The problem asks to describe the graph of the function $f(x, y)$ for each case from 7 to 16.

Okay, I will solve problem 7 through 16 in the image. Since the prompt requires only the solution without sketches, I will provide descriptions of the surfaces.

1. Problem Description

The problem asks to describe the graph of the function $f(x, y)$ for each case from 7 to

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6.

2. Solution Steps

7. $f(x, y) = 6$. This represents the plane $z = 6$, which is a plane parallel to the $xy$-plane and 6 units above it.

8. $f(x, y) = 6 - x$. This represents the plane $z = 6 - x$. This is a plane parallel to the $y$-axis. When $y = 0$, $z = 6 - x$, which is a line in the $xz$-plane.

9. $f(x, y) = 6 - x - 2y$. This represents the plane $z = 6 - x - 2y$, or $x + 2y + z = 6$. This is a plane that intersects the $x$-axis at $x = 6$, the $y$-axis at $y = 3$, and the $z$-axis at $z = 6$.

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0. $f(x, y) = 6 - x^2$. This represents the surface $z = 6 - x^2$. This is a parabolic cylinder parallel to the $y$-axis. The cross-section in the $xz$-plane is a parabola opening downward with vertex at $(0, 6)$.

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1. $f(x, y) = \sqrt{16 - x^2 - y^2}$. This represents the surface $z = \sqrt{16 - x^2 - y^2}$. Squaring both sides gives $z^2 = 16 - x^2 - y^2$, or $x^2 + y^2 + z^2 = 16$. Since $z \ge 0$, this is the upper hemisphere of a sphere with radius 4 centered at the origin.

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2. $f(x, y) = \sqrt{16 - 4x^2 - y^2}$. This represents the surface $z = \sqrt{16 - 4x^2 - y^2}$. Squaring both sides gives $z^2 = 16 - 4x^2 - y^2$, or $4x^2 + y^2 + z^2 = 16$. Dividing by 16, we get $\frac{x^2}{4} + \frac{y^2}{16} + \frac{z^2}{16} = 1$. Since $z \ge 0$, this is the upper half of an ellipsoid.

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3. $f(x, y) = 3 - x^2 - y^2$. This represents the surface $z = 3 - x^2 - y^2$, or $z = 3 - (x^2 + y^2)$. This is a paraboloid opening downward with vertex at $(0, 0, 3)$.

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4. $f(x, y) = 2 - x - y^2$. This represents the surface $z = 2 - x - y^2$. We can rewrite it as $x = 2 - z - y^2$. This is a parabolic cylinder.

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5. $f(x, y) = e^{-(x^2 + y^2)}$. This represents the surface $z = e^{-(x^2 + y^2)}$. This is a surface of revolution. As $x^2 + y^2$ increases, $z$ decreases. At $(0, 0)$, $z = 1$. This is a "bell-shaped" surface.

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6. $f(x, y) = \frac{x^2}{y}$, $y > 0$. This represents the surface $z = \frac{x^2}{y}$, $y > 0$.

3. Final Answer

Here are the descriptions of the surfaces:

7. Plane $z=6$

8. Plane $z=6-x$

9. Plane $z=6-x-2y$

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0. Parabolic cylinder $z=6-x^2$

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1. Upper hemisphere $x^2+y^2+z^2=16, z \ge 0$

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2. Upper half of an ellipsoid $\frac{x^2}{4} + \frac{y^2}{16} + \frac{z^2}{16} = 1, z \ge 0$

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3. Paraboloid $z=3-x^2-y^2$

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4. Parabolic cylinder $x = 2 - z - y^2$

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5. Bell-shaped surface $z=e^{-(x^2+y^2)}$

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6. Surface $z = \frac{x^2}{y}$, $y > 0$