Use the slicing method to verify that the volume of a right circular cone having radius $R$ and height $h$ is $V=\dfrac{1}{3}\pi R^{2}h.$

Solution We position the cone with its vertex at the origin and its axis on the $x$-axis, as shown in Figure 39.

The cone extends from $x=0$ to $x=h$. The cross section at any number $x$ is a circle. To obtain its area $A$, we need the radius $r$ of the circle. Embedded in Figure 39 are two similar triangles, one with sides $x$ and $r;$ the other with sides $h$ and $R,$ as shown in Figure 40. Because these triangles are similar (AAA), corresponding sides are in proportion. That is, $$\begin{eqnarray*} \dfrac{r}{x} &=&\dfrac{R}{h} \\[6pt] r &=&\dfrac{R}{h}x \end{eqnarray*}$$

So, $r$ is a function of $x$ and the area $A$ of the circular cross section is $$\begin{equation*} A=A(x)=\pi [ r(x)] ^{2}=\pi\! \left( \dfrac{R}{h}x\right) ^{2}= \dfrac{\pi R^{2}}{h^{2}}x^{2} \end{equation*}$$

Since $A$ is a continuous function of $x$ (where the slice was made), we can apply the slicing method. The volume $V$ of the right circular cone is $$\begin{equation*} V = \int_{a}^{b}A(x)~{\it dx}=\displaystyle \int_{0}^{h}\frac{\pi R^{2}}{h^{2}}x^{2}~{\it dx}=\frac{\pi R^{2}}{h^{2}}\!\displaystyle \int_{0}^{h}x^{2}~{\it dx}\\[11pt] = \frac{\pi R^{2}}{h^{2}}\left[ \frac{x^{3}}{ 3}\right] _{0}^{h}=\frac{\pi R^{2}h}{3} ~\hbox{cubic units} \end{equation*}$$