Calculus

Triple Integrals

Physical Applications of Triple Integrals

Page 1
Problem 1
Page 2
Problems 2-6

Mass and Static Moments of a Solid

Suppose we have a solid occupying a region \(U.\) Its volume density at a point \(M\left( {x,y,z} \right)\) is given by the function \(\rho\left( {x,y,z} \right).\) Then the mass of the solid \(m\) is expressed through 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
\[
{{M_{xy}} = \int\limits_U {z\rho \left( {x,y,z} \right)dxdydz} ,\;\;}\kern-0,3pt
{{M_{yz}} = \int\limits_U {x\rho \left( {x,y,z} \right)dxdydz} ,\;\;}\kern-0,3pt
{{M_{xz}} = \int\limits_U {y\rho \left( {x,y,z} \right)dxdydz} .}
\] The coordinates of the center of gravity of the solid are described by the expressions:
\[
{{\bar x = \frac{{{M_{yz}}}}{m} }={ \frac{{\iiint\limits_U {x\rho \left( {x,y,z} \right)dxdydz} }}{{\iiint\limits_U {\rho \left( {x,y,z} \right)dxdydz} }},\;\;}}\kern-0.3pt
{{\bar y = \frac{{{M_{xz}}}}{m} }={ \frac{{\iiint\limits_U {y\rho \left( {x,y,z} \right)dxdydz} }}{{\iiint\limits_U {\rho \left( {x,y,z} \right)dxdydz} }},\;\;}}\kern-0.3pt
{{\bar z = \frac{{{M_{xy}}}}{m} }={ \frac{{\iiint\limits_U {z\rho \left( {x,y,z} \right)dxdydz} }}{{\iiint\limits_U {\rho \left( {x,y,z} \right)dxdydz} }}.}}
\] 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
\[
{{{I_{xy}} }={ \iiint\limits_U {{z^2}\rho \left( {x,y,z} \right)dxdydz} ,\;\;}}\kern-0.3pt
{{{I_{yz}} }={ \iiint\limits_U {{x^2}\rho \left( {x,y,z} \right)dxdydz} ,\;\;}}\kern-0.3pt
{{{I_{xz}} }={ \iiint\limits_U {{y^2}\rho \left( {x,y,z} \right)dxdydz},}}
\] and the moments of inertia of a solid about the coordinate axes \(Ox, Oy, Oz\) are expressed by the formulas
\[
{{{I_x} = \iiint\limits_U {\left( {{y^2} + {z^2}} \right)\cdot}}\kern0pt{{\rho \left( {x,y,z} \right)dxdydz} ,\;\;}}\kern-0.3pt
{{{I_y} = \iiint\limits_U {\left( {{x^2} + {z^2}} \right)\cdot}}\kern0pt{{\rho \left( {x,y,z} \right)dxdydz} ,\;\;}}\kern-0.3pt
{{{I_z} = \iiint\limits_U {\left( {{x^2} + {y^2}} \right)\cdot}}\kern0pt{{\rho \left( {x,y,z} \right)dxdydz} .\;\;}}\kern-0.3pt
\] As seen, the following properties are valid:
\[
{{I_x} = {I_{xy}} + {I_{xz}},\;\;}\kern-0.3pt
{{I_y} = {I_{xy}} + {I_{yz}},\;\;}\kern-0.3pt
{{I_z} = {I_{xz}} + {I_{yz}}.}
\] The moment of inertia about the origin is called the integral
\[{I_0} = \iiint\limits_U {\left( {{x^2} + {y^2} + {z^2}} \right)}\cdot\kern0pt{\rho \left( {x,y,z} \right)dxdydz}.\] The moment of inertia about the origin can be expressed through the moments of inertia about the corodinate 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
\[{u\left( {x,y,z} \right) \text{ = }}\kern0pt{ \iiint\limits_U {\rho \left( {\xi ,\eta ,\zeta } \right)\frac{{d\xi d\eta d\zeta }}{r}} ,}\] where \({\rho \left( {\xi ,\eta ,\zeta } \right)}\) is the density of the body and
\[{r \text{ = }}\kern0pt{\sqrt {{{\left( {\xi – x} \right)}^2} + {{\left( {\eta – y} \right)}^2} + {{\left( {\zeta – z} \right)}^2}} }\] 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

Click on problem description to see solution.

 Example 1

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

 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 \(2\)).

 Example 3

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

 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 \(3\)).

 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 \(\left( {a \gt R} \right)?\)

 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.

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

A half-ball of radius R

Figure 1.

Since the half-ball is homogeneous, we set \(\rho \left( {x,y,z} \right) = {\rho _0}.\) Then
\[\require{cancel}
{\bar z = \frac{{{\bcancel{\rho _0}}\iiint\limits_U {zdxdydz} }}{{{\bcancel{\rho _0}}\iiint\limits_U {dxdydz} }} }
= {\frac{{\iiint\limits_U {zdxdydz} }}{{\iiint\limits_U {dxdydz} }} }
= {\frac{{\iiint\limits_U {zdxdydz} }}{V}.}
\] 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
\[
{\iiint\limits_U {zdxdydz} }
= {\iiint\limits_{U’} {r\cos \theta {r^2}\sin \theta drd\varphi d\theta } }
= {\int\limits_0^{2\pi } {d\varphi } \int\limits_0^R {{r^3}dr} \int\limits_0^{\large\frac{\pi }{2}\normalsize} {\cos \theta \sin \theta d\theta } }
= {\int\limits_0^{2\pi } {d\varphi } \int\limits_0^R {{r^3}dr} \int\limits_0^{\large\frac{\pi }{2}\normalsize} {\sin \theta d\left( {\sin \theta } \right)} }
= {\int\limits_0^{2\pi } {d\varphi } \int\limits_0^R {{r^3}dr} \cdot}\kern0pt{ \left[ {\left. {\left( {\frac{{{{\sin }^2}\theta }}{2}} \right)} \right|_{\theta = 0}^{\theta = \large\frac{\pi }{2}\normalsize}} \right] }
= {\int\limits_0^{2\pi } {d\varphi } \int\limits_0^R {{r^3}dr} \cdot \frac{1}{2} = \frac{1}{2}\int\limits_0^{2\pi } {d\varphi } \cdot \left[ {\left. {\left( {\frac{{{r^4}}}{4}} \right)} \right|_0^R} \right] }
= {\frac{{{R^4}}}{8}\int\limits_0^{2\pi } {d\varphi } }
= {\frac{{{R^4}}}{8} \cdot 2\pi }
= {\frac{{\pi {R^4}}}{4}.}
\] Thus, the coordinate \(\bar z\) of the center of gravity is
\[
{\bar z = \frac{{\iiint\limits_U {zdxdydz} }}{V} }
= {\frac{{\frac{1}{4}\pi {R^4}}}{{\frac{2}{4}\pi {R^3}}} }={ \frac{{3R}}{8}.}
\]

Page 1
Problem 1
Page 2
Problems 2-6