F(x,y) = -sqrt(16-(x-3)^2 - (y-3)^2) if (x-3)^2 + (y-3)^2 <= 16,
$$ \left\lbrace\right\rbrace $$sqrt(16-(x+3)^2 - (y+3)^2) if if (x+3)^2 + (y+3)^2 <= 16,
and 0 otherwise.
What is the domain and range of F(x,y)? can you graph it?

Here's the graph of the function $ F(x, y) $. It shows two hemispherical shapes:

- The negative hemisphere is centered at $ (3, 3) $, and the function takes values from $-4$ to $0$.
- The positive hemisphere is centered at $ (-3, -3) $, and the function takes values from $0$ to $4$.

The plot clearly illustrates how the function is defined over two distinct circular regions in the $xy$-plane, with values in the range $[-4, 4]$.

Would you like further details or have any questions? Here are some related topics you might be interested in:

1. How would the graph look if the hemispheres had different radii?
2. What is the effect of shifting the centers of the circles in $ F(x, y) $?
3. How do you compute the gradient of $ F(x, y) $ at different points?
4. Can we extend $ F(x, y) $ to three dimensions by introducing a new variable?
5. How would the range change if we modified the square root terms in $ F(x, y) $?

**Tip:** When graphing functions of two variables, always check the domain carefully to avoid undefined regions, which can lead to computational errors (as seen with the square root).

Explore the function F(x, y) = -sqrt(16-(x-3)^2 - (y-3)^2) for (x-3)^2 + (y-3)^2 <= 16 and sqrt(16-(x+3)^2 - (y+3)^2) for (x+3)^2 + (y+3)^2 <= 16, with a detailed analysis of its domain, range, and graphical representation. Understand how this piecewise function operates over circular regions in the xy-plane.

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