Chapter 7.3, Problem 42E

a.

To determine

To find:The volume of the given solid.

Answer to Problem 42E

The volume of the solidis $V=\pi \sqrt{3}-\frac{{\pi }^2}{6}$ .

Explanation of Solution

Given information:

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

Calculation:

The cross-sections are round discs with diameters extending from the curve, perpendicular to the $x$ -axis $y=\tan x$ to the curve $y=\sec x$ .

The formula to find the volume is,

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

The area of circle is given by,

  $A=\frac{\pi {\left(\text{diameter}\right)}^2}{4}$

Where,

  $\text{diameter}=\sec x-\tan x$

Here, $\sec x$ is greater than the $\tan x$ .

Now substituting the values to get,

  $\begin{array}{c}A=\frac{\pi {\left(\text{diameter}\right)}^2}{4}\\ =\frac{\pi {\left(\sec x-\tan x\right)}^2}{4}\\ =\frac{\pi \left({\sec }^{2}x-2\sec x\tan x+{\tan }^{2}x\right)}{4}\\ =\frac{\pi \left({\sec }^{2}x-2\sec x\tan x+{\sec }^{2}x-1\right)}{4}\\ A=\frac{\pi \left(2{\sec }^{2}x-2\sec x\tan x-1\right)}{4}\end{array}$

Now substituting the value of $A$ in volume formula to get,

  $\begin{array}{c}V={\displaystyle {\int }_a^{b}A\left(x\right)dx}\\ ={\displaystyle {\int }_a^b\frac{\pi \left(2{\sec }^{2}x-2\sec x\tan x-1\right)}{4}}dx\\ ={\displaystyle {\int }_{\frac{-\pi }{3}}^{\frac{\pi }{3}}\frac{\pi \left(2{\sec }^{2}x-2\sec x\tan x-1\right)}{4}dx}\end{array}$

Applying the distributing property,

  $\begin{array}{c}V={\displaystyle {\int }_{\frac{-\pi }{3}}^{\frac{\pi }{3}}\frac{\pi \left(2{\sec }^{2}x-2\sec x\tan x-1\right)}{4}dx}\\ ={\displaystyle {\int }_{\frac{-\pi }{3}}^{\frac{\pi }{3}}\frac{\pi }{2}\left({\sec }^{2}x\right)dx-{\displaystyle {\int }_{\frac{-\pi }{3}}^{\frac{\pi }{3}}\frac{\pi }{2}\left(\sec x\tan x\right)dx}}-{\displaystyle {\int }_{\frac{-\pi }{3}}^{\frac{\pi }{3}}\frac{\pi }{4}dx}\\ ={\left.\frac{\pi }{2}\left(\tan x\right)\right|}_{\frac{-\pi }{3}}^{\frac{\pi }{3}}-{\left.\frac{\pi }{2}\left(\sec x\right)\right|}_{\frac{-\pi }{3}}^{\frac{\pi }{3}}-{\left.\frac{\pi }{2}x\right|}_{\frac{-\pi }{3}}^{\frac{\pi }{3}}\\ =\frac{\pi }{2}\left(\sqrt{3}\right)+\frac{\pi }{2}\left(\sqrt{3}\right)-\left[\frac{\pi }{2}\left(2\right)-\frac{\pi }{2}\left(2\right)\right]-\left[\frac{{\pi }^2}{12}+\frac{{\pi }^2}{12}\right]\\ V=\pi \sqrt{3}-\frac{{\pi }^2}{6}\end{array}$

Therefore, the volume of the equilateral triangle is $V=\pi \sqrt{3}-\frac{{\pi }^2}{6}$ .

b.

To determine

To find: The volume of the square.

Answer to Problem 42E

The volume of the square is $V=4\sqrt{3}-\frac{2\pi }{3}$ .

Explanation of Solution

Given information:

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

Calculation:

The cross-sections are round discs with diameters extending from the curve, perpendicular to the $x$ -axis $y=\tan x$ to the curve $y=\sec x$ .

The formula to find the volume is,

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

The area of square is given by,

  $A={\left(\text{side}\right)}^2$

Where,

  $\text{side}=\sec x-\tan x$

Here, $\sec x$ is greater than the $\tan x$ .

Now substituting the values to get,

  $\begin{array}{c}A={\left(\text{side}\right)}^2\\ ={\left(\sec x-\tan x\right)}^2\\ =\left({\sec }^{2}x-2\sec x\tan x+{\tan }^{2}x\right)\\ =\left({\sec }^{2}x-2\sec x\tan x+{\sec }^{2}x-1\right)\\ A=\left(2{\sec }^{2}x-2\sec x\tan x-1\right)\end{array}$

Now substituting the value of $A$ in volume formula to get,

  $\begin{array}{c}V={\displaystyle {\int }_a^{b}A\left(x\right)dx}\\ ={\displaystyle {\int }_a^b\left(2{\sec }^{2}x-2\sec x\tan x-1\right)}dx\\ ={\displaystyle {\int }_{\frac{-\pi }{3}}^{\frac{\pi }{3}}\left(2{\sec }^{2}x-2\sec x\tan x-1\right)dx}\end{array}$

Applying the distributing property,

  $\begin{array}{c}V={\displaystyle {\int }_{\frac{-\pi }{3}}^{\frac{\pi }{3}}\left(2{\sec }^{2}x-2\sec x\tan x-1\right)dx}\\ =2{\displaystyle {\int }_{\frac{-\pi }{3}}^{\frac{\pi }{3}}\left({\sec }^{2}x\right)dx-2{\displaystyle {\int }_{\frac{-\pi }{3}}^{\frac{\pi }{3}}\left(\sec x\tan x\right)dx}}-{\displaystyle {\int }_{\frac{-\pi }{3}}^{\frac{\pi }{3}}1dx}\\ ={\left.2\left(\tan x\right)\right|}_{\frac{-\pi }{3}}^{\frac{\pi }{3}}-{\left.2\left(\sec x\right)\right|}_{\frac{-\pi }{3}}^{\frac{\pi }{3}}-{\left.x\right|}_{\frac{-\pi }{3}}^{\frac{\pi }{3}}\\ =2\left(\tan \frac{\pi }{3}\right)-2\left(\tan -\frac{\pi }{3}\right)-\left[2\left(\sec \frac{\pi }{3}\right)-2\left(\sec -\frac{\pi }{3}\right)\right]-\left[\left(\frac{\pi }{3}\right)-\left(-\frac{\pi }{3}\right)\right]\\ =2\left(\sqrt{3}\right)+2\left(\sqrt{3}\right)-\left[2-2\right]-\left[\frac{2\pi }{3}\right]\\ V=4\sqrt{3}-\frac{2\pi }{3}\end{array}$

Therefore, the volume of the square is $V=4\sqrt{3}-\frac{2\pi }{3}$ .