Problem 45 - Thomas Calculus

Question

Sketch the surfaces in Exercises \(13-76\) $$ x^{2}+y^{2}+z^{2}=4 $$

Step-by-Step Solution

Answer

The surface is a sphere centered at (0,0,0) with radius 2.

Step 1: Identify the Type of Surface

The given equation \(x^{2}+y^{2}+z^{2}=4\) is in the form \(x^2 + y^2 + z^2 = a^2\), which represents a sphere.

Step 2: Determine the Center and Radius

Comparing the equation \(x^{2}+y^{2}+z^{2}=4\) with the standard form of a sphere \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), we identify that \(h = 0\), \(k = 0\), \(l = 0\) and \(r^2 = 4\). Thus, the sphere is centered at (0, 0, 0) with a radius of \(\sqrt{4} = 2\).

Step 3: Sketch the Sphere

Since the sphere is centered at the origin (0,0,0) with a radius of 2, plot the sphere in three dimensions. Mark the points (2, 0, 0), (0, 2, 0), (0, 0, 2), (-2, 0, 0), (0, -2, 0), and (0, 0, -2) as these are intersections with the axes. Ensure the sketch represents a symmetric ball around the origin with surface all points equidistant (at 2 units) from the center.

Key Concepts

Sphere

A sphere is a perfectly round three-dimensional shape where every point on its surface is equidistant from its center. This distance is called the radius. In many ways, it resembles a ball or globe. Imagine a basketball or the planet Earth; these are physical representations of a sphere. Spheres are classified as one of the simplest types of shapes in three-dimensional geometry.
Spheres are unique because they have no faces, edges, or vertices, which distinguishes them from polygons or polyhedra. They are defined solely by their center and radius:

• The center of the sphere is a single point within the sphere.
• The radius extends from the center to any point on the surface.

Visualizing a sphere is easy; think of the shape you get when you perfectly inflate a beach ball, creating a flawless round object.

Equation of a Sphere

The equation of a sphere in a three-dimensional Cartesian coordinate system is derived from the distance formula. It can be written in its standard form as:\[(x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\]Here,

• \( (h, k, l) \) is the center of the sphere.
• \( r \) is the radius.

This standard equation implies that every point \((x, y, z)\) on the surface of the sphere maintains a constant distance from the center. In simpler terms, if you take any point on the surface and measure the distance to the center, it would always equal the radius \( r\).
Let's consider the example from the exercise, \(x^2 + y^2 + z^2 = 4\), which is a special case where the center is the origin \((0, 0, 0)\) and the radius squared is \(4\). This equation signifies a sphere at the origin with all points on the surface 2 units away from the center.

Radius and Center of a Sphere

Identifying the radius and center of a sphere from its equation is crucial for understanding and visualizing it in 3D space. For a sphere defined by the equation \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\):

• The center is represented by the coordinates \((h, k, l)\).
• The radius is \( r \), which you find by taking the square root of the value on the right-hand side of the equation \((r^2)\).

In cases where the equation is simplified, such as \(x^2 + y^2 + z^2 = 4\), it's easy to determine the center and radius:

• The center is \((0, 0, 0)\), since the terms do not include offsets \((h, k, l)\).
• The radius is \( \sqrt{4} \ = 2\).

This means every point on this sphere's surface is exactly 2 units away from the origin. Understanding this allows you to sketch or perceive the sphere mentally, identifying all intersections like those with the coordinate axes and knowing it maintains symmetry around its center.