# Thread: volume of a pyramid using integration

1. ## volume of a pyramid using integration

How would I find the volume of a pyramid, or any shape, using integration? Say I have a pyramid with height 500 and a base of side 750. How would I find the volume using integration? It needs to be in the form V = the summation from i=1 to n * V sub i, beofore converting to an integral and solved.

2. ## A picture is worth 1000 words, so here's 1000 words for you...

We use the concept of cross-sectional area: $\displaystyle V=\int_0^h A(x)dx$, where $\displaystyle A(x)$ gives the area of the cross-section created by slicing through the figure at $\displaystyle x$, and h is the height.

Picture a pyramid with its apex at the origin and the center of its base sitting on the point $\displaystyle (h,0,0)$ on the x-axis, where $\displaystyle h$ is of course its height. Now imagine cutting the pyramid with a vertical sheet parallel to the $\displaystyle yz$ plane passing through the point $\displaystyle (x,0,0)$ for $\displaystyle 0<x<h$. The cross sectional area is going to be a square. This square forms a "miniature" pyramid that is a scaled down model of the original. So, the ratio of x to s, the side of the cross sectional square, is the same as the ratio of h to b, the base side of the whole pyramid: $\displaystyle s=\frac bhx$. Thus, the area of this square is $\displaystyle A(x)=s^2=(\frac bhx)^2=\frac {b^2}{h^2}x^2$

Evaluating, $\displaystyle V=\int_0^hA(x)dx=\frac {b^2}{h^2}\int_0^hx^2dx=\frac {b^2}{h^2}\frac13x^3|_0^h=\frac13b^2h$

It is a known fact from the Greeks that the volume of any "pointed" object, that is, a two-dimensional shape as a base that comes to a point has an area that is one third its height times the area of its base.