Q 58 - Calculus

Question

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

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

Step-by-Step Solution

Answer

The required volume by using shells is $V\approx 12.5\mathrm{\pi }$.

Step 1. Given Information

We have given a function :-

$f\left(x\right)=2\ln x$

We have to find the volume of region of graph of this function and  $y=0, y=3 and x=0$, 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 :-

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

Here axis of revolution is  $x-axis$. So $r\left(y\right)=y$ and height is given by the function $h\left(y\right)=e^{\frac{y}{2}}$.

Then we get the volume as following :-

$$\begin{gathered} V=2\mathrm{\pi }{\int }_0^3\mathrm{y}\left({\mathrm{e}}^{\frac{\mathrm{y}}{2}}\right)\mathrm{dy} \\ \Rightarrow \mathrm{V}=2\mathrm{\pi }{\left[\frac{{\mathrm{y}}^2}{2}{\mathrm{e}}^{\frac{\mathrm{y}}{2}}-4{\mathrm{e}}^{\frac{\mathrm{y}}{2}}\right]}_0^3 \\ \Rightarrow \mathrm{V}=2\mathrm{\pi }\left[\frac{9}{2}{\mathrm{e}}^{\frac{3}{2}}-4{\mathrm{e}}^{\frac{3}{2}}-\left(0-4\right)\right] \\ \Rightarrow \mathrm{V}=2\mathrm{\pi }\left[\frac{1}{2}{\mathrm{e}}^{\frac{3}{2}}+4\right] \\ \Rightarrow \mathrm{V}=2\mathrm{\pi }\left[\frac{{\mathrm{e}}^{\frac{3}{2}}+8}{2}\right] \\ \Rightarrow \mathrm{V}\approx \mathrm{\pi }\left(4.5+8\right) \\ \Rightarrow \mathrm{V}\approx 12.5\mathrm{\pi } \end{gathered}$$