Chapter 8.3, Problem 40E

a.

To determine

To find: The find the volume of given solid.

Answer to Problem 40E

The volume is 3.79646 Unit².

Explanation of Solution

Given information :

The solid lies between planes perpendicular to the X-axis at $x=-\frac{\pi }{3}$ and $x=\frac{\pi }{3}$ .

The cross-section perpendicular to the X-axis and circular disks with diameters running from the curve $y=\tan x$ to the curve $y=\sec x$ .

Formula used : The volume of a solid of known integrable cross-section area $A\left(x\right)$ from $x=a$ to $x=b$ is the integral of $A$ from $a$ to $b$ ,

  $V={\displaystyle \underset{a}{\overset{b}{\int }}A\left(x\right)dx}$ .

The area of a circle of diameter “ $d$ ” is $\pi {\left(\frac{d}{2}\right)}^2$ Unit²

Calculation :

The cross-section perpendicular to X-axis is circular disk.

So, area is,

  $A\left(x\right)=\pi {\left(\frac{\sec x-\tan x}{2}\right)}^2$

Volume of the solid is,

So, the required volume is 3.79646 Unit².

b.

To determine

To find: The find the volume of given solid.

Answer to Problem 40E

The volume of the solid is approximately $4.8338$ Unit³.

Explanation of Solution

Given information : The solid lies between planes perpendicular to the X-axis at $x=-\frac{\pi }{3}$ and $x=\frac{\pi }{3}$ .

The cross-section perpendicular to the X-axis and circular disks with diameters running from the curve $y=\tan x$ to the curve $y=\sec x$ .

Formula used : The volume of a solid of known integrable cross-section area $A\left(x\right)$ from $x=a$ to $x=b$ is the integral of $A$ from $a$ to $b$ ,

  $V={\displaystyle \underset{a}{\overset{b}{\int }}A\left(x\right)dx}$ .

The area of the Square of side “ $a$ ” is $a^2$ Unit²

The cross section perpendicular to X-axis is a square.

So, the area $A\left(x\right)={\left(\sec x-\tan x\right)}^2$

Volume of the solid is,

The volume of the solid is approximately $4.8338$ Unit³.