We need to describe the domain of the following two functions geometrically: 27. $f(x, y, z) = \sqrt{x^2 + y^2 + z^2 - 16}$ 28. $f(x, y, z) = \sqrt{x^2 + y^2 - z^2 - 9}$

1. Problem Description

We need to describe the domain of the following two functions geometrically:

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7. $f(x, y, z) = \sqrt{x^2 + y^2 + z^2 - 16}$

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8. $f(x, y, z) = \sqrt{x^2 + y^2 - z^2 - 9}$

2. Solution Steps

For function 27: $f(x, y, z) = \sqrt{x^2 + y^2 + z^2 - 16}$, the expression inside the square root must be non-negative. Therefore, we have:

$x^2 + y^2 + z^2 - 16 \ge 0$

$x^2 + y^2 + z^2 \ge 16$

This represents all points $(x, y, z)$ that lie outside or on the sphere centered at the origin with radius $\sqrt{16} = 4$.

For function 28: $f(x, y, z) = \sqrt{x^2 + y^2 - z^2 - 9}$, the expression inside the square root must be non-negative. Therefore, we have:

$x^2 + y^2 - z^2 - 9 \ge 0$

$x^2 + y^2 - z^2 \ge 9$

$x^2 + y^2 \ge z^2 + 9$

This represents the points outside or on the hyperboloid of two sheets along the $z$-axis, $x^2 + y^2 - z^2 = 9$.

3. Final Answer

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7. The domain of $f(x, y, z) = \sqrt{x^2 + y^2 + z^2 - 16}$ is the set of all points $(x, y, z)$ that lie outside or on the sphere centered at the origin with radius 4, i.e., $x^2 + y^2 + z^2 \ge 16$.

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8. The domain of $f(x, y, z) = \sqrt{x^2 + y^2 - z^2 - 9}$ is the set of all points $(x, y, z)$ such that $x^2 + y^2 \ge z^2 + 9$, which represents the points outside or on the hyperboloid of two sheets along the $z$-axis, $x^2 + y^2 - z^2 = 9$.