Question

Consider the region between $f\left(x\right)=x-2$ and the x-axis on $[2,5]$. For each line of rotation given in Exercises 33–38, use the shell method to construct definite integrals to find the volume of the resulting solid.

Short answer

The volume is $18\pi$.

Step-by-step solution

Step 1. Given Information. 

We are given,

Step 2. Finding the Volume. 

As the region bounded by $f\left(x\right)=x-2$ and the x-axis from $x=2$, to $x=5$is rotated around the line $y=3$, so to find the volume by shell method shells of height given by $5-f^{-1}\left(y\right)$ are drawn parallel to the x-axis with average radius of $3-y$.

To find $f^{-1}\left(y\right)$solve $y=x-2$ for x gives

$x=y+2$

So $f^{-1}\left(y\right)=y+2$

So, the height of the shells is given by$5-(y+2)=3-y$

Therefore using the shell method,

$$\begin{gathered} \text{Volume}=2\pi {\int }_0^3(3-y)(3-y)dy \\ =2\pi {\left[\frac{{\displaystyle (3-y{)}^3}}{{\displaystyle 3}}\right]}_0^3 \\ =2\pi [-(0)+9] \\ =18\pi \end{gathered}$$

Hence, the volume is $18\pi$.