Using the Slicing Method to Find the Volume of a Cone

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*}