Q 22 - 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 $f\left(x\right)=\frac{1}{x}$ and the lines  $y=0, y=1, x=0 and x=2$, revolved around the y-axis.

Step-by-Step Solution

Answer

By using shells the volume is described as $\mathrm{V}=2\mathrm{\pi }{\int }_0^1\mathrm{xdx}+2\mathrm{\pi }{\int }_1^2\mathrm{dx}$

By using disks the volume is described as $V=\mathrm{\pi }{\int }_0^{\frac{1}{2}}1\mathrm{dy}+\mathrm{\pi }{\int }_{\frac{1}{2}}^1\frac{1}{{\mathrm{y}}^2}\mathrm{dy}$

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

Step 1. Given Information

We have given the following function :- 

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

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

Step 2. Volume by using Shells

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

The revolution is around y-axis. So that $r\left(x\right)=x$ and  from $x=0 to 1$height is given by $h\left(x\right)=1$ and from $x=1 to 2$ height is given by $h\left(x\right)=\frac{1}{x}$

So that volume is described as :-

$$\begin{gathered} V=2\mathrm{\pi }{\int }_0^1\mathrm{xdx}+2\mathrm{\pi }{\int }_1^2\mathrm{x}\times \frac{1}{\mathrm{x}}\mathrm{dx} \\ \Rightarrow \mathrm{V}=2\mathrm{\pi }{\int }_0^1\mathrm{xdx}+2\mathrm{\pi }{\int }_1^2\mathrm{dx} \end{gathered}$$

Step 3. Volume by using disks

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 :-

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

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

So that volume is described as :-

$$\begin{gathered} V=\mathrm{\pi }{\int }_0^{\frac{1}{2}}2\mathrm{dy}+\mathrm{\pi }{\int }_{\frac{1}{2}}^1{\left(\frac{1}{\mathrm{y}}\right)}^2\mathrm{dy} \\ \Rightarrow V=\mathrm{\pi }{\int }_0^{\frac{1}{2}}2\mathrm{dy}+\mathrm{\pi }{\int }_{\frac{1}{2}}^1\frac{1}{{\mathrm{y}}^2}\mathrm{dy} \end{gathered}$$      

Step 4. Easier method

There need less calculations in shells method then disks method.

So the shells method is easier than disks method.