Quadric Surfaces - Calculus 3

1. The equation $\frac{x^2}{9} + \frac{z^2}{4} = y$ represents a surface in 3D space. What happens to the cross-sectional area of this surface as y increases from 0 to 9?
   Answer: The area increases from 0 to $54\pi$ following the pattern $A = 6\pi y$ This is an elliptic paraboloid opening in the positive y-direction. Cross-sections at $y = k$ are ellipses with equation $\frac{x^2}{9k} + \frac{z^2}{4k} = 1$, giving semi-axes $a = 3\sqrt{k}$ and $b = 2\sqrt{k}$. The area is $\pi ab = \pi(3\sqrt{k})(2\sqrt{k}) = 6\pi k$. When $y = 9$, area = $54\pi$. Choice A is wrong (not cylindrical). Choice B has wrong coefficient. Choice D has wrong direction of change.
2. Consider the intersection of the surface $z = x^2 - y^2$ with the plane $z = 4$. What type of curve results, and how does this help classify the original surface?
   Answer: A rectangular hyperbola; this confirms the surface is a hyperbolic paraboloid with saddle-shaped characteristics Setting $z = 4$ gives $4 = x^2 - y^2$, or $\frac{x^2}{4} - \frac{y^2}{4} = 1$, which is a rectangular hyperbola. This confirms that $z = x^2 - y^2$ is a hyperbolic paraboloid (saddle surface). Choice B is wrong because the curve is hyperbolic, not elliptic. Choice C is wrong because $x^2 - y^2 = 4$ gives a hyperbola, not lines. Choice D is wrong because the equation doesn't represent a circle.
3. The surface $4x^2 + y^2 + 4z^2 = 16$ is rotated about which axis to generate a surface of revolution, and what type of curve generates this surface?
   Answer: Rotation about the y-axis; generated by rotating the ellipse $x^2 + 4z^2 = 4$ in the xz-plane The surface $\frac{x^2}{4} + \frac{y^2}{16} + \frac{z^2}{4} = 1$ is an ellipsoid with semi-axes 2, 4, 2 along x, y, z respectively. Since the x and z semi-axes are equal, this can be generated by rotating the ellipse $\frac{x^2}{4} + \frac{y^2}{16} = 1$ (or equivalently $x^2 + 4z^2 = 4$ when scaled) about the y-axis. Choice A has wrong axis and wrong ellipse. Choice C has wrong axis. Choice D is incorrect since two semi-axes are equal.
4. A quadric surface satisfies the equation $2x^2 + 2y^2 - z^2 = 8$. If this surface is intersected by planes of the form $z = c$ where $c$ is a constant, which statement correctly describes the resulting curves?
   Answer: Circles with radius $\sqrt{4 + \frac{c^2}{2}}$ exist for all real values of c The surface $\frac{x^2 + y^2}{4} - \frac{z^2}{8} = 1$ is a hyperboloid of one sheet. When intersected by $z = c$, we get $2x^2 + 2y^2 = 8 + c^2$, or $x^2 + y^2 = 4 + \frac{c^2}{2}$. This gives circles with radius $\sqrt{4 + \frac{c^2}{2}}$ for all real c (always positive). Choice B is wrong about restrictions on c. Choice C is wrong about curve type. Choice D is wrong about parabolic nature.
5. The surface defined by $\frac{x^2}{16} + \frac{y^2}{25} - \frac{z^2}{9} = -1$ has which of the following properties?
   Answer: It is a hyperboloid of two sheets with the gap occurring along the z-axis Rewriting as $\frac{z^2}{9} - \frac{x^2}{16} - \frac{y^2}{25} = 1$, this is a hyperboloid of two sheets opening along the z-axis. For real solutions, we need $|z| \geq 3$, creating a gap between $z = -3$ and $z = 3$. Choice A describes a hyperboloid of one sheet. Choice C would require all positive terms on the left. Choice D is incorrect because points exist when $|z| \geq 3$.
6. Consider the surface given by $x^2 - y^2 - z^2 = 0$. This surface has a distinctive geometric property that distinguishes it from paraboloids. What is this property?
   Answer: The surface contains straight lines passing through the origin in infinitely many directions When analyzing quadric surfaces in three dimensions, you need to recognize their defining characteristics by examining their equations and geometric properties. The surface $x^2 - y^2 - z^2 = 0$ can be rewritten as $x^2 = y^2 + z^2$, which is the equation of a double cone with vertex at the origin and axis along the x-direction. The key insight is that cones are ruled surfaces — they contain infinitely many straight lines. For this cone, any line from the origin to a point $(a, b, c)$ satisfying $a^2 = b^2 + c^2$ lies entirely on the surface. Since there are infinitely many such points forming circles in planes perpendicular to the x-axis, there are infinitely many straight lines through the origin contained in this surface. This makes answer (D) correct. Let's examine why the other options fail: (A) is incorrect because while some cross-sections are hyperbolas (not parabolas), they don't have uniform focal lengths. (B) is false since cones extend infinitely in both directions along their axis — this surface is unbounded and encloses no finite volume. (C) is wrong because cross-sections perpendicular to the x-axis are indeed circles, but their radius increases linearly (not quadratically) with distance: at $x = k$, the radius is $|k|$. Study tip: When identifying quadric surfaces, always check if you can rewrite the equation in a recognizable form. Cones have the pattern $\text{(one variable)}^2 = \text{(other variables)}^2$ and are the only quadric surfaces that are ruled through a single point.
7. Consider the equation $4x^2 + 9y^2 - 36z = 0$. After completing appropriate transformations, this surface can be classified as which type of quadric surface, and what is its orientation?
   Answer: Elliptic paraboloid opening in the positive z-direction Rewriting as $z = \frac{x^2}{9} + \frac{y^2}{4}$, this is an elliptic paraboloid opening upward (positive z-direction) since both coefficients are positive. The traces in planes $z = k > 0$ are ellipses, and traces in vertical planes through the z-axis are parabolas. Choice B is wrong because there's no $xy$ term. Choice C is wrong because this isn't homogeneous (no constant term makes it a cone). Choice D is wrong about the direction.
8. A surface is defined by the equation $y^2 + z^2 = 4x^2 + 9$. To determine the nature of traces in vertical planes, consider the intersection with the plane $x = k$ where $k$ is a positive constant. What can be concluded about the surface?
   Answer: It is a hyperboloid of one sheet since traces in planes $x = k$ are circles with radius increasing as $|k|$ increases Rewriting as $\frac{y^2 + z^2}{9} - \frac{4x^2}{9} = 1$, this is a hyperboloid of one sheet with axis along the x-direction. When $x = k$, we get $y^2 + z^2 = 4k^2 + 9$, giving circles of radius $\sqrt{4k^2 + 9}$. The minimum radius (when $k = 0$) is 3. Choice A is wrong about surface type. Choice B confuses terminology. Choice D is wrong because there's a constant term (not a cone).
9. A surface is described by the equation $x^2 + y^2 = 4z^2 + 16$. Which statement best characterizes this surface and its relationship to coordinate planes?
   Answer: It is a circular hyperboloid of one sheet with its axis along the z-direction and minimum radius of 4 When identifying quadric surfaces, you need to analyze the equation's form and understand how each variable contributes to the surface's shape. The equation $x^2 + y^2 = 4z^2 + 16$ can be rewritten in standard form by rearranging: $x^2 + y^2 - 4z^2 = 16$, or $\frac{x^2}{16} + \frac{y^2}{16} - \frac{z^2}{4} = 1$. This is the standard form of a hyperboloid of one sheet: $\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1$. Since the coefficients of $x^2$ and $y^2$ are equal, the cross-sections perpendicular to the z-axis are circles. At $z = 0$, we get $x^2 + y^2 = 16$, giving a circle of radius 4, which is the minimum radius of the surface. Answer A is wrong because a cone would have the form $x^2 + y^2 = k^2z^2$ with no constant term, and there's no vertex at (0,0,4). Answer B is incorrect because a cylinder would have no $z^2$ term at all. Answer C is wrong because a paraboloid would have $z$ to the first power, not $z^2$, and would look like $x^2 + y^2 = kz$. The key strategy is recognizing the standard forms of quadric surfaces. When you see two variables with the same sign and one with the opposite sign equal to a positive constant, think hyperboloid of one sheet. The constant term tells you about the minimum radius or "waist" of the surface.

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