Average Values

The average value of a function of one variable on the interval \([a,b]\) is defined by \[ [f]_{\rm av}= \frac{\displaystyle\int^b_a f(x){\it {\,d} x}}{b-a}. \]

Likewise, for functions of two variables, the ratio of the integral to the area of \(D\), \begin{equation*} [f]_{\rm av}=\frac{\displaystyle\intop\!\!\!\intop\nolimits_{D} f(x,y){\it {\,d} x}\, {\it {\,d} y}}{\displaystyle\intop\!\!\!\intop\nolimits_{D} {\it dx}\, {\it dy}},\tag{1} \end{equation*} is called the average value of \(f\) over \(D\). Similarly, the average value of a function \(f\) on a region \(W\) in 3-space is defined by \[ [f]_{\rm av}=\frac{\displaystyle\intop\!\!\!\intop\!\!\!\intop\nolimits_{W}f(x,y,z){\it {\,d} x}{\it {\,d} y}{\it {\,d} z}}{\displaystyle\intop\!\!\!\intop\!\!\!\intop\nolimits_{W} {\it dx}\,{\it dy}\,{\it dz}}. \]

example 1

Find the average value of \(f(x,y)=x\sin^2(xy)\) on the region \(D=[0,\pi]\) \(\times\) \([0,\pi]\).

solution First, we compute \begin{eqnarray*} \intop\!\!\!\intop\nolimits_{D} f(x,y)\,{\it dx}\, {\it dy} &=& \int^\pi_0 \int^\pi_0 x\sin^2(xy)\,{\it dx}\,{\it dy}\\[4pt] &=&\int^\pi_0\left[ \int^\pi_0\frac{1-\cos(2xy)}{2}x\, {\it dy}\right]{\it {\,d} x}\\[4pt] &=&\int^\pi_0\left[\frac{y}{2}-\frac{\sin(2xy)} {4x}\right]x\bigg|^\pi_{y=0} {\it {\,d} x}\\[4pt] &=&\int^\pi_0\left[\frac{\pi x}{2}-\frac{\sin(2\pi x)}{4}\right]{\it dx}=\left[\frac{\pi x^2}{4}+ \frac{\cos(2\pi x)}{8\pi}\right]\bigg|^\pi_0\\[4pt] &=&\frac{\pi^3}{4}+\frac{\cos(2\pi^2)-1}{8\pi}.\\[-32pt] \end{eqnarray*}

Thus, the average value of \(f\), by formula (1), is \[ \frac{\pi^3/4+[\cos(2\pi^2)-1]/8\pi}{\pi^2}=\frac{\pi}{4}+\frac{\cos(2\pi^2)-1}{8\pi^3}\approx 0.7839. \]

example 2

The temperature at points in the cube \(W=[-1,1]\) \(\times\) \([-1,1]\) \(\times\) \([-1,1]\) is proportional to the square of the distance from the origin.

• (a) What is the average temperature?
• (b) At which points of the cube is the temperature equal to the average temperature?

solution (a) Let \(c\) be the constant of proportionality, so \(T=c(x^2+y^2+z^2)\) and the average temperature is \([T]_{\rm av}=\frac{1}{8}{\intop\!\!\intop\!\!\intop}_WT{\it dx}\,{\it dy}\,{\it dz}\), because the volume of the cube is 8. Thus, \[ [T]_{\rm av}=\frac{c}{8}\int^1_{-1}\int^1_{-1} \int^1_{-1}(x^2+y^2+z^2)\,{\it dx}\,{\it dy}\,{\it dz}. \]

The triple integral is the sum of the integrals of \(x^2,y^2\), and \(z^2\). Because \(x,y\), and \(z\) enter symmetrically into the description of the cube, the three integrals will be equal, so that \[ [T]_{\rm av}=\frac{3c}{8}\int^1_{-1}\int^1_{-1}\int^1_{-1}z^2{\it dx}\,{\it dy}\,{\it dz}=\frac{3c}{8} \int^1_{-1}z^2\left(\int^1_{-1}\int^1_{-1} {\it dx}\,{\it dy}\right)\,{\it dz}. \]

The inner integral is equal to the area of the square \([-1,1]\times[-1,1]\). The area of that square is 4, and so \[ [T]_{\rm av}=\frac{3c}{8}\int^1_{-1} 4z^2 {\it dz}=\frac{3c}{2}\left(\frac{z^3}{3}\right)\bigg|^1_{-1}=c. \]

(b) The temperature is equal to the average temperature at all points satisfying \(c(x^2+y^2+z^2)\,{=}\,c\)—that is, at all points lying on the sphere \(x^2+y^2+z^2\,{=}\,1\).

The Center of Mass of Two-Dimensional Plates

\begin{equation*} \overline x= \frac{\displaystyle\intop\!\!\!\intop\nolimits_{D} x\delta(x,y)\,{\it dx}\,{\it dy}}{\displaystyle\intop\!\!\!\intop\nolimits_{D} \delta(x,y)\,{\it dx}\, {\it dy}} \qquad \hbox{and}\qquad \skew2\overline y=\frac{\displaystyle\intop\!\!\!\intop\nolimits_{D} y\delta(x,y)\,{\it dx}\,{\it dy}}{\displaystyle\intop\!\!\!\intop\nolimits_{D} \delta(x,y)\,{\it dx}\, {\it dy}},\tag{4} \end{equation*} where again \(\delta(x,y)\) is the mass density (see Figure 16.16).

example 3

Find the center of mass of the rectangle \([0,1]\times[0,1]\) if the mass density is \(e^{x+y}\).

solution First we compute the total mass: \begin{eqnarray*} \intop\!\!\!\intop\nolimits_{D} e^{x+y}{\it dx}\,{\it dy} &=&\int^1_0\int^1_0 e^{x+y}{\it dx}\,{\it dy}=\int^1_0 \big(e^{x+y}\big|^1_{x=0}\big)\,{\it dy} =\int^1_0 (e^{1+y}-e^y)\, {\it dy}\\[6pt] &=& (e^{1+y}-e^y) \big|^1_{y=0}=e^2-e-(e-1)=e^2-2e+1. \end{eqnarray*}

The numerator in formula (4) for \(\overline x\) is \begin{eqnarray*} \int^1_0\!\!\int^1_0\!x e^{x+y}{\it dx}\,{\it dy} &=&\int^1_0\!(xe^{x+y}-e^{x+y})\big|^1_{x=0}{\it dy} =\int^1_0\![e^{1+y}-e^{1+y}-(0e^y-e^y)]\,{\it dy}\\[6pt] &=&\int^1_0e^y {\it dy}=e^y\Big|^1_{y=0}=e-1,\\[-15pt] \end{eqnarray*} so that \[ \overline x=\frac{e-1}{e^2-2e+1}=\frac{e-1}{(e-1)^2}=\frac{1}{e-1}\approx0.582. \]

The roles of \(x\) and \(y\) may be interchanged in all these calculations, so that \(\skew2\overline y=1/(e-1)\approx 0.582\) as well.

Coordinates for the Center of Mass of Three-Dimensional Regions

\begin{equation*} \displaystyle \overline x=\frac{\displaystyle\intop\!\!\!\intop\!\!\!\intop\nolimits_{\! W}x\delta (x,y,z)\,{\it dx}\,{\it dy}\,{\it dz}}{{\rm mass}}\\[6pt] \displaystyle \skew2\overline y=\frac{\displaystyle\intop\!\!\!\intop\!\!\!\intop\nolimits_{\! W}y\delta (x,y,z)\,{\it dx}\,{\it dy}\,{\it dz}}{{\rm mass}}\\[6pt] \displaystyle \skew2\overline z=\frac{\displaystyle\intop\!\!\!\intop\!\!\!\intop\nolimits_{\! W}z\delta (x,y,z)\,{\it dx}\,{\it dy}\,{\it dz}}{{\rm mass}}\tag{7} \end{equation*}

example 4

The cube \([1,2]\times [1,2]\times [1,2]\) has mass density given by \(\delta(x,y,z)=(1+x)e^zy\). Find the total mass of the box.

solution The mass of the box is, by formula (6), \begin{eqnarray*} &&\int^2_1\int^2_1\int^2_1(1+x)e^zy\,{\it dx}\,{\it dy}\,{\it dz} =\int^2_1\int^2_1\left[\left(x+\frac{x^2}{2}\right)e^zy\right]^{x=2}_{x=1}{\it dy}\,{\it dz}\\[5pt] &&\quad =\int^2_1\int^2_1 \frac{5}{2}e^zy\,{\it dy}\,{\it dz} =\int^2_1\frac{15}{4}e^z{\it dz}=\left[\frac{15}{4}e^z\right]^{z=2}_{z=1} = \frac{15}{4}(e^2-e).\\[-36pt] \end{eqnarray*}

example 5

Find the center of mass of the hemispherical region \(W\) defined by the inequalities \(x^2+y^2+z^2\leq 1,z \geq 0\). (Assume that the density is unity.)

solution By symmetry, the center of mass must lie on the \(z\) axis, and so \(\overline x=\skew2\overline y=0\). To find \(\skew2\overline z\), we must compute, by formula (7), the numerator \(I={\intop\!\!\intop\!\!\intop}_W z\,{\it dx}\,{\it dy}\,{\it dz}\). The hemisphere is an elementary region, and thus the integral becomes \[ I=\int^1_0\int^{\sqrt{1-z^2}}_{-\sqrt{1-z^2}} \int^{\sqrt{1-y^2-z^2}}_{-\sqrt{1-y^2-z^2}}z\, {\it dx}\,{\it dy}\,{\it dz}. \]

Because \(z\) is a constant for the \(x\) and \(y\) integrations, we can remove it from the first two integral signs, to obtain \[ I=\int^1_0 z\left(\int^{\sqrt{1-z^2}}_{-\sqrt{1-z^2}} \int^{\sqrt{1-y^2-z^2}}_{-\sqrt{1-y^2-z^2}}{\it dx}\,{\it dy}\right)dz. \]

Instead of calculating the inner two integrals explicitly, we observe that they equal the double integral \({\intop\!\!\intop}_D {\it dx}\,{\it dy}\) over the disk \(x^2+y^2\leq 1-z^2\), considered as an \(x\)-simple region in the plane. The area of this disk is \(\pi(1-z^2)\), and so \[ I=\pi\int^1_0 z(1-z^2)\,{\it dz}=\pi\int^1_0(z-z^3)\,{\it dz}=\pi\left[\frac{z^2}{2}- \frac{z^4}{4}\right]^1_0=\frac{\pi}{4}. \]

The volume of the hemisphere is \(\frac{2}{3}\pi\), and so \(\skew2\overline z=(\pi/4)/( \frac{2}{3}\pi)=\frac{3}{8}\).

Historical Note

It is common knowledge that Archimedes observed the principle of the lever. Perhaps less known is that he was also responsible for discovering the concepts of center of mass and center of gravity. Only two of his works on mechanics have been handed down to us: On Floating Bodies and On the Equilibrium and Centers of Mass of Plane Figures. Both were translated into Latin by Niccolo Tartaglia, circa 1543.

In Equilibrium\(\ldots\), Archimedes began the field of applied mathematics, doing for mechanics what Euclid had accomplished for geometry. In this work he describes the principles behind all the machines of antiquity, including the lever, inclined plane, and pulley systems.

Surprisingly, Archimedes never carefully defined the center of mass; the first proper definition was given by Pappus of Alexandria in 340 C.E. The concept of equilibrium was to have a profound effect on the development of mechanical engineering (through the introduction of gears), architecture, and in art, permitting the construction of complex machines, large-scale buildings, and sculptures. Figure 16.17 shows sketches by Leonardo DaVinci, illustrating equilibrium positions of the human body.

Moments of Inertia About the Coordinate Axes

\begin{equation*} \begin{array}{c} I_x=\displaystyle\intop\!\!\!\intop\!\!\!\intop\nolimits_{W} (y^2+z^2)\,\delta\,dx{\it dy}\,{\it dz}, I_y=\intop\!\!\!\intop\!\!\!\intop\nolimits_{W} (x^2+z^2)\,\delta\,{\it dx}\,{\it dy}\,{\it dz},\\[9pt] I_z=\displaystyle\intop\!\!\!\intop\!\!\!\intop\nolimits_{W} (x^2+y^2)\, \delta\,{\it dx}\,{\it dy}\,{\it dz}. \end{array}\tag{8} \end{equation*}

example 6

Compute the moment of inertia \(I_z\) for the solid above the \(xy\) plane bounded by the paraboloid \(z=x^2+y^2\) and the cylinder \(x^2+y^2=a^2\), assuming \(a\) and the mass density to be constants.

solution The paraboloid and cylinder intersect at the plane \(z=a^2\). Using cylindrical coordinates, we find from equation (8), \[ I_z=\int^a_0\int^{2\pi}_0 \int^{r^2}_0\delta r^2\ {\cdot}\ r{\it dz}{\,d} \theta {\,d} r=\delta \int^a_0 \int^{2\pi}_0\int^{r^2}_0 r^3{\it dz}\,{\,d} \theta {\,d} r=\frac{\pi \delta a^6}{3}. \]

Historical Note

The theory of gravitational force fields and gravitational potentials was developed by Sir Isaac Newton (1642–1727). Newton withheld publication of his gravitational theories for quite some time. The result that a spherical planet has the same gravitational field as it would have if its mass were all concentrated at the planet’s center first appeared in his famous Philosophiae Naturalis Principia Mathematica, the first edition of which appeared in 1687. Using multiple integrals and spherical coordinates, we shall solve Newton’s problem here; remarkably, Newton’s published solution used only Euclidean geometry.

example 7

Let \(W\) be a region of constant density and total mass \(M\). Show that the gravitational potential is given by \[ V(x_1,y_1,z_1)=\left[\frac{1}{r}\right]_{\rm av} {\it GMm}, \] where \([1/r]_{\rm av}\) is the average over \(W\) of \[ f(x,y,z)=\frac{1}{\sqrt{(x-x_1)^2+( y-y_1)^2+(z-z_1)^2}}. \]

solution According to formula (9), \begin{eqnarray*} -V(x_1,y_1,z_1) & = & {\it Gm}\intop\!\!\!\intop\!\!\!\intop\nolimits_{W}\frac{\delta\, {\it dx}\,{\it dy}\,{\it dz}}{\sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}}\\ [10pt] & = & {\it Gm}\delta\intop\!\!\!\intop\!\!\!\intop\nolimits_{W}\frac{{\it dx}\,{\it dy}\,{\it dz}}{\sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}}\\[10pt] & = & {\it Gm}[\delta \hbox{ volume } (W)]\frac{ \displaystyle\intop\!\!\!\intop\!\!\!\intop\nolimits_{W} \displaystyle \frac{ {\it dx}\,{\it dy}\,{\it dz}}{\sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}}}{{\rm volume}\, (W)}\\ &=&{\it GmM}\left[\frac{1}{r}\right]_{\rm av} \end{eqnarray*} are required.

example 8

Find the gravitational potential acting on a unit mass of a spherical star with a mass \(M=3.02\times 10^{30}\) kg at a distance of \(2.25 \times 10^{11}\) m from its center \((G=6.67\times 10^{-11}\) N\(\,{\cdot}\,\)m\(^2\hbox{/}\)kg\(^2)\).

solution The negative potential is \[ -V=\frac{\it GM}{R}=\frac{6.67\times 10^{-11}\times 3.02\times 10^{30}} {2.25\times 10^{11}}=8.95\times 10^8\ \ {\rm m^2\hbox{/}s^2}. \]