Q 67 - Calculus

Question

Use a definite integral with the shell method to prove that the volume formula $V=\frac{4}{3}{\mathrm{\pi }}^3$ holds for a sphere of radius $3$. 

Step-by-Step Solution

Answer

By using the shells the volume of sphere of radius $3$ is given as :-

$$\begin{gathered} V=2\mathrm{\pi }{\int }_{-3}^{3}x\sqrt{9-x^2}dx \\ \Rightarrow V=4\mathrm{\pi }{\int }_0^{3}x\sqrt{9-x^2}dx \end{gathered}$$

By solving this integration we get $V=36\mathrm{\pi }$. This is the same as the volume by finding using volume formula.

So that the given volume formula $V=\frac{4}{3}{\mathrm{\pi r}}^3$ holds for sphere of radius $3$.

Step 1. Given Information

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

Step 2. Volume by using shells method

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

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

$$\begin{gathered} V=2\mathrm{\pi }{\int }_{-3}^{3}x\sqrt{9-x^2}dx \\ \Rightarrow V=4\mathrm{\pi }{\int }_0^{3}x\sqrt{9-x^2}dx \end{gathered}$$

Now put

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

Also when :-

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

and 

when :-

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

Then the integration for volume becomes :-

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

Step 3. Volume by using volume formula

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

$V=\frac{4}{3}{\mathrm{\pi r}}^3$

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

$$\begin{gathered} V=\frac{4}{3}\mathrm{\pi }\left(3^3\right) \\ \Rightarrow \mathrm{V}=\frac{4}{3}\mathrm{\pi }\left(27\right) \\ \mathrm{V}=36\mathrm{\pi } \end{gathered}$$

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

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