Problem 9
Question
Find the domain of the function. \(f(x, y, z)=\sqrt{1-x^{2}-y^{2}-z^{2}}\)
Step-by-Step Solution
Verified Answer
The domain is all points \((x, y, z)\) such that \(x^2 + y^2 + z^2 \leq 1\).
1Step 1: Understanding the Function
The function given is a multivariable function defined as \( f(x, y, z) = \sqrt{1 - x^2 - y^2 - z^2} \). The expression under the square root, \(1-x^2-y^2-z^2\), must be non-negative for the function to be defined in the real numbers.
2Step 2: Set Condition for Non-Negativity
To ensure the expression under the square root is non-negative, we must have: \[ 1 - x^2 - y^2 - z^2 \geq 0 \] Rearranging gives: \[ x^2 + y^2 + z^2 \leq 1 \].
3Step 3: Interpret the Condition
The condition \(x^2 + y^2 + z^2 \leq 1\) represents the set of all points \((x, y, z)\) that lie on or within the surface of a sphere of radius 1 centered at the origin in three-dimensional space. This is also known as a closed ball or a disk when in 3D.
4Step 4: Define the Domain
The domain of the function \( f(x, y, z) \) is all points \((x, y, z)\) such that they satisfy the condition \(x^2 + y^2 + z^2 \leq 1\). This can be expressed as: \[ \{(x, y, z) \mid x^2 + y^2 + z^2 \leq 1\} \].
Key Concepts
Domain of a FunctionInequalities in Three DimensionsSphere in Three-Dimensional Space
Domain of a Function
The domain of a function in mathematics represents all the possible input values that the function can accept without leading to any undefined behavior. For a multivariable function like the one given, \(f(x, y, z) = \sqrt{1-x^2-y^2-z^2}\), finding its domain requires analyzing the expression for any restrictions on the variables \(x\), \(y\), and \(z\).
In this function, the expression \(1 - x^2 - y^2 - z^2\) is under the square root. Hence, it needs to be non-negative, as the square root of a negative number is not defined within the realm of real numbers. This means:
Understanding the domain helps determine where the function can be applied, ensuring valid computations. This is particularly useful in multivariable calculus and in visualizing functions within a defined space.
In this function, the expression \(1 - x^2 - y^2 - z^2\) is under the square root. Hence, it needs to be non-negative, as the square root of a negative number is not defined within the realm of real numbers. This means:
- \(1 - x^2 - y^2 - z^2 \geq 0\)
- Rewriting this inequality, we have \(x^2 + y^2 + z^2 \leq 1\)
Understanding the domain helps determine where the function can be applied, ensuring valid computations. This is particularly useful in multivariable calculus and in visualizing functions within a defined space.
Inequalities in Three Dimensions
Inequalities in three-dimensional coordinates describe a variety of geometric shapes and regions in space. When dealing with inequalities involving three variables like \(x\), \(y\), and \(z\), such as \(x^2 + y^2 + z^2 \leq 1\) from our exercise, it defines a specific region in \(\mathbb{R}^3\).
In the context of calculus and physics, these inequalities help delineate regions where functions are applicable or where physical phenomena can occur.
- Here, \(x^2 + y^2 + z^2 \leq 1\) describes all the points inside or on the boundary of a sphere.
- The inequality describes a bounded region because of the less than or equal condition \(\leq\), encompassing everything from the center of the sphere up to its surface.
In the context of calculus and physics, these inequalities help delineate regions where functions are applicable or where physical phenomena can occur.
Sphere in Three-Dimensional Space
In three-dimensional space, a sphere represents a set of all points that are at the same distance from a central point, known as the radius. When defining a sphere using an inequality, as in the exercise, we are considering the area enclosed within the sphere.
- The standard equation for a sphere centered at the origin with radius \(r\) is \(x^2 + y^2 + z^2 = r^2\).
- In our scenario, the restriction \(x^2 + y^2 + z^2 \leq 1\) describes a solid sphere or a ball with a radius of 1, centered at the origin.
Other exercises in this chapter
Problem 9
Find the first partial derivatives of the function. $$ z=\sqrt{\left(1-x^{2 / 3}\right)^{3}-y^{2}} $$
View solution Problem 9
Evaluate the limit. $$ \lim _{(x, y, z) \rightarrow(\pi / 2,-\pi / 2,0)} \cos (x+y+z) $$
View solution Problem 10
Find the minimum value of \(f\) subject to the given constraint. In each case assume that the minimum value exists. $$ f(x, y, z)=3 z-x-2 y ; z=x^{2}+4 y^{2} $$
View solution Problem 10
Find all critical points. Determine whether each critical point yields a relative maximum value, a relative minimum value, or a saddle point. $$ f(u, v)=u^{3}+v
View solution