91Ó°ÊÓ

We have given the following function :-

$f\left(x\right)=\frac{1}{x}$

We have to describe the volume of the region between the graph of this function and the lines $y=0,y=1,x=0andx=2$ revolved around y-axis by using shells and disks both.

The given function is :-

$f\left(x\right)=\frac{1}{x}$

By using shells the volume between the graph and the $y-axis$is described as $V=2\mathrm{π}{∫}_{\mathrm{c}}^{\mathrm{d}}\mathrm{r}\left(\mathrm{x}\right)\mathrm{h}\left(\mathrm{x}\right)\mathrm{dx}$.

The revolution is around y-axis. So that $r\left(x\right)=x$and from localid="1651418625384" $x=0to1$height is given by localid="1651418166916" $h\left(x\right)=1$and from $x=1to2$height is given by $h\left(x\right)=\frac{1}{x}$

So that volume is described as :-

localid="1651418956369" $$\begin{gathered} V=2\mathrm{π}{∫}_0^1\mathrm{xdx}+2\mathrm{π}{∫}_1^2\mathrm{x}×\frac{1}{\mathrm{x}}\mathrm{dx} \\ ⇒\mathrm{V}=2\mathrm{π}{∫}_0^1\mathrm{xdx}+2\mathrm{π}{∫}_1^2\mathrm{dx} \end{gathered}$$

The given function is :-

$f\left(x\right)=\frac{1}{x}$

By using disks the volume between the graph and the $y-axis$is described as :-

localid="1651418479134" $V=\mathrm{π}{∫}_{\mathrm{a}}^{\mathrm{b}}{\left(\mathrm{r}\left(y\right)\right)}^{2}dy$

Here for localid="1651418798674" $y=0to\frac{1}{2}$, localid="1651419363234" $r\left(y\right)=2$and from$y=\frac{1}{2}to1$$r\left(y\right)=\frac{1}{y}$

So that volume is described as :-

localid="1651419377561" $$\begin{gathered} V=\mathrm{π}{∫}_0^{\frac{1}{2}}2\mathrm{dy}+\mathrm{π}{∫}_{\frac{1}{2}}^1{\left(\frac{1}{\mathrm{y}}\right)}^2\mathrm{dy} \\ ⇒V=\mathrm{π}{∫}_0^{\frac{1}{2}}2\mathrm{dy}+\mathrm{π}{∫}_{\frac{1}{2}}^1\frac{1}{{\mathrm{y}}^2}\mathrm{dy} \end{gathered}$$

There need less calculations in shells method then disks method.

So the shells method is easier than disks method.