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Let x = length of side of the square base of the pyramid and h its height.WebCalC On-line Calculus requires Scientific Notebook App (free viewer version available): Texas A&M U
At any height y, length of the square section
= x * (y/h)
so Area = [ x * (y/h) ]^2
Volume of a thin slice of pyramid at a height y and having thickness dy
= [ x * (y/h) ]^2 dy
Integrating from y = 0 to y = h,
Total volume
= (x/h)^2 ? y^2 dy [y=0 to y=h]
= (1/3) (x/h)^2 [ y^3 ] [y=0 to y=h]
= (1/3) (x/h)^2 * h^3
= (1/3) * h * x^2
i.e., one-third the height times area of the base.
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