Q. 36 - 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 $27\pi$. 

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 $x=5$ is rotated around the line $y=-2$, 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 $y-(-2)=y+2$.

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

$$\begin{gathered} \text{ Volume }=2\pi {\int }_0^3(y+2)(3-y)dy \\ =2\pi {\int }_0^3\left(y+6-y^2\right)dy \\ =2\pi {\left[\frac{y^2}{2}+6y-\frac{y^3}{3}\right]}_0^3 \\ =2\pi \left[\left(\frac{9}{2}+18-9\right)-\left(0\right)\right] \\ =27\pi \end{gathered}$$

Hence, the volume is $27\pi$.