Average Values over Solid Regions

From a Total to an Average

A triple integral accumulates the values of a function throughout a solid. Its result is a total: for example, total mass when the function is a density, or a total amount when the function measures an amount per unit volume. An average answers a different question: what single value would represent the function across the solid as a whole?

To get an average, divide the accumulated total by the amount of space over which it was accumulated. For a solid, that amount of space is its volume. This is the three-dimensional counterpart of finding the average value of a function on an interval by dividing its integral by the interval’s length.

Definition: Let \(E\) be a bounded solid with positive volume \(V(E)\), and let \(f\) be a scalar function that is integrable over \(E\). The average value of \(f\) over \(E\) is

f_{\mathrm{avg}} = \frac{1}{V(E)}\iiint_E f(x,y,z)\,dV.

Here \(V(E)=\iiint_E 1\,dV\). In Cartesian coordinates, \(dV=dx\,dy\,dz\), with the order chosen to match the bounds.

The requirement \(V(E)>0\) matters: division by zero is not defined, and a region with zero volume does not provide a three-dimensional solid over which to average. In the usual Calculus III setting, the solid has a piecewise-smooth boundary and the function is continuous on it, so the required volume and triple integrals exist.

Why the Formula Divides by Volume

Imagine splitting \(E\) into many small pieces, with the \(i\)-th piece having volume \(\Delta V_i\). Choose a sample point \((x_i,y_i,z_i)\) in each piece. The value \(f(x_i,y_i,z_i)\) represents the function there, so the sum \(f(x_i,y_i,z_i)\Delta V_i\) approximates that piece’s contribution to the total. Adding the contributions gives an approximation to the triple integral.

The corresponding weighted average of the sampled values is the approximate total divided by the approximate total volume. As the pieces become small, the sum in the numerator approaches the triple integral, while the sum of the piece volumes approaches \(V(E)\). Since \(V(E)>0\), the quotient approaches the integral divided by \(V(E)\).

Form a weighted sum.
Each sampled value is weighted by the volume of the small piece it represents: \(\sum_i f(x_i,y_i,z_i)\Delta V_i\).

Pass to the integral.
As the subdivision is refined, the weighted sums approach \(\iiint_E f\,dV\), and the sums \(\sum_i\Delta V_i\) approach \(V(E)\).

Normalize by the volume.
The weighted average is the approximate total divided by the approximate volume. Its limit is \(\frac{1}{V(E)}\iiint_E f\,dV\), which gives the stated average-value formula.

This reasoning also explains why the formula is a volume-weighted average, not an unweighted average of a few selected values. A larger portion of the solid contributes more to the total than a smaller portion.

Worked Examples

Worked Example: Average Value on a Rectangular Box

Find the average value of \(f(x,y,z)=x+2y+z\) on the box \(E=\{(x,y,z):0\le x\le2,\ 0\le y\le1,\ 0\le z\le3\}\).

First compute the volume. The side lengths are \(2\), \(1\), and \(3\), so

V(E)=2\cdot1\cdot3=6.

Now integrate the function using the stated bounds:

\begin{aligned} \iiint_E (x+2y+z)\,dV &=\int_0^2\int_0^1\int_0^3(x+2y+z)\,dz\,dy\,dx\\ &=\int_0^2\int_0^1\left[ (x+2y)z+\frac{z^2}{2}\right]_0^3\,dy\,dx\\ &=\int_0^2\int_0^1\left(3x+6y+\frac{9}{2}\right)\,dy\,dx\\ &=\int_0^2\left(3x+3+\frac{9}{2}\right)\,dx\\ &=\int_0^2\left(3x+\frac{15}{2}\right)\,dx\\ &=\left[\frac{3x^2}{2}+\frac{15x}{2}\right]_0^2 =6+15=21. \end{aligned}

Therefore,

f_{\mathrm{avg}}=\frac{21}{6}=\frac{7}{2}.

Check by multiplying the proposed average by the volume: \(\frac{7}{2}\cdot6=21\), the computed triple integral. Thus the average is \(\frac{7}{2}\).

Worked Example: Average Height on a Tetrahedron

Let \(E=\{(x,y,z):x\ge0,\ y\ge0,\ z\ge0,\ x+y+z\le2\}\). Find the average value of \(f(x,y,z)=z\) on \(E\). The bounds can be written as \(0\le z\le2\), \(0\le y\le2-z\), and \(0\le x\le2-y-z\).

First compute the volume using those bounds. Integrating in \(x\) gives the length \(2-y-z\) of the \(x\)-interval. Integrating in \(y\) gives the cross-sectional area \(\frac{(2-z)^2}{2}\). Hence

\begin{aligned} V(E) &=\int_0^2\int_0^{2-z}\int_0^{2-y-z}1\,dx\,dy\,dz\\ &=\int_0^2\int_0^{2-z}(2-y-z)\,dy\,dz\\ &=\int_0^2\frac{(2-z)^2}{2}\,dz\\ &=\frac12\int_0^2(4-4z+z^2)\,dz\\ &=\frac12\left[4z-2z^2+\frac{z^3}{3}\right]_0^2 =\frac12\left(8-8+\frac83\right)=\frac43. \end{aligned}

For the integral of \(z\), the same cross-sectional area applies, and \(z\) is constant over each cross-section:

\begin{aligned} \iiint_E z\,dV &=\int_0^2\int_0^{2-z}\int_0^{2-y-z}z\,dx\,dy\,dz\\ &=\int_0^2 z\frac{(2-z)^2}{2}\,dz\\ &=\frac12\int_0^2(4z-4z^2+z^3)\,dz\\ &=\frac12\left[2z^2-\frac{4z^3}{3}+\frac{z^4}{4}\right]_0^2\\ &=\frac12\left(8-\frac{32}{3}+4\right) =\frac12\cdot\frac43=\frac23. \end{aligned}

Divide by the volume:

f_{\mathrm{avg}}=\frac{2/3}{4/3}=\frac12.

Check: \(\frac12\cdot\frac43=\frac23\), which recovers the integral of \(z\) over \(E\). The average height coordinate is \(\frac12\), even though \(z\) ranges from \(0\) to \(2\) on the tetrahedron.

Worked Example: Average of a Function Beneath a Plane

Consider the solid \(E=\{(x,y,z):0\le x\le1,\ 0\le y\le1,\ 0\le z\le2-x-y\}\), which lies above the square \(0\le x,y\le1\) and below the plane \(z=2-x-y\). Find the average of \(f(x,y,z)=x+y+z\).

The height above each point \((x,y)\) is \(2-x-y\), so the volume is

\begin{aligned} V(E) &=\int_0^1\int_0^1(2-x-y)\,dy\,dx\\ &=\int_0^1\left(2-x-\frac12\right)\,dx\\ &=\left[\frac{3x}{2}-\frac{x^2}{2}\right]_0^1 =1. \end{aligned}

To integrate \(f\), first integrate with respect to \(z\). Write \(h=2-x-y\) for the upper \(z\)-bound. Then

\begin{aligned} \iiint_E(x+y+z)\,dV &=\int_0^1\int_0^1\int_0^h(x+y+z)\,dz\,dy\,dx\\ &=\int_0^1\int_0^1\left((x+y)h+\frac{h^2}{2}\right)\,dy\,dx. \end{aligned}

Compute the contribution from \((x+y)h\) by separating its two terms:

\begin{aligned} \int_0^1\int_0^1 x(2-x-y)\,dy\,dx &=\int_0^1\int_0^1(2x-x^2-xy)\,dy\,dx\\ &=1-\frac13-\frac14=\frac{5}{12},\\ \int_0^1\int_0^1 y(2-x-y)\,dy\,dx &=\int_0^1\int_0^1(2y-xy-y^2)\,dy\,dx\\ &=1-\frac14-\frac13=\frac{5}{12}. \end{aligned}

For the remaining term, expand the square and integrate each term over the unit square:

\begin{aligned} \int_0^1\int_0^1\frac{(2-x-y)^2}{2}\,dy\,dx &=\frac12\int_0^1\int_0^1 (4+x^2+y^2+2xy-4x-4y)\,dy\,dx\\ &=\frac12\left(4+\frac13+\frac13+\frac12-2-2\right)\\ &=\frac12\cdot\frac76=\frac{7}{12}. \end{aligned}

Adding the contributions gives

\iiint_E(x+y+z)\,dV =\frac{5}{12}+\frac{5}{12}+\frac{7}{12} =\frac{17}{12}.

Since \(V(E)=1\), the average is \(\frac{17}{12}\). Check: \(\frac{17}{12}\cdot1=\frac{17}{12}\), the computed integral. In this example the volume is one, so the average’s numerical value equals the integral’s; that equality is special to this volume, not a different average-value rule.

Interpreting the Result and Avoiding a Common Error

The integral and the average have different meanings. If \(f\) is measured in units of \(f\) per unit volume, then \(\iiint_E f\,dV\) has units of \(f\), while \(f_{\mathrm{avg}}\) has the same units as the function values themselves: the integral’s units are divided by units of volume. If \(f\) is a density, the integral is total mass and the average value is the mean density.

A common mistake is to stop after computing \(\iiint_E f\,dV\) and report that number as the average. In the first example, the integral is \(21\), but the box has volume \(6\). Reporting \(21\) as the average is wrong because it is the total accumulated value. The correction is

\text{average}=\frac{\text{integral}}{\text{volume}} =\frac{21}{6}=\frac72.

Always calculate or identify the volume before forming the quotient. Then check the result by multiplying the proposed average by the volume; the product must equal the triple integral. This is a direct check of the defining equation.

The average is a representative value for the solid, not necessarily a value attained at a particular point. For a constant function \(f=c\), the formula gives \(\frac{1}{V(E)}\iiint_E c\,dV=\frac{cV(E)}{V(E)}=c\), as expected. For a nonconstant function, the average summarizes its values across the entire solid, with larger portions of the solid contributing in proportion to their volume.

Check Your Understanding

For each question, identify the volume and the integral before calculating the average.

If \(V(E)=8\) and \(\iiint_E f\,dV=20\), what is the average value of \(f\) over \(E\)?
Why must the solid have positive volume for the average-value formula to apply?
If \(f=c\) everywhere on a solid of positive volume, show from the definition that its average is \(c\).
A function has units of kilograms per cubic meter. What are the units of its triple integral over a solid, and what are the units of its average value?
If a proposed average is \(4\) and the solid has volume \(3\), what value should the triple integral have for the proposed average to be correct?

Average Values over Solid Regions

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