Calculus

Line Integrals

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Physical Applications of Line Integrals

In physics, the line integrals are used, in particular, for computations of

Consider these applications in more details.

Mass of a Wire

Suppose that a piece of a wire is described by a curve \(C\) in three dimensions. The mass per unit length of the wire is a continuous function \(\rho \left( {x,y,z} \right).\) Then the total mass of the wire is expressed through the line integral of scalar function as

\[m = \int\limits_C {\rho \left( {x,y,z} \right)ds} .\]

If \(C\) is a curve parameterized by the vector function \(\mathbf{r}\left( t \right) =\) \(\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right),\) then the mass can be computed by the formula

\[m = \int\limits_\alpha ^\beta {\rho \left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right) \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2} + {{\left( {\frac{{dz}}{{dt}}} \right)}^2}} dt}.\]

If \(C\) is a curve in the \(xy\)-plane, then the mass of the wire is given by

\[m = \int\limits_C {\rho \left( {x,y} \right)ds},\]

or in parametric form

\[m = \int\limits_\alpha ^\beta {\rho \left( {x\left( t \right),y\left( t \right)} \right) \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} dt} .\]

Center of Mass and Moments of Inertia of a Wire

Let a wire is described by a curve \(C\) with a continuous density function \(\rho \left( {x,y,z} \right).\) The coordinates of the center of mass of the wire are defined as

\[\bar x = \frac{{{M_{yz}}}}{m},\;\; \bar y = \frac{{{M_{xz}}}}{m},\;\; \bar z = \frac{{{M_{xy}}}}{m},\]

where

\[{M_{yz}} = \int\limits_C {x\rho \left( {x,y,z} \right)ds} ,\;\; {M_{xz}} = \int\limits_C {y\rho \left( {x,y,z} \right)ds} ,\;\; {M_{xy}} = \int\limits_C {z\rho \left( {x,y,z} \right)ds}\]

are so-called first moments.

The moments of inertia about the \(x\)-axis, \(y\)-axis and \(z\)-axis are given by the formulas

\[{I_x} = \int\limits_C {\left( {{y^2} + {z^2}} \right)\rho \left( {x,y,z} \right)ds} ,\;\; {I_y} = \int\limits_C {\left( {{x^2} + {z^2}} \right)\rho \left( {x,y,z} \right)ds} ,\;\; {I_z} = \int\limits_C {\left( {{x^2} + {y^2}} \right)\rho \left( {x,y,z} \right)ds}.\]

Work

Work done by a force \(\mathbf{F}\) on an object moving along a curve \(C\) is given by the line integral

\[W = \int\limits_C {\mathbf{F} \cdot d\mathbf{r}} ,\]

where \(\mathbf{F}\) is the vector force field acting on the object, \(d\mathbf{r}\) is the unit tangent vector (Figure \(1\)).

Work done by a force F on an object moving along a curve C
Figure 1.

The notation \({\mathbf{F} \cdot d\mathbf{r}}\) means dot product of \(\mathbf{F}\) and \(d\mathbf{r}.\)

Note that the force field \(\mathbf{F}\) is not necessarily the cause of moving the object. It might be some other force acting to overcome the force field that is actually moving the object. In this case the work of the force \(\mathbf{F}\) could result in a negative value.

If a vector field is defined in the coordinate form

\[\mathbf{F} = \left( {P\left( {x,y,z} \right), Q\left( {x,y,z} \right), R\left( {x,y,z} \right)} \right),\]

then the work done by the force is calculated by the formula

\[W = \int\limits_C {\mathbf{F} \cdot d\mathbf{r}} = \int\limits_C {Pdx + Qdy + Rdz} .\]

If the object is moved along a curve \(C\) in the \(xy\)-plane, then the following formula is valid:

\[W = \int\limits_C {\mathbf{F} \cdot d\mathbf{r}} = \int\limits_C {Pdx + Qdy},\]

where \(\mathbf{F} \) \(= \left( {P\left( {x,y} \right),Q\left( {x,y} \right)} \right).\)

If a path \(C\) is specified by a parameter \(t\) (\(t\) often means time), the formula for calculating work becomes

\[W = \int\limits_\alpha ^\beta {\left[ {P\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)\frac{{dx}}{{dt}} + Q\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)\frac{{dy}}{{dt}} + R\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)\frac{{dz}}{{dt}}} \right]dt} ,\]

where \(t\) goes from \(\alpha\) to \(\beta.\)

If a vector field \(\mathbf{F}\) is conservative, then then the work on an object moving from \(A\) to \(B\) can be found by the formula

\[W = u\left( B \right) - u\left( A \right),\]

where \(u\left( {x,y,z} \right)\) is a scalar potential of the field.

Ampere's Law

The line integral of a magnetic field \(\mathbf{B}\) around a closed path \(C\) is equal to the total current flowing through the area bounded by the contour \(C\) (Figure \(2\)).

Ampere’s Law
Figure 2.

This is expressed by the formula

\[\int\limits_C {\mathbf{B} \cdot d\mathbf{r}} = {\mu _0}I,\]

where \({\mu _0}\) is the vacuum permeability constant, equal to \(1,26 \times {10^{ - 6}}\,\text{H/m}.\)

Faraday's Law

The electromotive force \(\varepsilon\) induced around a closed loop \(C\) is equal to the rate of the change of magnetic flux \(\psi\) passing through the loop (Figure \(3\)).

\[\varepsilon = \int\limits_C {\mathbf{E} \cdot d\mathbf{r}} = - \frac{{d\psi }}{{dt}}.\]
Faraday’s Law
Figure 3.

Solved Problems

Click or tap a problem to see the solution.

Example 1

Find the mass of a wire running along the plane curve \(C\) with the density \[\rho \left( {x,y} \right) = 3x + 2y.\] The curve \(C\) is the line segment from point \(A\left( {1,1} \right)\) to point \(B\left( {2,4} \right).\)

Example 2

Find the mass of a wire lying along the arc of the circle \[{x^2} + {y^2} = 1\] from \(A\left( {1,0} \right)\) to \(B\left( {0,1} \right)\) with the density \[\rho \left( {x,y} \right) = xy\] (Figure \(4\)).

Example 1.

Find the mass of a wire running along the plane curve \(C\) with the density \[\rho \left( {x,y} \right) = 3x + 2y.\] The curve \(C\) is the line segment from point \(A\left( {1,1} \right)\) to point \(B\left( {2,4} \right).\)

Solution.

We first find the parametric equation of the line \(AB:\)

\[ \frac{{x - {x_A}}}{{{x_B} - {x_A}}} = \frac{{y - {y_A}}}{{{y_B} - {y_A}}} = t,\;\; \Rightarrow \frac{{x - 1}}{{2 - 1}} = \frac{{y - 1}}{{4 - 1}} = t,\;\; \Rightarrow \frac{{x - 1}}{1} = \frac{{y - 1}}{3} = t\;\;\; \text{or}\;\;\left\{ {\begin{array}{*{20}{l}} {x = t + 1}\\ {y = 3t + 1} \end{array}} \right.,\]

where parameter \(t\) varies in the interval \(\left[ {0,1} \right].\) Then the mass of the wire is

\[ m = \int\limits_\alpha ^\beta {\rho \left( {x\left( t \right),y\left( t \right)} \right) \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} dt} = \int\limits_0^1 {\left( {3x\left( t \right) + 2y\left( t \right)} \right) \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} dt} = \int\limits_0^1 {\left( {9t + 5} \right)\sqrt {{1^2} + {3^2}} dt} = \sqrt {10} \int\limits_0^1 {\left( {9t + 5} \right)dt} = \sqrt {10} \left[ {\left. {\left( {\frac{{9{t^2}}}{2} + 5t} \right)} \right|_0^1} \right] = \frac{{19\sqrt {10} }}{2} \approx 30.\]

Example 2.

Find the mass of a wire lying along the arc of the circle \[{x^2} + {y^2} = 1\] from \(A\left( {1,0} \right)\) to \(B\left( {0,1} \right)\) with the density \[\rho \left( {x,y} \right) = xy\] (Figure \(4\)).

Solution.

A wire lying along the arc of the circle  x^2+y^2=10
Figure 4.

The circle with radius \(1\) and centered at the origin has parametric equations

\[x = \cos t,\;\;y = \sin t,\]

where the parameter \(t\) varies in the interval \(\left[ {0,{\frac{\pi }{2}}} \right].\) Then the mass of the wire is calculated as follows:

\[m = \int\limits_\alpha ^\beta {\rho \left( {x\left( t \right),y\left( t \right)} \right) \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} dt} = \int\limits_0^{\frac{\pi }{2}} {x\left( t \right)y\left( t \right) \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} dt} = \int\limits_0^{\frac{\pi }{2}} {\cos t\sin t \sqrt {{{\left( {\frac{{d\cos t}}{{dt}}} \right)}^2} + {{\left( {\frac{{d\sin t}}{{dt}}} \right)}^2}} dt} = \int\limits_0^{\frac{\pi }{2}} {\cos t\sin t \sqrt {{{\left( { - \sin t} \right)}^2} + {{\left( {\cos t} \right)}^2}} dt} = \int\limits_0^{\frac{\pi }{2}} {\cos t\sin tdt} = \frac{1}{2}\int\limits_0^{\frac{\pi }{2}} {\sin 2tdt} = \frac{1}{4}\left[ {\left. {\left( { - \cos 2t} \right)} \right|_0^{\frac{\pi }{2}}} \right] = \frac{1}{4}\left( { - \cos \pi + \cos 0} \right) = \frac{1}{2}.\]

See more problems on Page 2.

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