Calculus

Line Integrals

Line Integrals Logo

Physical Applications of Line Integrals

Solved Problems

Click or tap a problem to see the solution.

Example 3

Find the moment of inertia Ix of the circle \[{x^2} + {y^2} = {a^2}\] with the density ρ = 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 (0,0) to the point A (1,1) 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 3.

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

Solution.

The equation of the circle in parametric form is

\[\left\{ \begin{array}{l} x = a\cos t\\ y = a\sin t \end{array} \right.,\;\; 0 \le t \le 2\pi .\]

Then the moment of inertia \({I_x}\) about the \(x\)-axis can be calculated by the formula

\[{I_x} = \int\limits_C {{y^2}\rho ds} = \int\limits_0^{2\pi } {{{\left( {y\left( t \right)} \right)}^2}\rho \left( {x\left( t \right),y\left( t \right)} \right) \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} dt.}\]

So we have

\[{I_x} = \int\limits_0^{2\pi } {{{\left( {a\sin t} \right)}^2} \cdot 1 \cdot {{{\left( {\frac{{d\left( {a\cos t} \right)}}{{dt}}} \right)}^2} + {{\left( {\frac{{d\left( {a\sin t} \right)}}{{dt}}} \right)}^2}} dt} = \int\limits_0^{2\pi } {{a^2}{{\sin }^2}t \sqrt {{{\left( { - a\sin t} \right)}^2} + {{\left( {a\cos t} \right)}^2}} dt} = \int\limits_0^{2\pi } {{a^3}{{\sin }^2}t \sqrt {{{\sin }^2}t + {{\cos }^2}t} dt} = {a^3}\int\limits_0^{2\pi } {{{\sin }^2}t\,dt} = {a^3}\int\limits_0^{2\pi } {\frac{{1 - \cos 2t}}{2}dt} = \frac{{{a^3}}}{2}\int\limits_0^{2\pi } {\left( {1 - \cos 2t} \right)dt} = \frac{{{a^3}}}{2}\left[ {\left. {\left( {t - \frac{{\sin 2t}}{2}} \right)} \right|_0^{2\pi }} \right] = \frac{{{a^3}}}{2}\left( {2\pi - 0} \right) = \pi {a^3}.\]

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 \(y = \sqrt x.\)

Solution.

\(1.\;\) We first compute the work along the line segment \(y = x.\)

\[{W_1} = \int\limits_C {\mathbf{F} \cdot d\mathbf{r}} = \int\limits_C {Pdx + Qdy} = \int\limits_C {xydx + \left( {x + y} \right)dy} = \int\limits_0^1 {x \cdot xdx + \left( {x + x} \right)dx} = \int\limits_0^1 {\left( {{x^2} + 2x} \right)dx} = \left. {\left( {\frac{{{x^3}}}{3} + {x^2}} \right)} \right|_0^1 = \frac{1}{3} + 1 = \frac{4}{3}.\]

\(2.\;\) Now we find the work when the object moves along the curve \(y = \sqrt x.\)

\[{W_2} = \int\limits_C {\mathbf{F} \cdot d\mathbf{r}} = \int\limits_C {Pdx + Qdy} = \int\limits_C {xydx + \left( {x + y} \right)dy} = \int\limits_0^1 {x \cdot \sqrt x dx + \left( {x + \sqrt x } \right)\frac{{dx}}{{2\sqrt x }}} = \int\limits_0^1 {\left( {{x^{\frac{3}{2}}} + \frac{{{x^{\frac{1}{2}}}}}{2} + \frac{1}{2}} \right)dx} = \left. {\left( {\frac{{{x^{\frac{5}{2}}}}}{{\frac{5}{2}}} + \frac{{{x^{\frac{3}{2}}}}}{{2 \cdot \frac{3}{2}}} + \frac{x}{2}} \right)} \right|_0^1 = \frac{2}{5} + \frac{1}{3} + \frac{1}{2} = \frac{{37}}{{30}}.\]

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.

Solution.

An object is thrown under the angle alpha with the initial velocity v0
Figure 5.

We first find the equation of the path in parametric form (\(t\) is time):

\[x = {v_{0x}}t = {v_0}\cos \alpha \cdot t,\]
\[y = {v_{0y}}t - \frac{{g{t^2}}}{2} = {v_0}\sin\alpha \cdot t - \frac{{g{t^2}}}{2}.\]

At the moment of strike, \(y = 0,\) so that the fall time is

\[{v_0}\sin\alpha \cdot t - \frac{{g{t^2}}}{2} = 0,\;\; \Rightarrow t\left( {{v_0}\sin\alpha - \frac{{gt}}{2}} \right) = 0,\;\; \Rightarrow t = \frac{{2{v_0}\sin\alpha }}{g}.\]

The gravitational force can be written in the form \(\mathbf{F} = m\mathbf{g} \) \(= m\left( {0, - g} \right).\) Then the work done in moving the abject along the path is

\[W = \int\limits_\alpha ^\beta {\left( {P\frac{{dx}}{{dt}} + Q\frac{{dy}}{{dt}}} \right)dt} = \int\limits_0^{\frac{{2{v_0}\sin\alpha }}{g}} {\left( {0 \cdot \frac{{dx}}{{dt}} - g \cdot \frac{{dy}}{{dt}}} \right)dt} = - g\int\limits_0^{\frac{{2{v_0}\sin\alpha }}{g}} {\left( {\frac{{dy}}{{dt}}} \right)dt} = - g\int\limits_0^{\frac{{2{v_0}\sin\alpha }}{g}} {dy\left( t \right)} = - g\left[ {\left. {y\left( t \right)} \right|_{t = 0}^{\frac{{2{v_0}\sin\alpha }}{g}}} \right] = - g\left[ {\left. {\left( {{v_0}\sin \alpha t - \frac{{g{t^2}}}{2}} \right)} \right|_{t = 0}^{\frac{{2{v_0}\sin\alpha }}{g}}} \right] = - g\left( {\frac{{2v_0^2\,{{\sin }^2}\alpha }}{g} - \frac{{4gv_0^2\,{{\sin }^2}\alpha }}{{2{g^2}}}} \right) = 0.\]

This result can be easily explained. The gravitational field of Earth is conservative, since

\[\frac{{\partial Q}}{{\partial x}} = \frac{{\partial P}}{{\partial y}} = 0.\]

The scalar potential of the field can be written in the general form as

\[u\left( {x,y} \right) = \int {Pdx} + {C_1}\left( y \right) = \int {0dx} + {C_1}\left( y \right) = {C_0} + {C_1}\left( y \right).\]

By setting

\[\frac{{\partial u}}{{\partial y}} = Q\left( {x,y} \right) = - g,\]

we find that

\[\frac{d}{{dy}}{C_1}\left( y \right) = - g,\;\; \Rightarrow {C_1}\left( y \right) = - gy + {C_2}.\]

Thus, the potential of the gravitational field is

\[u\left( {x,y} \right) = {C_0} - gy + {C_2} = C - gy.\]

where \(C\) is a constant, which can be set zero. As a result, the potential of the field is

\[u\left( {x,y} \right) = - gy.\]

We see that the work done in moving the object from \(O\left( {0,0} \right)\) to the final point \(A\left( {L,0} \right)\) is

\[W = u\left( A \right) - u\left( O \right) = 0.\]

Example 6.

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

Solution.

To find the field a distance \(r\) from the wire, we consider a loop of radius \(r,\) centered on the wire with its plane perpendicular to the wire with the current \(I\) (Figure \(6\)).

Magnetic field B around a long straight wire carrying current I
Figure 6.

Since the field \(\mathbf{B}\) has a constant magnitude and the field is tangent to the loop everywhere, the dot product of the vectors \(\mathbf{B}\) and \(d\mathbf{r}\) is just \(Bdr.\) Then we can write

\[\oint\limits_C {\mathbf{B} \cdot d\mathbf{r}} = \oint\limits_C {Bdr} = B\oint\limits_C {dr} = 2\pi rB.\]

As a result we have

\[2\pi rB = {\mu _0}I\;\; \text{or}\;\;B = \frac{{{\mu _0}I}}{{2\pi r}}.\]

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 \(900\,\text{km/h}.\)

Solution.

According to Faraday's law,

\[\varepsilon = \oint\limits_C {E \cdot dr} = - \frac{{d\psi }}{{dt}}.\]

As the conducting ring moves through the Earth's magnetic field, there is a change in the magnetic flux \(\psi,\) passing through the ring.

Suppose that the magnetic field \(\mathbf{B}\) is perpendicular to the plane of the ring. Then change in the flux for the time \(\Delta t\) is

\[\Delta \psi = 2rBx = 2rBv\Delta t,\]

where \(x = v\Delta t,\) \(v\) is the velocity of the airplane, \(B\) is the magnetic field of the Earth. It follows from the last expression that

\[\varepsilon = - \frac{{d\psi }}{{dt}} = 2rBv.\]

Substituting the given values

\[v = 900\,\text{km/h} = 250\,\text{m/s},\;\; r = 1\,\text{cm} = 0,01\,\text{m},\;\; B = 5 \times {10^{ - 5}}\,\text{T},\]

we obtain the electromotive force:

\[\varepsilon = 2rBv = 2 \cdot 0,01 \cdot 5 \times {10^{ - 5}} \cdot 250 = 0,00025\,\text{V}.\]

As one can see, it's safe for human.

We can find the electric field in the conducting ring by the formula \(\varepsilon = \int\limits_C {\mathbf{E} \cdot d\mathbf{r}}.\)

By symmetry, the induced electric field will have a constant magnitude along the ring. Its direction will be tangential to the circle at every point. Hence, the line integral around the circle is

\[\varepsilon = \oint\limits_C {\mathbf{E} \cdot d\mathbf{r}} = \oint\limits_C {E \cdot dr \cdot \cos 0} = E\oint\limits_C {dr} = 2\pi rE.\]

Hence, the electric field strength is

\[E = \frac{\varepsilon }{{2\pi r}} = \frac{{0,00025}}{{2\pi \cdot 0,01}} = 0,004\,\text{V/m}.\]
Page 1 Page 2