Find The Volume of a Square Pyramid Using Integrals

Find the formula for the volume of a square pyramid using integrals in calculus.

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 volmes A dy that make the pyramid from y = 0 to y = H. Hence

Volume = _{0}^{H} A^{ 2} dy

= 4 _{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:

(a/2) / x = H / (H - y)

We now solve the above for x to obtain

x = a (H - y) / (2 H)

We now substitute x in the intergal that gives the volume to obtain

Volume = 4 (a/2H)^{2}_{0}^{H}(H - y)^{ 2} dy

Let us define t by

t = H - y and dt = - dy

The volume is now givwn by

Volume = 4 (a/2H)^{2}_{H}^{0}t^{ 2} (- dt)

Evaluate the integral and simplify

Volume = 4 (a/2H)^{2} [H^{3} / 3]

Volume = a^{2} H / 3

The volume of a square pyramid is given by the area of the base times the third of the height of the pyramid.
More references on
integrals and their applications in calculus.