Q. 63 - Calculus

Question

Use a definite integral 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

It has been proved that volume of the cone is $\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}$

Step 1. Given information

The height of the cone is $h$.

The base radius is $r$

The volume to be proved is $V=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}$

Step 2. Proof of the volume

Lets draw a cone having given dimensions:

Here a circular slice of thickness $dx$ has been cut at distance of $x$ from the base.

The radius of the slice is calculated as:

Since $∆ABC~∆AOE$

Hence, $$\begin{gathered} \frac{BC}{r}=\frac{h-x}{h} \\ BC=\frac{r}{h}(h-x) \\ =r-\frac{r}{h}x \end{gathered}$$

The volume of the slice is:

$$\begin{gathered} dv=\mathrm{\pi }{\left(\mathrm{BC}\right)}^2\mathrm{dx} \\ =\mathrm{\pi }{\left(\mathrm{r}-\frac{\mathrm{r}}{\mathrm{h}}\mathrm{x}\right)}^2\mathrm{dx} \\ =\mathrm{\pi }\left({\mathrm{r}}^2+\frac{{\mathrm{r}}^2}{{\mathrm{h}}^2}{\mathrm{x}}^2-2\frac{{\mathrm{r}}^2}{\mathrm{h}}\mathrm{x}\right)\mathrm{dx} \end{gathered}$$

The volume of the cone is:

$$\begin{gathered} V={\int }_0^{h}dv \\ ={\int }_0^h\mathrm{\pi }\left({\mathrm{r}}^2+\frac{{\mathrm{r}}^2}{{\mathrm{h}}^2}{\mathrm{x}}^2-2\frac{{\mathrm{r}}^2}{\mathrm{h}}\mathrm{x}\right)\mathrm{dx} \\ =\mathrm{\pi }{\int }_0^h\left({\mathrm{r}}^2+\frac{{\mathrm{r}}^2}{{\mathrm{h}}^2}{\mathrm{x}}^2-2\frac{{\mathrm{r}}^2}{\mathrm{h}}\mathrm{x}\right)\mathrm{dx} \\ =\mathrm{\pi }{\left[{\mathrm{r}}^2\mathrm{x}+\frac{{\mathrm{r}}^2}{{\mathrm{h}}^2}.\frac{{\mathrm{x}}^3}{3}-\frac{2{\mathrm{r}}^2}{2\mathrm{h}}{\mathrm{x}}^2\right]}_0^{\mathrm{h}} \\ =\mathrm{\pi }\left[\left({\mathrm{r}}^2\mathrm{h}+\frac{{\mathrm{r}}^2{\mathrm{h}}^3}{3{\mathrm{h}}^2}-\frac{{\mathrm{r}}^2{\mathrm{h}}^2}{\mathrm{h}}\right)-0\right] \\ =\mathrm{\pi }\left({\mathrm{r}}^2\mathrm{h}+\frac{{\mathrm{r}}^2\mathrm{h}}{3}-{\mathrm{r}}^2\mathrm{h}\right) \\ =\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h} \end{gathered}$$