From Polygonal Slices to Volume
A solid can be described by the shapes of its cross sections. If each cross section is a regular polygon, its area depends on two pieces of information: the number of polygon sides and the length of each side. Once that area is expressed in terms of position, the slice principle turns the cross-sectional areas into a volume integral.
This tutorial focuses on the area formula and its use when a slice’s side length is given as a function along the base. The key point is to use the correct formula for the polygon: the area of a square, for example, is not the area of a regular hexagon with the same side length.
Area of a Regular Polygon in Terms of Its Side
Let a regular polygon have \(n\) sides, each of length \(s\), where \(n\ge 3\). Join its center to each vertex. This divides the polygon into \(n\) congruent isosceles triangles. Draw a perpendicular from the center to the midpoint of one side. That perpendicular is the apothem, and it bisects both the side and the angle at the center.
The resulting right triangle has opposite side \(s/2\), adjacent side equal to the apothem \(r\), and angle \(\pi/n\) at the center. Therefore, by the definition of tangent, \(\tan(\pi/n)=(s/2)/r\), so \(r=s/(2\tan(\pi/n))\).
The area of a triangle is one half its base times its height. Each of the \(n\) triangles has base \(s\) and height \(r\). Adding their areas gives the following formula.
For a regular hexagon, \(n=6\) and \(\tan(\pi/6)=1/\sqrt{3}\). Thus \(A_6(s)=3\sqrt{3}s^2/2\). For other specified polygons, use their corresponding value of \(n\) in the general formula.
Why the Volume Integral Has This Form
Let \(A(x)\) denote the area of the cross section at position \(x\). The slice principle says that the volume is obtained by adding the volumes of thin slices across the base interval. A slice of thickness \(\Delta x\) has approximate volume \(A(x)\Delta x\). In the limit as the slice widths become arbitrarily small, this sum becomes the definite integral of \(A(x)\).
For regular \(n\)-gon slices with side length \(s(x)\), substitute \(s(x)\) into the polygon area formula. If \(s\) is continuous and nonnegative on \([a,b]\), the cross-sectional area is continuous there, and the volume is given by:
The polygon’s number of sides does not change along the base, so the factor \(n/(4\tan(\pi/n))\) is constant. The position-dependent part of the integrand is the square of the side length. If \(x\) is measured in metres, \(s(x)^2\) has units of square metres, and integrating with respect to \(x\) gives cubic metres.
Worked Examples
Worked Example: Hexagonal Slices Along a Two-Metre Base
A solid extends from \(x=0\) to \(x=2\) metres. Each cross section perpendicular to the \(x\)-axis is a regular hexagon whose side length is \(s(x)=2-x\) metres. Find the volume.
Step 1: Write the cross-sectional area. For a regular hexagon, \(A_6(s)=3\sqrt{3}s^2/2\). Substituting \(s(x)=2-x\) gives:
Step 2: Integrate over the base interval.
Check: The antiderivative is correct because the derivative of \(-(2-x)^3/3\) is \((2-x)^2\). At \(x=2\), the side length and area are zero; at \(x=0\), the side length is \(2\) metres and the area is \(6\sqrt{3}\) square metres. The positive integral is consistent with nonnegative slice areas throughout.
Worked Example: Pentagonal Slices with a Decreasing Side Length
A solid has regular pentagonal cross sections over \(0\le t\le 2\), with each side length \(s(t)=4-t\) centimetres. Find the volume.
Step 1: Apply the regular-pentagon area formula. Here \(n=5\) and \(\pi/n=36^\circ\), so the cross-sectional area is:
Step 2: Integrate the area. The factor \(5/(4\tan(36^\circ))\) is constant with respect to \(t\).
Check: Differentiating \(-5(4-t)^3/(12\tan(36^\circ))\) gives \(5(4-t)^2/(4\tan(36^\circ))\), the required area function. The volume is positive because the side length decreases from \(4\) centimetres to \(2\) centimetres, so every slice has positive area. Using \(\tan(36^\circ)\approx0.726543\), the volume is approximately \(32.116\ \text{cm}^3\), rounded to three decimal places.
Worked Example: Octagonal Slices with an Increasing Side Length
A solid extends over \(0\le y\le 1\) metre. Its cross sections are regular octagons, and each side has length \(s(y)=1+y\) metres. Find the volume.
Step 1: Simplify the octagon area factor. For \(n=8\), the general formula gives \(A_8(s)=2s^2/\tan(22.5^\circ)\). Since \(\tan(22.5^\circ)=\sqrt{2}-1\) and \(1/(\sqrt{2}-1)=\sqrt{2}+1\), this becomes:
Step 2: Substitute the side length and integrate.
Check: The derivative of \((1+y)^3/3\) is \((1+y)^2\), as required. At \(y=0\), the side is \(1\) metre; at \(y=1\), it is \(2\) metres. The area factor is positive, and the final volume is approximately \(11.266\ \text{m}^3\), rounded to three decimal places.
A Common Error: Using the Wrong Area
A frequent mistake is to use the square area \(s^2\) for every polygon. That ignores the shape of the cross section. For example, in the hexagonal example above, incorrectly setting \(A(x)=(2-x)^2\) would give:
The calculation integrates the side length squared correctly, but it uses the wrong cross-sectional area. The correct area of a regular hexagon with side \(s\) is \(3\sqrt{3}s^2/2\), not \(s^2\). Restoring that factor gives:
The error is not in the integration; it occurs before integration, when the polygon’s area formula is chosen. Always identify the number of sides and use the matching area formula.
A Reliable Setup
For any problem with regular polygonal slices, the setup can be organized into four steps. These steps keep the geometric information separate from the calculus.
Record its number of sides \(n\), and confirm that the given slice dimension is the side length \(s(x)\).
Use \(A(x)=n\,s(x)^2/(4\tan(\pi/n))\); for a hexagon, use the equivalent expression \(A(x)=3\sqrt{3}s(x)^2/2\).
Use the stated endpoints and integrate the cross-sectional area with respect to the coordinate along the base.
Verify the antiderivative, confirm that the volume is nonnegative, and make sure the units are cubic units.
Why the Formula Is Useful
The polygon area formula turns geometric information into an explicit function of position. That function is what the volume integral needs. Once the side length is known, the area varies as its square: doubling a polygon’s side length multiplies its area by four. The number of sides changes the constant factor, but not this squared dependence.
It is also important to distinguish the side length from other possible measures of a polygon, such as the distance across opposite vertices. The formula here takes \(s\) to be the length of one side. If a problem describes a different length, do not substitute it as \(s\) without first relating it to the side length.
Check Your Understanding
Use the regular-polygon area formula and the slice integral to answer these questions.
- What is the area of a regular hexagon with side length \(s\), and how does it follow from the formula for a regular \(n\)-gon?
- A solid has regular pentagonal slices with side length \(s(x)=3+x\) over \(0\le x\le 1\). Write, but do not evaluate, the volume integral.
- For regular octagonal slices, explain why the cross-sectional area is \(2(\sqrt{2}+1)s(x)^2\).
- What error is made if the cross-sectional area of a regular hexagon with side length \(s\) is set equal to \(s^2\)?
- If the base coordinate is measured in centimetres and the side length is also measured in centimetres, what units should the volume have?
