Calculus

Integration of Functions

Integration of Functions Logo

Force, Work and Energy

  • Work Done by a Force

    Work in physics is defined as the product of the force and displacement.

    When an object moves a distance \(\Delta x\) along a straight line as a result of action of a constant force \(F,\) the work done by the force is

    \[W = F\Delta x.\]

    This equation is called the constant force formula for work.

    If the force is measured in Newtons \(\left( N \right)\) and distance is in meters \(\left( m \right),\) then work is measured in Joules \(\left( J \right).\)

    If an object moves along a straight line from \(x = a\) to \(x = b\) under the influence of a variable force \(F\left( x \right),\) the work \(W\) done by the force is given by the definite integral

    \[W = \int\limits_a^b {F\left( x \right)dx} .\]

    Hooke’s Law

    Hooke’s Law says that the force it takes to stretch or compress a spring \(x\) units from its natural (unstressed) length is

    \[F = kx,\]

    where \(F\) is the applied force, \(k\) is the spring constant, \(x\) is the displacement from the original length.

    Hooke's Law
    Figure 1.

    Sometimes Hooke’s Law is written as

    \[F = – kx.\]

    In this expression \(F\) no longer means the applied force, but rather means the equal and oppositely directed restoring force.

    The work needed to stretch the spring from \(0\) to \(x\) is given by the integral

    \[{W = \int\limits_0^x {Fdx} }={ \int\limits_0^x {kxdx} }={ \frac{{k{x^2}}}{2}.}\]

    Newton’s Law of Gravitation

    According to Newton’s law of universal gravitation, the gravitational force acting between two objects is given by

    \[F = G\frac{{{m_1}{m_2}}}{{{x^2}}},\]

    where \({m_1}\) and \({m_2}\) ate the masses of the objects, \(x\) is the distance between the centers of their masses, and \(G\) is the gravitational constant.

    Coulomb’s Law

    Coulomb’s law states that the force with which stationary electrically charged particles repel or attract each other is given by

    \[F = k\frac{{{q_1}{q_2}}}{{{x^2}}},\]

    where \({q_1}\) and \({q_2}\) are the signed magnitudes of the charges, \(x\) is the distance between the charges, and \(k\) is Coulomb’s constant.

    Kinetic Energy

    The kinetic energy of a moving object is equal to

    \[{E_k} = \frac{1}{2}m{v^2},\]

    where \(m\) is the mass and \(v\) is the velocity of the object.

    If a rigid body is rotating about any line through the center of mass with an angular velocity \(\omega,\) then it has rotational kinetic energy, which is given by the equation

    \[{{E_r} = \int {\frac{{{v^2}dm}}{2}} }={ \int {\frac{{{{\left( {r\omega } \right)}^2}dm}}{2}} }={ \frac{{{\omega ^2}}}{2}\int {{r^2}dm} }={ \frac{1}{2}I{\omega ^2},}\]

    where \(I\) is the moment of inertia and the integration is performed over all mass elements of the body.

    Potential Energy

    Potential energy is defined as the energy stored in an object. There are different kinds of potential energy, depending on the type of force. For example, the potential energy associated with gravitational force is called gravitational potential energy.

    Assuming that the gravitational field near the Earth’s surface is constant, the gravitational potential energy of a body is given by the formula

    \[{E_p} = mgh,\]

    where \(m\) is the mass, \(g\) is the acceleration of gravity, and \(h\) is the height.


  • Solved Problems

    Click a problem to see the solution.

    Example 1

    A force of \(50\,\text{N}\) is required to stretch a spring \(10\,\text{cm}\) beyond its natural length. How much work will be done stretching the spring \(20\,\text{cm}\) from its natural length?

    Example 2

    Calculate the work that is required to pump the water through an upper opening out of a vertical cylindrical barrel with base radius \(R\) and altitude \(H.\)

    Example 3

    Calculate the work that has to be done to raise a body of mass \(m\) from the Earth’s surface to an altitude \(h.\) What is the work if the body is removed to infinity?

    Example 4

    A raindrop with initial mass \({m_0}\) starts falling from rest under the action of gravity and evenly evaporates losing every second mass equal to \(\mu.\) What is the work of gravity during the time from the beginning of the movement to the complete evaporation of the drop. Neglect air resistance.

    Example 5

    Find the force of gravity exerted by a semicircular ring of radius \(R\) and mass \(m\) on a particle of mass \({m_0}\) located at the centre of the ring.

    Example 6

    The dimensions of the Great Pyramid of Giza (also known as the Pyramid of Cheops) are as follows: height \(H = 140\,\text{m},\) side of the square base \(a = 200\,\text{m}.\) Assuming that the pyramid was built of limestone with density of \(\rho = 2500\,\large{\frac{{kg}}{{{m^3}}}}\normalsize,\) estimate the total work done in its building.

    Example 7

    Find the work required to produce a conical pile of sand with base radius \(R\) and altitude \(H.\)

    Example 8

    An infinite straight line is uniformly charged with positive electricity. The linear density of electricity is \(\mu.\) Calculate the force exerted by the straight line on a unit charge located in air at a distance \(\ell\) from the line.

    Example 9

    Find the kinetic energy of a homogeneous circular cylinder of density \(\rho\) with base radius \(R\) and altitude \(H\) rotating about its axis with angular velocity \(\omega.\)

    Example 10

    Calculate the kinetic energy of a homogeneous ball of density \(\rho\) and radius \(R\) rotating about its axis with angular velocity \(\omega.\)

    Example 1.

    A force of \(50\,\text{N}\) is required to stretch a spring \(10\,\text{cm}\) beyond its natural length. How much work will be done stretching the spring \(20\,\text{cm}\) from its natural length?

    Solution.

    According to Hooke’s law,

    \[F = kx.\]

    Hence, the spring constant is equal to

    \[{k = \frac{F}{x} = \frac{{50}}{{0.1}} }={ 500\,\left( {\frac{\text{N}}{\text{m}}} \right),}\]

    where the displacement \(x\) is measured in meters: \({x = 10\,\text{cm} = 0.1\,\text{m}}\)

    To find the work done by the external force, we integrate from \(x = 0\) to \(x = 20\,\text{cm} = 0.2\,\text{m}:\)

    \[{W = \int\limits_0^{0.2} {kxdx} }={ \int\limits_0^{0.2} {500xdx} }={ \left. {\frac{{500{x^2}}}{2}} \right|_0^{0.2} }={ \frac{{500 \times {{0.2}^2}}}{2} }={ 10\,\left( J \right).}\]

    Example 2.

    Calculate the work that is required to pump the water through an upper opening out of a vertical cylindrical barrel with base radius \(R\) and altitude \(H.\)

    Solution.

    Calculating work to pump the water out of a cylindrical barrel with base radius R and altitude H.
    Figure 2.

    To find the work, we take a thin representative slice with thickness \(dz\) at a height \(z\) from the bottom of the barrel. The mass of the slice is

    \[{dm = \rho dV = \rho Adz }={ \pi \rho {R^2}dz.}\]

    To pump out this volume of water out of the barrel, we need to raise it to the height \(H\). The work required for this is given by the expression

    \[{dW = dm \cdot g\left( {H – z} \right) }={ \pi \rho g{R^2}\left( {H – z} \right)dz.}\]

    The total amount of work is found by integration from \(z = 0\) to \(z = H:\)

    \[{W = \int\limits_0^H {dW} }={ \int\limits_0^H {\pi \rho g{R^2}\left( {H – z} \right)dz} }={ \pi \rho g{R^2}\int\limits_0^H {\left( {H – z} \right)dz} }={ \pi \rho g{R^2}\left. {\left( {Hz – \frac{{{z^2}}}{2}} \right)} \right|_0^H }={ \frac{{\pi \rho g{R^2}{H^2}}}{2}.}\]

    Example 3.

    Calculate the work that has to be done to raise a body of mass \(m\) from the Earth’s surface to an altitude \(h.\) What is the work if the body is removed to infinity?

    Solution.

    Assuming the radius of Earth is \(R,\) the mass of Earth is \(M,\) and acceleration due to gravity at its surface is \(g,\) we write the gravitational force acting on the body at the Earth’s surface in the form

    \[{F_0} = G\frac{{mM}}{{{R^2}}} = mg,\]

    where \(G\) is the gravitational constant, and \(g = \large{\frac{{GM}}{{{R^2}}}}\normalsize\) is the acceleration due to gravity.

    At a certain height \(x,\) the gravitational force is given by

    \[{F\left( x \right) = G\frac{{mM}}{{{{\left( {R + x} \right)}^2}}} }={ G\frac{{mM{R^2}}}{{{{\left( {R + x} \right)}^2}{R^2}}} }={ \frac{{mg{R^2}}}{{{{\left( {R + x} \right)}^2}}}.}\]

    The work to raise the body to an altitude \(h\) is determined through integration:

    \[\require{cancel}{W = \int\limits_0^h {F\left( x \right)dx} }={ \int\limits_0^h {\frac{{mg{R^2}}}{{{{\left( {R + x} \right)}^2}}}dx} }={ mg{R^2}\int\limits_0^h {\frac{{dx}}{{{{\left( {R + x} \right)}^2}}}} }={ mg{R^2}\left. {\left( { – \frac{1}{{R + x}}} \right)} \right|_0^h }={ mg{R^2}\left( { – \frac{1}{{R + h}} + \frac{1}{R}} \right) }={ mg\frac{{{R^\cancel{2}}h}}{{\cancel{R}\left( {R + h} \right)}} }={ mg\frac{{Rh}}{{R + h}}}.\]

    The work needed to move the body from the Earth’s surface to infinity is given by the limit

    \[{{W_\infty } = \lim\limits_{h \to \infty } W }={ \lim\limits_{h \to \infty } \left( {mg\frac{{Rh}}{{R + h}}} \right) }={ mg\lim\limits_{h \to \infty } \frac{{Rh}}{{R + h}} }={ mg\lim\limits_{h \to \infty } \frac{R}{{\frac{R}{h} + 1}} }={ mg\frac{R}{{0 + 1}} }={ mgR.}\]

    Example 4.

    A raindrop with initial mass \({m_0}\) starts falling from rest under the action of gravity and evenly evaporates losing every second mass equal to \(\mu.\) What is the work of gravity during the time from the beginning of the movement to the complete evaporation of the drop. Neglect air resistance.

    Solution.

    The mass of the raindrop varies according to the law

    \[m\left( t \right) = {m_0} – \mu t.\]

    So the complete evaporation time is

    \[T = \frac{{{m_0}}}{\mu }.\]

    Determine the elementary work over an infinitesimally small time interval \(\left[ {t,t + dt} \right].\) The force of gravity at the moment \(t\) is

    \[{F\left( t \right) = m(t)g }={ \left( {{m_0} – \mu t} \right)g.}\]

    For the time \(dt,\) the drop moves a distance equal to

    \[dx = v(t)dt = gtdt.\]

    Integrating from \(t = 0\) to \(t = T = \large{\frac{{{m_0}}}{\mu }}\normalsize\) gives the total work:

    \[{W = \int\limits_0^T {dW\left( t \right)} }={ \int\limits_0^T {F\left( t \right)} dx\left( t \right) }={ \int\limits_0^T {\left( {{m_0} – \mu t} \right){g^2}tdt} }={ {g^2}\int\limits_0^T {\left( {{m_0}t – \mu {t^2}} \right)dt} }={ {g^2}\left. {\left( {\frac{{{m_0}{t^2}}}{2} – \frac{{\mu {t^3}}}{3}} \right)} \right|_0^T }={ {g^2}\left( {\frac{{{m_0}{T^2}}}{2} – \frac{{\mu {T^3}}}{3}} \right) }={ {g^2}\left( {\frac{{m_0^3}}{{2{\mu ^2}}} – \frac{{\mu m_0^3}}{{3{\mu ^3}}}} \right) }={ \frac{{{g^2}m_0^3}}{{{\mu ^2}}}\left( {\frac{1}{2} – \frac{1}{3}} \right) }={ \frac{{{g^2}m_0^3}}{{6{\mu ^2}}}.}\]

    Example 5.

    Find the force of gravity exerted by a semicircular ring of radius \(R\) and mass \(m\) on a particle of mass \({m_0}\) located at the centre of the ring.

    Solution.

    Force of gravity exerted by a semicircular ring on a point mass.
    Figure 3.

    Consider a small arc of the ring cut off by an angle \(d\theta.\) The mass of this arc is \(dm = \large{\frac{{md\theta }}{\pi }}\normalsize.\) The force of gravity between the arc element \(dm\) and the central mass \({m_0}\) is given by the equation

    \[{dF = G\frac{{{m_0}dm}}{{{R^2}}} }={ G\frac{{{m_0}m}}{{\pi {R^2}}}d\theta ,}\]

    where \(G\) is the gravitational constant.

    Due to symmetry, the horizontal components \({dF_x}\) will cancel each other, so we will consider only the vertical components \({dF_y} = dF\sin\theta.\) The resulting force is found by integration from \(\theta = 0\) to \(\theta = \pi:\)

    \[{F = {F_y} = \int\limits_0^\pi {d{F_y}} }={ G\frac{{{m_0}m}}{{\pi {R^2}}}\int\limits_0^\pi {\sin \theta d\theta } }={ G\frac{{{m_0}m}}{{\pi {R^2}}}\left. {\left( { – \cos \theta } \right)} \right|_0^\pi }={ \frac{{2G{m_0}m}}{{\pi {R^2}}}.}\]

    Example 6.

    The dimensions of the Great Pyramid of Giza (also known as the Pyramid of Cheops) are as follows: height \(H = 140\,\text{m},\) side of the square base \(a = 200\,\text{m}.\) Assuming that the pyramid was built of limestone with density of \(\rho = 2500\,\large{\frac{{kg}}{{{m^3}}}}\normalsize,\) estimate the total work done in its building.

    Solution.

    Pyramid of Cheops
    Figure 4.

    Consider a thin slice in the pyramid with thickness \(dz\) drawn \(z\) units from the vertex.

    Calculation of total work done in building the Great Pyramid.
    Figure 5.

    The side length of the slice is \(\large{\frac{{az}}{H}}\normalsize.\) Hence, the mass of the slice is

    \[{dm = \rho dV = \rho {\left( {\frac{{az}}{H}} \right)^2}dz }={ \frac{{\rho {a^2}}}{{{H^2}}}{z^2}dz.}\]

    We assume that the work is done by gravity, so for our representative slice, the work is given by

    \[{dW = dm \cdot g\left( {H – z} \right) }={ \frac{{\rho g{a^2}}}{{{H^2}}}\left( {H – z} \right){z^2}dz.}\]

    The total work done in building the pyramid is

    \[{W = \int\limits_0^H {dW} }={ \int\limits_0^H {\frac{{\rho g{a^2}}}{{{H^2}}}\left( {H – z} \right){z^2}dz} }={ \frac{{\rho g{a^2}}}{{{H^2}}}\int\limits_0^H {\left( {H{z^2} – {z^3}} \right)dz} }={ \frac{{\rho g{a^2}}}{{{H^2}}}\left. {\left( {\frac{{H{z^3}}}{3} – \frac{{{z^4}}}{4}} \right)} \right|_0^H }={ \frac{{\rho g{a^2}}}{{{H^2}}}\left( {\frac{{{H^4}}}{3} – \frac{{{H^4}}}{4}} \right) }={ \frac{{\rho g{a^2}{H^2}}}{{12}}.}\]

    Substituting the values for the Great Pyramid yields

    \[{{W_{Cheops}} = \frac{{2500 \cdot 9.8 \cdot {{200}^2} \cdot {{140}^2}}}{{12}} }\approx{ 1.6 \times {10^{12}}\,\left( J \right)}\]

    Example 7.

    Find the work required to produce a conical pile of sand with base radius \(R\) and altitude \(H.\) The density of sand is \(\rho.\)

    Solution.

    Calculating work required to produce a conical pile of sand.
    Figure 6.

    Take a thin cylindrical slice with thickness \(dz\) at a height \(z\) from the base of the cone. Using the similarity of triangles, we can write the following proportion:

    \[{\frac{r}{R} = \frac{{H – z}}{H},}\;\; \Rightarrow {r = \frac{{R\left( {H – z} \right)}}{H}.}\]

    The mass of the slice is given by

    \[{dm = \rho dV = \pi \rho {r^2}dz }={ \frac{{\pi \rho {R^2}{{\left( {H – z} \right)}^2}}}{{{H^2}}}dz.}\]

    Assuming that the work is required to overcome gravity, we have

    \[{dW = dm \cdot gz }={ \frac{{\pi \rho g{R^2}}}{{{H^2}}}{\left( {H – z} \right)^2}zdz,}\]

    where \(dW\) denotes the elementary work necessary to pour the layer of sand at height \(z.\)

    To calculate the total work, we integrate from \(z = 0\) to \(z = H:\)

    \[{W = \int\limits_0^H {dW} }={ \frac{{\pi \rho g{R^2}}}{{{H^2}}}\int\limits_0^H {{{\left( {H – z} \right)}^2}zdz} }={ \frac{{\pi \rho g{R^2}}}{{{H^2}}}\int\limits_0^H {\left( {{H^2} – 2Hz + {z^2}} \right)zdz} }={ \frac{{\pi \rho g{R^2}}}{{{H^2}}}\int\limits_0^H {\left( {{H^2}z – 2H{z^2} + {z^3}} \right)dz} }={ \frac{{\pi \rho g{R^2}}}{{{H^2}}}\left. {\left( {\frac{{{H^2}{z^2}}}{2} – \frac{{2H{z^3}}}{3} + \frac{{{z^4}}}{4}} \right)} \right|_0^H }={ \pi \rho g{R^2}{H^2}\left( {\frac{1}{2} – \frac{2}{3} + \frac{1}{4}} \right) }={ \frac{{\pi \rho g{R^2}{H^2}}}{{12}}.}\]

    Example 8.

    An infinite straight line is uniformly charged with positive electricity. The linear density of electricity is \(\mu.\) Calculate the force exerted by the straight line on a unit charge located in air at a distance \(\ell\) from the line.

    Solution.

    Electrostatic force between an infinite charged straight line and a unit charge
    Figure 7.

    Let’s take a small segment of length \(dx\) on the straight line. It carries the charge \(dq=\mu dx.\) The force \(dF\) exerted by the elementary charge \(dq\) on a unit test charge is determined by Coulomb’s law:

    \[{dF = k\frac{{{q_1}{q_2}}}{{{r^2}}} }={ \frac{1}{{4\pi {\varepsilon _0}}}\frac{{1 \cdot \mu dx}}{{{r^2}}} }={ \frac{1}{{4\pi {\varepsilon _0}}}\frac{\mu }{{{r^2}}}dx.}\]

    Using the Pythagorean theorem, the distance \(r\) between the charges can be written as \(r = \sqrt {{l^2} + {x^2}} ,\) so

    \[dF = \frac{1}{{4\pi {\varepsilon _0}}}\frac{\mu }{{{l^2} + {x^2}}}dx.\]

    For integration over the infinite line it is convenient to change variable. We rewrite the expression for \(dF\) in terms of the angle \(\theta.\) Since

    \[x = l\tan \theta,\]

    then

    \[dx = l\,{\sec ^2}\theta \,d\theta.\]

    Thus, we obtain \(dF\) in the form

    \[{dF = \frac{1}{{4\pi {\varepsilon _0}}}\frac{\mu }{{{l^2} + {x^2}}}dx }={ \frac{1}{{4\pi {\varepsilon _0}}}\frac{\mu }{{{l^2} + {l^2}\,{{\tan }^2}\theta }}\,l\,{\sec ^2}\theta d\theta }={ \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\mu \cancel{l}\,{{\sec }^{2}}\theta }}{{{l^{\cancel{2}}}\left( {1 + {{\tan }^2}\theta } \right)}}d\theta }={ \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\mu \,{{\sec }^2}\theta }}{{l\left( {1 + {{\tan }^2}\theta } \right)}}d\theta .}\]

    Recall from trigonometry that

    \[1 + {\tan ^2}\theta = {\sec ^2}\theta .\]

    As a result, we get the following simple expression:

    \[dF = \frac{1}{{4\pi {\varepsilon _0}}}\frac{\mu }{l}d\theta .\]

    By symmetry, all horizontal components of the force cancel, so further we can consider only vertical components \(d{F_y} = {dF}\cos\theta.\) Hence, the total electrostatic force exerted on a unit test charge is given by

    \[{F = {F_y} }={ \int\limits_{ – \frac{\pi }{2}}^{\frac{\pi }{2}} {d{F_y}} }={ \int\limits_{ – \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{1}{{4\pi {\varepsilon _0}}}\frac{\mu }{l}\cos \theta d\theta } }={ \frac{\mu }{{4\pi {\varepsilon _0}l}}\int\limits_{ – \frac{\pi }{2}}^{\frac{\pi }{2}} {\cos \theta d\theta } }={ \frac{\mu }{{4\pi {\varepsilon _0}l}}\left. {\left( {\sin \theta } \right)} \right|_{ – \frac{\pi }{2}}^{\frac{\pi }{2}} }={ \frac{\mu }{{4\pi {\varepsilon _0}l}}\left[ {1 – \left( { – 1} \right)} \right] }={ \frac{\mu }{{2\pi {\varepsilon _0}l}}.}\]

    Example 9.

    Find the kinetic energy of a homogeneous circular cylinder of density \(\rho\) with base radius \(R\) and altitude \(H\) rotating about its axis with angular velocity \(\omega.\)

    Solution.

    Consider a thin hollow cylinder with inner radius \(r\) and wall thickness \(dr\) inside the initial cylinder.

    Homogeneous circular cylinder rotating around its axis with angular velocity omega.
    Figure 8.

    The mass of the the hollow cylinder is

    \[dm = \rho dV = 2\pi \rho Hrdr.\]

    The linear velocity of the layer is \(v = wr,\) so the kinetic energy of the elementary cylinder is given by

    \[{dE = \frac{{dm{v^2}}}{2} }={ \frac{{\cancel{2}\pi \rho Hrdr \cdot {{\left( {wr} \right)}^2}}}{\cancel{2}} }={ \pi \rho {w^2}H{r^3}dr.}\]

    The total kinetic energy of the initial cylinder can be found through integration:

    \[{E = \int\limits_0^R {dE\left( r \right)} }={ \int\limits_0^R {\pi \rho {w^2}H{r^3}dr} }={ \pi \rho {w^2}H\int\limits_0^R {{r^3}dr} }={ \left. {\frac{{\pi \rho {w^2}H{r^4}}}{4}} \right|_0^R }={ \frac{{\pi \rho {w^2}H{R^4}}}{4}}.\]

    Notice that the energy of rotation of the body can be expressed in terms of the moment of inertia. In our case,

    \[{E = \frac{{\pi \rho {w^2}H{R^4}}}{4} }={ \pi \rho {R^2}H \cdot \frac{{{R^2}}}{2} \cdot \frac{{{w^2}}}{2} }={ \frac{{m{R^2}}}{2} \cdot \frac{{{w^2}}}{2} }={ \frac{{I{w^2}}}{2},}\]

    where \(I = \large{\frac{{m{R^2}}}{2}}\normalsize\) is the moment of inertia of the solid circular cylinder of mass \(m\) and radius \(R.\)

    Example 10.

    Calculate the kinetic energy of a homogeneous ball of density \(\rho\) and radius \(R\) rotating about its axis with angular velocity \(\omega.\)

    Solution.

    Consider a thin disk of radius \(r\) and thickness \(dz\) inside the ball. The bottom side of the disk crosses the vertical axis at a distance of \(z\) units from the center of the ball.

    A homogeneous ball of radius R rotating about its axis with angular velocity omega.
    Figure 9.

    The disk has the shape of a cylinder. Therefore, its kinetic energy of rotation about the vertical axis is written as (see the previous Example \(9\))

    \[dE = \frac{{\pi \rho {\omega ^2}{r^4}}}{4}dz.\]

    The inner radius \(r\) is expressed in the form

    \[{r = \sqrt {{R^2} – {z^2}} ,}\;\; \Rightarrow {{r^4} = {\left( {{R^2} – {z^2}} \right)^2}.}\]

    To find the total kinetic energy of the ball we integrate the function \({dE\left( z \right)}\) from \(0\) to \(R\) and multiply the result by \(2.\) This yields:

    \[{E = 2\int\limits_0^R {dE\left( z \right)} }={ 2\int\limits_0^R {\frac{{\pi \rho {\omega ^2}{r^4}}}{4}dz} }={ \frac{{\pi \rho {\omega ^2}}}{2}\int\limits_0^R {{{\left( {{R^2} – {z^2}} \right)}^2}dz} }={ \frac{{\pi \rho {\omega ^2}}}{2}\int\limits_0^R {\left( {{R^4} – 2{R^2}{z^2} + {z^4}} \right)dz} }={ \frac{{\pi \rho {\omega ^2}}}{2}\left. {\left[ {{R^4}z – \frac{{2{R^2}{z^3}}}{3} + \frac{{{z^5}}}{5}} \right]} \right|_0^R }={ \frac{{\pi \rho {\omega ^2}}}{2}\left( {{R^5} – \frac{{2{R^5}}}{3} + \frac{{{R^5}}}{5}} \right) }={ \frac{{\pi \rho {\omega ^2}{R^5}}}{2}\left( {1 – \frac{2}{3} + \frac{1}{5}} \right) }={ \frac{{4\pi \rho {\omega ^2}{R^5}}}{{15}}.}\]