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

Let us find the volume of a pyramid of height ##h## with a ##btimes 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.