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 $f\left(x\right)=x^2+1$an the x-axis on $\left[0,1\right]$, revolved around the line $y=-2$.

Short answer

By using shells the volume is described as$V=2\mathrm{\pi }{\int }_0^1\left(-2-\mathrm{x}\right)\left({\mathrm{x}}^2+1\right)\mathrm{dx}$

By using disks the volume is described as$\mathrm{V}=\mathrm{\pi }{\int }_1^2\left({\left(2-\sqrt{\mathrm{y}-1}\right)}^2-2^2\right)\mathrm{dx}$

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

Step-by-step solution

Step 1. Given Information

We have given the following function :-

$f\left(x\right)=x^2+1$

We have to describe the volume of the region between the graph of this function and x-axis on $\left[0,1\right]$revolved around the line $y=-2$by using shells and disks both.

Step 2. Volume by using Shells

The given function is :-

$f\left(x\right)=x^2+1$

By using shells the volume between the graph and the x-axis is described as

$V=2\mathrm{\pi }{\int }_{\mathrm{c}}^{\mathrm{d}}\mathrm{r}\left(\mathrm{x}\right)\mathrm{h}\left(\mathrm{x}\right)\mathrm{dx}$

Here the revolution is around the line $x=-2$. So $r\left(x\right)=-2-x$.

Also the height is given by the function $h\left(x\right)=x^2+1$.

So that volume is described as :-

role="math" localid="1651586656308" $V=2\mathrm{\pi }{\int }_0^1\left(-2-\mathrm{x}\right)\left({\mathrm{x}}^2+1\right)\mathrm{dx}$

Step 3. Volume by using washers

The given function is :-

$f\left(x\right)=x^2+1$

By using the washers volume is described as

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

Where $R\left(x\right)$is outer radius and $r\left(x\right)$is inner radius.

From localid="1651645746182" $y=0to1$, $R\left(x\right)=2$and $r\left(x\right)=2$.

Also from $y=1to2$, $R\left(x\right)=2-\sqrt{y-1}$but $r\left(x\right)$is given by $2$.

So the required volume is described as :-

$$\begin{gathered} V=\mathrm{\pi }{\int }_0^1\left(2^2-2^2\right)\mathrm{dx}+\mathrm{\pi }{\int }_1^2\left({\left(2-\sqrt{\mathrm{y}-1}\right)}^2-2^2\right)\mathrm{dx} \\ \Rightarrow \mathrm{V}=0+\mathrm{\pi }{\int }_1^2\left({\left(2-\sqrt{\mathrm{y}-1}\right)}^2-2^2\right)\mathrm{dx} \\ \Rightarrow \mathrm{V}=\mathrm{\pi }{\int }_1^2\left({\left(2-\sqrt{\mathrm{y}-1}\right)}^2-2^2\right)\mathrm{dx} \end{gathered}$$

Step 4. Easier method

There need less calculations in shells method then disks method.

So the shells method is easier than disks method.