Q 21 - Calculus

Question

For each solid described in Exercises 21–24, set up volume integrals using both the shell and disk/washer methods. Which method produces an easier integral in each case, and why? Do not solve the integrals.

The region between the graph of $f\left(x\right)=\left|x\right|$ and the $x-axis$ on $\left[-2,2\right]$, revolved around the $x-axis$. 

Step-by-Step Solution

Answer

By using shells the volume is described as $2\left(2\mathrm{\pi }{\int }_0^2\mathrm{y}\left(2-\mathrm{y}\right)\mathrm{dy}\right)$.

By using disks the volume is described as $2\left(\mathrm{\pi }{\int }_0^2{\mathrm{x}}^2\mathrm{dx}\right)$.

The method of disks is easier here than the shells method.

Step 1. Given Information

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.

Step 2. Volume by using 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 $V=2\mathrm{\pi }{\int }_{\mathrm{c}}^{\mathrm{d}}\mathrm{r}\left(\mathrm{y}\right)\mathrm{h}\left(\mathrm{y}\right)\mathrm{dy}$

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 :-

$V=2\mathrm{\pi }{\int }_{-2}^2\mathrm{y}\left(2-\mathrm{y}\right)\mathrm{dy}$

We know that :-

${\int }_{-a}^{a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx$

Then we have :-

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

Step 3. Volume by using disks

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 $V=\mathrm{\pi }{\int }_{\mathrm{a}}^{\mathrm{b}}{\left(\mathrm{r}\left(\mathrm{x}\right)\right)}^2\mathrm{dx}$.

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

Then the volume is :-

$V=\mathrm{\pi }{\int }_{-2}^2{\mathrm{x}}^2\mathrm{dx}$

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

Then we have :-

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

Step 4. Easier method

By using shells volume is described as :-

$2\left(2\mathrm{\pi }{\int }_0^2\mathrm{y}\left(2-\mathrm{y}\right)\mathrm{dy}\right)$

Also by using disks volume is described as :-

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

There need less calculations in disks method then shells method.

So the disks method is easier than shells method.