Problem 9 - Thomas Calculus

Question

In Exercises \(1-16,\) give a geometric description of the set of points in space whose coordinates satisfy the given pairs of equations. $$x^{2}+y^{2}+z^{2}=1, \quad x=0$$

Step-by-Step Solution

Answer

The set of points is a circle of radius 1 in the yz-plane.

Step 1: Identify the First Equation

The first equation given is \( x^2 + y^2 + z^2 = 1 \), which represents the equation of a sphere in three-dimensional space, centered at the origin \((0,0,0)\) with a radius of 1.

Step 2: Identify the Second Equation

The second equation is \( x = 0 \), which describes a plane that is parallel to the \( yz \)-plane and passing through the origin.

Step 3: Intersect the Sphere and the Plane

To find the geometric description of the set of points, consider the intersection of the sphere and the plane \( x = 0 \). Substitute \( x = 0 \) into the sphere's equation, which results in \( y^2 + z^2 = 1 \). This equation represents a circle with radius 1 centered at the origin in the \( yz \)-plane.

Step 4: Describe the Geometric Set

The set of points in space described by these equations is a circle of radius 1 centered at \((0,0,0)\) and lying entirely on the \( yz \)-plane.

Key Concepts

Sphere Equation

The sphere equation is a mathematical representation of all the points in three-dimensional space that are equidistant from a central point. The general form of a sphere's equation is \[ (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2 \] where

• \((h, k, l)\) represents the center of the sphere.
• \(r\) is the radius of the sphere, representing the constant distance from the center to any point on the sphere's surface.

In our exercise, the given sphere is described by the equation \( x^2 + y^2 + z^2 = 1 \). This indicates that the sphere is centered at the origin, \((0, 0, 0)\), and has a radius of 1. This means every point on the sphere is 1 unit away from the origin. This fundamental understanding of the sphere helps in visualizing and interpreting intersections with other geometric objects such as planes.

Plane Equation

In geometry, a plane equation describes a flat, two-dimensional surface extending indefinitely in three-dimensional space. The general equation for a plane in three-dimensional space is \( ax + by + cz = d \), where

• \(a, b, \text{ and } c\) are constants that determine the plane's orientation.
• \(d\) is the distance from the plane to the origin when the normal vector is not normalized.

In our particular problem, we have the equation \( x = 0 \). This simplifies conventionally as there is no coefficient for \(y\) and \(z\), implicating that the plane is actually parallel to the \( yz \)-plane and passes through every point where the \( x \) coordinate is zero. This means that for any point on this plane, moving solely in the \( x \)-axis direction won't change the position on the plane. Understanding this concept helps when considering how this plane might intersect with other geometric entities.

Intersection of Sphere and Plane

When a sphere and a plane intersect, they usually form a shape such as a circle or an ellipse depending on the relative positioning. By finding out how two geometric shapes intersect, we can understand more about their spatial relationships. In this scenario, we have a sphere centered at the origin with a radius of 1, described by \( x^2 + y^2 + z^2 = 1 \) and a plane \( x = 0 \), which is essentially the \( yz \)-plane. To find the intersection, substitute the equation of the plane into the equation of the sphere, resulting in \( y^2 + z^2 = 1 \). This is the equation of a circle with radius 1 centered at the origin in the plane. Thus, when the plane intersects the sphere, the resulting geometric figure is a circle situated on the \( yz \)-plane. This forms a perfect ring, highlighting the beautiful symmetry inherent in spherical geometries. Understanding these intersections can be pivotal in fields such as computer graphics, physics simulations, and even constructing architectural structures.