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Triple Integrals and Solid Regions · Tutorial 624 of 999

A Volume and Average Value Modeling Workshop

Build a three-dimensional region from its geometric conditions, then use its bounds to calculate and check volume and average value.

Intermediate 10 min read

What You'll Learn

  • Translate inequalities and bounding surfaces into iterated bounds for a solid.
  • Calculate a region’s volume from a triple integral of one.
  • Find a scalar field’s average by dividing its integral by the actual volume.
  • Check a volume using a second cross-sectional description.
  • Use range estimates and endpoint checks to identify modeling errors.

From Geometric Conditions to a Model

A three-dimensional region is often described in words or by inequalities rather than by a ready-to-use integral. To calculate its volume or the average value of a scalar field over it, first translate those conditions into bounds that describe every point in the region exactly once.

The modeling choices matter as much as the integration. A lower surface must be below its corresponding upper surface, projected bounds must match the solid, and an average must be divided by the volume of the solid itself—not by the volume of a convenient box around it.

Definition: Let \(D\) be a bounded solid region whose boundary is sufficiently well behaved for triple integration, and let \(f\) be an integrable scalar field on \(D\). The volume of \(D\) is \(\iiint_D 1\,dV\). If this volume is positive, the average value of \(f\) over \(D\) is \(\frac{1}{\operatorname{Vol}(D)}\iiint_D f\,dV\).

In a Cartesian iterated integral, the bounds must describe \(D\), and the differential order must agree with the order of integration. The earlier material on bounds and integration order provides the setup tools; here the emphasis is on connecting the model, the calculation, and a useful check.

A Reliable Modeling Workflow

1
Identify the solid’s constraints.
Record the inequalities and bounding surfaces, including which surface is lower and which is upper.
2
Choose a projection and order.
Determine the projected region for the outer variables, then express the remaining variable between its two bounding surfaces.
3
Calculate volume and field integral.
Integrate \(1\) for volume and \(f\) for the field total, using the same region description.
4
Check the result.
Test the bounds at representative points, compare volume with a geometric or alternate-slice calculation, and check whether the average lies within the field’s range.

Worked Example: A Sloping Roof Over a Triangular Base

Worked Example: A Sloping Roof Over a Triangular Base

Let \(D\) lie above the triangular base \(x\geq 0\), \(y\geq 0\), \(x+y\leq 2\), and between the planes \(z=0\) and \(z=4-x-y\). Find the volume and the average value of \(f(x,y,z)=z\).

The base constraints give \(0\leq x\leq 2\) and \(0\leq y\leq 2-x\). The vertical bounds are \(0\leq z\leq 4-x-y\). The upper bound is at least \(2\) on this base, so it never falls below the lower bound.

Integrating \(1\) first with respect to \(z\), the volume is

$$ \begin{aligned} \operatorname{Vol}(D) &=\int_0^2\int_0^{2-x}\int_0^{4-x-y}1\,dz\,dy\,dx\\ &=\int_0^2\int_0^{2-x}(4-x-y)\,dy\,dx\\ &=\int_0^2\left[(4-x)(2-x)-\frac{(2-x)^2}{2}\right]dx\\ &=\int_0^2\left(6-4x+\frac{x^2}{2}\right)dx\\ &=\left[6x-2x^2+\frac{x^3}{6}\right]_0^2 =12-8+\frac{8}{6} =\frac{16}{3}. \end{aligned} $$

For the field integral, integrating \(z\) over its vertical bounds gives \(\frac{1}{2}(4-x-y)^2\). Thus

$$ \begin{aligned} \iiint_D z\,dV &=\frac12\int_0^2\int_0^{2-x}(4-x-y)^2\,dy\,dx\\ &=\frac16\int_0^2\big((4-x)^3-8\big)\,dx\\ &=\frac16\left(\frac{4^4-2^4}{4}-16\right)\\ &=\frac16(60-16) =\frac{22}{3}. \end{aligned} $$

Therefore the average value is

$$ f_{\mathrm{avg}} =\frac{\iiint_D z\,dV}{\operatorname{Vol}(D)} =\frac{22/3}{16/3} =\frac{11}{8}. $$

As a volume check, slice horizontally. At height \(z\) between \(0\) and \(2\), the entire triangular base remains under the roof, so the cross-sectional area is \(2\). For \(2\leq z\leq 4\), the cross-section is \(x+y\leq 4-z\), a right triangle of area \(\frac{(4-z)^2}{2}\). Hence

$$ \operatorname{Vol}(D) =\int_0^2 2\,dz+\int_2^4\frac{(4-z)^2}{2}\,dz =4+\frac{4}{3} =\frac{16}{3}, $$

agreeing with the iterated integral. Also \(0\leq z\leq 4\) throughout \(D\), and the average \(11/8\) lies in that range.

Why the Average Formula Works

The average over a solid is a volume-weighted mean. Imagine partitioning \(D\) into small pieces. On a piece with volume \(\Delta V_i\), choose a representative point \(P_i\). The field total is approximated by adding the value on each piece multiplied by its volume, while the total volume is approximated by adding the piece volumes. The ratio of those sums is therefore

$$ \frac{\sum_i f(P_i)\,\Delta V_i}{\sum_i\Delta V_i}. $$

As the partition is refined, the numerator approaches \(\iiint_D f\,dV\) and the denominator approaches \(\iiint_D 1\,dV=\operatorname{Vol}(D)\). Provided the field is integrable and the solid has positive volume, the limiting ratio is precisely the stated average-value formula. This also explains why each part of the solid is weighted according to its volume rather than counted as one equally weighted sample.

Worked Example: A Prism and a Linear Field

Worked Example: A Prism and a Linear Field

Let \(D\) be given by \(0\leq x\leq 1\), \(0\leq y\leq 1-x\), and \(0\leq z\leq 2\). Find its volume and the average of \(f(x,y,z)=x+2y+z\).

The triangular projection in the \(xy\)-plane and constant vertical bounds describe the solid, so

$$ \begin{aligned} \operatorname{Vol}(D) &=\int_0^1\int_0^{1-x}\int_0^2 1\,dz\,dy\,dx\\ &=\int_0^1\int_0^{1-x}2\,dy\,dx\\ &=2\int_0^1(1-x)\,dx =2\left[x-\frac{x^2}{2}\right]_0^1 =1. \end{aligned} $$

For the field integral, integrate each term in \(f\) with respect to \(z\). Since \(\int_0^2(x+2y+z)\,dz=2x+4y+2\),

$$ \begin{aligned} \iiint_D f\,dV &=\int_0^1\int_0^{1-x}(2x+4y+2)\,dy\,dx\\ &=\int_0^1\left[2x(1-x)+2(1-x)^2+2(1-x)\right]dx\\ &=\int_0^1(4-4x)\,dx\\ &=[4x-2x^2]_0^1 =2. \end{aligned} $$

Thus \(f_{\mathrm{avg}}=2/1=2\). To check the field total directly, the average coordinates over this prism are \(x_{\mathrm{avg}}=1/3\), \(y_{\mathrm{avg}}=1/3\), and \(z_{\mathrm{avg}}=1\), as follows by integrating each coordinate and dividing by the volume \(1\). Linearity gives \(1/3+2(1/3)+1=2\), matching the result. The field ranges from \(0\) to \(4\) on the solid, so the computed average is also within its possible range.

Worked Example: Mixed Vertical Bounds and an Alternate Setup

Worked Example: Mixed Vertical Bounds and an Alternate Setup

Consider the solid \(D\) defined by \(0\leq x\leq 1\), \(0\leq y\leq 1\), and \(x\leq z\leq 2-y\). Find its volume. Then find the average of \(f(x,y,z)=z\).

Over the unit square, \(x\leq 1\leq 2-y\), so the lower surface \(z=x\) does not exceed the upper surface \(z=2-y\). The volume is

$$ \begin{aligned} \operatorname{Vol}(D) &=\int_0^1\int_0^1\int_x^{2-y}1\,dz\,dy\,dx\\ &=\int_0^1\int_0^1(2-y-x)\,dy\,dx\\ &=\int_0^1\left(\frac32-x\right)dx\\ &=\left[\frac32x-\frac{x^2}{2}\right]_0^1 =1. \end{aligned} $$

The field total is

$$ \begin{aligned} \iiint_D z\,dV &=\int_0^1\int_0^1\int_x^{2-y}z\,dz\,dy\,dx\\ &=\frac12\int_0^1\int_0^1\big((2-y)^2-x^2\big)\,dy\,dx\\ &=\frac12\left(\frac73-\frac13\right) =1. \end{aligned} $$

Consequently, the average value of \(z\) is \(1/1=1\).

An alternate volume check slices by \(z\). For \(0\leq z\leq 1\), the conditions imply \(0\leq x\leq z\) and \(0\leq y\leq 1\), giving cross-sectional area \(z\). For \(1\leq z\leq 2\), they imply \(0\leq x\leq 1\) and \(0\leq y\leq 2-z\), giving area \(2-z\). Therefore

$$ \operatorname{Vol}(D) =\int_0^1z\,dz+\int_1^2(2-z)\,dz =\frac12+\frac12 =1, $$

which confirms the original setup. Since \(0\leq z\leq 2\), the average value \(1\) is also consistent with the field’s range.

A Common Mistake: Dividing by the Wrong Volume

The field integral is not itself an average. It is the total accumulated value, with each point weighted by volume. To obtain an average, divide by the volume of the region being integrated over.

For the first example, the field integral was \(22/3\), and the solid’s volume was \(16/3\). A tempting but incorrect calculation is to divide by the volume \(16\) of the enclosing box \(0\leq x\leq 2\), \(0\leq y\leq 2\), \(0\leq z\leq 4\), giving \((22/3)/16=11/24\). That is not the average over \(D\): much of the box lies outside the solid. The correct calculation is \((22/3)/(16/3)=11/8\). The denominator must match the region in the numerator.

Checks That Strengthen a Model

A correct-looking integral is not enough: test whether its bounds match the geometry. Useful checks include:

  • Check lower and upper surfaces. At points in the projected region, verify that the proposed lower bound does not exceed the upper bound.
  • Check representative endpoints. Substitute boundary values into the bounds to see whether they reproduce the described edges, corners, or intersections.
  • Compare volumes. If practical, compute the volume from a different cross-section or use a familiar geometric formula.
  • Check the average against the field’s range. If \(m\leq f\leq M\) throughout a positive-volume solid, then \(m\leq f_{\mathrm{avg}}\leq M\). Indeed, integrating the inequalities gives \(m\operatorname{Vol}(D)\leq\iiint_D f\,dV\leq M\operatorname{Vol}(D)\); division by the positive volume proves the claim.

These checks do not replace a correct setup, but they can expose reversed bounds, a mistaken projection, or an incorrect denominator before the error is carried through a calculation.

Check Your Understanding

Use the definitions and modeling checks from this workshop to answer the following.

  1. For a solid under \(z=3-x-y\) and above \(z=0\), what condition must the projected region satisfy for the vertical bounds to be valid?
  2. Why must the volume be positive before the average value over a solid is defined by a volume ratio?
  3. A field satisfies \(2\leq f\leq 7\) throughout a solid of positive volume. What range must its average value lie in, and why?
  4. In the mixed-bounds example, which inequalities describe the cross-section when \(1\leq z\leq 2\)?
  5. Why is the volume of an enclosing box generally not the correct denominator for an average over a smaller solid?

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