We need to sketch the graph of $f(x, y)$ for the following functions: 7. $f(x, y) = 6$ 8. $f(x, y) = 6 - x$ 9. $f(x, y) = 6 - x - 2y$ 10. $f(x, y) = 6 - x^2$ 11. $f(x, y) = \sqrt{16 - x^2 - y^2}$ 12. $f(x, y) = \sqrt{16 - 4x^2 - y^2}$ 13. $f(x, y) = 3 - x^2 - y^2$ 14. $f(x, y) = 2 - x - y^2$

1. Problem Description

We need to sketch the graph of $f(x, y)$ for the following functions:

7. $f(x, y) = 6$

8. $f(x, y) = 6 - x$

9. $f(x, y) = 6 - x - 2y$

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0. $f(x, y) = 6 - x^2$

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

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2. $f(x, y) = \sqrt{16 - 4x^2 - y^2}$

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3. $f(x, y) = 3 - x^2 - y^2$

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4. $f(x, y) = 2 - x - y^2$

2. Solution Steps

We need to describe the surface for each function.

7. $f(x, y) = 6$. This is equivalent to $z = 6$. This is a horizontal plane parallel to the $xy$-plane, intersecting the $z$-axis at $z = 6$.

8. $f(x, y) = 6 - x$. This is equivalent to $z = 6 - x$. This is a plane parallel to the $y$-axis. In the $xz$-plane, it's the line $z = 6 - x$, or $x + z = 6$.

9. $f(x, y) = 6 - x - 2y$. This is equivalent to $z = 6 - x - 2y$, or $x + 2y + z = 6$. This is a plane. The intercepts are $(6, 0, 0)$, $(0, 3, 0)$, and $(0, 0, 6)$.

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0. $f(x, y) = 6 - x^2$. This is equivalent to $z = 6 - x^2$. This is a cylinder parallel to the $y$-axis. In the $xz$-plane, it's the parabola $z = 6 - x^2$, or $x^2 = 6 - z$.

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1. $f(x, y) = \sqrt{16 - x^2 - y^2}$. This is equivalent to $z = \sqrt{16 - x^2 - y^2}$, where $z \geq 0$. Squaring both sides, we get $z^2 = 16 - x^2 - y^2$, or $x^2 + y^2 + z^2 = 16$. This is a sphere with radius 4 centered at the origin. Since $z \geq 0$, we only have the upper hemisphere.

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2. $f(x, y) = \sqrt{16 - 4x^2 - y^2}$. This is equivalent to $z = \sqrt{16 - 4x^2 - y^2}$, where $z \geq 0$. Squaring both sides, we get $z^2 = 16 - 4x^2 - y^2$, or $4x^2 + y^2 + z^2 = 16$. Dividing by 16, we get $\frac{x^2}{4} + \frac{y^2}{16} + \frac{z^2}{16} = 1$. This is an ellipsoid. Since $z \geq 0$, we only have the upper half.

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3. $f(x, y) = 3 - x^2 - y^2$. This is equivalent to $z = 3 - x^2 - y^2$, or $x^2 + y^2 = 3 - z$. This is a paraboloid opening downwards, with its vertex at $(0, 0, 3)$.

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4. $f(x, y) = 2 - x - y^2$. This is equivalent to $z = 2 - x - y^2$, or $x = 2 - z - y^2$. This is a parabolic cylinder.

3. Final Answer

7. Horizontal plane at $z = 6$.

8. Plane $z = 6 - x$.

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

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0. Cylinder $z = 6 - x^2$.

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

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

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

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