You need to consider a cone, centered on y axis, whose top is in origin.
You should use the following formula to find the volume of the cone using integral such that:
`V = int_a^b A(x) dx`
Notice that the cross section of the cone is a triangle, whose base is 2R and height is h.
Using similar triangles yields the following ratios such that:
`R/h = r/(h-x) => r = (R/h)(h-x)`
Evaluating the area of the cross section yields:
`A(x) = pi*r^2 => A(x) = pi*(R/h)^2(h-x)^2`
You should substitute `pi*(R/h)^2(h-x)^2` for `A(x)` in formula of volume such that:
`V = int_0^hpi*(R/h)^2(h-x)^2 dx = pi*(R/h)^2*int_0^h(h-x)^2 dx `
You should come up with the following substitution `h-x = t => -dx = dt => dx = -dt` `V = -pi*(R/h)^2*int_h^0 t^2dt => V = -pi*(R/h)^2*(t^3/3)|_h^0`
`V = -pi*(R/h)^2*(-h^3/3)`
`V = (pi*R^2*h)/3`
Substituting 18 for h and 6 for R yields:
`V = pi*36*18/3 => V = 216 pi`
Hence, evaluating the volume of cone, using integral, yields `V = 216 pi.`