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 $9\pi$.

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 x-axis, 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 y.

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

$x=y+2$

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

Therefore using the shell method ,

$$\begin{gathered} \text{Volume}=2\pi {\int }_0^3\left(y\right)[5-(y+2\left)\right]dy \\ =2\pi {\int }_0^3\left(3y-y^2\right)dy \\ =2\pi {\left(\frac{{\displaystyle 3}}{{\displaystyle 2}}{y}^2-\frac{{\displaystyle y^3}}{{\displaystyle 3}}\right)}_0^3 \\ =2\pi \left(\left[\frac{{\displaystyle 27}}{{\displaystyle 2}}-9\right]-\left[0\right]\right) \\ =9\pi \end{gathered}$$

Hence, the volume is $9\pi$.