Question

Using the method of integration, show that the volume of a right circular cone of base radius $r$ and height $h$ is $V=\frac{1}{3}\pi r^{2}h$.

Answer 1

Let $ABC$ be the right circular cone.
Consider a circular section 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 $A{O}^{\prime }E$, we have
$\frac{y}{x}=\frac{r}{h}\Rightarrow y=\frac{rx}{h}$
Therefore, area of the circular section $DE=\pi y^2=\frac{\pi r^2{x}^2}{h^2}$
Therefore, volume of the circular section $DE=\frac{\pi r^2{x}^{2}dx}{h^2}$
i.e., $dV=\frac{\pi r^2{x}^{2}dx}{h^2}$
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,
$V=\sum dV,={\int }_0^h\frac{\pi r^2{x}^{2}dx}{h^2}=\frac{\pi r^2}{h^2}{\left[\frac{x^3}{3}\right]}_0^h$
$=\frac{\pi r^2}{3h^2}[h^3-0]=\frac{\pi r^{2}h}{3}$.