denotes a "force field", which returns the force on an object as a function of position, the work performed on a point mass that traverses the oriented curve $C$ is $W=\int _{\mathbf {q} \in C}\mathbf {F} (\mathbf {q} )\cdot d\mathbf {q}$

Example 2: Consider the gravitational field that surrounds a point mass of $M$ located at the origin: $\mathbf {g} (\mathbf {q} )=-{\frac {GM}{|\mathbf {q} |^{2}}}{\frac {\mathbf {q} }{|\mathbf {q} |}}$ using Newton's inverse square law. The force acting on a point mass of $m$ at position $\mathbf {q}$ is $\mathbf {F} (\mathbf {q} )=m\mathbf {g} (\mathbf {q} )=-{\frac {GMm}{|\mathbf {q} |^{2}}}{\frac {\mathbf {q} }{|\mathbf {q} |}}$. In spherical coordinates the force is $\mathbf {F} (r,\theta ,\phi )=-{\frac {GMm}{r^{2}}}{\hat {\mathbf {r} }}$ (note that ${\hat {\mathbf {r} }},{\hat {\mathbf {\theta } }},{\hat {\mathbf {\phi } }}$ are the unit length mutually orthogonal basis vectors for spherical coordinates).

Consider an arbitrary path $C$ that $m$ traverses that starts at an altitude of $r=r_{1}$ and ends at an altitude of $r=r_{2}$. The work done by the gravitational field is:

$W=\int _{\mathbf {q} \in C}\mathbf {F} (\mathbf {q} )\cdot d\mathbf {q}$

The infinitesimal displacement $d\mathbf {q}$ is equivalent to the displacement expressed in spherical coordinates: $dr\cdot {\hat {\mathbf {r} }}+r\cdot d\theta \cdot {\hat {\mathbf {\theta } }}+r\sin \theta \cdot d\phi \cdot {\hat {\mathbf {\phi } }}$.

$W=\int _{\mathbf {q} \in C}-{\frac {GMm}{r^{2}}}{\hat {\mathbf {r} }}\cdot (dr\cdot {\hat {\mathbf {r} }}+r\cdot d\theta \cdot {\hat {\mathbf {\theta } }}+r\sin \theta \cdot d\phi \cdot {\hat {\mathbf {\phi } }})=\int _{r=r_{1}}^{r_{2}}-{\frac {GMm}{r^{2}}}dr={\frac {GMm}{r}}{\bigg |}_{r=r_{1}}^{r_{2}}=GMm\left({\frac {1}{r_{2}}}-{\frac {1}{r_{1}}}\right)$

The work is equal to the amount of gravitational potential energy lost, so one possible function for the gravitational potential energy is $\phi (r,\theta ,\phi )=-{\frac {GMm}{r}}$ or equivalently, $\phi (\mathbf {q} )=-{\frac {GMm}{|\mathbf {q} |}}$.

Example 3: Consider the spiral $C$ parameterized with respect to $t$ in cylindrical coordinates by $\mathbf {q} _{C}(t)=(\rho =R,\phi =t,z=k{\frac {t}{2\pi }})$. Consider the problem of determining the spiral's length with $t$ restricted to the range $[0,2\pi ]$. An infinitesimal change of $dt$ in $t$ results in the infinitesimal displacement:

$d\mathbf {q} ={\frac {d\mathbf {q} _{C}}{dt}}dt=\left({\frac {d\rho }{dt}}{\hat {\mathbf {\rho } }}+\rho {\frac {d\phi }{dt}}{\hat {\mathbf {\phi } }}+{\frac {dz}{dt}}{\hat {\mathbf {z} }}\right)dt=\left(R{\hat {\mathbf {\phi } }}+{\frac {k}{2\pi }}{\hat {\mathbf {z} }}\right)dt$

The length of the infinitesimal displacement is $|d\mathbf {q} |={\sqrt {R^{2}+\left({\frac {k}{2\pi }}\right)^{2}}}\cdot dt$.

The length of the spiral is therefore: $\int _{\mathbf {q} \in C}|d\mathbf {q} |=\int _{t=0}^{2\pi }{\sqrt {R^{2}+\left({\frac {k}{2\pi }}\right)^{2}}}\cdot dt=2\pi {\sqrt {R^{2}+\left({\frac {k}{2\pi }}\right)^{2}}}={\sqrt {(2\pi R)^{2}+k^{2}}}$

Path Independence

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

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

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

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

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

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

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.

Surface Integrals

Given any oriented surface $\sigma$ (oriented means that the there is a preferred direction to pass through the surface), an infinitesimal portion of the surface is defined by an infinitesimal area $|dS|$, and a unit length outwards oriented normal vector $\mathbf {n}$. $\mathbf {n}$ has a length of 1 and is perpendicular to the surface of $\sigma$, while penetrating $\sigma$ in the preferred direction. The infinitesimal portion of the surface is denoted by the infinitesimal "surface vector": $\mathbf {dS} =|dS|\mathbf {n}$. If a vector field $\mathbf {F} :\mathbb {R} ^{3}\to \mathbb {R} ^{3}$ denotes a flow density, then the flow through the infinitesimal surface portion in the preferred direction is $\mathbf {F} (\mathbf {q} )\cdot \mathbf {dS}$.

The infinitesimal "surface vector" $\mathbf {dS} =\mathbf {n} |dS|$ describes the infinitesimal surface element in a manner similar to how the infinitesimal displacement $d\mathbf {q}$ describes an infinitesimal portion of a path. More specifically, similar to how the interior points on a path do not affect the total displacement, the interior points on a surface to not affect the total surface vector.

The displacement between two points is independent of the path that connects them.

Consider for instance two paths $C_{1}$ and $C_{2}$ that both start at point $A$, and end at point $B$. The total displacements, $\int _{\mathbf {q} \in C_{1}}d\mathbf {q}$ and $\int _{\mathbf {q} \in C_{2}}d\mathbf {q}$, are both equivalent and equal to the displacement between $A$ and $B$. Note however that the total lengths $\int _{\mathbf {q} \in C_{1}}|d\mathbf {q} |$ and $\int _{\mathbf {q} \in C_{2}}|d\mathbf {q} |$ are not necessarily equivalent.

Similarly, given two surfaces $S_{1}$ and $S_{2}$ that both share the same counter-clockwise oriented boundary $C$, the total surface vectors $\int _{\mathbf {q} \in S_{1}}\mathbf {dS}$ and $\int _{\mathbf {q} \in S_{2}}\mathbf {dS}$ are both equivalent and are a function of the boundary $C$. This implies that a surface can be freely deformed within its boundaries without changing the total surface vector. Note however that the surface areas $\int _{\mathbf {q} \in S_{1}}|\mathbf {dS} |$ and $\int _{\mathbf {q} \in S_{2}}|\mathbf {dS} |$ are not necessarily equivalent.

The fact that the total surface vectors of $S_{1}$ and $S_{2}$ are equivalent is not immediately obvious. To prove this fact, let $\mathbf {F}$ be a constant vector field. $S_{1}$ and $S_{2}$ share the same boundary, so the flux/flow of $\mathbf {F}$ through $S_{1}$ and $S_{2}$ is equivalent. The flux through $S_{1}$ is $\Phi _{1}=\int _{\mathbf {q} \in S_{1}}\mathbf {F} \cdot \mathbf {dS} =\mathbf {F} \cdot \int _{\mathbf {q} \in S_{1}}\mathbf {dS}$, and similarly for $S_{2}$ is $\Phi _{2}=\int _{\mathbf {q} \in S_{2}}\mathbf {F} \cdot \mathbf {dS} =\mathbf {F} \cdot \int _{\mathbf {q} \in S_{2}}\mathbf {dS}$. Since $\mathbf {F} \cdot \int _{\mathbf {q} \in S_{1}}\mathbf {dS} =\mathbf {F} \cdot \int _{\mathbf {q} \in S_{2}}\mathbf {dS}$ for every choice of $\mathbf {F}$, it follows that $\int _{\mathbf {q} \in S_{1}}\mathbf {dS} =\int _{\mathbf {q} \in S_{2}}\mathbf {dS}$.

The geometric significance of the total surface vector is that each component measures the area of the projection of the surface onto the plane formed by the other two dimensions. Let $\sigma$ be a surface with surface vector $\mathbf {S} =S_{x}\mathbf {i} +S_{y}\mathbf {j} +S_{z}\mathbf {k}$. It is then the case that: $S_{x}$ is the area of the projection of $\sigma$ onto the yz-plane; $S_{y}$ is the area of the projection of $\sigma$ onto the xz-plane; and $S_{z}$ is the area of the projection of $\sigma$ onto the xy-plane.

The boundary $\partial \Sigma$ of $\Sigma$ is counter-clockwise oriented.

Given an oriented surface $\Sigma$, another important concept is the oriented boundary. The boundary of $\Sigma$ is an oriented curve $\partial \Sigma$ but how is the orientation chosen? If the boundary is "counter-clockwise" oriented, then the boundary must follow a counter-clockwise direction when the oriented surface normal vectors point towards the viewer. The counter-clockwise boundary also obeys the "right-hand rule": If you hold your right hand with your thumb in the direction of the surface normals (penetrating the surface in the "preferred" direction), then your fingers will wrap around in the direction of the counter-clockwise oriented boundary.

Example 1: Consider the Cartesian points $(0,0,0)$; $(1,0,0)$; $(0,1,0)$; and $(0,0,1)$.

Let $\sigma _{1}$ be the surface formed by the triangular planes $\{(0,1,0),(0,0,0),(1,0,0)\}$; $\{(0,0,1),(0,0,0),(0,1,0)\}$; and $\{(1,0,0),(0,0,0),(0,0,1)\}$ where the vertices are listed in a counterclockwise direction relative to the surface normal directions. The surface vectors of each plane are respectively ${\frac {1}{2}}\mathbf {k}$; ${\frac {1}{2}}\mathbf {i}$; and ${\frac {1}{2}}\mathbf {j}$ respectively which add to a total surface vector of $\mathbf {S} _{1}={\frac {1}{2}}(\mathbf {i} +\mathbf {j} +\mathbf {k} )$.

Let $\sigma _{2}$ be the surface formed by the single triangular plane $\{(1,0,0),(0,1,0),(0,0,1)\}$ where the vertices are listed in a counterclockwise direction relative to the normal direction. It can be seen that $\sigma _{1}$ and $\sigma _{2}$ share a the common counter clockwise boundary $(1,0,0)\to (0,1,0)\to (0,0,1)$The surface vector is $\mathbf {S} _{2}={\frac {1}{2}}(\mathbf {j} -\mathbf {i} )\times (\mathbf {k} -\mathbf {i} )={\frac {1}{2}}(\mathbf {i} +\mathbf {j} +\mathbf {k} )$ which is equivalent to $\mathbf {S} _{1}$.

Example 2: This example will show how moving a point that is in the interior of a "triangular mesh" does not affect the total surface vector. Consider the points $P_{0},P_{1},P_{2},\dots ,P_{n}$ where $n\geq 3$. Let the closed path $C$ be defined by the cycle $P_{1}\to P_{2}\to \dots \to P_{n}\to P_{1}$. For simplicity, $P_{n+1}=P_{1}$. For each $i=1,2,\dots ,n$, $\mathbf {v} _{i}$ will denote the displacement of $P_{i}$ relative to $P_{0}$. Like with $P_{n+1}$, $\mathbf {v} _{n+1}=\mathbf {v} _{1}$.

Let $\sigma$ denote a surface that is a "triangular mesh" comprised of the closed fan of triangles: $\{P_{2},P_{0},P_{1}\}$; $\{P_{3},P_{0},P_{2}\}$; ...; $\{P_{n},P_{0},P_{n-1}\}$; $\{P_{1},P_{0},P_{n}\}$ where the vertices of each triangle are listed in a counterclockwise direction. It can be seen that the counterclockwise boundary of $\sigma$ is $C$ and does not depend on the location of $P_{0}$. The total surface vector for $\sigma$ is:

$\mathbf {S} ={\frac {1}{2}}(\mathbf {v} _{1}\times \mathbf {v} _{2})+{\frac {1}{2}}(\mathbf {v} _{2}\times \mathbf {v} _{3})+\dots +{\frac {1}{2}}(\mathbf {v} _{n-1}\times \mathbf {v} _{n})+{\frac {1}{2}}(\mathbf {v} _{n}\times \mathbf {v} _{n+1})={\frac {1}{2}}\sum _{i=1}^{n}(\mathbf {v} _{i}\times \mathbf {v} _{i+1})$

Now displace $P_{0}$ by $\mathbf {w}$ to get $P'_{0}$. The displacement vector of $P_{i}$ relative to $P_{0}'$ becomes $\mathbf {v} '_{i}=\mathbf {v} _{i}-\mathbf {w}$. The counterclockwise boundary is unaffected. The total surface vector is:

$\mathbf {S} '={\frac {1}{2}}\sum _{i=1}^{n}(\mathbf {v} '_{i}\times \mathbf {v} '_{i+1})={\frac {1}{2}}\sum _{i=1}^{n}((\mathbf {v} _{i}-\mathbf {w} )\times (\mathbf {v} _{i+1}-\mathbf {w} ))={\frac {1}{2}}\sum _{i=1}^{n}((\mathbf {v} _{i}\times \mathbf {v} _{i+1})-(\mathbf {v} _{i}\times \mathbf {w} )-(\mathbf {w} \times \mathbf {v} _{i+1})+(\mathbf {w} \times \mathbf {w} ))$

$={\frac {1}{2}}\sum _{i=1}^{n}((\mathbf {v} _{i}\times \mathbf {v} _{i+1})-\mathbf {w} \times (\mathbf {v} _{i+1}-\mathbf {v} _{i}))=\mathbf {S} -{\frac {1}{2}}\mathbf {w} \times (\mathbf {v} _{n+1}-\mathbf {v} _{1})=\mathbf {S} -{\frac {1}{2}}\mathbf {w} \times \mathbf {0} =\mathbf {S}$

Therefore moving the interior point $P_{0}$ neither affects the boundary, nor the total surface vector.

Calculating Surface Integrals

To calculate a surface integral, the oriented surface $\sigma$ must be parameterized. Let $\mathbf {q} _{\sigma }(u,v)$ be a continuous function that maps each point $(u,v)$ from a two-dimensional domain $D_{u,v}$ to a point in $\sigma$. $\mathbf {q} _{\sigma }(u,v)$ must be continuous and onto. While $\mathbf {q} _{\sigma }(u,v)$ does not necessarily have to be one to one, the parameterization should never "fold back" on itself. The infinitesimal increases in $u$ and $v$ are respectively $du$ and $dv$. These respectively give rise to the displacements ${\frac {\partial \mathbf {q} _{\sigma }}{\partial u}}du$ and ${\frac {\partial \mathbf {q} _{\sigma }}{\partial v}}dv$. Assuming that the surface's orientation follows the right hand rule with respect to the displacements ${\frac {\partial \mathbf {q} _{\sigma }}{\partial u}}du$ and ${\frac {\partial \mathbf {q} _{\sigma }}{\partial v}}dv$, the surface vector that arises is $\mathbf {dS} =({\frac {\partial \mathbf {q} _{\sigma }}{\partial u}}\times {\frac {\partial \mathbf {q} _{\sigma }}{\partial v}})dudv$.

In the surface integral $\iint _{\mathbf {q} \in \sigma }\mathbf {F} (\mathbf {q} )\cdot \mathbf {dS}$, the differential $\mathbf {dS}$ can be replaced with $({\frac {\partial \mathbf {q} _{\sigma }}{\partial u}}\times {\frac {\partial \mathbf {q} _{\sigma }}{\partial v}})dudv$ to get $\iint _{(u,v)\in D_{u,v}}\mathbf {F} (\mathbf {q} _{\sigma }(u,v))\cdot ({\frac {\partial \mathbf {q} _{\sigma }}{\partial u}}\times {\frac {\partial \mathbf {q} _{\sigma }}{\partial v}})dudv$.

Example 3: Consider the problem of computing the surface area of a sphere of radius $R$.

Center the sphere $\sigma$ on the origin, and using $u$ and $v$ as the parameter variables, the sphere can be parameterized in spherical coordinates via $\mathbf {q} _{\sigma }(u,v)=(r=R,\theta =u,\phi =v)$ where $u\in [0,\pi ]$ and $v\in [-\pi ,+\pi ]$. The infinitesimal displacements from small changes in the parameters are:

$du$ causes ${\frac {\partial \mathbf {q} _{\sigma }}{\partial u}}du=({\frac {\partial r}{\partial u}}{\hat {\mathbf {r} }}+r{\frac {\partial \theta }{\partial u}}{\hat {\mathbf {\theta } }}+r\sin \theta {\frac {\partial \phi }{\partial u}}{\hat {\mathbf {\phi } }})du=(R{\hat {\mathbf {\theta } }})du$

$dv$ causes ${\frac {\partial \mathbf {q} _{\sigma }}{\partial v}}dv=({\frac {\partial r}{\partial v}}{\hat {\mathbf {r} }}+r{\frac {\partial \theta }{\partial v}}{\hat {\mathbf {\theta } }}+r\sin \theta {\frac {\partial \phi }{\partial v}}{\hat {\mathbf {\phi } }})dv=(R\sin(u){\hat {\mathbf {\phi } }})dv$

The infinitesimal surface vector is hence $\mathbf {dS} =(R{\hat {\mathbf {\theta } }})du\times (R\sin(u){\hat {\mathbf {\phi } }})dv=(R^{2}\sin(u){\hat {\mathbf {r} }})dudv$. While not important to this example, note how the parameterization was chosen so that the surface vector points outwards. The area is $|\mathbf {dS} |=R^{2}\sin(u)dudv$.

The total surface area is hence:

$\iint _{\mathbf {q} \in \sigma }|\mathbf {dS} |=\int _{u=0}^{\pi }\int _{v=-\pi }^{+\pi }R^{2}\sin(u)dvdu=2\pi R^{2}\int _{u=0}^{\pi }\sin(u)du=2\pi R^{2}(-\cos(u){\bigg |}_{u=0}^{\pi })=4\pi R^{2}$

More on Surface Integrals and Area

Resources

• Line Integrals, Mathematics LibreTexts
• Vector Calculus, Wikibooks: Calculus
• Line integral, Wikipedia
• Introduction to The Line Integral, Khan Academy

Licensing

Content obtained and/or adapted from:

• Vector calculus, Wikibooks: Calculus under a CC BY-SA license