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## Homework Statement

There is no specific problem, I'm just confused after reading the chapter.

Consider a pyramid 3 m high with a square base that is 3 m on a side. The cross section of the pyramid perpendicular to the altidude x m down from the vertex is a square x m on a side.

Now I understand the volume is,

[tex]

\int_{a}^{b} A(x) dx = \int_{0}^{3} x^2 dx = 9

[/tex]

but then I get confused. How would the volume of a rectangle with the same square base, using the same method, be any different?

**edit: Thanks, it makes much more sense to me now.**

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