Chapter 8.3, Problem 2E

a.

To determine

To find: The formula for the area $A\left(x\right)$ of the cross-0section of the solid that are perpendicular to the X-axis.

Answer to Problem 2E

The formula for the cross-section area is $A\left(x\right)=\pi x$

Explanation of Solution

Given information :

The solid lies between planes perpendicular to the X-axis at $x=0$ and $x=4$ .

The cross-section perpendicular to the X-axis between these planes runs from $y=-\sqrt{x}$ to $y=\sqrt{x}$ .

The cross-section are circular disk with diameter in the $xy$ plane.

  

Formula used :

Area of a circular disk of diameter $d=\pi {\left(\frac{d}{2}\right)}^2$

Calculation :

The cross-section perpendicular to the X-axis run from $y=-\sqrt{x}$ to $y=\sqrt{x}$ .

So the diameter of the circular disk be $\sqrt{x}-\left(-\sqrt{x}\right)=2\sqrt{x}$

So, the cross-section area,

  $A\left(x\right)=\pi {\left(\frac{2\sqrt{x}}{2}\right)}^2=\pi x$

Thus, the formula for the cross-section area is $A\left(x\right)=\pi x$

b.

To determine

To find: The formula for the area $A\left(x\right)$ of the cross-0section of the solid that are perpendicular to the X-axis

Answer to Problem 2E

The formula for the cross-section area is $A\left(x\right)=4x$ .

Explanation of Solution

Given information :

The solid lies between planes perpendicular to the X-axis at $x=0$ and $x=4$ .

The cross-section perpendicular to the X-axis between these planes runs from $y=-\sqrt{x}$ to $y=\sqrt{x}$ .

The cross-section are squares with base in $xy$ planes.

Formula used :

Area of a square of side $a=a^2$ .

Calculation:

The cross-section perpendicular to the X-axis run from $y=-\sqrt{x}$ to $y=\sqrt{x}$ .

So the length of the side of square be $\sqrt{x}-\left(-\sqrt{x}\right)=2\sqrt{x}$

The cross-section area is $A\left(x\right)={\left(2\sqrt{x}\right)}^2=4x$

Thus, the formula for the cross-section area is $A\left(x\right)=4x$ .

c.

To determine

To find: The formula for the area $A\left(x\right)$ of the cross-0section of the solid that are perpendicular to the X-axis.

Answer to Problem 2E

The formula for the cross-section area is $A\left(x\right)=2x$ .

Explanation of Solution

Given information :

The solid lies between planes perpendicular to the X-axis at $x=0$ and $x=4$ .

The cross-section perpendicular to the X-axis between these planes runs from $y=-\sqrt{x}$ to $y=\sqrt{x}$ .

The cross-section are square with diameter in the $xy$ plane.

Concept used :

The length of square of side “ $a$ ” be $2\sqrt{a}$ .

So, the area of the square is given by

  $\frac{2\sqrt{a}\times 2\sqrt{a}}{2}=\frac{{\left(Diagonal\right)}^2}{2}$

Calculation :

The cross-section perpendicular to the X-axis run from $y=-\sqrt{x}$ to $y=\sqrt{x}$ .

So the length of the side of square be $\sqrt{x}-\left(-\sqrt{x}\right)=2\sqrt{x}$

The cross-section area $A\left(x\right)=\frac{{\left(2\sqrt{x}\right)}^2}{2}=2x$

Thus, the formula for the cross-section area is $A\left(x\right)=2x$ .

d.

To determine

To find: The formula for the area $A\left(x\right)$ of the cross-0section of the solid that are perpendicular to the X-axis.

Answer to Problem 2E

The formula for the cross-section area is $A\left(x\right)=\sqrt{3}x$ .

Explanation of Solution

Given information :

The solid lies between planes perpendicular to the X-axis at $x=0$ and $x=4$ .

The cross-section perpendicular to the X-axis between these planes runs from $y=-\sqrt{x}$ to $y=\sqrt{x}$ .

The cross-section are equilateral triangles with diameter in the $xy$ plane.

Formula used :

The area of the equilateral triangle of side “ $a$ ” is $\frac{\sqrt{3}}{4}{a}^2$

The cross-section perpendicular to the X-axis run from $y=-\sqrt{x}$ to $y=\sqrt{x}$ .

So the length of the side of square be $\sqrt{x}-\left(-\sqrt{x}\right)=2\sqrt{x}$

The cross-section area $A\left(x\right)=\frac{\sqrt{3}}{4}{\left(2\sqrt{x}\right)}^2=\sqrt{3}x$

Thus, the formula for the cross-section area is $A\left(x\right)=\sqrt{3}x$ .