Question

how to prove the volume formula of a pyramid is 1/3xhx(area of the surface)?

Answer 1

Answer on Question #48087 – Math – Geometry

How to prove the volume formula of a pyramid is $\frac{1}{3} h$ (area of the surface)?

Solution

On the left, $ACB$ is a section of a pyramid. $AC$ is a line through the base, and represents one dimension of the area, $A$, of the base. $DE$ is any section through the pyramid at height $y$, and is parallel to the base, and represents an area $A_i$, $h$ is the height of the pyramid.

Because the area section at $DE$, $A_i$, is proportional to the area of the base of the pyramid, $A$ (at $AC$) and these areas, $A$ and $A_i$, are proportional to their heights squared, then:

$$\frac{A_i}{A} = \frac{y^2}{h^2}.$$

So, $A_i$ is:

$$A_i = A \frac{y^2}{h^2}.$$

As each component of the total volume is:

$$dV = A_i dy.$$

Substituting the formula for the area:

$$dV = A \frac{y^2}{h^2} dy.$$

We now have the means of integrating to get the volume, $V$, between the limits 0 and $h$:

$$V = \int_0^h A \frac{y^2}{h^2} dy.$$

And by integrating:

$$V = \left(A \frac{y^3}{3h^2}\right)_0^h = \frac{1}{3} Ah.$$

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