Chapter 8.3, Problem 1E

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 1E

The formula for the cross-section area is $A\left(x\right)=\pi \left(1-x^2\right)$

Explanation of Solution

Given information :

The solid lies between planes perpendicular to the X-axis at $x=-1$ and $x=1$ .

The cross-section perpendicular to the X-axis between these planes runs from $y=-\sqrt{1-x^2}$ to $y=\sqrt{1-x^2}$ .

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 $y=-\sqrt{1-x^2}$ to $y=\sqrt{1-x^2}$ .

So the diameter of the circular disk be $\sqrt{1-x^2}-\left(-\sqrt{1-x^2}\right)=2\sqrt{1-x^2}$

So, the cross-section area,

  $A\left(x\right)=\pi {\left(\frac{2\sqrt{1-x^2}}{2}\right)}^2=\pi \left(1-x^2\right)$

Thus, the formula for the cross-section area is $A\left(x\right)=\pi \left(1-x^2\right)$

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 1E

The formula for the cross-section area is $A\left(x\right)=4\left(1-x^2\right)$ .

Explanation of Solution

Given information :

The solid lies between planes perpendicular to the X-axis at $x=-1$ and $x=1$ .

The cross-section perpendicular to the X-axis between these planes runs from $y=-\sqrt{1-x^2}$ to $y=\sqrt{1-x^2}$ .

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

  

Formula used :

Area of a square of side “ $a$ ” is $a^2$

Calculation:

The cross-section perpendicular to the X-axis run from $y=-\sqrt{1-x^2}$ to $y=\sqrt{1-x^2}$ .

So the diameter of the circular disk be $\sqrt{1-x^2}-\left(-\sqrt{1-x^2}\right)=2\sqrt{1-x^2}$

So, the cross-section area,

  $A\left(x\right)={\left(2\sqrt{1-x^2}\right)}^2=4\left(1-x^2\right)$

Thus, the formula for the cross-section area is $A\left(x\right)=4\left(1-x^2\right)$

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 1E

The formula for the cross-section area is $A\left(x\right)=2\left(1-x^2\right)$ .

Explanation of Solution

Given information :

The solid lies between planes perpendicular to the X-axis at $x=-1$ and $x=1$ .

The cross-section perpendicular to the X-axis between these planes runs from $y=-\sqrt{1-x^2}$ to $y=\sqrt{1-x^2}$ .

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

  

Concept used :

The length of square’s diagonal is $\sqrt{2}$ times the length of its side.

So, the formula for area of a square can be written as $\frac{{\left(Diagonal\right)}^2}{2}$ .

Calculation :

The cross-section perpendicular to the X-axis run from $y=-\sqrt{1-x^2}$ to $y=\sqrt{1-x^2}$ .

So, length of the diagonal of the square be $\sqrt{1-x^2}-\left(-\sqrt{1-x^2}\right)=2\sqrt{1-x^2}$

So, the cross-section area,

  $A\left(x\right)=\frac{{\left(2\sqrt{1-x^2}\right)}^2}{2}=2\left(1-x^2\right)$

Thus, the formula for the cross-section area is $A\left(x\right)=2\left(1-x^2\right)$

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 1E

The formula for the cross-section area is $A\left(x\right)=\frac{\sqrt{3}}{2}\sqrt{1-x^2}$ .

Explanation of Solution

Given information :

The solid lies between planes perpendicular to the X-axis at $x=-1$ and $x=1$ .

The cross-section perpendicular to the X-axis between these planes runs from $y=-\sqrt{1-x^2}$ to $y=\sqrt{1-x^2}$ .

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 $y=-\sqrt{1-x^2}$ to $y=\sqrt{1-x^2}$ .

So, length of the diagonal of the square be $\sqrt{1-x^2}-\left(-\sqrt{1-x^2}\right)=2\sqrt{1-x^2}$

So, the cross-section area,

  $A\left(x\right)=\frac{\sqrt{3}}{4}\left(2\sqrt{1-x^2}\right)=\frac{\sqrt{3}}{2}\sqrt{1-x^2}$

Thus, the formula for the cross-section area is $A\left(x\right)=\frac{\sqrt{3}}{2}\sqrt{1-x^2}$