Problem 17
Question
In Exercises \(13-18\) , describe the sets of points in space whose coordinates satisfy the given inequalities or combinations of equations and inequalities. $$ \begin{array}{l}{\text { a. } x^{2}+y^{2}+z^{2}=1, \quad z \geq 0} \\ {\text { b. } x^{2}+y^{2}+z^{2} \leq 1, \quad z \geq 0}\end{array} $$
Step-by-Step Solution
Verified Answer
Part a is the upper hemisphere surface; part b is the solid upper hemisphere.
1Step 1: Understand the Equation
For part a, the equation \(x^2 + y^2 + z^2 = 1\) represents a sphere in 3D space with a radius of 1, centered at the origin \((0, 0, 0)\). The condition \(z \geq 0\) means we are only considering the upper hemisphere of this sphere where the z-coordinates are non-negative.
2Step 2: Analyze Part a
Combining both the equation and the condition \(z \geq 0\), the set of points is the upper hemisphere of the unit sphere, which includes all points on the sphere's surface that rest in the half-space above the xy-plane.
3Step 3: Understand the Inequality
For part b, the inequality \(x^2 + y^2 + z^2 \leq 1\) describes a solid sphere (including its interior) with radius 1 centered at the origin. The condition \(z \geq 0\) limits this region to the top half of the sphere, similar to the restriction in part a.
4Step 4: Analyze Part b
The combination of the inequality and the condition \(z \geq 0\) results in a solid (filled) upper hemisphere. This includes all points with \(z \geq 0\) that lie inside or on the surface of the unit sphere.
Key Concepts
3D Coordinate SystemUnit SphereInequalities in Space
3D Coordinate System
When we consider the 3D coordinate system, we visualize it as a space where each point is defined by three coordinates: \(x\), \(y\), and \(z\). Unlike the 2D coordinate system, which is flat and consists of the axes \(x\) and \(y\), the 3D system introduces an additional dimension: height or depth, which is the \(z\)-axis.
In this system, the origin is at the point \((0, 0, 0)\), where all three axes intersect.
In this system, the origin is at the point \((0, 0, 0)\), where all three axes intersect.
- The \(x\)-axis allows movement left or right.
- The \(y\)-axis allows movement forward or backward.
- The \(z\)-axis allows movement up or down.
Unit Sphere
A unit sphere in a 3D coordinate system is a special sphere. It is centered at the origin \((0, 0, 0)\) and has a radius of 1. The equation of a unit sphere is \[ x^2 + y^2 + z^2 = 1\]This equation tells us that the sum of the squares of the coordinates \(x\), \(y\), and \(z\) always equals 1. This sum ensures that every point on the surface of this sphere is exactly one unit away from the origin.
The unit sphere is a fundamental concept in geometry because it provides a simple model for understanding spherical shapes and distances from a central point. It's frequently used in exercises to build an intuition about spheres and the concept of radius. When we look at subsets of a unit sphere, like only the upper hemisphere where \(z \geq 0\), we apply additional conditions to focus our view on specific sections of this sphere.
The unit sphere is a fundamental concept in geometry because it provides a simple model for understanding spherical shapes and distances from a central point. It's frequently used in exercises to build an intuition about spheres and the concept of radius. When we look at subsets of a unit sphere, like only the upper hemisphere where \(z \geq 0\), we apply additional conditions to focus our view on specific sections of this sphere.
Inequalities in Space
Inequalities are essential for describing regions in space rather than just surfaces or lines. For example, the inequality \[x^2 + y^2 + z^2 \leq 1\]defines the volume of a sphere, including all points inside it along with its surface. This inequality specifies that the distance from the origin to any point \((x, y, z)\) is less than or equal to 1.
When combined with other conditions, such as \(z \geq 0\), we can describe even more specific areas in space. Here, the inequality and the condition come together to define a solid upper hemisphere. We include both the interior volume and the surface of the hemisphere, allowing for the description of a more complex, filled 3D region.
Such inequalities are vital tools in geometry and calculus, as they let us express and solve for bounded regions in space, enabling a deeper understanding of spatial volumes and surfaces.
When combined with other conditions, such as \(z \geq 0\), we can describe even more specific areas in space. Here, the inequality and the condition come together to define a solid upper hemisphere. We include both the interior volume and the surface of the hemisphere, allowing for the description of a more complex, filled 3D region.
Such inequalities are vital tools in geometry and calculus, as they let us express and solve for bounded regions in space, enabling a deeper understanding of spatial volumes and surfaces.
Other exercises in this chapter
Problem 17
In Exercises \(15-18\) a. Find the area of the triangle determined by the points \(P, Q,\) and \(R .\) b. Find a unit vector perpendicular to plane \(P Q R .\)
View solution Problem 17
In Exercises \(17-19\) , write \(u\) as the sum of a vector parallel to \(v\) and a vector orthogonal to \(v\) . $$ \mathbf{u}=3 \mathbf{j}+4 \mathbf{k}, \quad
View solution Problem 17
In Exercises \(17-22,\) express each vector in the form \(\mathbf{v}=v_{1} \mathbf{i}+\) \(v_{2} \mathbf{j}+v_{3} \mathbf{k} .\) \(\overrightarrow{P_{1} P_{2}}\
View solution Problem 18
In Exercises \(15-18\) a. Find the area of the triangle determined by the points \(P, Q,\) and \(R .\) b. Find a unit vector perpendicular to plane \(P Q R .\)
View solution