Quadric Surfaces - Multivariable Calculus

A solid region in $\mathbb{R}^3$ is bounded below by the surface $S_1$ given by $z = \sqrt{x^2+y^2}$ and bounded above by the surface $S_2$ given by $z = 6 - x^2 - y^2$ . What types of quadric surfaces are $S_1$ and $S_2$ ?
Answer: $S_1$ is an elliptic cone and $S_2$ is a circular paraboloid. First, consider the surface $S_1$ given by $z = \sqrt{x^2+y^2}$ . Since $z$ must be non-negative, this is the upper nappe of a cone. Squaring both sides gives $z^2 = x^2+y^2$ , which is the equation of a double cone. Because of the principal square root, we only have the part with $z \ge 0$ . This is the upper half of a circular (or elliptic, but here circular) cone. Next, consider the surface $S_2$ given by $z = 6 - x^2 - y^2$ . This can be rewritten as $z = 6 - (x^2+y^2)$ . This is the equation of a circular paraboloid that opens downwards along the $z$ -axis, with its vertex at $(0, 0, 6)$ . So, the solid is bounded by a cone and a circular paraboloid.
Consider the surface $z^2 = x^2 + 4y^2$ . If this surface is intersected by the plane $x + z = 6$ , what type of curve results?
Answer: A parabola in the plane x + z = 6 with vertex at (3, 0, 3) The surface z² = x² + 4y² is a circular cone. From the plane equation x + z = 6, we get x = 6 - z. Substituting into the cone equation: z² = (6 - z)² + 4y², which gives z² = 36 - 12z + z² + 4y². Simplifying: 0 = 36 - 12z + 4y², so 12z = 36 + 4y², or z = 3 + y²/3. Since x = 6 - z, we have x = 6 - (3 + y²/3) = 3 - y²/3. The parametric equations are x = 3 - y²/3, z = 3 + y²/3, showing a parabola. When y = 0, x = 3 and z = 3, giving the vertex at (3, 0, 3). Options B and C are incorrect because the intersection of a cone and a plane typically gives a conic section, and our algebra shows a parabolic relationship. Option D is incorrect as this would require a linear relationship in y.
The surface $x^2 + 4y^2 - 8z = 16$ intersects the coordinate planes in three curves. What types of curves are these intersections?
Answer: xy-plane: ellipse; xz-plane: parabola; yz-plane: parabola For the xy-plane (z = 0): x² + 4y² = 16, which gives x²/16 + y²/4 = 1. This is an ellipse. For the xz-plane (y = 0): x² - 8z = 16, so z = (x² - 16)/8. This is a parabola opening upward. For the yz-plane (x = 0): 4y² - 8z = 16, so z = (y² - 4)/2. This is also a parabola opening upward. The surface is an elliptic paraboloid, which characteristically has elliptical cross-sections parallel to one coordinate plane and parabolic cross-sections in the other two coordinate planes.
A surface is formed by rotating the curve $x=2z^2$ in the $xz$ -plane about the $x$ -axis. What is the equation of the resulting surface?
Answer: $x = 2y^2 + 2z^2$ When a curve in a coordinate plane is rotated about an axis, any point $(x, 0, z)$ on the curve $x=2z^2$ in the $xz$ -plane traces a circle in a plane perpendicular to the axis of rotation (the $x$ -axis). The radius of this circle is the distance from the axis, which is $z$ . For a general point $(x, y, z')$ on the surface, the radius of rotation is $\sqrt{y^2+(z')^2}$ . This radius must be equal to the original distance $z$ from the axis. So, $z = \sqrt{y^2+(z')^2}$ , which means $z^2 = y^2+(z')^2$ . We replace $z^2$ in the original equation with this expression. The original equation is $x=2z^2$ . Replacing $z^2$ with $(y^2+z^2)$ —dropping the prime from $z'$ for the general coordinate—we get $x = 2(y^2+z^2)$ , which is $x = 2y^2 + 2z^2$ . This surface is an elliptic paraboloid (specifically, a circular paraboloid) opening along the positive $x$ -axis.
Which of the following equations represents a quadric surface that is bounded in all three coordinate directions?
Answer: $\frac{x^2}{25} + \frac{y^2}{9} + \frac{z^2}{4} = 1$ A quadric surface is bounded in all directions only if it's an ellipsoid. Option C represents an ellipsoid since all terms are positive and equal to 1. For any fixed value of one variable within the appropriate range, the other two variables are constrained to an ellipse, making the entire surface bounded. Option A is a hyperboloid of one sheet (unbounded in x and y), option B is a circular cone (unbounded in all directions), and option D is a circular paraboloid (unbounded in the x direction). Only ellipsoids are bounded in all three coordinate directions.
Consider the surface defined by $y^2 = x^2 + z^2$ . Which statement correctly describes the behavior of this surface along the y-axis?
Answer: The surface intersects the y-axis only at the origin, and cross-sections perpendicular to the y-axis are circles with radius |y| When analyzing surfaces defined by equations, you need to examine two key behaviors: where the surface intersects coordinate axes and what its cross-sections look like at different values. To find where this surface intersects the y-axis, set $x = 0$ and $z = 0$ in the equation $y^2 = x^2 + z^2$ . This gives you $y^2 = 0$ , so $y = 0$ . The surface only touches the y-axis at the origin $(0,0,0)$ . For cross-sections perpendicular to the y-axis, fix $y$ at some constant value and see what shape results. The equation becomes $x^2 + z^2 = y^2$ . Since $y^2$ is constant, this describes a circle in the $xz$ -plane with radius $\sqrt{y^2} = |y|$ . As $|y|$ increases, these circular cross-sections get larger. Looking at the wrong answers: Choice A incorrectly claims the surface intersects the y-axis at all points, which contradicts our calculation showing only the origin works. It also gives the wrong radius formula. Choice C correctly identifies the y-axis intersection but wrongly describes the cross-sections as ellipses—they're actually circles since the coefficients of $x^2$ and $z^2$ are equal. Choice D gets the y-axis intersection right but incorrectly describes square cross-sections instead of circular ones. This surface is actually a circular cone opening along the y-axis, which you can visualize by recognizing the standard form $y^2 = x^2 + z^2$ . Study tip: When analyzing surfaces, always check axis intersections by setting other coordinates to zero, and find cross-sections by holding one variable constant.
The equation $x^2 + z^2 = 1$ defines a surface in $\mathbb{R}^3$ . Which of the following best describes this surface?
Answer: A circular cylinder of radius 1 with its central axis along the $y$ -axis. In $\mathbb{R}^2$ (the $xz$ -plane), the equation $x^2 + z^2 = 1$ represents a circle of radius 1. However, in $\mathbb{R}^3$ , the equation describes the set of all points $(x, y, z)$ that satisfy the condition. Since the variable $y$ is not present in the equation, there is no restriction on its value. This means that for any point $(x, z)$ that satisfies $x^2 + z^2 = 1$ , the point $(x, y, z)$ will be on the surface for any real number $y$ . Geometrically, this means we take the circle of radius 1 in the $xz$ -plane and extend it infinitely in both the positive and negative $y$ -directions. This creates a circular cylinder of radius 1 whose central axis is the $y$ -axis.
The quadric surface defined by $4x^2 - y^2 + z^2 + 8x + 6y - 4z - 5 = 0$ is which of the following?
Answer: A hyperboloid of one sheet centered at $(-1, 3, 2)$ . To identify the surface, we must complete the square for each variable. Start by grouping terms: $(4x^2 + 8x) - (y^2 - 6y) + (z^2 - 4z) - 5 = 0$ . Factor out the leading coefficients: $4(x^2 + 2x) - (y^2 - 6y) + (z^2 - 4z) - 5 = 0$ . Complete the square inside each parenthesis: $4(x^2 + 2x + 1) - (y^2 - 6y + 9) + (z^2 - 4z + 4) - 5 - 4(1) - (-1)(9) - (1)(4) = 0$ . Simplify the equation: $4(x+1)^2 - (y-3)^2 + (z-2)^2 - 5 - 4 + 9 - 4 = 0$ . This simplifies to $4(x+1)^2 - (y-3)^2 + (z-2)^2 - 4 = 0$ , or $4(x+1)^2 - (y-3)^2 + (z-2)^2 = 4$ . Dividing by 4 gives the standard form: $\frac{(x+1)^2}{1} - \frac{(y-3)^2}{4} + \frac{(z-2)^2}{4} = 1$ . This is the equation of a hyperboloid of one sheet because it has the form $+X^2 - Y^2 + Z^2 = 1$ . The center is at $(-1, 3, 2)$ .
A quadric surface has the equation $Ax^2 + By^2 + Cz^2 + Dx + Ey + Fz + G = 0$ where all of A, B, C are positive. After completing the square, the resulting equation could represent any of the following EXCEPT:
Answer: A hyperboloid of one sheet, if the constant term is negative after completing the square When you encounter quadric surfaces with positive coefficients for all squared terms, you're dealing with elliptic-type surfaces. The key insight is that after completing the square, you'll get an equation of the form $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} + \frac{(z-l)^2}{c^2} = K$ where $K$ is the constant term and $a^2, b^2, c^2$ are all positive. Since all coefficients $A$ , $B$ , and $C$ are positive, the left side of this equation represents the sum of three non-negative terms (squared distances). This sum can only equal zero or positive values. Options A, B, and C are all possible outcomes. If $K > 0$ (option B), you get an ellipsoid. If $K = 0$ (option C), the only solution is the single point $(h,k,l)$ where all three squared terms equal zero. If $K < 0$ (option A), there's no solution since you can't have a sum of non-negative terms equal a negative number, giving you an empty set. However, option D is impossible. A hyperboloid of one sheet requires the general form $\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1$ where at least one coefficient must be negative. Since all of $A$ , $B$ , and $C$ are positive in this problem, you cannot obtain a hyperboloid of any type. Study tip: Remember that when all squared-term coefficients have the same sign, you get elliptic surfaces (ellipsoids, points, or empty sets). Mixed signs are required for hyperbolic surfaces.
Consider the two surfaces $S_1: x^2 + y^2 - z^2 = 1$ and $S_2: -x^2 - y^2 + z^2 = 1$ . What is the most significant topological difference between these two surfaces?
Answer: $S_1$ is a connected surface, while $S_2$ is composed of two disconnected components. $S_1$ is the equation of a hyperboloid of one sheet, with the $z$ -axis as its axis of symmetry. This surface is a single, connected piece. $S_2$ can be rewritten as $x^2 + y^2 = z^2 - 1$ . For a real solution to exist, we must have $z^2 - 1 \ge 0$ , which means $|z| \ge 1$ . This condition splits the surface into two separate components (or sheets), one for $z \ge 1$ and one for $z \le -1$ . This is a hyperboloid of two sheets. The most fundamental topological difference is that a hyperboloid of one sheet is connected, while a hyperboloid of two sheets is not.
A surface is given by the equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1$ where $a, b, c > 0$ . For which values of the parameters does the surface have exactly two connected components?
Answer: For all positive values of a, b, and c, since this always represents a hyperboloid of two sheets When you encounter a quadratic surface equation, the key is recognizing the standard forms and understanding what the signs tell you about the surface's geometry. The equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1$ is the standard form of a hyperboloid of two sheets. The crucial insight is that this surface type is determined entirely by the pattern of positive and negative terms, not by the specific values of the parameters. Since we have one positive term ( $x^2$ ) and two negative terms ( $y^2$ and $z^2$ ), this will always create a hyperboloid with two separate sheets opening along the x-axis, regardless of what positive values $a$ , $b$ , and $c$ take. Choice A is incorrect because there's no meaningful "discriminant" $b^2 + c^2 - a^2$ that determines connectivity for quadratic surfaces. This confuses quadratic surface theory with quadratic equation solutions. Choice B is wrong because the condition $a > b + c$ doesn't control the number of components. The parameter values only affect the shape and scale of the hyperboloid, not whether it has one or two sheets. Choice D makes the right conclusion but for the wrong reason. While the components are indeed separated along the x-axis, this isn't a special property that depends on parameter values—it's inherent to this surface type. The correct answer is C. For any positive $a$ , $b$ , and $c$ , this equation always represents a hyperboloid of two sheets. Study tip: Memorize the standard forms of quadratic surfaces. The sign pattern of the squared terms immediately tells you the surface type, while parameter values only affect size and proportions.
For the family of surfaces defined by $x^2 + y^2 - z^2 = k$ , where $k$ is a real parameter, how does the geometry of the surface change as $k$ decreases from a large positive value to a large negative value?
Answer: It transitions from a hyperboloid of one sheet to a cone, and then to a hyperboloid of two sheets. Let's analyze the equation $x^2 + y^2 - z^2 = k$ . When $k > 0$ , this is the standard equation for a hyperboloid of one sheet, with the $z$ -axis as its axis. As $k$ decreases towards 0, the radius of the circular cross-section at $z=0$ (given by $x^2+y^2=k$ ) shrinks. When $k=0$ , the equation becomes $x^2 + y^2 = z^2$ , which is the equation of a double cone. When $k < 0$ , let $k = -c^2$ for some $c>0$ . The equation becomes $x^2 + y^2 - z^2 = -c^2$ , which can be rewritten as $z^2 - x^2 - y^2 = c^2$ . This is the standard equation for a hyperboloid of two sheets. Therefore, as $k$ decreases from positive to negative, the surface transitions from a hyperboloid of one sheet to a cone, and then to a hyperboloid of two sheets.
Which statement accurately describes a geometric property of the hyperboloid of one sheet defined by the equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1$ ?
Answer: Through every point on the surface, there pass two distinct straight lines that lie entirely on the surface. A key property of a hyperboloid of one sheet is that it is a doubly ruled surface. This means that for any point on the surface, there are exactly two distinct straight lines that pass through that point and are entirely contained within the surface. Choice A describes a hyperboloid of two sheets. Choice B describes a paraboloid. Choice C is incorrect; the intersection with a plane $z=k$ is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 + \frac{k^2}{c^2}$ , which is an ellipse for any real $k$ .
A surface is described by the vector parameterization $\vec{r}(u, v) = \langle 2u\cos(v), 3u\sin(v), u \rangle$ for $u \ge 0$ and $0 \le v \le 2\pi$ . Which quadric surface does this parameterization represent?
Answer: An elliptic cone Let the components of $\vec{r}(u,v)$ be $x$ , $y$ , and $z$ . So, $x = 2u\cos(v)$ , $y = 3u\sin(v)$ , and $z = u$ . To find the Cartesian equation, we need to eliminate the parameters $u$ and $v$ . From the equations for $x$ and $y$ , we can write $\frac{x}{2} = u\cos(v)$ and $\frac{y}{3} = u\sin(v)$ . Squaring and adding these gives $(\frac{x}{2})^2 + (\frac{y}{3})^2 = u^2\cos^2(v) + u^2\sin^2(v) = u^2$ . Since $z=u$ , we can substitute $z$ for $u$ in this equation to get $\frac{x^2}{4} + \frac{y^2}{9} = z^2$ . This is the standard equation for an elliptic cone with its axis along the $z$ -axis. The condition $u \ge 0$ means we only get the upper nappe of the cone ( $z \ge 0$ ).
A quadric surface has the property that every horizontal cross-section (z = constant) is a circle, and every vertical cross-section through the z-axis is a parabola opening upward. Which equation could represent this surface?
Answer: $x^2 + y^2 = 2z + 8$ For horizontal cross-sections to be circles, we need x² + y² to appear together with the same coefficients. For vertical cross-sections through the z-axis to be parabolas, z must appear linearly (not quadratically). Option B: x² + y² = 2z + 8 gives circles x² + y² = 2c + 8 for z = c (horizontal sections), and parabolas z = (x² - 8)/2 or z = (y² - 8)/2 for vertical sections through the axes. This describes a circular paraboloid. Option A gives circular hyperboloids, C gives an ellipsoid, and D gives a sphere.
Which statement provides the most accurate geometric description of the surface defined by the equation $z^2 = 4y^2$ in three-dimensional space, $\mathbb{R}^3$ ?
Answer: Two intersecting planes with the line of intersection being the $x$ -axis. The equation $z^2 = 4y^2$ can be rewritten as $z^2 - 4y^2 = 0$ . Factoring this difference of squares gives $(z - 2y)(z + 2y) = 0$ . For this product to be zero, one of the factors must be zero. This means the surface is the union of the points satisfying $z - 2y = 0$ and the points satisfying $z + 2y = 0$ . The equations $z=2y$ and $z=-2y$ each define a plane in $\mathbb{R}^3$ . Since the variable $x$ is missing from both equations, these planes are parallel to the $x$ -axis. Both planes contain the origin $(0,0,0)$ and thus intersect. Their line of intersection is where both equations hold, which requires $y=0$ and $z=0$ . This is precisely the $x$ -axis. Therefore, the surface consists of two planes intersecting at the $x$ -axis.
An elliptic paraboloid is given by the equation $z = \frac{x^2}{4} + \frac{y^2}{9}$ . Which of the following best describes the intersection of this surface with the plane $y=2$ ?
Answer: A parabola in the plane $y=2$ opening in the positive $z$ -direction. To find the intersection, substitute $y=2$ into the equation of the paraboloid: $z = \frac{x^2}{4} + \frac{2^2}{9} = \frac{x^2}{4} + \frac{4}{9}$ . This equation relates $z$ and $x$ . Since $y$ is fixed at 2, this describes a curve in the plane $y=2$ . The equation is of the form $z = ax^2 + k$ with $a>0$ , which is the equation of a parabola opening in the positive $z$ -direction. Its vertex is at $(0, 2, 4/9)$ .
The equation $\frac{(x-2)^2}{9} + \frac{(y+1)^2}{4} + \frac{(z-3)^2}{16} = 0$ represents what type of geometric object in three-dimensional space?
Answer: An ellipsoid with center (2, -1, 3) that has been degenerated to a single point When you encounter an equation that looks like the standard form of an ellipsoid but equals zero instead of one, you need to think about what mathematical conditions could make this true. The given equation has the form of an ellipsoid: $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} + \frac{(z-l)^2}{c^2} = 0$ with center at $(h,k,l) = (2,-1,3)$ . Since we're adding three squared terms (which are always non-negative), the only way their sum can equal zero is if each term individually equals zero. This happens when $(x-2)^2 = 0$ , $(y+1)^2 = 0$ , and $(z-3)^2 = 0$ , which means $x = 2$ , $y = -1$ , and $z = 3$ . The equation is satisfied by exactly one point: $(2,-1,3)$ . Answer A incorrectly describes this as having "zero volume" - while technically true, this misses the key insight that it's a degenerate case. Answer B is wrong because the equation does have a real solution - the single point $(2,-1,3)$ . Answer C calls it a "point ellipsoid" but then mentions semi-axes lengths, which is contradictory since a single point has no meaningful axes. Answer D correctly identifies this as a degenerate ellipsoid that has collapsed to a single point. Remember: when you see a sum of squared terms equal to zero, each squared term must be zero individually. This creates degenerate cases where curves become points, or surfaces become curves or points.

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