91Ó°ÊÓ

We have to prove that the volume formula of sphere $V=\frac{4}{3}{\mathrm{Ï€°ù}}^3$is holds for a volume of sphere of radius$3$.

A sphere of radius $3$can by rotating the region $y=\sqrt{3^2-x^2}ory=\sqrt{9--x^2}$around x-axis on $\left[-3,3\right]$.

Then by using shells method the volume is given by :-

role="math" localid="1651763409717" $$\begin{gathered} V=2\mathrm{π}{∫}_{-3}^{3}x\sqrt{9-x^2}dx \\ ⇒V=4\mathrm{π}{∫}_0^{3}x\sqrt{9-x^2}dx \end{gathered}$$

Now put

$$\begin{gathered} 9-x^2=t^2 \\ ⇒-2xdx=2tdt \\ ⇒xdx=-tdt \end{gathered}$$

Also when :-

$$\begin{gathered} x=0, \\ {t}^2=9 \\ ⇒t=3 \end{gathered}$$

and

when :-

$$\begin{gathered} x=3, \\ {t}^2=0 \\ ⇒t=0 \end{gathered}$$

Then the integration for volume becomes :-

$$\begin{gathered} V=-4\mathrm{π}{∫}_3^0\mathrm{t}\sqrt{{\mathrm{t}}^2}\mathrm{dt} \\ ⇒\mathrm{V}=-4\mathrm{π}{∫}_3^0{\mathrm{t}}^2\mathrm{dt} \\ ⇒\mathrm{V}=-4\mathrm{π}{\left[\frac{{\mathrm{t}}^3}{3}\right]}_3^0 \\ ⇒\mathrm{V}=-4\mathrm{π}\left[0-\frac{27}{3}\right] \\ ⇒\mathrm{V}=-4\mathrm{π}\left(-9\right) \\ ⇒\mathrm{V}=36\mathrm{π} \end{gathered}$$

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

$V=\frac{4}{3}{\mathrm{Ï€°ù}}^3$

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

role="math" $$\begin{gathered} V=\frac{4}{3}\mathrm{π}\left(3^3\right) \\ ⇒\mathrm{V}=\frac{4}{3}\mathrm{π}\left(27\right) \\ \mathrm{V}=36\mathrm{π} \end{gathered}$$

So the volume of sphere of radius $3$is $36\mathrm{Ï€}$. This is the same as the volume finding by using shells method.

So we can conclude that the given volume formula of sphere $V=\frac{4}{3}{\mathrm{Ï€°ù}}^3$holds for sphere of radius$r=3$.