Problem 51 - University Calculus: Early Transcendentals

Question

Find the center \(C\) and the radius \(a\) for the spheres. $$(x+2)^{2}+y^{2}+(z-2)^{2}=8$$

Step-by-Step Solution

Answer

Center is \((-2, 0, 2)\); radius is \(2\sqrt{2}\).

Step 1: Identify the standard form of a sphere's equation

The general form of a sphere's equation is \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\), where \((h, k, l)\) is the center and \(r\) is the radius of the sphere.

Step 2: Extract the center from the equation

Compare the given equation \((x+2)^{2}+y^{2}+(z-2)^{2}=8\) to the standard form \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\). This provides the center as \((-2, 0, 2)\).

Step 3: Calculate the radius

The right side of the equation \((x+2)^{2}+y^{2}+(z-2)^{2}=8\) is \(8\), representing \(r^2\). Take the square root of 8 to find \(r = \sqrt{8} = 2\sqrt{2}\).

Key Concepts

Geometry of a Sphere

A sphere is a three-dimensional object where all points on its surface are equidistant from a central point. This characteristic provides unique properties that are fundamental in geometry.
Spheres are the 3D counterparts of circles in 2D geometry. When working with spheres, we often encounter the need to determine their size and location within a coordinate system, usually described by an equation.

• A sphere's surface is completely smooth, without any edges or vertices.
• The diameter of a sphere is the longest distance between any two points on the surface, passing through the center.
• Every point on the sphere's surface shares the same distance from the center, known as the radius.
• The volume and surface area can be calculated if the radius is known, using the formulas:
  • Volume: \( \frac{4}{3} \pi r^3 \)
  • Surface Area: \( 4\pi r^2 \)

Understanding these properties of spheres can greatly simplify calculations involving other geometric figures and help students connect the concepts of plane and solid geometry.

Center of a Sphere

The center of a sphere is a crucial component because it serves as the reference point for defining the sphere's position in space. In the standard form equation of a sphere, the center is denoted by the coordinates \((h, k, l)\).
Identifying the center from an equation means comparing the equation to the standard sphere form, \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\). Looking at this form, we see that:

• The values \(h\), \(k\), and \(l\) represent the sphere's center on the x-, y-, and z-axes, respectively.
• For example, in an equation such as \((x+2)^2 + y^2 + (z-2)^2 = 8\), the center is \((-2, 0, 2)\).This is determined because:
  • \((x+2)^2\) becomes \(x - (-2)\), showing the x-coordinate is -2.
  • \(y^2\) means \(y-0\), indicating the y-coordinate is 0.
  • \((z-2)^2\) indicates \(z - 2\), thus the z-coordinate is 2.

By locating the center, we grasp how the sphere is situated in this virtual 3D space.

Radius of a Sphere

The radius of a sphere is a measure of the distance from its center to any point on its surface. It is the constant distance creating the spherical surface and is denoted as \(r\) in the sphere's equation, \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\).
Finding this involves understanding the relationship presented in the equation:

• The right side of the equation relates directly to the square of the radius, \(r^2\).
• For example, in the equation \((x+2)^2 + y^2 + (z-2)^2 = 8\), the number 8 represents \(r^2\).
• To find \(r\), take the square root of this value. Here, \(r = \sqrt{8} = 2\sqrt{2}\).

Knowing the radius is essential as it defines the size of the sphere. It helps compute other properties like volume and surface area. Remember, any change in radius affects these attributes, which further impacts calculations in geometric problem-solving.