Find the formula for the volume of a square pyramid using integrals in calculus.
In what follows, \( C \) represents the constant of integration.
Problem & Derivation
Problem: A pyramid is shown in the figure below. Its base is a square of side \( a \) and is orthogonal to the y-axis. The height of the pyramid is \( H \). Use integrals and their properties to find the volume of the square pyramid in terms of \( a \) and \( H \).
Solution to the problem:
Let us first position the pyramid so that two opposite sides of the square base are perpendicular to the x-axis and the center of its base is at the origin of the x-y system of axes. If we look at the pyramid in a direction orthogonal to the x-y plane, it will look like a two-dimensional shape as shown below. \( AC \) is the slant height.
Let \( x = A'B' \) be the length of half of the side of the square at height \( y \). The area \( A \) of the square at height \( y \) is given by:
\[ A(x) = (2x)^2 \]The volume is found by adding all the volumes \( A \, dy \) that make the pyramid from \( y = 0 \) to \( y = H \). Hence:
\[ \text{Volume} = \int_{0}^{H} A(x) \, dy \] \[ = 4 \int_{0}^{H} x^2 \, dy \]We now use the fact that triangles \( ABC \) and \( AB'C' \) are similar and therefore the lengths of their corresponding sides are proportional to write:
\[ \frac{a/2}{x} = \frac{H}{H - y} \]We now solve the above for \( x \) to obtain:
\[ x = \frac{a (H - y)}{2H} \]We now substitute \( x \) in the integral that gives the volume to obtain:
\[ \text{Volume} = 4 \left(\frac{a}{2H}\right)^2 \int_{0}^{H} (H - y)^2 \, dy \]Let us define \( t \) by:
\[ t = H - y \quad \text{and} \quad dt = -dy \]Substitute and change the limits of integration to write the volume as follows:
\[ \text{Volume} = 4 \left(\frac{a}{2H}\right)^2 \int_{H}^{0} t^2 (-dt) \] \[ = 4 \left(\frac{a}{2H}\right)^2 \int_{0}^{H} t^2 \, dt \]Evaluate the integral and simplify:
\[ \text{Volume} = 4 \left(\frac{a}{2H}\right)^2 \left[\frac{H^3}{3}\right] \] \[ \Large \displaystyle \boxed{\text{Volume} = \frac{a^2 H}{3}} \]The volume of a square pyramid is given by the area of the base times one-third of the height of the pyramid.