Question

Describe the region \(R\) in the \(x y\) -plane that corresponds to the domain of the function, and find the range of the function. $$ f(x, y)=\sqrt{16-x^{2}-y^{2}} $$

Short answer

The region in the xy-plane that corresponds to the domain of the function \(f(x, y)=\sqrt{16-x^{2}-y^{2}}\) is the closed disk of radius 4 centered at the origin, including its boundary. The range of the function is the interval [0, 4].

Step-by-step solution

Define the Domain

The domain of \(f(x, y)\) is given by the inequality \(16 - x^{2} - y^{2} >= 0\). This is the equation of a circle centered at the origin with radius 4 in the xy-plane. The values inside the square root must be nonnegative since square root of negative number is undefined in real numbers. Hence, the domain of the function is the closed disk of radius 4 centered at the origin, including its boundary.

Solve the Inequality

We solve the inequality \(16 - x^{2} - y^{2} >= 0\) to find the values of \(x\) and \(y\). Rearranging the terms, we get \(x^{2} + y^{2} <= 16\). This means that both \(x\) and \(y\) should satisfy the condition \((-4 <= x <= 4)\) and \((-4 <= y <= 4)\) respectively. The region within and on the circle represent all the possible values of \(x\) and \(y\) that make the function defined.

Find the Range

The range of the function \(f(x, y)\) is the set of all possible values of \(f\). The minimum value occurs when the value inside the square root is at its minimum (i.e., \(x^{2} + y^{2} = 16\)), which gives \(f(x, y) = 0\). The maximum value occurs when \(x = y = 0\) (as we can't exceed the radius of 4), which gives \(f(x, y) = \sqrt{16} = 4\). Hence, the range of \(f(x, y)\) is the interval \([0, 4]\) on the real number line.

Key concepts

Domain of a Multivariable Function

When we talk about the domain of a multivariable function such as \( f(x, y) \), we're referring to the set of all possible pairs \( (x, y) \) that we can plug into the function so that the function 'makes sense' or is well defined. In other words, the domain is all the input values that won't cause the function to output errors or undefined values.

For the function \( f(x, y) = \sqrt{16 - x^2 - y^2} \), we need to ensure that the expression under the square root is nonnegative, because the square root of a negative number is not a real number. Therefore, the function is only defined when \( 16 - x^2 - y^2 \geq 0 \). This corresponds to the set of points \( (x, y) \) inside or on the boundary of a circle with radius 4 on the xy-plane. Imagine shading this circular region; all shaded (x,y) points form the domain of our function.

Range of a Multivariable Function

The range of a multivariable function, on the other hand, consists of all the possible output values the function can produce using the input values from the domain. For the function \( f(x, y) \), which outputs real numbers, the range is particularly interesting because it's governed by the maximum and minimum values attainable within the domain.

The minimum value of \( f(x, y) \) is zero, which occurs when the expression under the square root equals zero – meaning we are on the circle's edge \( x^2 + y^2 = 16 \). The maximum value, attained when \( x \) and \( y \) are both zero (this point is the circle's center), is 4, which is the square root of 16. Therefore, we can see that the range of \( f(x, y) \) is the interval \([0, 4]\), including the endpoints.

Solving Inequalities

Solving inequalities is a foundational skill needed to determine the domain of functions. In our function, we're presented with the inequality \( 16 - x^2 - y^2 \geq 0 \). To solve this, you want to think about the relationship between \( x \) and \( y \) that would make this inequality true.

Rearrange terms to isolate \( x^2 + y^2 \) on one side, yielding \( x^2 + y^2 \leq 16 \). What this is telling us is that all the x and y coordinate pairs that lie within a circle of radius 4 in the xy-plane will satisfy this inequality. Graphing systems of inequalities, like these, shows us the domain or feasible region visually.

Circle Equation in XY-Plane

A circle in the xy-plane can be described with the equation \( x^2 + y^2 = r^2 \), where \( r \) is the radius of the circle and the center is at the origin (0,0). In our example, our circle has a radius of 4, so the equation is \( x^2 + y^2 = 16 \).

The domain of our function contains all the points that would fall within the circle's boundary when graphed on the xy-plane. In other words, if you draw this circle on a graph and color inside it, you've drawn all the possible \( (x, y) \) that the function could accept as inputs—a great visualization for understanding domain!