2.3: Solid Geometry - The Volume of a Truncated Pyramid

The Gaussian-integral example (Section 2.1) and the ellipse-area example (Section 2.2) showed easy cases as a method of analysis: for checking whether formulas are correct. The next level of sophistication is to use easy cases as a method of synthesis: for constructing formulas.

As an example, take a pyramid with a square base and slice a piece from its top using a knife parallel to the base. This truncated pyramid (called the frustum) has a square base and square top parallel to the base. Let h be its vertical height, \(b\) be the side length of its base, and \(a\) be the side length of its top.

Question

What is the volume of the truncated pyramid?

Let’s synthesize the formula for the volume. It is a function of the three lengths \(h, a, \text{ and } b\). These lengths split into two kinds: height and base lengths. For example, flipping the solid on its head interchanges the meanings of \(a \text{ and } b\) but preserves \(h\); and no simple operation interchanges height with \(a \text{ or } b\). Thus the volume probably has two factors, each containing a length or lengths of only one kind:

\[V(h, a, b) = f(h) × g(a, b). \label{2.9} \]

Proportional reasoning will determine \(f; a\) bit of dimensional reasoning and a lot of easy-cases reasoning will determine \(g\).

Question

What is \(f\) : How should the volume depend on the height?

To find \(f\), use a proportional-reasoning thought experiment. Chop the solid into vertical slivers, each like an oil-drilling core; then imagine doubling \(h\). This change doubles the volume of each sliver and therefore doubles the whole volume \(V\). Thus \(f ∼ h\) and \(V ∝ h\):

\[V = h \times g(a, b). \label{2.10} \]

Question

What is \(g\) : How should the volume depend on \(a \text{ and } b\)?

Because V has dimensions of \(L^{3}\), the function \(g(a, b)\) has dimensions of \(L^{2}\). That constraint is all that dimensional analysis can say. Further constraints are needed to synthesize \(g\), and these constraints are provided by the method of easy cases.

Easy cases

Question

What are the easy cases of \(a \text{ and } b\)?

The easiest case is the extreme case \(a = 0\) (an ordinary pyramid). The symmetry between a and b suggests two further easy cases, namely \(a = b\) and the extreme case \(b = 0\). The easy cases are then threefold:

\(a = 0\) \(b = 0\) \(c = 0\)

When \(a = 0\), the solid is an ordinary pyramid, and \(g\) is a function only of the base side length \(b\). Because \(g\) has dimensions of \(L^{2}\), the only possibility for \(g\) is \(g ∼ b\); in addition, \(V ∝ h\); so, \(V ∼ hb^{2}\). When \(b = 0\), the solid is an upside-down version of the \(b = 0\) pyramid and therefore has volume \(V ∼ ha^{2}\). When a = b, the solid is a rectangular prism having volume \(V = ha^{2}\) (or \(hb^{2}\)).

Question

Is there a volume formula that satisfies the three easy-cases constraints?

The \(a = 0\) and \(b = 0\) constraints are satisfied by the symmetric sum \(V ∼ h(a^{2} + b^{2})\). If the missing dimensionless constant is 1/2, making \(V = h(a^{2} + b^{2})/2\), then the volume also satisfies the \(a = b\) constraint, and the volume of an ordinary pyramid \((a = 0)\) would be \(hb^{2}/2\).

Question

When \(a = 0\), is the prediction \(V = hb^{2}/2\) correct?

Testing the prediction requires finding the exact dimensionless constant in \(V ∼ hb^{2}\). This task looks like a calculus problem: Slice a pyramid into thin horizontal sections and add (integrate) their volumes. However, a simple alternative is to apply easy cases again.

The easy case is easier to construct after we solve a similar but simpler problem: to find the area of a triangle with base \(b\) and height \(h\). The area satisfies \(A ∼ hb\), but what is the dimensionless constant? To find it, choose b and h to make an easy triangle: a right triangle with \(h = b\). Two such triangles make 2 an easy shape: a square with area \(b\). Thus each right triangle has area \(A = b^{2}/2\); the dimensionless constant is 1/2. Now extend this reasoning to three dimensions find an ordinary pyramid (with a square base) that combines with itself to make an easy solid.

Question

What is the easy solid?

A convenient solid is suggested by the pyramid’s square base: Perhaps each base is one face of a cube. The cube then requires six pyramids whose tips meet in the center of the cube; thus the pyramids have the aspect ratio \(h = b/2\). For numerical simplicity, let’s meet this condition with \(b = 2\) and \(h = 1\).

Six such pyramids form a cube with volume \(b^{3} = 8\), so the volume of one pyramid is 4/3. Because each pyramid has volume \(V ∼ hb^{2}\), and \(hb^{2} = 4\) for these pyramids, the dimensionless constant in \(V ∼ hb^{2}\) must be 1/3. The volume of an ordinary pyramid (a pyramid with \(a = 0\)) is therefore \(V = hb^{2}/3\).

Problem 2.7 Triangular base

Guess the volume of a pyramid with height h and a triangular base of area A. Assume that the top vertex lies directly over the centroid of the base. Then try Problem 2.8.

Problem 2.8 Vertex location

The six pyramids do not make a cube unless each pyramid’s top vertex lies directly above the center of the base. Thus the result \(V = hb^{2}/3\) might apply only with this restriction. If instead the top vertex lies above one of the base vertices, what is the volume?

The prediction from the first three easy-cases tests was \(V = hb^{2}/2\) (when \(a = 0\)), whereas the further easy case \(h = b/2\) alongside \(a = 0\) just showed that \(V = hb^{2}/3\). The two methods are making contradictory predictions.

Question

How can this contradiction be resolved ?

The contradiction must have snuck in during one of the reasoning steps. To find the culprit, revisit each step in turn. The argument for \(V ∝ h\) looks correct. The three easy-case requirements that \(V ∼ hb^{2}\) when \(a = 0\), that \(V ∼ ha^{2}\) when \(b = 0\), and that \(V = h(a^{2} + b^{2}/2\) when \(a = b\) also look correct. The mistake was leaping from these constraints to the prediction \(V ∼ h(a^{2} + b^{2})\) for any \(a \text{ or } b\).

Instead let’s try the following general form that includes an \(ab\) term:

\[V = h(αa^{2} + βab + γb^{2}). \label{2.11} \]

Then solve for the coefficients \(α, β, \text{ and } γ\) by reapplying the easy-cases requirements.

The \(b = 0\) test along with the \(h = b/2\) easy case, which showed that \(V = hb^{2}/3\) for an ordinary pyramid, require that \(α = 1/3\). The \(a = 0\) test similarly requires that \(γ = 1/3\). And the \(a = b\) test requires that \(α + β + γ = 1\). Therefore \(β = 1/3\) and voilà,

\[V = \frac{1}{3}h(a^{2} + ab + b^{2}). \label{2.12} \]

This formula, the result of proportional reasoning, dimensional analysis, and the method of easy cases, is exact (Problem 2.9)!

Multiple problems

Problem 2.9 Integration

Use integration to show that \(V = h(a^{2} + ab + b^{2})/3\).

Problem 2.10 Truncated triangular pyramid

Instead of a pyramid with a square base, start with a pyramid with an equilateral triangle of side length b as its base. Then make the truncated solid by slicing a piece from the top using a knife parallel to the base. In terms of the height h and the top and bottom side lengths \(a \text{ and } b\), what is the volume of this solid? (See also Problem 2.7.)

Problem 2.11 Truncated cone

What is the volume of a truncated cone with a circular base of radius \(r_{1}\) and circular top of radius \(r_{2}\) (with the top parallel to the base)? Generalize your formula to the volume of a truncated pyramid with height \(h\), a base of an arbitrary shape and area \(A_{base}\), and a corresponding top of area \(A_{top}\).