Problem 49 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the Sphere Equation Format

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

Step 2: Identify the Center from the Equation

Compare the given equation to the standard form:- From \( (x-\sqrt{2})^2 \), we identify \(h = \sqrt{2}\).- From \( (y-\sqrt{2})^2 \), we identify \(k = \sqrt{2}\).- From \( (z+\sqrt{2})^2 \), rewrite as \((z-(-\sqrt{2}))^2\) to identify \(l = -\sqrt{2}\).Thus, the center \((h, k, l) = (\sqrt{2}, \sqrt{2}, -\sqrt{2})\).

Step 3: Determine the Radius

The right side of the equation is 2, which is equal to \(r^2\) in the standard form. To find the radius, take the square root: \[ r = \sqrt{2} \] Thus, the radius of the sphere is \(\sqrt{2}\).

Step 4: State the Answer

The center and radius of the sphere have been determined. The center of the sphere is \((\sqrt{2}, \sqrt{2}, -\sqrt{2})\) and the radius is \(\sqrt{2}\).

Key Concepts

Standard Form of a Sphere

When dealing with spheres in a coordinate system, understanding the equation for a sphere is important. The standard form of a sphere's equation is given by:

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

Here, \((h, k, l)\) represents the coordinates of the center of the sphere. The variable \(r\) stands for the radius of the sphere, and each term in the equation corresponds to one of the three axes (x, y, and z).

The equation is set up so that for any point \((x, y, z)\) on the sphere, the sum of squared distances from the center equals the square of the radius. This form ensures that all points satisfy the condition to lie on the surface of the sphere. If you break it down, every part of the equation indicates how far in each respective direction a point can be from the center while still lying on the sphere's surface.

Center of a Sphere

The center of a sphere is the point that is equidistant from all points on the surface of the sphere. Given our standard sphere equation:

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

in this form, the center can be easily identified as \((h, k, l)\).

For instance, in the equation \((x-\sqrt{2})^2 + (y-\sqrt{2})^2 + (z+\sqrt{2})^2 = 2\), we find the center by comparing to the standard form:
- The \(h\) value comes from \((x - h)^2\), giving us \(h = \sqrt{2}\).
- The \(k\) value is found by \((y - k)^2\), which also results in \(k = \sqrt{2}\).
- The \(l\) value arises from \((z + \sqrt{2})^2\), which can be rewritten as \((z - (-\sqrt{2}))^2\), hence \(l = -\sqrt{2}\).

Consequently, the sphere's center is at \((\sqrt{2}, \sqrt{2}, -\sqrt{2})\), clearly setting a point in 3D space from which all surface points are measured.

Radius of a Sphere

The radius of a sphere is the constant distance from its center to any point on the sphere's surface. In the standard equation

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

the value \(r^2\) is the square of the radius.

To find \(r\), we take the square root of the constant on the right side of the equation. For example, in our exercise equation:

• \((x-\sqrt{2})^2 + (y-\sqrt{2})^2 + (z+\sqrt{2})^2 = 2\)

The right side of this equation equals 2. Thus, we determine the radius by calculating the square root:

• \(r = \sqrt{2}\)

This simple conversion tells us that every point on the sphere's surface is \(\sqrt{2}\) units away from the center, thereby confirming its size and position in space.