Q. 33 - Calculus

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. 

Step-by-Step Solution

Answer

The volume is $36\pi$. 

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 $x=2$ is rotated around the y-axis, so to find the volume by shell method shells are drawn parallel to y-axis with average radius equal to x and the height of the shell is given by $y=f\left(x\right)=x-2$

Therefore using the shell method

$$\begin{gathered} \text{ Volume }=2\pi {\int }_2^5\left(x\right)(x-2)dx \\ =2\pi {\int }_2^5\left(x^2-2x\right)dx \\ =2\pi {\left[\frac{{\displaystyle x^3}}{{\displaystyle 3}}-x^2\right]}_2^5 \\ =2\pi \left(\left[\frac{{\displaystyle 125}}{{\displaystyle 3}}-25\right]-\left[\frac{{\displaystyle 8}}{{\displaystyle 3}}-4\right]\right) \\ =36\pi \end{gathered}$$

Hence, the volume is $36\pi$.