Problem 55 - University Calculus: Early Transcendentals

Question

Find the center \(C\) and the radius \(a\) for the spheres. $$x^{2}+y^{2}+z^{2}+4 x-4 z=0$$

Step-by-Step Solution

Answer

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

Step 1: Write the equation of the sphere

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

Step 2: Rearrange the terms

Rewrite the given equation: \(x^2 + y^2 + z^2 + 4x - 4z = 0\) so you can group the \(x\) terms, \(y\) terms, and \(z\) terms. It remains \(x^2 + 4x + y^2 + z^2 - 4z = 0\).

Step 3: Complete the square for x terms

Complete the square for \(x\) by taking half of the coefficient of \(x\), squaring it, and adding inside the equation. The coefficient of \(x\) is 4, so half of it is 2, and the square is 4. Thus, add 4 inside the equation: \((x^2 + 4x + 4)\).

Step 4: Complete the square for z terms

Complete the square for \(z\) by taking half of the coefficient of \(z\), squaring it, and adding inside the equation. The coefficient of \(z\) is -4, so half of it is -2, and the square is 4. Thus, add 4 inside the equation: \((z^2 - 4z + 4)\).

Step 5: Rewrite the equation as completed squares

Write the equation as \((x + 2)^2 + y^2 + (z - 2)^2 = 4 + 4\), and simplify it to \((x + 2)^2 + y^2 + (z - 2)^2 = 8\).

Step 6: Identify the center and radius

From \((x + 2)^2 + y^2 + (z - 2)^2 = 8\), the center \(C\) of the sphere is \((-2, 0, 2)\) and the radius \(a\) is \(\sqrt{8} = 2\sqrt{2}\).

Key Concepts

Center of a Sphere

In the context of a sphere, the center is a point in three-dimensional space that is equidistant from every point on the surface of the sphere. Understanding the center is crucial because it helps define the overall position and symmetry of the sphere.

When dealing with the equation of a sphere in the standard form \[(x - h)^2 + (y - k)^2 + (z - l)^2 = a^2\]it is easy to identify the center by observing the values of \(h\), \(k\), and \(l\). These constants represent the coordinates \((h, k, l)\) for the center of the sphere. The equation reveals how each axis \(x\), \(y\), and \(z\) is offset from these values.

In our original problem, after completing the square for the \(x\) and \(z\) terms, we transformed the equation to \[(x + 2)^2 + y^2 + (z - 2)^2 = 8\]This implies that the center \(C\) of the sphere is

• \(x\)-coordinate: \(-2\) (since \(x + 2 \Rightarrow h = -2\))
• \(y\)-coordinate: \(0\) (since \(y^2\) lacks an internal shift \(y\Rightarrow k = 0\))
• \(z\)-coordinate: \(2\) (since \(z - 2 \Rightarrow l = 2\))

Thus the center of the sphere is given as \((-2, 0, 2)\).

Radius of a Sphere

The radius of a sphere is the distance from the center to any point on the surface of the sphere. It is a crucial measurement that defines the size of the sphere and influences the volume and surface area.

In the equation of a sphere \[(x - h)^2 + (y - k)^2 + (z - l)^2 = a^2\]\(a\) represents the radius. However, in this squared form, \(a^2\) is given, meaning the actual radius is the square root of the term on the right side of the equation.

For the provided problem, our rewritten equation becomes \[(x + 2)^2 + y^2 + (z - 2)^2 = 8\]Since the right side is 8, the radius \(a\) is found by taking the square root:

• \(a = \sqrt{8}\)
• \(a = 2\sqrt{2}\)

Hence, the radius of the sphere is \(2\sqrt{2}\). This reveals the distance from the center \((-2, 0, 2)\) to any point on the sphere's surface.

Completing the Square

Completing the square is a vital algebraic technique used to convert a quadratic equation into a perfect square trinomial. This form makes it easier to identify components such as the center and radius in the sphere's equation.

When you "complete the square", you adjust the equation to reveal squared binomials. In the context of a sphere, it helps us reframe terms on the left into the standard sphere equation form, making properties like center and radius accessible.

To complete the square, follow these general steps:

• Take the coefficient of the linear term, halve it, and square it.
• Add and subtract this squared value inside the equation to maintain equality.
• Rewrite the trinomial as a squared binomial.

For the exercise, we completed the square for the \(x\) and \(z\) variables.

• For \(x\): Given \(x^2 + 4x\), half of 4 is 2, and \(2^2 = 4\). Thus, \(x^2 + 4x + 4 = (x + 2)^2\).
• For \(z\): Given \(z^2 - 4z\), half of -4 is -2, and \(-2^2 = 4\). Thus, \(z^2 - 4z + 4 = (z - 2)^2\).

This adjustment helped convert our original problem's equation into a readily interpretable form where \( (x + 2)^2 + y^2 + (z - 2)^2 = 8 \), revealing the sphere's core properties.