# 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 volumes 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 integral 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 given 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.

Area under a curve .

Area between two curves .

Find The Volume of a Solid of Revolution .

Volume by Cylindrical Shells Method .