Problem 47 - Thomas Calculus

Question

Find the centers and radii of the spheres in Exercises \(47-50\) $$(x+2)^{2}+y^{2}+(z-2)^{2}=8$$

Step-by-Step Solution

Answer

Center: (-2, 0, 2); Radius: \(2\sqrt{2}\)

Step 1: Identify the equation of the sphere

The given equation of the sphere is \[ (x+2)^{2}+y^{2}+(z-2)^{2}=8 \] This is in the standard form of the equation of a sphere, which is \[ (x-h)^{2}+(y-k)^{2}+(z-l)^{2}=r^{2} \] where \((h, k, l)\) is the center of the sphere and \(r\) is the radius.

Step 2: Determine the center of the sphere

Compare each term in the given equation \[ (x+2)^{2}+y^{2}+(z-2)^{2} = 8 \] with the standard sphere equation to identify the center:- \((x+2)^{2}\) suggests \(h = -2\),- \(y^{2}\) suggests \(k = 0\),- \((z-2)^{2}\) suggests \(l = 2\).Therefore, the center of the sphere is \((-2, 0, 2)\).

Step 3: Calculate the radius of the sphere

In the general form of the sphere equation, the right-hand side represents \(r^{2}\), where \(r\) is the radius. From the equation \[ (x+2)^{2}+y^{2}+(z-2)^{2}=8 \] we can see that \(r^{2} = 8\). Therefore, the radius \(r\) is the square root of 8, which simplifies to \(2\sqrt{2}\).

Key Concepts

Center of a Sphere

Understanding the center of a sphere is an essential step in analyzing any sphere given by its equation. The equation of a sphere often comes in the standard form: \[ (x-h)^2 + (y-k)^2 + (z-l)^2 = r^2 \] This form is neat because it allows you to easily identify the sphere's center, given by the coordinate \((h, k, l)\). The center acts like an anchor point.
In simpler terms, it defines where the sphere is located in a 3D space. For example, in the equation \((x+2)^2 + y^2 + (z-2)^2 = 8\), you can find the center by comparing each term to the standard form. Start with the \((x+2)^2\). It shows \(h = -2\).
Next, \(y^2\) indicates \(k = 0\).Finally, \((z-2)^2\) states \(l = 2\). These values tell us that the center of the sphere is at \((-2, 0, 2)\). Recognizing this center helps visualize and understand the sphere's position, making it easier to connect mathematical equations to real-world 3D shapes.

Radius of a Sphere

The radius of a sphere is just as crucial as its center. It tells you the size or how big the sphere is, extending evenly in all directions from the center. Within the sphere equation, \[ (x-h)^2 + (y-k)^2 + (z-l)^2 = r^2 \] The term \(r^2\) on the right-hand side represents the square of the radius. To get the actual radius, you simply take the square root of this term. For example, consider our sphere's equation \((x+2)^2 + y^2 + (z-2)^2 = 8\). Here, \(r^2 = 8\). Thus, \(r\) itself is \(\sqrt{8}\), which simplifies to \(2\sqrt{2}\).
The radius \(2\sqrt{2}\) offers a clear picture of how far the surface of the sphere is from the center \((-2, 0, 2)\). Understanding the radius is fundamental as it not only describes the size but also plays a role in calculations for volume and surface area when delving deeper into geometry.

Standard Form of a Sphere

The standard form of a sphere's equation is a powerful tool in geometry and mathematics. It offers a convenient way to read key properties of a sphere, namely the center and the radius.
The equation looks like this:\[ (x-h)^2 + (y-k)^2 + (z-l)^2 = r^2 \]Each part of this equation holds significant information:

• The terms \((x-h)^2, (y-k)^2,\) and \((z-l)^2\) are tied directly to the sphere's center at \((h, k, l)\).
• The term \(r^2\) reveals the squared radius, indicating the size of the sphere.

Determining the center and radius is straightforward thanks to this form. You're basically matching each part of a given sphere equation to the standard form to deduce \(h, k, and l\) for the center and \(r\) for the radius.
This ease of use is why understanding the standard form is important for students, simplifying more complex geometry tasks. Whether you’re doing homework or engaging in more advanced studies, remembering this form helps solve sphere-related problems efficiently.