Q 23 - 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 $f\left(x\right)=x^2+1$ an the x-axis on $\left[0,1\right]$, revolved around the line $x=3$.

Step-by-Step Solution

Answer

By using shells the volume is described as $\mathrm{V}=2\mathrm{\pi }{\int }_0^1(3-\mathrm{x})\left({\mathrm{x}}^2+1\right)\mathrm{dx}$

By using disks the volume is described as $V=\mathrm{\pi }{\int }_0^1\left(3^2-2^2\right)\mathrm{dy}+\mathrm{\pi }{\int }_1^2\left({\left(3-\sqrt{\mathrm{y}-1}\right)}^2-2^2\right)\mathrm{dy}$

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

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 $x=3$ 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 $\mathrm{x}=3$. So $\mathrm{r}\left(\mathrm{x}\right)=3-\mathrm{x}$.

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

So that volume is described as :-

$\mathrm{V}=2\mathrm{\pi }{\int }_0^1(3-\mathrm{x})\left({\mathrm{x}}^2+1\right)\mathrm{dx}$

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 $y=0 to 1$, $R\left(x\right)=3$ and $r\left(x\right)=2$.

Also from $y=1 to 2$, $R\left(x\right)=3-\sqrt{y-1}$ but $r\left(x\right)$ is given by $2$.

So the required volume is described as :-

$V=\mathrm{\pi }{\int }_0^1\left(3^2-2^2\right)\mathrm{dy}+\mathrm{\pi }{\int }_1^2\left({\left(3-\sqrt{\mathrm{y}-1}\right)}^2-2^2\right)\mathrm{dy}$

Step 4. Easier method

There need less calculations in shells method then disks method.

So the shells method is easier than disks method.