Physical Applications of Line Integrals

Solved Problems

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

\(C\) is the line segment \(y = x;\)
\(C\) 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

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