Calculus

Line Integrals

Physical Applications of Line Integrals

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Problem 1
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Problems 2-7

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

  • mass of a wire;
  • center of mass and moments of inertia of a wire;
  • work done by a force on an object moving in a vector field;
  • magnetic field around a conductor (Ampere’s Law);
  • voltage generated in a loop (Faraday’s Law of magnetic induction).

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
\[{m \text{ = }}\kern0pt
{\int\limits_\alpha ^\beta {\rho \left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)\cdot }}\kern0pt{{ \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2} + {{\left( {\frac{{dz}}{{dt}}} \right)}^2}} dt}}
\] 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
\[{m \text{ = }}\kern0pt
{\int\limits_\alpha ^\beta {\rho \left( {x\left( t \right),y\left( t \right)} \right) \cdot }}\kern0pt{{ \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} dt} .}
\]

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
\[{\bar x = \frac{{{M_{yz}}}}{m},\;\;\;}\kern0pt{\bar y = \frac{{{M_{xz}}}}{m},\;\;\;}\kern0pt{\bar z = \frac{{{M_{xy}}}}{m},}\] where
\[
{{M_{yz}} = \int\limits_C {x\rho \left( {x,y,z} \right)ds} ,\;\;\;}\kern-0.3pt
{{M_{xz}} = \int\limits_C {y\rho \left( {x,y,z} \right)ds} ,\;\;\;}\kern-0.3pt
{{M_{xy}} = \int\limits_C {z\rho \left( {x,y,z} \right)ds} }
\] are so-called first moments.

The moments of inertia about the \(x\)-axis, \(y\)-axis and \(z\)-axis are given by the formulas
\[
{{I_x} = \int\limits_C {\left( {{y^2} + {z^2}} \right)\rho \left( {x,y,z} \right)ds} ,\;\;}\kern-0.3pt
{{I_y} = \int\limits_C {\left( {{x^2} + {z^2}} \right)\rho \left( {x,y,z} \right)ds} ,\;\;}\kern-0.3pt
{{I_z} = \int\limits_C {\left( {{x^2} + {y^2}} \right)\rho \left( {x,y,z} \right)ds}}
\]

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 \(1\)).

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

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

Figure 1.

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
\[
{\mathbf{F} \text{ = }}\kern0pt
{ \left( {P\left( {x,y,z} \right), Q\left( {x,y,z} \right), }\right.}\kern0pt{\left.{ R\left( {x,y,z} \right)} \right),}
\] 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
\[
{W \text{ = }}\kern0pt
{\int\limits_\alpha ^\beta {\left[ {P\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)\frac{{dx}}{{dt}} }\right.}}+{{\left.{ Q\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)\frac{{dy}}{{dt}} }\right.}}+{{\left.{ R\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)\frac{{dz}}{{dt}}} \right]dt} ,}
\] 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

Figure 2.

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 \(2\)). 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 on problem description to see 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 3

Find the moment of inertia \({I_x}\) of the circle \({x^2} + {y^2} = {a^2}\) with the density \(\rho = 1.\)

 Example 4

Find the work done by the force field \(\mathbf{F}\left( {x,y} \right) \) \(= \left( {xy,x + y} \right)\) on an object moving from the origin \(O\left( {0,0} \right)\) to the point \(A\left( {1,1} \right)\) along the path \(C,\) where

  1. \(C\) is the line segment \(y = x;\)
  2. \(C\) is the curve \(y = \sqrt x.\)

 Example 5

An object with a mass of \(m\) is thrown under the angle \(\alpha\) with the initial velocity \({v_0}\) (Figure \(5\)). Calculate the work performed by the gravitational force \(\mathbf{F} = m\mathbf{g}\) while the object moves until the moment it strikes the ground.

 Example 6

Find the magnetic field in vacuum a distance \(r\) from the axis of a long straight wire carrying current \(I.\)

 Example 7

Evaluate the maximum electromotive force \(\varepsilon\) and the electric field \(E\) induced in a finger ring of radius \(1\,\text{cm}\) when the passenger flies on an airplane in the magnetic field of the Earth with the velocity of \(900\,\text{km/h}.\)

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 \(AB:\)
\[
{{\frac{{x – {x_A}}}{{{x_B} – {x_A}}} = \frac{{y – {y_A}}}{{{y_B} – {y_A}}} }={ t,\;\;}}\Rightarrow
{{\frac{{x – 1}}{{2 – 1}} = \frac{{y – 1}}{{4 – 1}} }={ t,\;\;}}\Rightarrow
{{\frac{{x – 1}}{1} = \frac{{y – 1}}{3} }={ t\;\;\;}}\kern0pt
{\text{or}\;\;\left\{ {\begin{array}{*{20}{l}}
{x = t + 1}\\
{y = 3t + 1}
\end{array}} \right.,}
\] where parameter \(t\) varies in the interval \(\left[ {0,1} \right].\) Then the mass of the wire is
\[
{m \text{ = }}\kern0pt{ \int\limits_\alpha ^\beta {\rho \left( {x\left( t \right),y\left( t \right)} \right) \cdot}}\kern0pt{{ \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} dt} }
= {{\int\limits_0^1 {\left( {3x\left( t \right) + 2y\left( t \right)} \right) \cdot}}\kern0pt{{ \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} dt} }}
= {\int\limits_0^1 {\left( {9t + 5} \right)\sqrt {{1^2} + {3^2}} dt} }
= {\sqrt {10} \int\limits_0^1 {\left( {9t + 5} \right)dt} }
= {\sqrt {10} \left[ {\left. {\left( {\frac{{9{t^2}}}{2} + 5t} \right)} \right|_0^1} \right] }
= {\frac{{19\sqrt {10} }}{2} }\approx{ 30.}
\]

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Problem 1
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Problems 2-7