Volume of a Frustrum of a Pyramid

[Ghostscythe]

Homework Statement

Find the volume, using only the variables a, b, and h in your answer.

A frustum of a pyramid with square base of side b, square top of side a, and height h:

Homework Equations

Area = length*width.

Length of side S = ??

The Attempt at a Solution

I know that generally when finding volume, it's going to be the integral from a[bottom] to b[top] of a cross-section's area. usually that'd just be integral[pi*r^2] a..b for a cylinder.

For this, obviously, I need to find an equation for the length side S that is a function of h, so S(h) = SL.

My first thought was (b-a)h, but that wouldn't change as the integral changes, so it's out. I can't think of a dynamic equation to suit the problem, and would appreciate help pushing me in that direction.

 

[Dick]

Let x be the height of your cross section. So x ranges from 0 to h. Let S(x) be the side length as a function of x. So S(0)=b and S(h)=a and S varies linearly in between. Now can you write an explicit form for S(x)?

 

[Ghostscythe]

Ahh, soo..S(x) = b + [(a-b)x]/hSo it would be the integral from 0..h of ( b + [(a-b)x]/h )^2 dx..

And the volume is:

V = [PLAIN]http://www4a.wolframalpha.com/Calculate/MSP/MSP105419c6ggha31a9g60d000062bg6e58533dg93g?MSPStoreType=image/gif&s=16&w=117&h=36

 

[Ghostscythe]

Update - got this problem right, then did another frustrum (of a cone) and got that right as well. Thanks for the pointer, appreciate it! :D

I love how this parallels the derivative formula in the easy/hard way...(1/3)*h*(a-b)^2 every time, give or take a Pi...lol.