Using the method of integration, show that the volume of a right circular cone of base radius r and height h is `V=1/3pir^2h`.

Explanation

Let ABC be the right circular cone. Consider a circular secition of the cone DE, with plane parallel to its base, of thickness `dx`, at a distance of x from the apex A.
If y be its radius, then from similar triangles, AOC and `AO^'E`, we have
`y/x=r/himpliesy=(rx)/(h)`
Therefore, area of the circular section `DE=piy^2=(pir^2x^2)/(h^2)`
Therefore, volume of the circular section `DE=(pir^2x^2dx)/(h^2)`
i.e., `dV=(pir^2x^2dx)/(h^2)`
Now, the total volume of the cone can be obtained as the summation (integration) of the volumes of each circular sections such as `DE`, i.e.,
`V=sumdV=underset0oversethint(pir^2x^2dx)/(h^2)=(pir^2)/(h^2)[(x^3)/(3)]_0^h`
`=(pir^2)/(3h^2)[h^3-0]=(pir^2h)/(3)`