91Ó°ÊÓ

We have given the following function :-

$f\left(x\right)=\left|x\right|$

We have to describe the volume of the region between the graph of this function and x-axis on $\left[-2,2\right]$, revolved around the x-axis by using both methods disks and shells.

The given function is :-

$f\left(x\right)=\left|x\right|$

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

The revolution is around x-axis. So that $r\left(y\right)=y$.

Also the we need to find volume between the given function $\left|x\right|$and x-axis on $\left[-2,2\right]$.

So the height of shell is given by $h\left(x\right)=2-y$.

Then the volume is described as :-

role="math" $V=2\mathrm{π}{∫}_{-2}^2\mathrm{y}\left(2-\mathrm{y}\right)\mathrm{dy}$

We know that :-

role="math" localid="1651416867862" ${∫}_{-a}^{a}f\left(x\right)dx=2{∫}_0^{a}f\left(x\right)dx$

Then we have :-

$V=2\left(2\mathrm{π}{∫}_0^2\mathrm{y}\left(2-\mathrm{y}\right)\mathrm{dy}\right)$

The given function is $f\left(x\right)=\left|x\right|$.

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

Here $r\left(x\right)=x$and the limits will be $-2and2$.

Then the volume is :-

$V=\mathrm{π}{∫}_{-2}^2{\mathrm{x}}^2\mathrm{dx}$

We know that ${∫}_{-a}^{a}f\left(x\right)dx=2{∫}_0^{a}f\left(x\right)dx$.

Then we have :-

role="math" localid="1651416376954" $V=2\left(\mathrm{π}{∫}_0^2{\mathrm{x}}^2\mathrm{dx}\right)$

By using shells volume is described as :-

$2\left(2\mathrm{π}{∫}_0^2\mathrm{y}\left(2-\mathrm{y}\right)\mathrm{dy}\right)$

Also by using disks volume is described as :-

$V=2\left(\mathrm{π}{∫}_0^2{\mathrm{x}}^2\mathrm{dx}\right)$

There need less calculations in disks method then shells method.

So the disks method is easier than shells method.