Q 68 - Calculus

Question

Use a definite integral with the shell method to prove that a cone of radius $r$ and height $h$ has volume given by the formula $V=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}$. 

Step-by-Step Solution

Answer

By using shells method the volume of the cone is given by :-

$$\begin{gathered} V=2\mathrm{\pi }{\int }_{-r}^{r}x\left[\frac{h}{2}\left(1-\frac{x}{r}\right)\right]dx \\ \Rightarrow V=4\mathrm{\pi }{\int }_0^{r}x\left[\frac{h}{2}\left(1-\frac{x}{r}\right)\right]dx \\ \Rightarrow V=\frac{4\mathrm{\pi }}{2}{\int }_0^r\left(xh-\frac{x^{2}h}{r}\right)dx \end{gathered}$$

By solving this integration we get :-

$V=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}$.

Step 1. Given Information

By using shells method we have to prove the formula of volume of cone.

That is we have to prove that volume of cone of radius $r$ and height $h$ is $\frac{1}{3}{\mathrm{\pi }}^2\mathrm{h}$ by using shells method.

Step 2. Required proof

We know that a cone of radius $r$ can be obtained  by the rotating the graph of function $y=\frac{h}{2}\left(1-\frac{x}{r}\right)$ and x-axis on $\left[-r,r\right]$. The region obtained by this rotation give us a cone.

Then by using shells method volume of region between a graph of a function and x-axis is :-

$V=2\mathrm{\pi }{\int }_{\mathrm{c}}^{\mathrm{d}}\mathrm{r}\left(\mathrm{x}\right)\mathrm{h}\left(\mathrm{x}\right)\mathrm{dx}$.

Here $r\left(x\right)=x$ and $h\left(x\right)=\frac{h}{2}\left(1-\frac{x}{r}\right)$ and rotation is from $\left[-r,r\right]$. So the volume is given by :-

$$\begin{gathered} V=2\mathrm{\pi }{\int }_{-r}^{r}x\left[\frac{h}{2}\left(1-\frac{x}{r}\right)\right]dx \\ \Rightarrow V=4\mathrm{\pi }{\int }_0^{r}x\left[\frac{h}{2}\left(1-\frac{x}{r}\right)\right]dx \\ \Rightarrow V=\frac{4\mathrm{\pi }}{2}{\int }_0^r\left(xh-\frac{x^{2}h}{r}\right)dx \\ \Rightarrow V=\frac{4\mathrm{\pi }}{2}{\left[\frac{x^{2}h}{2}-\frac{x^{3}h}{3r}\right]}_0^r \\ \Rightarrow V=\frac{4\mathrm{\pi }}{2}\left[\left(\frac{r^{2}h}{2}-\frac{r^{3}h}{3r}\right)-0\right] \\ \Rightarrow V=\frac{4\mathrm{\pi }}{2}\left[\frac{r^{2}h}{2}-\frac{r^{2}h}{3}\right] \\ \Rightarrow V=\frac{4\mathrm{\pi }}{2}\left(\frac{3r^{2}h-2r^{2}h}{6}\right) \\ \Rightarrow V=2\mathrm{\pi }\left(\frac{{\mathrm{r}}^2\mathrm{h}}{6}\right) \\ \Rightarrow \mathrm{V}=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h} \end{gathered}$$