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

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. |