Use triple integrals to derive the volume formula of a pyramidal frustum. This f

Question

Use triple integrals to derive the volume formula of a pyramidal frustum. This frustum has the height of h whose ends are squares with lengths of the sides a and A

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Explanation / Answer

I will give the solution for a triple integral Let the bottom base of the pyramid be of square of size A and the top base a square of side a , such that A> a now let us take the origin at the center of the bottom face. at a height z, we have a square of let's say side l which we have to determine . at a height z+dz, we will also have a square of approximately side l . so the voulme of this element will be given by the V = (area of this square)dz now area of this square = double integral (dx dy ) the limits will be determined by the size of this square l now if the extend the frustum to complete the pyramid ,, then by using geometry and the concept of similar triangles ,, we can see that l = A - z(A-a)/h so the volume of the pyramid = triple integral (dx dy dz) where the limits on x and y are from -l/2 to l/2 and the limits on z are from o to h now the double integral (dx dy ) = x(from -l/2 to l/2)* y (from -l/2 to l/2) = l^2 =[A - z(A-a)/h]^(2) = [A^(2) + ((A-a)^2* z^2)/h^2 - 2A(A-a)z/h ] now integration of this term over z from 0 to h will give = A^2.h + (A-a)^2.h/3 - (A-a)/h = h/3 [ 3A^2 + (A-a)^ - 3A(A-a)] = h/3 [3A^2 + A^2 + a^2 -2Aa -3A^2 +3Aa] = h/3[A^2 + Aa + a^2 ] which is the required volume

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