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 $x=2$, so to find the volume by shell method shells are drawn parallel to y-axis with average radius equal to $x-2$and the height of the shell is given by,

$$\begin{gathered} y=f\left(x\right) \\ =x-2 \end{gathered}$$

Therefore using the shell method

$$\begin{gathered} \text{Volume}=2\pi {\int }_2^5(x-2)(x-2)dx \\ =2\pi {\int }_2^5(x-2{)}^{2}dx \\ =2\pi {\left[\frac{{\displaystyle (x-2{)}^3}}{{\displaystyle 3}}\right]}_2^5 \\ =\frac{{\displaystyle 2\pi }}{{\displaystyle 3}}(\left[27\right]-\left[0\right]) \\ =18\pi \end{gathered}$$

Hence, the volume is $18\pi$.