Chapter 7, Problem 25RE

To determine

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

Answer to Problem 25RE

The volume of the solid generated is $V=\pi \left(2-\ln 3\right)$ .

Explanation of Solution

Given information:

  $\begin{array}{c}y=e^{x/2}\\ y=1\\ x=\ln 3\end{array}$

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 $x$ -axis and it utilizes the cylindrical shell method.

The limits of the integration are $x=0\text{ and }x=\ln 3$ .

Finding the area $A\left(x\right)$ of the cross section perpendicular to $x$ -axis on height $x\in \left[0,\ln 3\right]$ .

Since, the area $A\left(x\right)$ of the cross section perpendicular to $x$ -axis on height $x\in \left[0,\ln 3\right]$ is a circular region of radius $e^{x/2}$ with the circular region of the radius $1$ , cut from its center.

  $A\left(x\right)=\left(e^x-1\right)\pi$

The volume of the solid is given by,

  $\begin{array}{c}V={\displaystyle {\int }_0^{\ln 3}\left(e^x-1\right)\pi }dx\\ ={\left.\pi \left(e^x-x\right)\right|}_0^{\ln 3}\\ =\pi \left(e^{\ln 3}-\ln 3-1+0\right)\\ V=\pi \left(2-\ln 3\right)\end{array}$

Therefore, the volume of the solid generated is $V=\pi \left(2-\ln 3\right)$ .