Chapter 9.4, Problem 55E

a.

To determine

Tofind: the area of a typical cross section.

Answer to Problem 55E

the area of a typical cross section is $A=\frac{\pi }{4}\left(e^{2x}\right)$ .

Explanation of Solution

Given information:

The figure provided in the question:

  

Calculation: to find the area of the cross section area, the radius of the cross section is,

  $r=\frac{e^x}{2}$

Area of the cross section is :

  $\begin{array}{l}A=\pi {\left(r\right)}^2\\ A=\pi {\left(\frac{e^x}{2}\right)}^2\text{ [substitute]}\\ A=\frac{\pi }{4}\left(e^{2x}\right)\end{array}$

Thus, the area of the cross section is $A=\frac{\pi }{4}\left(e^{2x}\right)$ .

b.

To determine

To find: the expression volume of the horn as an improper integral.

Answer to Problem 55E

the volume expression of the horn is $V={\displaystyle {\int }_{-\infty }^{\ln 2}\frac{\pi }{4}\left(e^{2x}\right)dx}$ .

Explanation of Solution

Given information:

  

Calculation:to find the volume expression, integrate the area of the cross section area from $-\infty \text{ to }\ln 2$ ,

  $\begin{array}{l}V={\displaystyle {\int }_{-\infty }^{\ln 2}A\left(x\right)dx}\\ V={\displaystyle {\int }_{-\infty }^{\ln 2}\frac{\pi }{4}\left(e^{2x}\right)dx\text{ [substitute]}}\end{array}$

Thus, the expression is $V={\displaystyle {\int }_{-\infty }^{\ln 2}\frac{\pi }{4}\left(e^{2x}\right)dx}$ .

c.

To determine

To find: the volume of the horn.

Answer to Problem 55E

The volume of the horn is $\frac{\pi }{2}$ .

Explanation of Solution

Given information:

  

Calculation: to find the volume of the horn, evaluate the expression of the volume,

  $\begin{array}{l}V={\displaystyle {\int }_{-\infty }^{\ln 2}\frac{\pi }{4}\left(e^{2x}\right)dx}\\ V=\frac{\pi }{4}{\left[\frac{e^{2x}}{2}\right]}_{-\infty }^{\ln 2}\text{ [integrate and simplify]}\\ V=\frac{\pi }{4}\left(\frac{e^{2\left(\ln 2\right)}}{2}-\frac{e^{2\left(-\infty \right)}}{2}\right)\\ V=\frac{\pi }{4}\left(2-0\right)\text{ [simplify]}\\ V=\frac{\pi }{2}\end{array}$

Thus, the volume is $\frac{\pi }{2}$ .