From Polygonal Slices to Pyramid Volume
A pyramid has a polygonal base and triangular sides meeting at one apex. Its volume can be found by treating it as a stack of thin cross sections parallel to the base. The key geometric fact is that these cross sections shrink in a predictable way: their corresponding lengths are proportional to their distance from the apex.
Once the length scale is known, the area of each slice follows. Integrating those areas over the pyramid’s height derives the volume formula, rather than requiring it as a memorized rule.
How the Cross Sections Scale
Take a pyramid with apex \(A\), and a plane through its base. A second plane parallel to the base cuts the pyramid at distance \(x\) from the apex. The perpendicular distance from the apex to the base plane is \(H\).
Each point on the cross section lies on a line from the apex to a corresponding point on the base. Because the two planes are parallel, the cross section is a scaled copy of the base. The scale factor is the fraction of the full perpendicular height reached by the slice:
This holds for a pyramid whose apex is not directly above the center of its base, as well as for a right pyramid. The cross section may be shifted relative to the base, but its shape and scale are determined by the same ratio.
Lengths scale by \(x/H\). Areas scale by the square of the length scale, so the cross-sectional area at distance \(x\) from the apex is
This formula has the expected endpoint values. At the apex, \(x=0\), the cross-sectional area is zero. At the base, \(x=H\), it is \(B\).
Deriving the Volume Formula
The slice principle gives the volume as the integral of the cross-sectional area with respect to distance along the pyramid’s height. Substitute the area function and integrate from the apex to the base:
Here \(B\) and \(H\) are constant, so they can be taken outside the integral. Evaluating the remaining power integral gives
The factor \(1/3\) comes from integrating the quadratic change in slice area over the height. It is not a factor from the base shape: the argument uses only the base area and the fact that parallel sections scale uniformly. The base can be any polygon.
Why the Scaling Argument Works
The relation between distance and scale can be made precise. Set the apex’s perpendicular coordinate to zero, and choose the coordinate direction toward the base. The base plane then has coordinate \(H\), while a parallel slice at distance \(x\) has coordinate \(x\).
Join the apex to any point \(P\) on the base. The point where this line meets the slice is a fraction \(t\) of the way from the apex to \(P\). Its perpendicular coordinate is \(tH\). Since it lies in the slice plane, that coordinate is \(x\), so \(tH=x\), and therefore \(t=x/H\). This same value of \(t\) applies to every line from the apex through the base.
Consequently, every segment in the base connecting two points corresponds to a segment in the slice whose length is multiplied by \(x/H\). A polygon’s area is multiplied by the square of this length factor. Thus \(A(x)=B(x/H)^2\), which justifies the area function used in the integral.
Worked Examples
Worked Example: A Square Pyramid
A square pyramid has base side length \(6\) cm and perpendicular height \(10\) cm. Find its volume by integrating the areas of its parallel square slices.
The base area is \(B=6^2=36\text{ cm}^2\). At distance \(x\) cm from the apex, the slice side length is \(6(x/10)=3x/5\) cm. Therefore its area is
Integrate from \(x=0\) to \(x=10\):
Check the integration: the derivative of \(\frac{9x^3}{75}\) is \(\frac{9x^2}{25}\), the stated slice-area function. Also, the formula gives \(\frac13(36)(10)=120\text{ cm}^3\), in agreement. The units are area times length, hence cubic centimetres.
Worked Example: An Oblique Pyramid With a Rectangular Base
An oblique pyramid has a rectangular base measuring \(8\) m by \(3\) m and perpendicular height \(9\) m. Find its volume using slices parallel to the base.
The base area is \(B=8\cdot3=24\text{ m}^2\). Even though the apex need not be above the rectangle’s center, each parallel slice is a translated, scaled copy of the base. Its side lengths are \(8x/9\) m and \(3x/9=x/3\) m. Thus
Integrating over the height yields
For a slice at \(x=3\) m, the two side lengths are \(8(3/9)=8/3\) m and \(3(3/9)=1\) m, giving area \(8/3\text{ m}^2\). The formula gives \(8(3^2)/27=8/3\text{ m}^2\) as well. Finally, the volume formula gives \(\frac13(24)(9)=72\text{ m}^3\). Both checks agree.
Worked Example: A Regular Hexagonal Pyramid
A regular hexagonal pyramid has base side length \(4\) cm and perpendicular height \(6\) cm. Find its volume by integrating the cross-sectional areas.
The area of a regular hexagon with side length \(s\) is \(\frac{3\sqrt{3}}{2}s^2\). Thus the base area is
At distance \(x\) from the apex, each slice side length is \(4(x/6)=2x/3\) cm. Its area is therefore
Integrate over \(0\le x\le6\):
The antiderivative differentiates back to \(\frac{2\sqrt{3}}{3}x^2\). The general formula gives \(\frac13(24\sqrt{3})(6)=48\sqrt{3}\text{ cm}^3\), confirming the result. At half the height, \(x=3\), the side length is \(2\) cm and the slice area is \(6\sqrt{3}\text{ cm}^2\); substituting \(x=3\) into \(A(x)\) gives the same value.
A Common Scaling Mistake
A frequent error is to scale area by the same factor as length. For the square pyramid with base area \(36\text{ cm}^2\) and height \(10\) cm, this mistake would assign the slice area \(36(x/10)\), rather than \(36(x/10)^2\). Integrating that incorrect model gives
which is not the correct volume. The error is visible at \(x=5\): the side length is half the base side, so the slice area is \(36(1/2)^2=9\text{ cm}^2\), not \(36(1/2)=18\text{ cm}^2\). Always square the scale factor when converting a length ratio into an area ratio.
Also keep track of where distance is measured from. In this tutorial, \(x=0\) is at the apex and \(x=H\) is at the base. A coordinate measured from the base would require a different expression for the slice scale, so its formula should not be used without changing the bounds and interpretation consistently.
When This Derivation Is Useful
This method shows how a geometric volume can emerge from the changing area of its cross sections. It applies regardless of whether the base is a triangle, rectangle, or another polygon, provided the slices are parallel to that base and the object is a pyramid with a single apex.
It also gives a practical check on a model: the slice area must approach zero at the apex and equal the base area at the base. The volume must have units of area times length. These checks help catch incorrect scaling, height, or units before relying on a final calculation.
Check Your Understanding
Use the distance \(x\) measured from the apex in each question.
- A pyramid has base area \(50\text{ cm}^2\) and height \(12\) cm. What is the linear scale factor of a slice at \(x=3\) cm?
- For the pyramid in Question 1, write a formula for the area of the slice at distance \(x\).
- Why is the area scale factor the square of the length scale factor?
- Derive the volume of a pyramid with base area \(B\) and height \(H\) by integrating its slice areas.
- A pyramid has a regular polygonal base with area \(18\text{ m}^2\) and height \(5\) m. Find its volume.
