Problem 48 - Calculus Early Transcendentals

Question

Find the domain of the following functions. If possible, give a description of the domains (for example, all points outside a sphere of radius 1 centered at the origin ).$$p(x, y, z)=\sqrt{x^{2}+y^{2}+z^{2}-9}.$$

Step-by-Step Solution

Answer

$$ Answer: The domain of the function is the set of all points \((x, y, z)\) that lie on or outside a sphere of radius 3 centered at the origin.

Step 1: Identify the condition for the function to be defined

Since we are dealing with a square root function, the term under the square root (also called the radicand) should be non-negative (equal to or greater than zero) for the function to be real and defined. So, we must determine the values of \(x\), \(y\), and \(z\) that satisfy the inequality: $$x^{2}+y^{2}+z^{2}-9 \ge 0$$

Step 2: Rearrange the inequality

Let's add 9 to both sides of the inequality to have the more recognizable form of the equation of a sphere: $$x^{2}+y^{2}+z^{2} \ge 9$$

Step 3: Interpret the inequality geometrically

The inequality $$x^{2}+y^{2}+z^{2} \ge 9$$ can be recognized as the equation of a sphere in three dimensions. This inequality represents the set of all points \((x, y, z)\) for which their distance squared from the origin \((0, 0, 0)\) is equal to or greater than 9. In other words, the points lie on or outside a sphere of radius 3 centered at the origin.

Step 4: Describe the domain

Now that we have found out the condition for the function to be defined, we can describe the domain as the set of all points \((x, y, z)\) that lie outside a sphere of radius 3 centered at the origin. This will be the domain of the function \(p(x, y, z)\).

Key Concepts

Sphere

When we talk about a sphere in three dimensions, we are referring to a perfectly round shape, much like a 3D circle. For any sphere, you need a center point, and a radius, which is the distance from the center to any point on the surface. The general formula for a sphere centered at the origin \(0, 0, 0\) is \[ x^2 + y^2 + z^2 = r^2 \]where \(r\) is the radius.
This formula tells us the location of all points that are exactly \(r\) units away from the origin.

• If the points satisfy \(x^2 + y^2 + z^2 = 9\), they lie exactly on a sphere with radius 3.
• If \(x^2 + y^2 + z^2 > 9\), the points lie outside this sphere.

Understanding this helps us visualize domains for functions involving spheres.

Square root function

The square root function, denoted as \sqrt{ \ }\, has specific requirements to be defined. Specifically, the expression inside the square root (called the radicand) must be non-negative. In mathematical terms, the radicand should be \(\ge 0 \). This ensures that the output of the function is a real number.
For the function \(p(x, y, z) = \sqrt{x^2 + y^2 + z^2 - 9}\):

• The radicand is \(x^2 + y^2 + z^2 - 9 \).
• It should satisfy \(x^2 + y^2 + z^2 - 9 \ge 0\).

This leads us to the inequality \(x^2 + y^2 + z^2 \ge 9\), meaning our function is only defined for values of \(x\), \(y\), and \(z\) that satisfy this condition. Understanding this is key to finding the domain of square root functions in three dimensions.

Inequality in three dimensions

In 3D geometry, inequalities help define regions of space rather than specific points. The inequality \(x^2 + y^2 + z^2 \ge 9\) represents all points \(x, y, z\) that lay on or outside a sphere of radius 3. Such problems are often discussed in physics and engineering when defining boundaries.

• The boundary is the surface of the sphere, where \(x^2 + y^2 + z^2 = 9\).
• The region includes external points where \(x^2 + y^2 + z^2 > 9\).

This kind of inequality helps visualize regions like areas of influence, or where particular actions or conditions are applicable. It's a powerful concept in analyzing the topology of spaces, especially within calculus and optimization problems.