Problem 12 - 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. $$x^{2}+(y-1)^{2}+z^{2}=4, \quad y=0$$

Step-by-Step Solution

Answer

A circle in the xz-plane with center at \((0, 0)\) and radius \(\sqrt{3}\).

Step 1: Understand the first equation

The first equation \( x^2 + (y - 1)^2 + z^2 = 4 \) describes a sphere in 3D space. The sphere has a center at the point \((0, 1, 0)\) and a radius of 2, because the general form of a sphere's equation is \((x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\). Here, \(h = 0\), \(k = 1\), \(l = 0\), and \(r = 2\).

Step 2: Interpret the second equation

The second equation \( y = 0 \) represents a plane that is parallel to the xz-plane but shifted downward in the y-direction by 1 unit. This plane cuts through the 3D space at \( y = 0 \).

Step 3: Find the intersection

To find the geometric description of where the sphere and plane intersect, substitute \( y = 0 \) in the first equation: \( x^2 + (0 - 1)^2 + z^2 = 4 \), simplifying to \( x^2 + 1 + z^2 = 4 \) or \( x^2 + z^2 = 3 \).

Step 4: Analyze the intersection equation

The simplified equation \( x^2 + z^2 = 3 \) is a circle in the xz-plane. This circle is centered at \((0, 0)\) with a radius of \(\sqrt{3}\). Because it results from the intersection with \( y = 0 \), it exists only in this slice of the sphere.

Key Concepts

Sphere Equation

A sphere in 3D geometry is a beautifully symmetric shape. It includes all points that are equidistant from a central point, much like a circle in 2D space. One way to represent a sphere mathematically is through its equation:

• The standard form of a sphere's equation is \[(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\]

Here,

• \( (h, k, l) \) is the center of the sphere, determining the position in 3D space.
• \( r \) is the radius, indicating how large the sphere is from its center point.

This equation essentially sums the squared distances from each of the point's coordinates to the center's coordinates, ensuring the point is always exactly \( r \) units from the center. By understanding this, we can easily identify its center and radius just like in the equation \( x^2 + (y - 1)^2 + z^2 = 4 \), where the center is at \( (0, 1, 0) \) and the radius is \( 2 \). This means the sphere is centered slightly above the xz-plane, at y = 1.

Plane Intersection

When dealing with 3D geometry, often a shape like a sphere will intersect a plane. A plane equation in the form \( y = c \) is straightforward and represents a flat, two-dimensional slice of space, remaining parallel to the xz-plane. In our exercise, the plane described by \( y = 0 \) is horizontal and intersects the space at the y-coordinate zero. To determine where it intersects with a sphere, replace \( y \) in the sphere's equation with the plane's constant (here \( y = 0 \)). The new equation will describe the intersection line or shape. For the provided equations, substituting \( y = 0 \) into the sphere's equation gives us: \[x^2 + (0 - 1)^2 + z^2 = 4\]This simplifies to \( x^2 + z^2 = 3 \), identifying another geometric shape at this slice.

Circle in Space

In three-dimensional space, a circle can appear in various contexts, particularly when a sphere intersects with a plane. The resulting intersection can be a perfect circle. When we plug \( y = 0 \) into our sphere’s equation and simplify, we find the intersection described by \( x^2 + z^2 = 3 \). This represents a circle centered at \( (0,0) \) in the xz-plane with a radius given by \( \sqrt{3} \). Important aspects of this circle:

• This circle lies entirely within the plane \( y = 0 \).
• Its size is dictated by the formula for a circle’s radius (in this case, \( r = \sqrt{3} \)).

Being aware of how spheres and planes interact to form circles can deepen your understanding of geometry. It highlights how 2D shapes originate from 3D intersections, providing insight on visual and spatial reasoning.