# Finding The Domain of $f(x,y) = \sqrt{4-x^2-y^2}$.

The function is $f(x,y) = \sqrt{4-x^2-y^2}$.

I know that the only allowable $x$ and $y$ are those that yield either zero under the square root symbol, or some positive number. With that in mind,

$$4=x^2-y^2 \ge 0 \rightarrow x^2 + y^2 \le 4$$

So, I solved for each variable, finding the domains of each one, both of which have the same one, $[-2,2]$; I then wrote the domain of $f(x,y)$ as $\{(x,y) \mid x,y \in [-2,2]\}$. In the answer key, it is written as $\{(x,y) \mid x^2 + y^2 \le 4\}$. Is the way I wrote proper?

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Your version is not correct. what you have described is a square (and its interior) while the correct answer is a disk.

For example, the point $(1.5,1.5)$ is in the set described by your answer, but the function is not defined there.

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No.

Notice that

$\{ (x,y) \mid x, y \in [-2, 2] \}$ is a square in a plane but

$\{(x,y) \mid x^2+y^2 \leq 4\}$ is a disk with radius 2.

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No, it is not. Graph your solution and the one in the answer key: yours is a square of side $4$ centred on the origin, and the one in the answer key is a circle of radius $2$ centred on the origin. Your region includes too much: $\langle 2,2\rangle$, for instance, is clearly not in the domain of $f$.

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No, for the same reason we discussed in your previous question. Your description includes $(2,2)$ and $(\frac 32,\frac 32)$, which are not in the domain as the sum of squares is greater than $4$.

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Your answer is not correct because it contains points which don't satisfy the constraint $x^2+y^2\leq 4$. For example, for $(x, y) = (2, 2)$, $x^2 + y^2 = 2^2 + 2^2 = 4 + 4 = 8\not\leq 4$.

Geometrically, the set you wrote down describes a square of side length $4$ centred at the origin, while the correct set describes a disk of radius $2$ centred at the origin. The latter is strictly contained within the former.

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