How do you find the volume of a pyramid using integrals?

Let us find the volume of a pyramid of height ##h## with a ##b\times b## square base.

If ##y## is the vertical distance from the top of the pyramid, then the square cross-sectional area ##A(y)## can be expressed as

##A(y)=(b/hy)^2=b^2/h^2y^2##.

So, the volume ##V## can be found by the integral

##V=int_0^hA(y) dy=b^2/h^2int_0^hy^2 dy=b^2/h^2[y^3/3]_0^h =1/3b^2h##.

I hope that this was helpful.