# Frustrum of a pyramid

Find the volume of frustum of a pyramid with square base of side b, square top of side a, and height h.

Usually when I do these volume problems, I treat them as an equation rotating around an axis, but this object has flat sides so I don't know how to begin.

The solid looks like that.

I know I need to make an integral from 0 to h of the area of the square but while I usually replace the radius in the formula with an equation, I cannot do so here.

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I know I need to make an integral from 0 to h of the area of the square but while I usually replace the radius in the formula with an equation, I cannot do so here.
Yes you can: Let s(y) be the length of the side of the square at height y. You know that s(0) = b and s(h) = a. You also know that s(y) has to a linear function of y. (Why?) Use these facts to find s(y).

So S(y) is (a - b)x / h ?

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I think you meant (a - b)y / h. You know this is not it because when y = h, it yields a - b. You want it to yield just a.

Oh whoop, I should have seen my linear equation needed a "+ b" on the end based on the 0,b point. So the equation should look like s(y) = (a-b)y/h + b ?

You got it.