### 4 Applications of triple and higher integrals

The integral $\int \int \int f\left(x,y,z\right)\phantom{\rule{0.3em}{0ex}}dzdydx$ (or ${\int }_{V}f\left(x,y,z\right)\phantom{\rule{0.3em}{0ex}}dV$ ) may represent many physical quantities depending on the function $f\left(x,y,z\right)$ and the limits used.

#### 4.1 Volume

The integral ${\int }_{V}1\phantom{\rule{0.3em}{0ex}}dV$ (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 $\int \int \int \rho \left(x,y,z\right)\phantom{\rule{0.3em}{0ex}}dzdydx$ (or ${\int }_{V}\rho \left(x,y,z\right)\phantom{\rule{0.3em}{0ex}}dV$ ), 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 $\rho \left(z\right)=a-b×z$ 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 $2{0}^{\circ }$ 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 $\rho \left(z\right)=a-b×z$ . 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

$\phantom{\rule{2em}{0ex}}M={\int }_{0}^{100}{\int }_{0}^{400}{\int }_{y∕10-40}^{0}\left(a-bz\right)\phantom{\rule{0.3em}{0ex}}dzdydx$

which can be evaluated as follows

$\begin{array}{rcll}M& =& {\int }_{0}^{100}{\int }_{0}^{400}{\int }_{y∕10-40}^{0}\left(a-bz\right)\phantom{\rule{0.3em}{0ex}}dzdydx& \text{}\\ & =& {\int }_{0}^{100}{\int }_{0}^{400}{\left[az-\frac{b}{2}{z}^{2}\right]}_{y∕10-40}^{0}\phantom{\rule{0.3em}{0ex}}dydx& \text{}\\ & =& {\int }_{0}^{100}{\int }_{0}^{400}\left[0-a\left(y∕10-40\right)+\frac{b}{2}{\left(y∕10-40\right)}^{2}\right]\phantom{\rule{0.3em}{0ex}}dydx& \text{}\\ & =& {\int }_{0}^{100}{\int }_{0}^{400}\left[40a-\frac{a}{10}y+\frac{b}{200}{y}^{2}-4by+800b\right]\phantom{\rule{0.3em}{0ex}}dydx& \text{}\\ & =& {\int }_{0}^{100}{\left[40ay-\frac{a}{20}{y}^{2}+\frac{b}{600}{y}^{3}-2b{y}^{2}+800by\right]}_{0}^{400}\phantom{\rule{0.3em}{0ex}}dx& \text{}\\ & =& {\int }_{0}^{100}\left[16000a-8000a+\frac{320000}{3}b-320000b+320000b\right]\phantom{\rule{0.3em}{0ex}}dx& \text{}\\ & =& {\int }_{0}^{100}\left[8000a+\frac{320000}{3}b\right]\phantom{\rule{0.3em}{0ex}}dx& \text{}\\ & =& 8×1{0}^{5}a+\frac{3.2}{3}×1{0}^{7}b=7.984×1{0}^{8}+\frac{0.16}{3}×1{0}^{7}=7.989×1{0}^{8}\phantom{\rule{1em}{0ex}}kg\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

So the mass of water in the reservoir is $7.989×1{0}^{8}$ 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 $7.984×1{0}^{8}$ kg while assuming a uniform density of $\rho =1000\text{kg}{\text{m}}^{-3}$ gives a total mass of $8×1{0}^{8}$ kg.

#### 4.4 Centre of mass

The expressions for the centre of mass $\left(\stackrel{¯}{x},\phantom{\rule{1em}{0ex}}\stackrel{¯}{y},\phantom{\rule{1em}{0ex}}\stackrel{¯}{z}\right)$ of a solid of density $\rho \left(x,y,z\right)$ are given below

$\begin{array}{rcll}\stackrel{¯}{x}=\frac{\int \rho \left(x,y,z\right)x\phantom{\rule{0.3em}{0ex}}dV}{\int \rho \left(x,y,z\right)\phantom{\rule{0.3em}{0ex}}dV}\phantom{\rule{2em}{0ex}}\stackrel{¯}{y}=\frac{\int \rho \left(x,y,z\right)y\phantom{\rule{0.3em}{0ex}}dV}{\int \rho \left(x,y,z\right)\phantom{\rule{0.3em}{0ex}}dV}\phantom{\rule{2em}{0ex}}\stackrel{¯}{z}=\frac{\int \rho \left(x,y,z\right)z\phantom{\rule{0.3em}{0ex}}dV}{\int \rho \left(x,y,z\right)\phantom{\rule{0.3em}{0ex}}dV}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& & & \text{}\end{array}$

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

$\phantom{\rule{2em}{0ex}}\stackrel{¯}{x}=\frac{\int x\phantom{\rule{0.3em}{0ex}}dV}{\int \phantom{\rule{0.3em}{0ex}}dV}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\stackrel{¯}{y}=\frac{\int y\phantom{\rule{0.3em}{0ex}}dV}{\int \phantom{\rule{0.3em}{0ex}}dV}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\stackrel{¯}{z}=\frac{\int z\phantom{\rule{0.3em}{0ex}}dV}{\int \phantom{\rule{0.3em}{0ex}}dV}$

##### 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 ${\int }_{V}f\left(x,y,z\right)\phantom{\rule{0.3em}{0ex}}dV$ becomes ${\int }_{x=0}^{4}{\int }_{y=0}^{4-x}{\int }_{z=0}^{4-x-y}f\left(x,y,z\right)\phantom{\rule{0.3em}{0ex}}dzdydx$ . The volume is given by $\begin{array}{rcll}V& =& {\int }_{V}\phantom{\rule{0.3em}{0ex}}dV={\int }_{x=0}^{4}{\int }_{y=0}^{4-x}{\int }_{z=0}^{4-x-y}\phantom{\rule{0.3em}{0ex}}dzdydx& \text{}\\ & =& {\int }_{x=0}^{4}{\int }_{y=0}^{4-x}{\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}z\right]}_{z=0}^{4-x-y}\phantom{\rule{0.3em}{0ex}}dydx& \text{}\\ & =& {\int }_{x=0}^{4}{\int }_{y=0}^{4-x}\left(4-x-y\right)\phantom{\rule{0.3em}{0ex}}dydx& \text{}\\ & =& {\int }_{x=0}^{4}{\left[4y-xy-\frac{1}{2}{y}^{2}\right]}_{y=0}^{4-x}\phantom{\rule{0.3em}{0ex}}dx& \text{}\\ & =& {\int }_{x=0}^{4}\left[8-4x+\frac{1}{2}{x}^{2}\right]\phantom{\rule{0.3em}{0ex}}dx& \text{}\\ & =& {\left[8x-2{x}^{2}+\frac{1}{6}{x}^{3}\right]}_{0}^{4}=32-32+\frac{64}{6}=\frac{32}{3}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

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

2. The $x$ coordinate of the centre of mass i.e. $\stackrel{¯}{x}$ is given by $\stackrel{¯}{x}=\frac{\int x\phantom{\rule{0.3em}{0ex}}dV}{\int \phantom{\rule{0.3em}{0ex}}dV}$ .

The denominator $\int \phantom{\rule{0.3em}{0ex}}dV$ is the formula for the volume i.e. $\frac{32}{3}$ while the numerator $\int x\phantom{\rule{0.3em}{0ex}}dV$ was calculated in an earlier Example to be $\frac{32}{3}$ .

Thus $\stackrel{¯}{x}=\frac{\int x\phantom{\rule{0.3em}{0ex}}dV}{\int \phantom{\rule{0.3em}{0ex}}dV}=\frac{32∕3}{32∕3}=1$ .

By symmetry (or by evaluating relevant integrals), it can be shown that $\stackrel{¯}{y}=\stackrel{¯}{z}=1$ 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

$\phantom{\rule{2em}{0ex}}I=\text{Mass}×{\text{Distance}}^{2}\phantom{\rule{2em}{0ex}}\text{or}\phantom{\rule{2em}{0ex}}I=M{d}^{2}$

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

$\phantom{\rule{2em}{0ex}}{I}_{z}={\int }_{V}\rho \left(x,y,z\right)\left({x}^{2}+{y}^{2}\right)\phantom{\rule{0.3em}{0ex}}dz$

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

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

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,

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

$\phantom{\rule{2em}{0ex}}{I}_{x}=\rho {\int }_{V}\left({y}^{2}+{z}^{2}\right)\phantom{\rule{0.3em}{0ex}}dx=\rho {\int }_{0}^{1}{\int }_{0}^{1}{\int }_{0}^{1}\left({y}^{2}+{z}^{2}\right)\phantom{\rule{0.3em}{0ex}}dzdydx$

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

$\phantom{\rule{2em}{0ex}}{I}_{x}=\frac{2}{3}\rho =\frac{2}{3}M$

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