Chapter 7, Problem 45RE

To determine

To find:The volume of the solid generated by resolving the region bounded by the curves.

Answer to Problem 45RE

The volume of the solid generated is $V=3\pi \text{ cu}\text{. unit}$ .

Explanation of Solution

Given information:

  $\begin{array}{c}y=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x$}\right.\\ y=0\\ x=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ x=2\end{array}$

Formula used:

The formula for cylindrical shell is,

  $V={\displaystyle {\int }_a^{b}2\pi xf\left(x\right)dx}$

Calculation:

Let's graph thegiven area and draw some conclusions based on it.

  

Taking into consideration the integration with respect to $x$ because the region revolves around the $y$ -axis and it utilizes the cylindrical shell method.

The limits of the integration are $x=\frac{1}{2}\text{ and }x=2$ .

The volume of the solid is given by,

  $\begin{array}{c}V={\displaystyle {\int }_a^{b}2\pi x}f\left(x\right)dx\\ ={\displaystyle {\int }_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{2}2\pi x\cdot \frac{1}{x}dx}\\ =2\pi {\displaystyle {\int }_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{2}dx}\\ ={\left.2\pi x\right|}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^2\\ V=3\pi \end{array}$

Therefore, the volume of the solid generated is $V=3\pi \text{ cu}\text{. unit}$ .