4 Applications of triple and higher integrals

The integral $\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->f\left(x,y,z\right)dzdydx$ (or ${\int \; <!--nolimits\; -->}_{V}f\left(x,y,z\right)dV$ ) may represent many physical quantities depending on the function $f\left(x,y,z\right)$ and the limits used.

4.1 Volume

The integral ${\int \; <!--nolimits\; -->}_{V}1dV$ (i.e. the integral of the function $f\left(x,y,z\right)=1$ ) with appropriate limits gives the volume of the solid described by $V$ . This is sometimes more convenient than finding the volume by means of a double integral.

4.2 Mass

The integral $\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->\rho \left(x,y,z\right)dzdydx$ (or ${\int \; <!--nolimits\; -->}_V\rho \left(x,y,z\right)dV$ ), with appropriate limits, gives the mass of the solid bounded by $V$ .

4.3 Mass of water in a reservoir

The introduction to this Section concerned the mass of water in a reservoir. Imagine that the reservoir is rectangular in profile and that the width along the dam (i.e. measured in the $x$ direction) is 100 m. Imagine also that the length of the reservoir (measured away from the dam i.e. in the $y$ direction) is 400 m. The depth of the reservoir is given by $40-y∕10$ m i.e. the reservoir is 40 m deep along the dam and the depth reduces to zero at the end away from the dam.

The density of the water can be approximated by $\rho \left(z\right)=a-b\times z$ where $a=998\text{kg}{\text{m}}^{-3}$ and $b=0.05\text{kg}{\text{m}}^{-4}.$ I.e. at the surface ( $z=0$ ) the water has density $998\text{kg}{\text{m}}^{-3}$ (corresponding to a temperature of $20^{\circ }$ C) while $40$ m down i.e. $z=-40$ , the water has a density of $1000\text{kg}{\text{m}}^{-3}$ (corresponding to the lower temperature of $4^{\circ }$ C).

Figure 27

The mass of water in the reservoir is given by the integral of the function $\rho \left(z\right)=a-b\times z$ . For each value of $x$ and $y$ , the limits on $z$ will be from $y∕10-40$ (bottom) to $0$ (top). Limits on $y$ will be $0$ to $400$ m while the limits of $x$ will be $0$ to $100$ m. The mass of water is therefore given by the integral

$M={\int \; <!--nolimits\; -->}_0^{100}{\int \; <!--nolimits\; -->}_0^{400}{\int \; <!--nolimits\; -->}_{y∕10-40}^0\left(a-bz\right)dzdydx$

which can be evaluated as follows

So the mass of water in the reservoir is $7.989\times 10^8$ kg.

Notes :

1. In practice, the profile of the reservoir would not be rectangular and the depth would not vary so smoothly.
2. The variation of the density of water with height is only a minor factor so it would only be taken into account when a very exact answer was required. Assuming that the water had a uniform density of $\rho =998\text{kg}{\text{m}}^{-3}$ would give a total mass of $7.984\times 10^8$ kg while assuming a uniform density of $\rho =1000\text{kg}{\text{m}}^{-3}$ gives a total mass of $8\times 10^8$ kg.

4.4 Centre of mass

The expressions for the centre of mass $\left(\bar{x},\bar{y},\bar{z}\right)$ of a solid of density $\rho \left(x,y,z\right)$ are given below

In the (fairly common) case where the density $\rho$ does not vary with position and is constant, these results simplify to

$\bar{x}=\frac{\int \; <!--nolimits\; -->xdV}{\int \; <!--nolimits\; -->dV}\bar{y}=\frac{\int \; <!--nolimits\; -->ydV}{\int \; <!--nolimits\; -->dV}\bar{z}=\frac{\int \; <!--nolimits\; -->zdV}{\int \; <!--nolimits\; -->dV}$

Example 22

A tetrahedron is enclosed by the planes $x=0$ , $y=0$ , $z=0$ and $x+y+z=4$ . Find

1. the volume of this tetrahedron,  
2. the position of the centre of mass.

Solution

1. Note that this tetrahedron was considered in Example 18, see Figure 24. It was shown that in this case the volume integral ${\int \; <!--nolimits\; -->}_{V}f\left(x,y,z\right)dV$ becomes ${\int \; <!--nolimits\; -->}_{x=0}^4{\int \; <!--nolimits\; -->}_{y=0}^{4-x}{\int \; <!--nolimits\; -->}_{z=0}^{4-x-y}f\left(x,y,z\right)dzdydx$ . The volume is given by $$\begin{array}{rcll}V& =& {\int \; <!--nolimits\; -->}_{V}dV={\int \; <!--nolimits\; -->}_{x=0}^4{\int \; <!--nolimits\; -->}_{y=0}^{4-x}{\int \; <!--nolimits\; -->}_{z=0}^{4-x-y}dzdydx& \\ & =& {\int \; <!--nolimits\; -->}_{x=0}^4{\int \; <!--nolimits\; -->}_{y=0}^{4-x}{\left[z\right]}_{z=0}^{4-x-y}dydx& \\ & =& {\int \; <!--nolimits\; -->}_{x=0}^4{\int \; <!--nolimits\; -->}_{y=0}^{4-x}\left(4-x-y\right)dydx& \\ & =& {\int \; <!--nolimits\; -->}_{x=0}^4{\left[4y-xy-\frac{1}{2}{y}^2\right]}_{y=0}^{4-x}dx& \\ & =& {\int \; <!--nolimits\; -->}_{x=0}^4\left[8-4x+\frac{1}{2}{x}^2\right]dx& \\ & =& {\left[8x-2x^2+\frac{1}{6}{x}^3\right]}_0^4=32-32+\frac{64}{6}=\frac{32}{3}& \end{array}$$

   Thus the volume of the tetrahedron is $\frac{32}{3}\approx 10.3$

2. The $x$ coordinate of the centre of mass i.e. $\bar{x}$ is given by $\bar{x}=\frac{\int \; <!--nolimits\; -->xdV}{\int \; <!--nolimits\; -->dV}$ .

   The denominator $\int \; <!--nolimits\; -->dV$ is the formula for the volume i.e. $\frac{32}{3}$ while the numerator $\int \; <!--nolimits\; -->xdV$ was calculated in an earlier Example to be $\frac{32}{3}$ .

   Thus $\bar{x}=\frac{\int \; <!--nolimits\; -->xdV}{\int \; <!--nolimits\; -->dV}=\frac{32∕3}{32∕3}=1$ .
   
   By symmetry (or by evaluating relevant integrals), it can be shown that $\bar{y}=\bar{z}=1$ i.e. the centre of mass is at $\left(1,1,1\right)$ .

4.5 Moment of inertia

The moment of inertia $I$ of a particle of mass $M$ about an axis $PQ$ is defined as

$I=\text{Mass}\times {\text{Distance}}^2\text{or}I=M{d}^2$

where $d$ is the perpendicular distance from the particle to the axis.

To find the moment of inertia of a larger object, it is necessary to carry out a volume integration over all such particles. The distance of a particle at $\left(x,y,z\right)$ from the $z$ -axis is given by $\sqrt{x^2+y^2}$ so the moment of inertia of an object about the $z$ -axis is given by

$I_z={\int \; <!--nolimits\; -->}_V\rho \left(x,y,z\right)\left(x^2+y^2\right)dz$

Similarly, the moments of inertia about the $x$ -axis and $y$ -axis are given by

$I_x={\int \; <!--nolimits\; -->}_V\rho \left(x,y,z\right)\left(y^2+z^2\right)dx\text{ and }{I}_y={\int \; <!--nolimits\; -->}_V\rho \left(x,y,z\right)\left(x^2+z^2\right)dy$

In the case where the density is constant over the object, so $\rho \left(x,y,z\right)=\rho$ , these formulae reduce to

$I_x=\rho {\int \; <!--nolimits\; -->}_V\left(y^2+z^2\right)dx\text{ , }{I}_y=\rho {\int \; <!--nolimits\; -->}_V\left(x^2+z^2\right)dy\text{ and }{I}_z=\rho {\int \; <!--nolimits\; -->}_V\left(x^2+y^2\right)dz$

When possible, the moment of inertia is expressed in terms of $M$ , the mass of the object.

Example 23

Find the moment of inertia (about the $x$ -axis) of the cube of side $1$ , mass $M$ and density $\rho$ shown in Example 16, page 43.

Solution

For the cube,

$\text{Mass = Volume}\times \text{Density i.e. }M=1^3\times \rho =\rho$

The moment of inertia (about the $x$ -axis) is given by

$I_x=\rho {\int \; <!--nolimits\; -->}_V\left(y^2+z^2\right)dx=\rho {\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^1\left(y^2+z^2\right)dzdydx$

This integral was shown to equal $\frac{2}{3}$ in Example 16. Thus

$I_x=\frac{2}{3}\rho =\frac{2}{3}M$

By applying symmetry, it can also be shown that the moments of inertia about the $y$ - and $z$ -axes are also equal to $\frac{2}{3}M$ .