The volume of a cone is one third the volume of the cylinder that contains the cone. How can we prove it?

i think it would depend on how big the cone and the cylinder are, we would need dimensions to solve this.

You need to know some calculus, more specifically integration.

If you know integration, you'll know that the area bounded by a curve, the x axis, the line y=a, and the line y=b is expressed as an integral, which is derived from finding a limiting sum of increasingly small rectangular strips that approximate the area of the function.

What we want this time is to find the volume of a solid, which can be done in a similar way, this time taking the limiting sum of increasingly thin cylinders, and then convert that into an integral.

So, we let f(x)=(r/h)x, where r is the radius of the base of the cone, and h is the height. We want, instead of finding the sum of rectangles with height f(x), we want to revolve those rectangles around the x axis to form thin cylinders, with radius f(x).

Notice that the area of the rectangles are f(x)*dx (d meaning delta, i.e. very small change in x), whereas the cylinders are pi*(f(x))^2*dx, so in stead of an integral of the function f(x), we have an integral of the function pi*(f(x))^2.

So, we want to find the integral from x=0 to x=h of pi*((r/h)x)^2, which will give you the result 1/3*pi*r^2*h, which is 1/3rd the volume of a the cylinder with the same dimensions.

I hope it helps :/