Problem 36
Question
Find the domain of the following functions. $$f(x, y, z)=\sqrt{49-x^{2}-y^{2}-z^{2}}$$
Step-by-Step Solution
Verified Answer
The domain is the set of all points \((x, y, z)\) such that \(x^2 + y^2 + z^2 \leq 49\).
1Step 1: Identify the function type
The function given is a square root function. This means we need the expression inside the square root to be greater than or equal to zero in order for the function to have real values.
2Step 2: Set the expression under the root to be non-negative
To find the domain of the function, we need to solve the inequality: \[49-x^2-y^2-z^2 \geq 0\] This ensures that the value under the square root is non-negative.
3Step 3: Rearrange the inequality
Rearrange the inequality to simplify it: \[x^2 + y^2 + z^2 \leq 49\] This new form represents a solid sphere in three-dimensional space.
4Step 4: Describe the domain
The inequality \(x^2 + y^2 + z^2 \leq 49\) represents all points \((x, y, z)\) that lie inside or on the surface of the sphere centered at the origin with a radius of 7.
Key Concepts
Understanding the Square Root FunctionSolving InequalitiesExploring Three-Dimensional SpaceConcept of a Solid Sphere
Understanding the Square Root Function
The square root function is a mathematical function that extracts the non-negative root of a number or expression. In formula terms, \(\sqrt{x}\) represents a number, which, when multiplied by itself, results in \(x\). It is denoted as \(\sqrt{ \; }\).
When dealing with square roots, the value or expression under the square root must be non-negative to produce real numbers. This is because square roots of negative numbers are not real. Therefore, for any function that involves a square root, solving for the domain begins with ensuring the expression inside the root is at least zero.
In the problem given, the function is \(f(x, y, z)=\sqrt{49-x^{2}-y^{2}-z^{2}}\). To find its domain, we need to find when \(49-x^{2}-y^{2}-z^{2} \) is greater than or equal to zero.
When dealing with square roots, the value or expression under the square root must be non-negative to produce real numbers. This is because square roots of negative numbers are not real. Therefore, for any function that involves a square root, solving for the domain begins with ensuring the expression inside the root is at least zero.
In the problem given, the function is \(f(x, y, z)=\sqrt{49-x^{2}-y^{2}-z^{2}}\). To find its domain, we need to find when \(49-x^{2}-y^{2}-z^{2} \) is greater than or equal to zero.
Solving Inequalities
Inequalities are mathematical statements that describe a range of values rather than an exact number. Here's a quick breakdown of inequalities:
By rearranging \(49-x^{2}-y^{2}-z^{2} \geq 0\) we get \(x^2 + y^2 + z^2 \leq 49\), an inequality demonstrating all valid \(x, y, z\) that our function can accept. Solving inequalities effectively defines the domain of the function in this context.
- The symbol \(>\) means 'greater than.'
- \(<\) signifies 'less than.'
- \(\geq\) represents 'greater than or equal to.'
- \(\leq\) stands for 'less than or equal to.'
By rearranging \(49-x^{2}-y^{2}-z^{2} \geq 0\) we get \(x^2 + y^2 + z^2 \leq 49\), an inequality demonstrating all valid \(x, y, z\) that our function can accept. Solving inequalities effectively defines the domain of the function in this context.
Exploring Three-Dimensional Space
Three-dimensional space refers to the physical universe that we exist in, characterized by three dimensions—length, width, and height. This concept allows us to represent objects and distances in a space where each point is defined by coordinates \(x, y, z\).
The function \(f(x, y, z)\) involves three variables, indicating that it represents a shape or volume in three-dimensional space. One common three-dimensional shape we deal with in math is a sphere, which ties directly into our solution. By understanding that our inequality \(x^2 + y^2 + z^2 \leq 49\) represents a sphere, we can appreciate how this equation describes a volume in space and its boundaries.
The function \(f(x, y, z)\) involves three variables, indicating that it represents a shape or volume in three-dimensional space. One common three-dimensional shape we deal with in math is a sphere, which ties directly into our solution. By understanding that our inequality \(x^2 + y^2 + z^2 \leq 49\) represents a sphere, we can appreciate how this equation describes a volume in space and its boundaries.
Concept of a Solid Sphere
A solid sphere in mathematics includes all points inside a sphere as well as those on its surface. Unlike a hollow sphere or shell, a solid sphere encompasses everything within its boundary. This set includes the internal volume and the sphere's outer "skin."
The domain of the function \(f(x, y, z)=\sqrt{49-x^{2}-y^{2}-z^{2}}\) can be visualized as a solid sphere. Here, the sphere is centered at the origin \((0, 0, 0)\) with a radius of \(7\) (since \(49\) is the square of \(7\)).
The inequality \(x^2 + y^2 + z^2 \leq 49\) describes every valid point \(x, y, z\) inside or on this sphere. Points \( (x, y, z) \) that satisfy this inequality are part of the domain, offering a clear spatial interpretation.
The domain of the function \(f(x, y, z)=\sqrt{49-x^{2}-y^{2}-z^{2}}\) can be visualized as a solid sphere. Here, the sphere is centered at the origin \((0, 0, 0)\) with a radius of \(7\) (since \(49\) is the square of \(7\)).
The inequality \(x^2 + y^2 + z^2 \leq 49\) describes every valid point \(x, y, z\) inside or on this sphere. Points \( (x, y, z) \) that satisfy this inequality are part of the domain, offering a clear spatial interpretation.
Other exercises in this chapter
Problem 34
Find the domain of the following functions. $$z=\ln \left(x-y^{2}\right)$$
View solution Problem 35
Find the domain of the following functions. $$f(x, y, z)=\frac{1}{\sqrt{36-4 x^{2}-9 y^{2}-z^{2}}}$$
View solution Problem 37
Find the domain of the following functions. $$f(x, y, z)=\sqrt[3]{16-x^{2}-y^{2}-z^{2}}$$
View solution Problem 38
Find the domain of the following functions. $$f(x, y)=\cos \sqrt{x^{2}+y^{2}}$$
View solution