Problem 33 - Thomas Calculus

Question

Write inequalities to describe the sets in Exercises \(29-34\) The (a) interior and (b) exterior of the sphere of radius 1 centered at the point \((1,1,1)\)

Step-by-Step Solution

Answer

Interior: \((x-1)^2 + (y-1)^2 + (z-1)^2 < 1\); Exterior: \((x-1)^2 + (y-1)^2 + (z-1)^2 > 1\).

Step 1: Understanding the Problem

We are asked to describe the interior and exterior of a sphere using inequalities. The sphere has a radius of 1 and is centered at the point \((1, 1, 1)\). The interior is the space inside the sphere, excluding the surface, and the exterior is the space outside, excluding the surface.

Step 2: Equation of the Sphere

The equation of a sphere with radius \(r\) centered at \((h, k, l)\) is given by:\[(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\]Substituting \(r = 1\) and \((h, k, l) = (1, 1, 1)\), we find the equation of the sphere:\[(x - 1)^2 + (y - 1)^2 + (z - 1)^2 = 1\]

Step 3: Describing the Interior of the Sphere

The interior of the sphere includes all points where the sum of the squared differences from the center is less than the square of the radius. Therefore, the inequality describing the interior is:\[(x - 1)^2 + (y - 1)^2 + (z - 1)^2 < 1\]

Step 4: Describing the Exterior of the Sphere

The exterior includes all points where that sum is greater than the square of the radius. The inequality for the exterior is:\[(x - 1)^2 + (y - 1)^2 + (z - 1)^2 > 1\]

Key Concepts

Inequalities

Inequalities are a key concept in mathematics that allow us to express a range of values or describe regions where certain conditions are met. In the context of a sphere, inequalities help delineate the boundaries of its interior and exterior. For example, when dealing with a sphere centered at \(1,1,1\) with a radius of 1, we can use inequalities to identify all points inside (interior) or outside (exterior) the sphere.
In the given exercise, inequalities are used in two specific ways:

• Interior of the sphere: The inequality \( (x - 1)^2 + (y - 1)^2 + (z - 1)^2 < 1 \) describes points whose Euclidean distance from the center of the sphere is less than 1.
• Exterior of the sphere: The inequality \( (x - 1)^2 + (y - 1)^2 + (z - 1)^2 > 1 \) encompasses all points located further from the center than the sphere's surface.

This method of using inequalities allows us to work with complex three-dimensional spaces in a manageable and understandable way.

Interior and Exterior

The concepts of interior and exterior are crucial when working with three-dimensional objects like spheres. The interior refers to the space where points are located inside the sphere but not on its surface. Meanwhile, the exterior is the space where points lie outside the sphere.
Let's delve a bit deeper into each concept:

• Interior of the Sphere: For any sphere, its interior is the region containing all points with a distance from the center that is less than the radius. This is visualized as the volume within the sphere. In this exercise, this region is described using the inequality \( (x - 1)^2 + (y - 1)^2 + (z - 1)^2 < 1 \), ensuring points are strictly inside and not on the boundary.
• Exterior of the Sphere: The exterior surrounds the sphere, encompassing all points farther from the center than those precisely on the sphere. For our scenario, the inequality \( (x - 1)^2 + (y - 1)^2 + (z - 1)^2 > 1 \) excludes both the interior and the sphere’s surface.

Understanding these definitions helps clarify how we can partition space into meaningful categories according to distance, which is particularly useful in fields such as physics and engineering.

Three-Dimensional Geometry

Three-dimensional geometry involves the study of shapes and figures in a space that has three dimensions: length, width, and height. Unlike two-dimensional geometry, which deals with flat shapes like circles and squares, three-dimensional geometry encompasses objects like spheres, cubes, and pyramids.
A sphere, one of the simplest three-dimensional objects, is perfectly symmetrical and has all points on its surface equidistant from a central point, its center. This exercise provides a practical application of three-dimensional geometry through understanding and visualizing the properties of a sphere centered at \(1,1,1\) with a radius of 1.
Three-dimensional geometry is important because:

• It allows us to represent and calculate the properties of real-world objects, such as volume and surface area.
• It is fundamental for various scientific and engineering disciplines, where modeling objects in space is essential.
• It provides a foundation for more advanced topics in mathematics such as topology and complex systems analysis.

Through exercises like this, students can better grasp how mathematical principles apply to the world beyond two-dimensional constraints, offering insights into how geometry functions in more realistic settings.