Physical Applications of Triple Integrals

Solved Problems

Example 3.

Find the mass of a ball of radius \(R\) whose density \(\gamma\) is proportional to the squared distance from the center.

Solution.

According to the condition of the problem, the density \(\gamma\) is given by \(\gamma = a{r^2},\) where \(a\) is a constant, \(r\) is the distance from the center. It is convenient to calculate the mass of the ball in spherical coordinates:

\[m = \iiint\limits_U {\gamma \left( {r,\varphi ,\theta } \right){r^2}\sin \theta drd\varphi d\theta } = \iiint\limits_U {a{r^2}{r^2}\sin \theta drd\varphi d\theta } = a\iiint\limits_U {{r^4}\sin \theta drd\varphi d\theta } = a\int\limits_0^{2\pi } {d\varphi } \int\limits_0^R {{r^4}dr} \int\limits_0^\pi {\sin \theta d\theta } = a\int\limits_0^{2\pi } {d\varphi } \int\limits_0^R {{r^4}dr} \cdot \left[ {\left. {\left( { - \cos \theta } \right)} \right|_0^\pi } \right] = a\int\limits_0^{2\pi } {d\varphi } \int\limits_0^R {{r^4}dr} \cdot \left( { - \cos \pi + \cos 0} \right) = 2a\int\limits_0^{2\pi } {d\varphi } \int\limits_0^R {{r^4}dr} = 2a\int\limits_0^{2\pi } {d\varphi } \cdot \left[ {\left. {\left( {\frac{{{r^5}}}{5}} \right)} \right|_0^R} \right] = \frac{{2a{R^5}}}{5}\int\limits_0^{2\pi } {d\varphi } = \frac{{2a{R^5}}}{5} \cdot 2\pi = \frac{{4a\pi {R^5}}}{5}.\]

Example 4.

Find the moment of inertia of a right circular homogeneous cone about its axis. The cone has base radius \(R,\) height \(H\) and the total mass \(m\) (Figure \(3\)).

Solution.

The moment of inertia of a body about the \(z\)-axis is expressed by the formula

\[{I_z} = \iiint\limits_U {\gamma \left( {x,y,z} \right)\left( {{x^2} + {y^2}} \right)dxdydz} .\]

Because the cone is homogeneous, then the density \(\gamma \left( {x,y,z} \right) = {\gamma _0}\) can be taken outside the integral sign:

\[{I_z} = {\gamma _0}\int\limits_U {\left( {{x^2} + {y^2}} \right)dxdydz} .\]

We pass to cylindrical coordinates by replacing

\[x = r\cos \varphi ,\;\; y = r\sin \varphi ,\;\; \Rightarrow {x^2} + {y^2} = {r^2},\;\; dxdydz = rdrd\varphi dz.\]

The new variables range within

\[0 \le r \le R,\;\; 0 \le \varphi \le 2\pi ,\;\; r\frac{H}{R} \le z \le H.\]

Then the moment of inertia is

\[ {I_z} = {\gamma _0}\int\limits_0^{2\pi } {d\varphi } \int\limits_0^R {{r^3}dr} \int\limits_{r\frac{H}{R}}^H {dz} = {\gamma _0}\int\limits_0^{2\pi } {d\varphi } \int\limits_0^R {{r^3}dr} \cdot \left[ {\left. z \right|_{r\frac{H}{R}}^H} \right] = {\gamma _0}\int\limits_0^{2\pi } {d\varphi } \int\limits_0^R {{r^3}\left( {H - r\frac{H}{R}} \right)dr} = {\gamma _0}H\int\limits_0^{2\pi } {d\varphi } \int\limits_0^R {\left( {{r^3} - \frac{{{r^4}}}{R}} \right)dr} = {\gamma _0}H\int\limits_0^{2\pi } {d\varphi } \cdot \left[ {\left. {\left( {\frac{{{r^4}}}{4} - \frac{{{r^5}}}{{5R}}} \right)} \right|_0^R} \right] = {\gamma _0}H\int\limits_0^{2\pi } {\left( {\frac{{{R^4}}}{4} - \frac{{{R^4}}}{5}} \right)d\varphi } = \frac{{{\gamma _0}H{R^4}}}{{20}}\int\limits_0^{2\pi } {d\varphi } = \frac{{{\gamma _0}H{R^4}}}{{20}} \cdot 2\pi = \frac{{{\gamma _0}\pi H{R^4}}}{{10}}.\]

We express the density \({\gamma _0}\) in terms of the known mass of the cone \(m.\) Since

\[m = {\gamma _0}V,\;\; V = \frac{1}{3}\pi {R^2}H,\]

hence,

\[{\gamma _0} = \frac{m}{V} = \frac{m}{{\frac{1}{3}\pi {R^2}H}} = \frac{{3m}}{{\pi {R^2}H}}.\]

Finally we obtain

\[{I_z} = \frac{{{\gamma _0}\pi H{R^4}}}{{10}} = \frac{{3m}}{{\bcancel{\pi} {R^2}\bcancel{H}}} \cdot \frac{{\bcancel{\pi} \bcancel{H}{R^4}}}{{10}} = \frac{{3m{R^2}}}{{10}}.\]

It is interesting that the moment of inertia of the cone does not depend on its height.

Example 5.

With what force does a homogeneous ball of mass \(M\) attract a material point of mass \(m,\) located at distance \(a\) from the center of the ball \(\left( {a \gt R} \right)?\)

Solution.

Without sacrificing generality, the material point can be placed on the \(z\)-axis (Figure \(4\)), so that its coordinate is \(\left( {0,0,a} \right).\)

A material point attracted by a ball — Figure 4.

We solve the problem as follows. First, we calculate the potential of the ball and then find the force of attraction of the point and the ball.

To find the potential of the ball, it is more convenient to first determine the potential of the sphere (using the surface integral) instead of calculating the triple integral, and then get the result for the ball (by performing one more integration).

So, calculate the potential of the sphere of arbitrary radius \(r\) \(\left( {r \le R} \right).\) Consider a small area element \(dS\) on the sphere as shown in Figure \(5.\)

Attraction of a material point P and a small surface element dS — Figure 5.

The mass of this area element is

\[dM = \rho \left( r \right)drdS,\]

where \(\rho \left( r \right)\) is the sphere density and \(dr\) is its width. This sphere creates the potential at the point \(P,\) which is equal to

\[du = \rho \left( r \right)dr\iint\limits_S {\frac{{dS}}{\delta }} = \rho \left( r \right)dr \cdot \iint\limits_S {\frac{{dS}}{{\sqrt {{a^2} + {r^2} - 2ar\cos \theta } }}} ,\]

where the distance \(\delta\) from the area element \(dS\) to the point \(P\) is expressed in terms of \(a, r, \theta.\)

Taking into account that the area element is

\[dS = {r^2}\sin \theta d\theta d\varphi,\]

we get

\[du = \rho \left( r \right)dr \cdot \int\limits_S {\frac{{dS}}{{\sqrt {{a^2} + {r^2} - 2ar\cos \theta } }}} = \rho \left( r \right)dr\int\limits_0^{2\pi } {d\varphi } \int\limits_0^\pi {\frac{{{r^2}\sin \theta d\theta }}{{\sqrt {{a^2} + {r^2} - 2ar\cos \theta } }}} = \rho \left( r \right)dr \cdot 2\pi {r^2} \int\limits_0^\pi {\frac{{\sin \theta d\theta }}{{\sqrt {{a^2} + {r^2} - 2ar\cos \theta } }}} .\]

Calculate separately the integral over the variable \(\theta.\) We make the following substitution: Let

\[v = {a^2} + {r^2} - 2ar\cos \theta .\]

Then

\[dv = 2ar\sin \theta d\theta ,\;\; \Rightarrow \sin \theta d\theta = \frac{{dv}}{{2ar}}.\]

As a result, we find the integral

\[I = \int {\frac{{\sin \theta d\theta }}{{\sqrt {{a^2} + {r^2} - 2ar\cos \theta } }}} = \int {\frac{{dv}}{{2ar\sqrt v }}} = \frac{1}{{ar}}\int {\frac{{dv}}{{2\sqrt v }}} = \frac{{\sqrt v }}{{ar}} = \frac{1}{{ar}}\sqrt {{a^2} + {r^2} - 2ar\cos \theta } .\]

Thus, the potential of the sphere of radius \(r\) is given by

\[du = \rho \left( r \right)dr \cdot 2\pi {r^2} \cdot \frac{1}{{ar}} \cdot \left[ {\left. {\left( {\sqrt {{a^2} + {r^2} - 2ar\cos \theta } } \right)} \right|_{\theta = 0}^{\theta = \pi }} \right] = \frac{{2\pi r\rho \left( r \right)dr}}{a} \left[ {\sqrt {{a^2} + {r^2} + 2ar} - \sqrt {{a^2} + {r^2} - 2ar} } \right] = \frac{{2\pi r\rho \left( r \right)dr}}{a} \left[ {\sqrt {{{\left( {a + r} \right)}^2}} - \sqrt {{{\left( {a - r} \right)}^2}} } \right] = \frac{{2\pi r\rho \left( r \right)dr}}{a} \cdot 2r = \frac{{4\pi {r^2}\rho \left( r \right)dr}}{a}.\]

Now we can calculate the potential of the ball of radius \(R.\) We suppose for simplicity that the density of the ball is constant and equal to \({\rho_0}.\) This yields

\[u = \int\limits_0^R {du} = \int\limits_0^R {\frac{{4\pi {r^2}{\rho _0}dr}}{a}} = \frac{{4\pi {\rho _0}}}{a}\int\limits_0^R {{r^2}dr} = \frac{{4\pi {\rho _0}}}{a}\left[ {\left. {\left( {\frac{{{r^3}}}{3}} \right)} \right|_0^R} \right] = \frac{{4\pi {\rho _0}{R^3}}}{{3a}}.\]

In this expression \({\frac{4}{3}} \pi {R^3} = V\) is the volume of the ball, and \({\rho_0}V = M\) is the mass of the ball. Thus, we have proved that the gravitational potential of the ball at the distance \(a\) from the center of the ball \(\left( {a \gt R} \right)\) is expressed by the formula

\[u = \frac{M}{a}.\]

Next, it is easy to find the force of attraction between the ball and the material point. Since

\[\mathbf{F} = - Gm\,\mathbf{\text{grad}}\,u,\]

the force is

\[F = Gm\frac{{\partial u}}{{\partial z}} = Gm{\left. {\left[ {\frac{\partial }{{\partial z}}\left( {\frac{M}{z}} \right)} \right]} \right|_{z = a}} = - G\frac{{mM}}{{{a^2}}}.\]

The "minus" sign means that the force is directed opposite to the \(z\)-axis, that is, it is the force of attraction.

As it can be seen, the attractive force of the ball and the point has the same form as the force of attraction between two point masses! This is one of the fundamental results in astrophysics and celestial mechanics. Because of this, the planets and stars can often be considered as material points in the description of their movement. To prove this result, Isaac Newton was even forced to postpone publication of his masterpiece on astronomy. Perhaps the difficulties were related to the fact that he did not use spherical coordinates to solve this problem...

Example 6.

Suppose that a planet has a radius \(R\) and its density is expressed by the formula

\[\gamma \left( r \right) = \frac{{R + r}}{{2r}}{\gamma _0}.\]

Find the mass of the planet.

Solution.

Consider in detail the law of density variation. If \(r = R,\) then

\[\gamma \left( R \right) = \frac{{R + R}}{{2R}}{\gamma _0} = {\gamma _0},\]

where \({\gamma_0}\) is a surface density of the planet. Here \(\gamma \to \infty\) as \(r \to 0\) (Figure \(6\)).

The density of a planet depending on the distance from the center — Figure 6.

We calculate the mass of the planet using triple integral by the formula:

\[M = \iiint\limits_U {dV} .\]

By converting to spherical coordinates, we have

\[M = \iiint\limits_U {dV} = \iiint\limits_U {\gamma \left( r \right){r^2}\sin \theta drd\varphi d\theta } = \iiint\limits_U {\frac{{R + r}}{{2r}}{\gamma _0}{r^2}\sin \theta drd\varphi d\theta } = \frac{{{\gamma _0}}}{2}\int\limits_U {\left( {R + r} \right)r\sin \theta drd\varphi d\theta } = \frac{{{\gamma _0}}}{2}\int\limits_0^R {\left( {R + r} \right)rdr} \int\limits_0^{2\pi } {d\varphi } \int\limits_0^\pi {\sin \theta d\theta } = \frac{{{\gamma _0}}}{2}\int\limits_0^R {\left( {R + r} \right)rdr} \int\limits_0^{2\pi } {d\varphi } \cdot \left[ {\left. {\left( { - \cos \theta } \right)} \right|_0^\pi } \right] = \frac{{{\gamma _0}}}{2}\int\limits_0^R {\left( {R + r} \right)rdr} \int\limits_0^{2\pi } {d\varphi } \cdot \left( { - \cos \pi + \cos 0} \right) = {\gamma _0}\int\limits_0^R {\left( {R + r} \right)rdr} \int\limits_0^{2\pi } {d\varphi } = {\gamma _0}\int\limits_0^R {\left( {R + r} \right)rdr} \cdot 2\pi = 2\pi {\gamma _0}\int\limits_0^R {\left( {Rr + {r^2}} \right)dr} = 2\pi {\gamma _0}\left[ {\left. {\left( {\frac{{R{r^2}}}{2} + \frac{{{r^3}}}{3}} \right)} \right|_0^R} \right] = 2\pi {\gamma _0}\left( {\frac{{{R^3}}}{2} + \frac{{{R^3}}}{3}} \right) = 2\pi {\gamma _0}\frac{{5{R^3}}}{6} = \frac{{5\pi {\gamma _0}{R^3}}}{3}.\]

Since the volume of the planet is \({\frac{4}{3}} \pi {R^3},\) the answer can be written in the form:

\[M = \frac{{5\pi {\gamma _0}{R^3}}}{3} = \frac{5}{4}{\gamma _0}V.\]

As it can be seen, the mass of the planet is \(25\%\) more compared with the case when the density is distributed uniformly.