Physical Applications of Triple Integrals

Mass and Static Moments of a Solid

Suppose we have a solid occupying a region U. Its volume density at a point M (x, y, z) is given by the function ρ (x, y, z). Then the mass of the solid m is expressed by the triple integral as

\[m = \iiint\limits_U {\rho \left( {x,y,z} \right)dxdydz} .\]

The static moments of the solid about the coordinate planes \(Oxy, Oxz, Oyz\) are given by the formulas

\[{M_{xy}} = \int\limits_U {z\rho \left( {x,y,z} \right)dxdydz} ,\;\;{M_{yz}} = \int\limits_U {x\rho \left( {x,y,z} \right)dxdydz} ,\;\;{M_{xz}} = \int\limits_U {y\rho \left( {x,y,z} \right)dxdydz} .\]

The coordinates of the center of gravity of the solid are described by the expressions:

\[\bar x = \frac{{{M_{yz}}}}{m} = \frac{{\iiint\limits_U {x\rho \left( {x,y,z} \right)dxdydz} }}{{\iiint\limits_U {\rho \left( {x,y,z} \right)dxdydz} }},\;\; \bar y = \frac{{{M_{xz}}}}{m} = \frac{{\iiint\limits_U {y\rho \left( {x,y,z} \right)dxdydz} }}{{\iiint\limits_U {\rho \left( {x,y,z} \right)dxdydz} }},\;\; \bar z = \frac{{{M_{xy}}}}{m} = \frac{{\iiint\limits_U {z\rho \left( {x,y,z} \right)dxdydz} }}{{\iiint\limits_U {\rho \left( {x,y,z} \right)dxdydz} }}.\]

If a solid is homogeneous with density \({\rho \left( {x,y,z} \right)} = 1\) for all points \({M\left( {x,y,z} \right)}\) in the region \(U,\) then the center of gravity of the solid is determined only by the shape of the solid and is called the centroid.

Moments of Inertia of a Solid

The moments of inertia of a solid about the coordinate planes \(Oxy, Oxz, Oyz\) are given by

\[{I_{xy}} = \iiint\limits_U {{z^2}\rho \left( {x,y,z} \right)dxdydz} ,\;\; {I_{yz}} = \iiint\limits_U {{x^2}\rho \left( {x,y,z} \right)dxdydz} ,\;\; {I_{xz}} = \iiint\limits_U {{y^2}\rho \left( {x,y,z} \right)dxdydz},\]

and the moments of inertia of a solid about the coordinate axes \(Ox, Oy, Oz\) are expressed by the formulas

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

As it can be seen, the following properties are valid:

\[{I_x} = {I_{xy}} + {I_{xz}},\;\; {I_y} = {I_{xy}} + {I_{yz}},\;\; {I_z} = {I_{xz}} + {I_{yz}}.\]

The moment of inertia about the origin is called the integral

\[{I_0} = \iiint\limits_U {\left( {{x^2} + {y^2} + {z^2}} \right) \rho \left( {x,y,z} \right)dxdydz}.\]

The moment of inertia about the origin can be expressed through the moments of inertia about the coordinate planes as follows:

\[{I_0} = {I_{xy}} + {I_{yz}} + {I_{xz}}.\]

Tensor of Inertia

Using the \(6\) numbers considered above: \({I_x},{I_y},{I_z},{I_{xy}},{I_{xz}},{I_{yz}},\) we can construct the so-called matrix of inertia or the tensor of inertia of the solid:

\[I = \left( {\begin{array}{*{20}{c}} {{I_x}}&{ - {I_{xy}}}&{ - {I_{xz}}}\\ { - {I_{xy}}}&{{I_y}}&{ - {I_{yz}}}\\ { - {I_{xz}}}&{ - {I_{yz}}}&{{I_z}} \end{array}} \right).\]

This tensor is symmetric and, hence, it can be transformed to a diagonal view by choosing the appropriate coordinate axes \(Ox', Oy', Oz'.\) The values of the diagonal elements (after transforming the tensor to a diagonal form) are called the main moments of inertia, and the indicated directions of the axes are called the eigenvalues or the principal axes of inertia of the body.

If a body rotates about an axis which does not coincide with a principal axis of inertia, it will experience vibrations at the high rotation speeds. Therefore, when designing such devices it is necessary the axis of rotation to be coinciding with one of the principal axes of inertia. For example, when replacing car tires, it's often necessary to balance the wheels by attaching small lead weights to ensure the coincidence of the rotation axis with the principal axis of inertia and to eliminate vibration.

Gravitational Potential and Attraction Force

The Newton potential of a body at a point \(P\left( {x,y,z} \right)\) is called the integral

\[u\left( {x,y,z} \right) = \iiint\limits_U {\rho \left( {\xi ,\eta ,\zeta } \right)\frac{{d\xi d\eta d\zeta }}{r}} ,\]

where \({\rho \left( {\xi ,\eta ,\zeta } \right)}\) is the density of the body and

\[r = \sqrt {{{\left( {\xi - x} \right)}^2} + {{\left( {\eta - y} \right)}^2} + {{\left( {\zeta - z} \right)}^2}} .\]

The integration is performed over the whole volume of the body. Knowing the potential, one can calculate the force of attraction of the material point of mass \(m\) and the distributed body with the density \({\rho \left( {\xi ,\eta ,\zeta } \right)}\) by the formula

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

where \(G\) is the gravitational constant.

Solved Problems

Example 1.

Find the centroid of a homogeneous half-ball of radius \(R.\)

Solution.

We introduce the system of coordinates in such a way that the half-ball is located at \(z \ge 0\) and centered at the origin (Figure \(1\text{).}\)

Using this system of coordinates, we find the centroid (the center of gravity) of the solid. Obviously, by symmetry,

\[\bar x = \bar y = 0.\]

Calculate the coordinate \(\bar z\) of the centroid by the formula

\[\bar z = \frac{{{M_{xy}}}}{m} = \frac{{\iiint\limits_U {z\rho \left( {x,y,z} \right)dxdydz} }}{{\iiint\limits_U {\rho \left( {x,y,z} \right)dxdydz} }}.\]

Since the half-ball is homogeneous, we set \(\rho \left( {x,y,z} \right) = {\rho _0}.\) Then

\[\bar z = \frac{{{\bcancel{\rho _0}}\iiint\limits_U {zdxdydz} }}{{{\bcancel{\rho _0}}\iiint\limits_U {dxdydz} }} = \frac{{\iiint\limits_U {zdxdydz} }}{{\iiint\limits_U {dxdydz} }} = \frac{{\iiint\limits_U {zdxdydz} }}{V}.\]

The symbol \(V\) in the denominator denotes the volume of the solid, which is equal to

\[V = \frac{1}{2}\left( {\frac{4}{3}\pi {R^3}} \right) = \frac{2}{3}\pi {R^3}.\]

It remains to compute the triple integral \({\iiint\limits_U {zdxdydz} }.\) For this, we pass to spherical coordinates. In this case, the radial coordinate is denoted by \(r\) in order not to be confused with the density \(\rho.\) As a result, we have

\[\iiint\limits_U {zdxdydz} = \iiint\limits_{U'} {r\cos \theta {r^2}\sin \theta drd\varphi d\theta } = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^R {{r^3}dr} \int\limits_0^{\frac{\pi }{2}} {\cos \theta \sin \theta d\theta } = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^R {{r^3}dr} \int\limits_0^{\frac{\pi }{2}} {\sin \theta d\left( {\sin \theta } \right)} = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^R {{r^3}dr} \cdot \left[ {\left. {\left( {\frac{{{{\sin }^2}\theta }}{2}} \right)} \right|_{\theta = 0}^{\theta = \frac{\pi }{2}}} \right] = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^R {{r^3}dr} \cdot \frac{1}{2} = \frac{1}{2}\int\limits_0^{2\pi } {d\varphi } \cdot \left[ {\left. {\left( {\frac{{{r^4}}}{4}} \right)} \right|_0^R} \right] = \frac{{{R^4}}}{8}\int\limits_0^{2\pi } {d\varphi } = \frac{{{R^4}}}{8} \cdot 2\pi = \frac{{\pi {R^4}}}{4}.\]

Thus, the coordinate \(\bar z\) of the center of gravity is

\[\bar z = \frac{{\iiint\limits_U {zdxdydz} }}{V} = \frac{{\frac{1}{4}\pi {R^4}}}{{\frac{2}{4}\pi {R^3}}} = \frac{{3R}}{8}.\]

Example 2.

Determine the mass and coordinates of the center of gravity of the unit cube with the density

\[\rho \left( {x,y,z} \right) = x + 2y + 3z\]

(Figure \(2\)).

Solution.

The unit cube with non-homogeneous density — Figure 2.

First we calculate the mass of the cube:

\[ m = \iiint\limits_U {\rho \left( {x,y,z} \right)dxdydz} = \int\limits_0^1 {dx} \int\limits_0^1 {dy} \int\limits_0^1 {\left( {x + 2y + 3z} \right)dz} = \int\limits_0^1 {dx} \int\limits_0^1 {dy} \cdot \left[ {\left. {\left( {xz + 2yz + \frac{{3{z^2}}}{2}} \right)} \right|_{z = 0}^{z = 1}} \right] = \int\limits_0^1 {dx} \int\limits_0^1 {\left( {x + 2y + \frac{3}{2}} \right)dy} = \int\limits_0^1 {dx} \cdot \left[ {\left. {\left( {xy + {y^2} + \frac{3}{2}y} \right)} \right|_{y = 0}^{y = 1}} \right] = \int\limits_0^1 {\left( {x + 1 + \frac{3}{2}} \right)dx} = \int\limits_0^1 {\left( {x + \frac{5}{2}} \right)dx} = \left. {\left( {\frac{{{x^2}}}{2} + \frac{5}{2}x} \right)} \right|_0^1 = \frac{1}{2} + \frac{5}{2} = 3.\]

Now we calculate the static moments \({M_{xy}},{M_{xz}},{M_{yz}}.\)

\[ {M_{xy}} = \iiint\limits_U {z\rho \left( {x,y,z} \right)dxdydz} = \int\limits_0^1 {dx} \int\limits_0^1 {dy} \int\limits_0^1 {z\left( {x + 2y + 3z} \right)dz} = \int\limits_0^1 {dx} \int\limits_0^1 {dy} \int\limits_0^1 {\left( {xz + 2zy + 3{z^2}} \right)dz} = \int\limits_0^1 {dx} \int\limits_0^1 {dy} \cdot \left[ {\left. {\left( {\left( {x + 2y} \right)\frac{{{z^2}}}{2} + {z^3}} \right)} \right|_{z = 0}^{z = 1}} \right] = \int\limits_0^1 {dx} \int\limits_0^1 {\left( {\frac{1}{2}\left( {x + 2y} \right) + 1} \right)dy} = \frac{1}{2}\int\limits_0^1 {dx} \int\limits_0^1 {\left( {x + 2y + 2} \right)dy} = \frac{1}{2}\int\limits_0^1 {dx} \cdot \left[ {\left. {\left( {xy + {y^2} + 2y} \right)} \right|_{y = 0}^{y = 1}} \right] = \frac{1}{2}\int\limits_0^1 {\left( {x + 3} \right)dx} = \frac{1}{2}\left[ {\left. {\left( {\frac{{{x^2}}}{2} + 3x} \right)} \right|_0^1} \right] = \frac{1}{2}\left( {\frac{1}{2} + 3} \right) = \frac{7}{4}.\]

Similarly, we find the moments \({M_{xz}}\) and \({M_{yz}}:\)

\[ {M_{xz}} = \iiint\limits_U {y\rho \left( {x,y,z} \right)dxdydz} = \int\limits_0^1 {dx} \int\limits_0^1 {dy} \int\limits_0^1 {y\left( {x + 2y + 3z} \right)dz} = \int\limits_0^1 {dx} \int\limits_0^1 {dy} \int\limits_0^1 {\left( {yx + 2{y^2} + 3yz} \right)dz} = \int\limits_0^1 {dx} \int\limits_0^1 {dy} \cdot \left[ {\left. {\left( {yxz + 2{y^2}z + \frac{{3y{z^2}}}{2}} \right)} \right|_{z = 0}^{z = 1}} \right] = \int\limits_0^1 {dx} \int\limits_0^1 {\left( {yx + 2{y^2} + \frac{{3y}}{2}} \right)dy} = \int\limits_0^1 {dx} \cdot \left[ {\left. {\left( {\frac{{x{y^2}}}{2} + \frac{{2{y^3}}}{3} + \frac{{3{y^2}}}{4}} \right)} \right|_{y = 0}^{y = 1}} \right] = \int\limits_0^1 {\left( {\frac{x}{2} + \frac{2}{3} + \frac{3}{4}} \right)dx} = \int\limits_0^1 {\left( {\frac{x}{2} + \frac{{17}}{{12}}} \right)dx} = \left. {\left( {\frac{{{x^2}}}{4} + \frac{{17x}}{{12}}} \right)} \right|_0^1 = \frac{1}{4} + \frac{{17}}{{12}} = \frac{{20}}{{12}} = \frac{5}{3}.\]

\[ {M_{yz}} = \iiint\limits_U {x\rho \left( {x,y,z} \right)dxdydz} = \int\limits_0^1 {dx} \int\limits_0^1 {dy} \int\limits_0^1 {x\left( {x + 2y + 3z} \right)dz} = \int\limits_0^1 {dx} \int\limits_0^1 {dy} \int\limits_0^1 {\left( {{x^2} + 2xy + 3xz} \right)dz} = \int\limits_0^1 {dx} \int\limits_0^1 {dy} \cdot \left[ {\left. {\left( {{x^2}z + 2xyz + \frac{{3x{z^2}}}{2}} \right)} \right|_{z = 0}^{z = 1}} \right] = \int\limits_0^1 {dx} \int\limits_0^1 {\left( {{x^2} + 2xy + \frac{{3x}}{2}} \right)dy} = \int\limits_0^1 {dx} \cdot \left[ {\left. {\left( {{x^2}y + x{y^2} + \frac{{3xy}}{2}} \right)} \right|_{y = 0}^{y = 1}} \right] = \int\limits_0^1 {\left( {{x^2} + x + \frac{{3x}}{2}} \right)dx} = \int\limits_0^1 {\left( {{x^2} + \frac{{5x}}{2}} \right)dx} = \left. {\left( {\frac{{{x^3}}}{3} + \frac{{5{x^2}}}{4}} \right)} \right|_0^1 = \frac{1}{3} + \frac{5}{4} = \frac{{19}}{{12}}.\]

Calculate the coordinates of the center of gravity of the cube:

\[\bar x = \frac{{{M_{yz}}}}{m} = \frac{{\frac{{19}}{{12}}}}{3} = \frac{{19}}{{36}},\;\; \bar y = \frac{{{M_{xz}}}}{m} = \frac{{\frac{5}{3}}}{3} = \frac{5}{9} = \frac{{20}}{{36}},\;\; \bar z = \frac{{{M_{xy}}}}{m} = \frac{{\frac{7}{4}}}{3} = \frac{7}{{12}} = \frac{{21}}{{36}}.\]

See more problems on Page 2.