91Ó°ÊÓ

We have to prove that the volume formula for cone $V=\frac{1}{3}{\mathrm{Ï€°ù}}^2\mathrm{h}$holds for the volume of cone of radius $3$and height $5$.

A cone of radius $3$and height $5$can obtained by rotating the region $y=\frac{h}{2}\left(1-\frac{x}{r}\right)ory=\frac{5}{2}\left(1-\frac{x}{3}\right)$around x-axis on $\left[-3,3\right]$.

Then by using shells method volume is given by :-

$$\begin{gathered} V=\frac{2\mathrm{π}}{2}{∫}_{-3}^3\mathrm{x}\left[5\left(1-\frac{\mathrm{x}}{3}\right)\right]\mathrm{dx} \\ ⇒\mathrm{V}=\mathrm{π}{∫}_{-3}^3\left(5\mathrm{x}-\frac{5{\mathrm{x}}^2}{3}\right)\mathrm{dx} \\ ⇒\mathrm{V}=2\mathrm{π}{∫}_0^3\left(5\mathrm{x}-\frac{5{\mathrm{x}}^2}{3}\right)\mathrm{dx} \\ ⇒\mathrm{V}=2\mathrm{π}{\left[\frac{5{\mathrm{x}}^2}{2}-\frac{5{\mathrm{x}}^3}{9}\right]}_0^3 \\ ⇒\mathrm{V}=2\mathrm{π}\left[\left(\frac{45}{2}-15\right)-\left(0+0\right)\right] \\ ⇒\mathrm{V}=2\mathrm{π}\left[\frac{45-30}{2}\right] \\ ⇒\mathrm{V}=2\mathrm{π}×\frac{15}{2} \\ ⇒\mathrm{V}=15\mathrm{π} \end{gathered}$$

The volume of cone is given by the following formula :-

$V=\frac{1}{3}{\mathrm{Ï€°ù}}^2\mathrm{h}$

Put $r=3,h=5$, then we have :-

$$\begin{gathered} V=\frac{1}{3}\mathrm{π}{\left(3\right)}^2×5 \\ ⇒\mathrm{V}=15\mathrm{π} \end{gathered}$$

So the volume of cone of radius $3$and height $5$is $15\mathrm{Ï€}$. This is the same as the volume finding by using shells method.

So we can conclude that the given volume formula of cone $V=\frac{1}{3}{\mathrm{Ï€°ù}}^2\mathrm{h}$holds for cone of radius $3$and height $5$.