Find the domain and range of f(x,y)=root of 9-xsquare minus ysquare

The given function is:

\[
f(x, y) = \sqrt{9 - x^2 - y^2}
\]

To find the **domain** and **range**, let's examine the expression in detail.

### Domain:
The domain consists of all values of \(x\) and \(y\) for which the function is defined. Since the square root function is only defined for non-negative values, we need:

\[
9 - x^2 - y^2 \geq 0
\]

Rearranging this inequality:

\[
x^2 + y^2 \leq 9
\]

This inequality represents the equation of a circle with radius 3 centered at the origin \((0, 0)\) in the \(xy\)-plane. Therefore, the domain is all points \((x, y)\) inside or on this circle, mathematically written as:

\[
D = \{(x, y) \in \mathbb{R}^2 : x^2 + y^2 \leq 9\}
\]

### Range:
The function \(f(x, y) = \sqrt{9 - x^2 - y^2}\) represents the square root of a non-negative quantity. The maximum value of the expression under the square root occurs when \(x = 0\) and \(y = 0\), which gives:

\[
f(0, 0) = \sqrt{9 - 0^2 - 0^2} = \sqrt{9} = 3
\]

The minimum value of the square root function is 0, which happens when \(x^2 + y^2 = 9\), i.e., on the boundary of the circle.

Thus, the range of the function is:

\[
R = [0, 3]
\]

### Summary:
- **Domain**: \( x^2 + y^2 \leq 9 \) (a disk of radius 3 centered at the origin).
- **Range**: \( [0, 3] \) (the function takes values between 0 and 3).

Would you like to explore this further, or have any specific questions? Here are some related questions to think about:

1. What is the geometric meaning of the domain in terms of circles?
2. How would the domain change if the function was \( \sqrt{16 - x^2 - y^2} \)?
3. What happens if the square root function was replaced by a cube root?
4. How does the range change if the expression inside the square root includes other terms like \( +1 \)?
5. How would the domain be affected if this was a 3-dimensional function?

**Tip**: Always check for values that make the expression inside a square root negative when finding domains!

Find the domain and range of f(x, y) = √(9 - x² - y²)

Explore how to determine the domain and range of the function f(x, y) = √(9 - x² - y²). This multivariable calculus problem explains key concepts like the geometry of circles and inequalities, making it ideal for students in Grades 10-12.

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