Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 27-32, describe geometrically the domain of each of the indicated functions of three variables. $$ f(x, y, z)=\ln \left(x^{2}+y^{2}+z^{2}\right) $$

Step-by-Step Solution

Answer

The domain is all space except the origin.

Step 1: Understanding the Function

The function given is \( f(x, y, z) = \ln(x^2 + y^2 + z^2) \). This is a natural logarithm function applied to the sum of squares of three variables \( x, y, \text{ and } z \).

Step 2: Determine Where the Function is Defined

The natural logarithm, \( \ln(t) \), is only defined for \( t > 0 \). Therefore, for the function \( f(x, y, z) \) to be defined, the argument of the logarithm, \( x^2 + y^2 + z^2 \), must be greater than zero, i.e., \( x^2 + y^2 + z^2 > 0 \).

Step 3: Geometrical Interpretation of the Domain

The condition \( x^2 + y^2 + z^2 > 0 \) essentially describes all points in three-dimensional space, except the origin \((0, 0, 0)\). This is because \( x^2 + y^2 + z^2 \) is the squared distance from the origin to the point \((x,y,z)\), which is zero only at the origin. Therefore, the geometrical domain is the entire space minus the origin.

Key Concepts

Geometric Interpretation

When discussing the geometric interpretation of a function's domain in multivariable calculus, it's about understanding how the function behaves spatially. Consider the function given: \[f(x, y, z) = \ln(x^2 + y^2 + z^2)\]This function will only have real and defined values if the expression inside the logarithm, \(x^2 + y^2 + z^2\), is greater than zero. In three-dimensional space, the expression \(x^2 + y^2 + z^2\) represents the square of the distance from a point \((x, y, z)\) to the origin.

• If this expression equals zero, the point is at the origin at \((0, 0, 0)\).
• If greater than zero, it marks all points other than the origin.

Thus, the geometric interpretation of the domain is that every point in space is part of the domain except the origin. Think of it as an entire universe of points, with just one tiny speck not included.

Domain of a Function

The domain of a function refers to all the possible input values the function can accept. For our function \(f(x, y, z) = \ln(x^2 + y^2 + z^2)\), determining the domain involves figuring out when the expression within the natural logarithm is valid. Natural logarithms, represented as \(\ln(t)\), are defined only for positive numbers. Hence, \(x^2 + y^2 + z^2\) must be more than zero for the logarithm to function correctly. If this sum is zero, it would imply the input is invalid.

• Solve for positive inputs: \(x^2 + y^2 + z^2 > 0\).
• The equation represents all points except where \(x = y = z = 0\).

In practical terms, this means the domain excludes the origin, allowing all other combinations of \(x\), \(y\), and \(z\). This clarifies the safe zone for our function to operate on valid inputs.

Natural Logarithm

A natural logarithm is a fundamental concept in calculus, denoted as \( \ln(x)\), representing the logarithm to the base \( e\), where \( e \approx 2.71828\). This special logarithm is widely used in mathematical applications due to its unique properties and relationships.The key aspect when dealing with natural logs is ensuring their argument is positive:

• The formula is valid and provides meaningful results when \(x > 0\).
• When \(x \leq 0\), the logarithm is undefined in real numbers, leading to non-real solutions.

In the given function \(f(x, y, z) = \ln(x^2 + y^2 + z^2)\), the natural logarithm emphasizes the condition \(x^2 + y^2 + z^2 > 0\) is necessary. This enforcement ensures the result is within the real number scope, maintaining mathematical functions' integrity and avoiding undefined or non-real outputs.