Problem 53 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the Standard Equation of a Sphere

The standard equation for a sphere is \((x-h)^2 + (y-k)^2 + (z-l)^2 = a^2\), where \((h, k, l)\) is the center and \(a\) is the radius of the sphere.

Step 2: Compare with Given Equation

The given equation is \((x-\sqrt{2})^2 + (y-\sqrt{2})^2 + (z+\sqrt{2})^2 = 2\). Here, we will identify the terms corresponding to \((x-h)\), \((y-k)\), and \((z-l)\) and compare them to the standard equation.

Step 3: Determine the Center of the Sphere

Comparing terms in the equation, we see that:- \(x-h = x - \sqrt{2}\) implies \(h = \sqrt{2}\)- \(y-k = y - \sqrt{2}\) implies \(k = \sqrt{2}\)- \(z-l = z + \sqrt{2}\) implies \(l = -\sqrt{2}\)Thus, the center \(C\) of the sphere is \((\sqrt{2}, \sqrt{2}, -\sqrt{2})\).

Step 4: Determine the Radius of the Sphere

The right side of the equation is equal to \(a^2\). Thus, \(a^2 = 2\). Therefore, the radius \(a\) is \(\sqrt{2}\).

Step 5: Verify the Solution

Rewriting the equation with the identified parameters: \((x-\sqrt{2})^2 + (y-\sqrt{2})^2 + (z+\sqrt{2})^2 = 2\).We have found the center \((\sqrt{2}, \sqrt{2}, -\sqrt{2})\) and the radius \(\sqrt{2}\), which satisfies all parts of the equation.

Key Concepts

Geometry

In the world of geometry, spheres are fascinating objects. A sphere is a perfectly symmetrical three-dimensional shape, similar to a ball, where every point on its surface is equidistant from its center.
This fundamental structure can represent various natural and man-made entities such as planets, bubbles, and marbles.
To delve deep, the equation of a sphere is crucial. This equation helps determine the key features of a sphere, such as its center and its radius. In mathematical terms, this is usually presented in the form:

• \((x-h)^2 + (y-k)^2 + (z-l)^2 = a^2\)

This equation reveals that geometry is not just lines and angles. It's also about visualizing forms like spheres in a mathematical sense.
Studying this can improve spatial understanding and problem-solving skills, making geometry a vital part of both math and the wider world.

Center of a Sphere

The center of a sphere is one of the most important concepts in understanding its structure and position in space.
In our example scenario, the center is given by the coordinates

• \(h, k, l\)

This is a point in a 3D space from which all surface points of the sphere are at equal distance.
Identifying this center involves comparing the given equation segments with a standard equation of a sphere's form. For instance:

• Comparing terms such as \(x-h = x - \sqrt{2}\) gives \(h = \sqrt{2}\).
• Similarly, \(y-k = y - \sqrt{2}\) gives \(k = \sqrt{2}\).
• And \(z-l = z + \sqrt{2}\) results in \(l = -\sqrt{2}\).

Therefore, the center \(C\) is calculated to be \((\sqrt{2}, \sqrt{2}, -\sqrt{2})\).
Understanding the center helps in grasping how spheres are aligned in a three-dimensional environment, adding context to their geometric properties.

Radius of a Sphere

The radius of a sphere is the line from its center to any point on its surface.
It is a constant value across the sphere and is crucial for determining the size of a sphere.
In the equation

• \((x-h)^2 + (y-k)^2 + (z-l)^2 = a^2\)

'a' denotes the radius. To find the radius, you equate the right side of the given sphere's equation with \(a^2\).
For the equation provided:

• We have: \(a^2 = 2\).
• Finding 'a' involves taking the square root which provides: \(a = \sqrt{2}\).

Thus, the radius of this sphere is \(\sqrt{2}\).
Understanding the radius concept helps in visualizing the sphere's magnitude and its spread within the surrounding space, allowing further exploration of properties such as surface area and volume.