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 = ∫0H A 2 dy
= 4 ∫0H 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∫0H(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∫H0t 2 (- dt)
Evaluate the integral and simplify
Volume = 4 (a/2H)2 [H3 / 3]
Volume = a2 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.