Problem 11
Question
In Exercises 11-14, describe the region in space defined by the inequalities. \(x^{2}+y^{2}+z^{2}<1\)
Step-by-Step Solution
Verified Answer
The region is an open ball (interior of a sphere) of radius 1 centered at the origin.
1Step 1: Analyzing the Inequality
The inequality \(x^2 + y^2 + z^2 < 1\) represents a geometric shape. We begin by comparing it to the equation \(x^2 + y^2 + z^2 = 1\), which defines a sphere centered at the origin (0, 0, 0) with a radius of 1.
2Step 2: Understanding the Sphere's Properties
The inequality uses the spherical equation \(x^2 + y^2 + z^2 = 1\) to describe a complete sphere. Therefore, for the inequality \(x^2 + y^2 + z^2 < 1\), it defines the interior of this sphere. This means it includes all the points inside the sphere but excludes those on the surface.
3Step 3: Defining the Region in Space
The inequality \(x^2 + y^2 + z^2 < 1\) describes a region in space that contains all points within a sphere of radius 1, centered at the origin. It is not just the surface; it includes everything inside up to just before the boundary. Hence, the region is an open ball with radius 1 centered at the origin.
Key Concepts
SpheresInequalitiesOpen Ball
Spheres
A sphere is a perfectly symmetrical geometric shape. It is defined as the set of all points in three-dimensional space that are at a fixed distance, known as the radius, from a given point called the center. Imagine the globe we live on; it's an example of a three-dimensional sphere. In mathematical terms, if the sphere's center is at the origin
- the center of the coordinate system (0, 0, 0),
- the equation would be \[ x^2 + y^2 + z^2 = r^2 \]
- where \(r\) is the radius.
Inequalities
When dealing with mathematical expressions, inequalities are statements that one quantity is less than or greater than another. They help in defining ranges and boundaries within mathematical spaces. Let's say you have an inequality like \[ x^2 + y^2 + z^2 < 1 \]. This inequality means that we are looking at the set of all points that are strictly inside a unit sphere, not touching its surface. In other words, all combinations of \(x, y, z\) such that
- their squared sum is less than 1
- fit inside this region.
Open Ball
An open ball is a concept that describes a region in space, particularly within the field of metric spaces, without including the boundary. Think of it as a sphere that lacks its outer shell. In three-dimensional geometry, if you have a center (say at the origin) and a radius, the space inside this boundary is termed an 'open ball'. Consider the inequality \[ x^2 + y^2 + z^2 < 1 \], which defines an open ball around the origin with a radius of 1. It encompasses
- every point inside the spherical space,
- omitting those exactly on its boundary.
Other exercises in this chapter
Problem 11
Create your own vectors \(\vec{u}, \vec{v}\) and \(\vec{w}\) in \(\mathbb{R}^{2}\) and show that \(\vec{u} \cdot(\vec{v}+\vec{w})=\vec{u} \cdot \vec{v}+\vec{u}
View solution Problem 11
Let \(\vec{u}=\langle 1,-2\rangle\) and \(\vec{v}=\langle 1,1\rangle .\) (a) Find \(\vec{u}+\vec{v}, \vec{u}-\vec{v}, 2 \vec{u}-3 \vec{v} .\) (b) Sketch the abo
View solution Problem 12
Give the equation of the described plane in standard and general forms. Contains the intersecting lines \(\vec{\ell}_{1}(t)=\langle 5,0,3\rangle+t\langle-1,1,1\
View solution Problem 12
Write the vector, parametric and symmetric equations of the lines described. Passes through the point of intersection of \(\ell_{1}(t)\) and \(\ell_{2}(t)\) and
View solution