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 xy system of axes. If we look at the pyramid in a direction orthogonal to the xy 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.
