Problem 19
Question
In Exercises \(17-24,\) describe the sets of points in space whose coordinates satisfy the given inequalities or combinations of equations and inequalities. $$ \text { a. } x^{2}+y^{2}+z^{2} \leq 1 \quad \text { b. } x^{2}+y^{2}+z^{2}>1 $$
Step-by-Step Solution
Verified Answer
a: Solid sphere of radius 1 at origin; b: Exterior of sphere with radius 1.
1Step 1: Analyze the Equation for Part (a)
The inequality \( x^2 + y^2 + z^2 \leq 1 \) represents the space of points whose distance from the origin (0,0,0) is less than or equal to 1. Geometrically, this is described as a solid sphere including the surface, centered at the origin with a radius of 1.
2Step 2: Analyze the Equation for Part (b)
The inequality \( x^2 + y^2 + z^2 > 1 \) represents the space of points whose distance from the origin exceeds 1. This describes the exterior of a sphere centered at the origin, where points do not include those on the sphere's surface.
Key Concepts
InequalitiesSolid SphereSphere Exterior
Inequalities
Inequalities are mathematical expressions involving the symbols \(<\), \(>\), \(\leq\), or \(\geq\) to compare two values. They represent a range of possibilities rather than a fixed value, allowing us to express conditions where solutions can vary.
When dealing with inequalities in three-dimensional geometry, especially involving quadratic terms, we are often exploring shapes like spheres, cylinders, and cones. For example, the inequality \(x^2 + y^2 + z^2 \leq 1\) describes a set of points within a certain boundary. In this case, it describes all points inside or on the surface of a sphere with a radius of 1, centered at the origin.
When dealing with inequalities in three-dimensional geometry, especially involving quadratic terms, we are often exploring shapes like spheres, cylinders, and cones. For example, the inequality \(x^2 + y^2 + z^2 \leq 1\) describes a set of points within a certain boundary. In this case, it describes all points inside or on the surface of a sphere with a radius of 1, centered at the origin.
- The "\(\leq\)" symbol indicates that the points on the boundary are included in the solution set.
- Alternatively, the inequality \(x^2 + y^2 + z^2 > 1\) uses a ">" symbol, excluding boundary points and representing the exterior region beyond the sphere's surface.
Solid Sphere
A solid sphere is the set of all points in space that lie inside or on the surface of a sphere. Mathematically, it can be described using an inequality like \(x^2 + y^2 + z^2 \leq r^2\), where \(r\) is the radius of the sphere and \((0,0,0)\) is the center of the sphere.
For example:
For example:
- The inequality \(x^2 + y^2 + z^2 \leq 1\) suggests a solid sphere with a radius of 1 centered at the origin.
- This includes every point within this sphere, like points at \((0,0,1)\), \((0.5, 0.5, 0.5)\), and so on, as long as their combined squared distances to the origin do not exceed 1.
Sphere Exterior
The sphere exterior is the region in three-dimensional space that lies outside a sphere. This region can be described by an inequality, such as \(x^2 + y^2 + z^2 > r^2\), indicating that the distance from any point in this region to the sphere's center exceeds its radius. Hence, none of the points on the sphere's surface are included.
- Consider the inequality \(x^2 + y^2 + z^2 > 1\). This describes the exterior of a sphere of radius 1, centered at the origin. Points like \((1,0,0)\) are just outside and meet this condition.
- The region consists entirely of points that stay beyond the reach of the sphere’s outer surface.
Other exercises in this chapter
Problem 19
Sketch the surfaces in Exercises \(13-44.\) ELLIPSOIDS $$4 x^{2}+9 y^{2}+4 z^{2}=36$$
View solution Problem 19
Express each vector in the form \(\mathbf{v}=v_{1} \mathbf{i}+\) \(v_{2} \mathbf{j}+v_{3} \mathbf{k}.\) \(\overrightarrow{A B}\) if \(A\) is the point \((-7,-8,
View solution Problem 20
Perpendicular diagonals Show that squares are the only rectangles with perpendicular diagonals.
View solution Problem 20
Sketch the surfaces in Exercises \(13-44.\) ELLIPSOIDS $$9 x^{2}+4 y^{2}+36 z^{2}=36$$
View solution