Q 69 - Calculus

Question

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

Step-by-Step Solution

Answer

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

$$\begin{gathered} V=2\mathrm{\pi }{\int }_{-r}^r\mathrm{x}\left(\sqrt{{\mathrm{r}}^2-{\mathrm{x}}^2}\right)\mathrm{dx} \\ \Rightarrow \mathrm{V}=4\mathrm{\pi }{\int }_0^r\mathrm{x}\left(\sqrt{{\mathrm{r}}^2-{\mathrm{x}}^2}\right)\mathrm{dx} \end{gathered}$$

By solving this integration we get :-

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

Step 1. Given Information

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

That is we have to prove that volume of sphere of radius $r$ is $V=\frac{4}{3}{\mathrm{\pi r}}^3$ by using shells method.

Step 2. Required proof

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

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

$$\begin{gathered} V=2\mathrm{\pi }{\int }_{-r}^r\mathrm{x}\left(\sqrt{{\mathrm{r}}^2-{\mathrm{x}}^2}\right)\mathrm{dx} \\ \Rightarrow \mathrm{V}=4\mathrm{\pi }{\int }_0^r\mathrm{x}\left(\sqrt{{\mathrm{r}}^2-{\mathrm{x}}^2}\right)\mathrm{dx} \end{gathered}$$

Now put

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

Also when

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

and when 

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

Then we have :-

$$\begin{gathered} V=-4\mathrm{\pi }{\int }_{\mathrm{r}}^0\sqrt{{\mathrm{t}}^2}·\mathrm{tdt} \\ \Rightarrow \mathrm{V}=-4\mathrm{\pi }{\int }_{\mathrm{r}}^0\mathrm{t}·\mathrm{tdt} \\ \Rightarrow \mathrm{V}=-4\mathrm{\pi }{\int }_{\mathrm{r}}^0{\mathrm{t}}^2\mathrm{dt} \\ \Rightarrow \mathrm{V}=-4\mathrm{\pi }{\left[\frac{{\mathrm{t}}^3}{3}\right]}_{\mathrm{r}}^0 \\ \Rightarrow \mathrm{V}=-4\mathrm{\pi }\left[0-\frac{{\mathrm{r}}^3}{3}\right] \\ \Rightarrow \mathrm{V}=\frac{4}{3}{\mathrm{\pi r}}^3 \end{gathered}$$
Hence proved.