We are asked to describe the graphs of several functions of two variables, $f(x, y)$.
2025/6/3
1. Problem Description
We are asked to describe the graphs of several functions of two variables, .
2. Solution Steps
7. $f(x, y) = 6$. This is a plane parallel to the $xy$-plane, located at $z = 6$.
8. $f(x, y) = 6 - x$. This is a plane. We can rewrite it as $z = 6 - x$ or $x + z = 6$. This is a plane parallel to the $y$-axis, intersecting the $xz$-plane in the line $x + z = 6$.
9. $f(x, y) = 6 - x - 2y$. This is a plane. We can rewrite it as $z = 6 - x - 2y$ or $x + 2y + z = 6$.
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0. $f(x, y) = 6 - x^2$. This is a cylinder. We can rewrite it as $z = 6 - x^2$. This is a parabolic cylinder parallel to the $y$-axis.
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1. $f(x, y) = \sqrt{16 - x^2 - y^2}$. Since $z = f(x, y)$, we have $z = \sqrt{16 - x^2 - y^2}$. Squaring both sides, we have $z^2 = 16 - x^2 - y^2$, or $x^2 + y^2 + z^2 = 16$. Since $z = \sqrt{16 - x^2 - y^2}$, we must have $z \ge 0$. Thus, this is the upper hemisphere of a sphere with radius $\sqrt{16} = 4$ centered at the origin.
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2. $f(x, y) = \sqrt{16 - 4x^2 - y^2}$. Since $z = f(x, y)$, we have $z = \sqrt{16 - 4x^2 - y^2}$. 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 $\frac{x^2}{4} + \frac{y^2}{16} + \frac{z^2}{16} = 1$. Since $z = \sqrt{16 - 4x^2 - y^2}$, we must have $z \ge 0$. Thus, this is the upper half of an ellipsoid.
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3. $f(x, y) = 3 - x^2 - y^2$. Since $z = f(x, y)$, we have $z = 3 - x^2 - y^2$, or $x^2 + y^2 = 3 - z$. This is a paraboloid opening downwards with vertex at $(0, 0, 3)$.
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4. $f(x, y) = 2 - x - y^2$. Since $z = f(x, y)$, we have $z = 2 - x - y^2$, or $x = 2 - y^2 - z$. This is a parabolic cylinder.
3. Final Answer
7. Plane at $z=6$.
8. Plane $x+z=6$.
9. Plane $x+2y+z=6$.
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0. Parabolic cylinder $z=6-x^2$.
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1. Upper hemisphere of $x^2+y^2+z^2=16$.
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2. Upper half of ellipsoid $\frac{x^2}{4} + \frac{y^2}{16} + \frac{z^2}{16} = 1$.
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3. Paraboloid $x^2+y^2 = 3-z$.
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