Chapter 13.5, Problem 48E

To determine

To graph: The system of inequalities $\{\begin{array}{c}{x}^2+y^2\le 81\\ x^2+y^2\ge 64\end{array}$ .

Explanation of Solution

Given information:

The inequalities $\{\begin{array}{c}{x}^2+y^2\le 81\\ x^2+y^2\ge 64\end{array}$ .

Graph:

Consider the inequalities $\{\begin{array}{c}{x}^2+y^2\le 81\\ x^2+y^2\ge 64\end{array}$ .

Take the first inequality $x^2+y^2\le 81$ .

Take the equality and construct the graph

  $x^2+y^2=81$ .

This represent a circle with center $\left(0,0\right)$ and radius 9.

Take a point say $\left(0,0\right)$ that does not lie on boundary of circle.

Substitute it in inequality $x^2+y^2\le 81$ .

  $\begin{array}{c}{0}^2+0^2\le 81\\ 0\le 81\end{array}$

This is true so shade the part that is toward the point $\left(0,0\right)$ .

That is shade the inner part of the circle and boundary will be solid.

Take the second inequality $x^2+y^2\ge 64$ .

Take the equality and construct the graph $x^2+y^2=64$ .

This represent a circle with center $\left(0,0\right)$ and radius 8.

Take a point say $\left(0,0\right)$ that does not lie on boundary of circle.

Substitute it in inequality $x^2+y^2\ge 64$ .

  $\begin{array}{c}{0}^2+0^2\ge 64\\ 0\ge 64\end{array}$

This is false so shade the part that is away from the point $\left(0,0\right)$ .

That is shade the outer part of the circle and boundary will be solid.

  

The blue color represents the inequality $x^2+y^2\ge 64$ , green color represents the inequality $x^2+y^2\le 81$ . The dark region represents the common region.

Now, take the common region to both the inequalities.

  

The red color region depicts the common region to inequalities $\{\begin{array}{c}{x}^2+y^2\le 81\\ x^2+y^2\ge 64\end{array}$ .

Interpretation:

The common region is a circle as it the common intersection of two circles $x^2+y^2=81$ and $x^2+y^2=64$ . The width of circle so obtained is 1 unit because outer radius is 9 and inner radius is 8.