Problem 11 - APEX Calculus

Question

(a) Find the domain \(D\) of the given function. (b) State whether \(D\) is an open or closed set. (c) State whether \(D\) is bounded or unbounded. $$ f(x, y)=\sqrt{9-x^{2}-y^{2}} $$

Step-by-Step Solution

Answer

Domain: \( D = \{(x, y) \mid x^2 + y^2 \leq 9\} \); Closed and Bounded.

Step 1: Determine where the function is well-defined

The function given is \( f(x, y) = \sqrt{9 - x^2 - y^2} \). Since it involves a square root, the expression inside the square root must be non-negative for \( f(x, y) \) to be real and defined. Therefore, we require \( 9 - x^2 - y^2 \geq 0 \). Rearranging this gives us \( x^2 + y^2 \leq 9 \). This inequality describes a circle centered at the origin with radius 3. The domain \( D \) consists of all points \((x, y)\) where \( x^2 + y^2 \leq 9 \).

Step 2: Identify the nature of the set

The domain \( D \) is the set of all points \( (x, y) \) where \( x^2 + y^2 \leq 9 \). This includes all points on the boundary \( x^2 + y^2 = 9 \) and all points inside this circle. Therefore, \( D \) is a closed set because it includes its boundary.

Step 3: Determine boundedness

A set is bounded if it can be enclosed within some circle of finite radius. Here, \( D \) is enclosed within a circle of radius 3 (as given by \( x^2 + y^2 \leq 9 \)), which means it is bounded. Therefore, \( D \) is a bounded set.

Key Concepts

Domain of a Function

When we talk about the "domain of a function" in multivariable calculus, we're referring to all possible input values for which the function is defined. Consider the function \( f(x, y) = \sqrt{9 - x^2 - y^2} \). Because this function involves a square root, the expression under the square root sign must be non-negative.

This means for our function to have real and defined outputs, the inequality \( 9 - x^2 - y^2 \geq 0 \) must hold true. When we rearrange this inequality, we get \( x^2 + y^2 \leq 9 \).

• It describes a circle in the xy-plane.
• The circle is centered at the origin \((0, 0)\).
• It has a radius of 3.

The domain \( D \) therefore, is the set of all points \((x, y)\) within this circle, including those exactly on the circle's boundary. Understanding the domain as a geometric region helps visualize all possible \( (x, y) \) inputs that the function can 'accept' and produce real outputs for.

Closed Set

A set is described as 'closed' if it contains all its boundary points. So, to check if a domain is closed, we've got to see if it includes all the 'edge' points. For the domain \( D = \{ (x, y) \mid x^2 + y^2 \leq 9 \} \), it includes all points \( (x, y) \) where the distance from the origin is less than or equal to 3.

Importantly, it also includes points where \( x^2 + y^2 = 9 \), which lie exactly on the circle's edge. Therefore, since the boundary defined by \( x^2 + y^2 = 9 \) is part of \( D \), the domain is indeed a closed set. Closed sets are often represented by solid circles or disks (including their boundaries), helping us understand the full extent of the set.

Bounded Set

A set in multivariable calculus is considered 'bounded' if there's a way to enclose it within a finite region, often described as a circle or sphere with a certain radius. With our domain \( D \), the inequality \( x^2 + y^2 \leq 9 \) specifies a circle with a center at the origin and a radius of 3.

Since \( D \) is entirely contained within this circle, it's easily encased within a finite area of the plane.

• There are no points in \( D \) that have values outside a fixed-distance circle.
• This means \( D \) does not extend infinitely in any direction.

In summary, the domain \( D \) being bounded tells us that it’s constrained to a specific, limited region in the xy-plane, supporting the concept that it is a bounded set.