Q 59 - Calculus

Question

Use definite integrals to find the volume of each solid of revolution described in Exercises $49-61$. (It is your choice whether to use disks/washers or shells in these exercises.)

The region bounded the graph of  $f\left(x\right)=\frac{2}{x}, y=0, y=3, x=0 and x=2$, revolved around the x-axis.

Step-by-Step Solution

Answer

The required volume by using shells is $V=10\mathrm{\pi }$.

Step 1. Given Information

We have given a function :- 
$f\left(x\right)=\frac{2}{x}$

We have to find the volume of region of graph of this function and $y=0,y=3,x=0 and x=2$, revolved around the x-axis.

Step 2. Find the integral and evaluate it to calculate volume

We know that by using shells the volume is given by :-

$V=2\mathrm{\pi }{\int }_c^d\mathrm{r}\left(\mathrm{y}\right)\mathrm{h}\left(\mathrm{y}\right)\mathrm{dy}$

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

From $y=0 to 1$ the height is $y=2$ and from $y=1 to 3$ the height is given by $\frac{2}{y}$.

So the volume is given by following :-

$$\begin{gathered} V=2\mathrm{\pi }{\int }_0^{1}2\mathrm{ydy}+2\mathrm{\pi }{\int }_1^3\mathrm{y}\times \frac{2}{\mathrm{y}}\mathrm{dy} \\ \Rightarrow \mathrm{V}=2\mathrm{\pi }{\int }_0^{1}2\mathrm{ydy}+2\mathrm{\pi }{\int }_1^{3}2\mathrm{dy} \\ \Rightarrow \mathrm{V}=2\mathrm{\pi }{\left[\frac{2{\mathrm{y}}^2}{2}\right]}_0^1+2\mathrm{\pi }{\left[2\mathrm{y}\right]}_1^3 \\ \Rightarrow \mathrm{V}=2\mathrm{\pi }\left[1-0\right]+2\mathrm{\pi }\left[6-2\right] \\ \Rightarrow \mathrm{V}=2\mathrm{\pi }+8\mathrm{\pi } \\ \Rightarrow \mathrm{V}=10\mathrm{\pi } \end{gathered}$$