Answer: Circle of radius 1; circle of radius $\sqrt{5}$; hyperbola

Trace at z = 0: Setting $z = 0$: $x^2 + y^2 = 1$. This is a circle of radius 1. Trace at z = 2: Setting $z = 2$: $x^2 + y^2 = 1 + 4 = 5$. This is a circle of radius $\sqrt{5}$. Trace at y = 0: Setting $y = 0$: $x^2 - z^2 = 1$. This is a hyperbola in the $xz$-plane. Why distractors fail: Option B incorrectly identifies horizontal traces as ellipses and the vertical trace as a parabola. Option C gives the wrong radius at $z = 2$. Option D incorrectly says the $z = 0$ trace is a hyperbola.

Answer: The angle measured from the positive $z$-axis to the line segment from the origin to the point

Identify what each spherical coordinate measures: In the convention $(\rho, \phi, \theta)$: $\rho$ is the distance from the origin, $\phi$ is the polar angle from the positive $z$-axis, and $\theta$ is the azimuthal angle in the $xy$-plane. Why Option B is correct: $\phi$ is the angle between the positive $z$-axis and the line from the origin to the point, with $0 \le \phi \le \pi$. Why distractors fail: Option A describes $\rho$. Option C describes $\theta$. Option D describes $r$ in cylindrical coordinates.

Answer: An ellipse

Set z = 0 in the equation: Substituting $z = 0$ gives $\frac{x^2}{4} + \frac{y^2}{9} = 1$. Identify the resulting curve: The equation $\frac{x^2}{4} + \frac{y^2}{9} = 1$ is the standard form of an ellipse with semi-axes $a = 2$ and $b = 3$. Why distractors fail: A hyperbola (A) would require a minus sign between the two terms. A circle (B) would require equal denominators. A parabola (D) would require one variable to be linear, not squared.

Answer: $x = r\cos\theta$, $y = r\sin\theta$

Recall cylindrical coordinate definitions: Cylindrical coordinates extend polar coordinates into 3D. In the $xy$-plane, the polar relationships are $x = r\cos\theta$ and $y = r\sin\theta$, with $z$ unchanged. Why Option B is correct: This matches the standard definition: $x = r\cos\theta$ and $y = r\sin\theta$. Why distractors fail: Option A swaps cosine and sine. Options C and D use the same trigonometric function for both $x$ and $y$, which cannot correctly parameterize a circle.

Answer: $\frac{x^2}{4} + \frac{y^2}{9} + \frac{z^2}{16} = 1$

Recall the standard form of an ellipsoid: An ellipsoid has the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$, where all three squared terms are positive and summed to equal 1. Why Option A is correct: Option A has three positive squared terms summing to 1, which matches the ellipsoid standard form with $a=2$, $b=3$, $c=4$. Why distractors fail: Option B has a negative $z^2$ term, making it a hyperboloid of one sheet. Option C equals $z$ (not 1), making it an elliptic paraboloid. Option D has two negative terms, making it a hyperboloid of two sheets.

Answer: Cylindrical coordinates with $0 \le r \le \sqrt{3}$, $0 \le \theta \le 2\pi$, $1 \le z \le \sqrt{4 - r^2}$

Determine the geometry: The region is a spherical cap: inside the sphere $x^2+y^2+z^2=4$ (radius 2) and above the plane $z=1$. The sphere and plane intersect where $x^2+y^2 = 4-1 = 3$, giving a maximum radial distance of $r = \sqrt{3}$ in the $xy$-plane. Set up cylindrical bounds: In cylindrical coordinates: $r$ ranges from $0$ to $\sqrt{3}$ (the radius of the intersection circle), $\theta$ ranges from $0$ to $2\pi$, and for each fixed $r$, $z$ runs from the plane $z = 1$ up to the sphere $z = \sqrt{4 - r^2}$. These bounds are correct and clean. Why Option B is incorrect: Using $0 \le \rho \le 2$ and $0 \le \phi \le \frac{\pi}{3}$ describes a full spherical sector down to the origin — it does not enforce the flat lower boundary $z = 1$. Points with small $\rho$ near the origin would be incorrectly included. Why Option C is incorrect: Using $r \le 2$ is too large — at $z = 1$, the sphere only reaches $r = \sqrt{3} \approx 1.73$, not $r = 2$. Using $z \le 2$ as a flat upper bound ignores the curved spherical cap shape, incorrectly including points outside the sphere. Why Option D is incorrect: While the $\phi$ and $\rho$ bounds in Option D are set up to account for both boundaries, the lower bound $\rho = \frac{1}{\cos\phi} = \sec\phi$ (from $z = \rho\cos\phi = 1$) is a $\phi$-dependent lower limit, making the bounds significantly more complex to work with. Option A in cylindrical coordinates is simpler, has constant limits on $r$ and $ heta$, and is the most straightforward and least error-prone setup for this region.

Answer: The $x$-axis

Identify the positive term: In a hyperboloid of two sheets, the variable with the positive coefficient determines the axis along which the two sheets open. Why Option C is correct: The $x^2$ term is the only positive one. Setting $y = z = 0$ gives $\frac{x^2}{4} = 1$, so $x = \pm 2$. The surface exists only where $|x| \ge 2$, forming two sheets separated along the $x$-axis. Why distractors fail: Options A and B name axes whose variables have negative signs. Option D is incorrect because a hyperboloid of two sheets always opens along a single axis.

Answer: The conversion is correct, but the surface is a cylinder (not a sphere) because there is no $z$-dependence

Verify the conversion: Multiplying both sides of $r = 6\sin\theta$ by $r$: $r^2 = 6r\sin\theta$. Since $r^2 = x^2+y^2$ and $r\sin\theta = y$, we get $x^2+y^2 = 6y$. The conversion is correct. Classify the surface in 3D: Completing the square: $x^2 + (y-3)^2 = 9$. This is a circle of radius 3 centered at $(0,3)$ in the $xy$-plane. Crucially, the equation places no restriction on $z$, so the surface extends infinitely along the $z$-axis — making it a circular cylinder, not a sphere. Why a sphere requires z-dependence: A sphere centered at $(0, 3, 0)$ with radius 3 would require the equation $x^2 + (y-3)^2 + z^2 = 9$, which involves $z$. The absence of $z$ in the original equation is the key reason the student's classification fails. Why distractors fail: Option A is incorrect — $r\sin\theta = y$, not $x$, so the conversion to $x^2+y^2=6y$ is valid. Option C is wrong because the lack of $z$-dependence makes it a cylinder, not a sphere. Option D is wrong because cones require $z$ to appear explicitly as a squared or linear term.

Answer: A cone; $x^2 + y^2 = \frac{z^2}{3}$

Interpret φ = π/6: In spherical coordinates, a fixed $\phi$ means the angle from the positive $z$-axis is constant. This generates a half-cone (for $\rho \ge 0$) with vertex at the origin. Derive the Cartesian equation: From $\cos\phi = \frac{\sqrt{3}}{2}$ and $\sin\phi = \frac{1}{2}$: $\tan\phi = \frac{1}{\sqrt{3}}$. Since $\tan\phi = \frac{\sqrt{x^2+y^2}}{z}$, we get $\frac{x^2+y^2}{z^2} = \frac{1}{3}$, i.e., $x^2 + y^2 = \frac{z^2}{3}$. Why distractors fail: Option A confuses a constant angle with a constant radius. Option B confuses $\phi = \text{const}$ with $z = \text{const}$ (a horizontal plane). Option D describes a cylinder, which has no dependence on $z$.

Answer: A sphere of radius 2 centered at $(0, 0, 2)$

Multiply both sides by ρ: Multiplying by $\rho$: $\rho^2 = 4\rho\cos\phi$. Since $\rho^2 = x^2+y^2+z^2$ and $\rho\cos\phi = z$, this becomes $x^2+y^2+z^2 = 4z$. Complete the square: $x^2 + y^2 + z^2 - 4z = 0 \Rightarrow x^2 + y^2 + (z-2)^2 = 4$. This is a sphere of radius 2 centered at $(0,0,2)$. Why distractors fail: Option A: a cone would have the form $\phi = \text{const}$. Option B: a sphere of radius 4 at the origin would be $\rho = 4$. Option D: a paraboloid is not obtained from this type of equation.

Answer: Elliptic paraboloid

Identify the surface structure: The equation $z = x^2 + y^2$ has two squared variables on one side and a single linear variable on the other. This is the hallmark of a paraboloid. Why Option C is correct: Since both squared terms are positive and have the same sign, the cross-sections parallel to the $xy$-plane are circles (a special case of ellipses), making this an elliptic paraboloid opening upward. Why distractors fail: An ellipsoid (A) requires all three variables squared and set equal to 1. A hyperboloid of one sheet (B) has one negative squared term. An elliptic cone (D) has the form $z^2 = x^2 + y^2$, with $z$ also squared.

Answer: $(3, \frac{\pi}{2}, 4)$

Compute r: $r = \sqrt{x^2 + y^2} = \sqrt{0 + 9} = 3$. Compute θ: Since $x = 0$ and $y = 3 > 0$, the point lies on the positive $y$-axis, so $\theta = \frac{\pi}{2}$. Identify z: In cylindrical coordinates, $z$ remains unchanged, so $z = 4$. Why distractors fail: Option A uses $\theta = 0$, which corresponds to the positive $x$-axis. Option C uses $r = 5$, which is $\sqrt{x^2+y^2+z^2}$ (the spherical $\rho$). Option D uses $\theta = \pi$, which is the negative $x$-axis.

Answer: Elliptic paraboloid

Rewrite in standard form: Rearranging: $4x^2 + y^2 = 8z$, or equivalently $\frac{x^2}{2} + \frac{y^2}{8} = z$. Classify the surface: This has the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = z$, which is an elliptic paraboloid opening in the positive $z$-direction. Why distractors fail: An elliptic cone (A) would require $z^2$ on the right or a degenerate equation equal to 0 with all squared terms. A hyperboloid of one sheet (C) requires opposite signs on squared terms equal to a nonzero constant. A hyperbolic paraboloid (D) requires one positive and one negative squared term.

Answer: Because the right-hand side is 0 instead of 1, causing the surface to degenerate to a point at the origin

Compare cone and hyperboloid forms: Both the cone and the hyperboloid of one sheet have the pattern of two positive and one negative squared term. The key difference is that the cone has $= 0$ on the right, while the hyperboloid has $= 1$. Why Option A is correct: Setting the equation equal to 0 means the surface passes through the origin and degenerates: at $z = 0$, only the point $(0,0,0)$ satisfies it. The cross-sections for $z = k \neq 0$ are ellipses that grow linearly with $|k|$, forming a cone. Why distractors fail: Option B is insufficient — hyperboloids also have a negative squared term. Option C is irrelevant — both cones and hyperboloids have three squared terms. Option D is false — $a$, $b$, $c$ can differ for hyperboloids.

Answer: $z = 5 - x^2 - y^2$

Check circular cross-sections: For cross-sections parallel to the $xy$-plane to be circles, the coefficients of $x^2$ and $y^2$ must be equal in magnitude. Option A: $z = 5 - x^2 - y^2$ gives $x^2 + y^2 = 5 - z$ (circles for $z < 5$). Check the point (0, 0, 5): Substituting $(0,0,5)$ into $z = 5 - x^2 - y^2$: $5 = 5 - 0 - 0 = 5$. ✓ Check downward opening: The negative coefficients on $x^2$ and $y^2$ mean $z$ decreases as $x$ or $y$ move away from zero, so the paraboloid opens downward. ✓ Why distractors fail: Option B opens upward (positive coefficients on $x^2$ and $y^2$) and passes through $(0,0,-5)$. Option C is a sphere, not a paraboloid — it doesn't open in one direction. Option D has $-x^2 + y^2$, giving hyperbolic (not circular) cross-sections — it is a hyperbolic paraboloid.

Answer: $x^2 + y^2 + z^2 = 16$

Identify rotational symmetry about the z-axis: A surface is independent of $\theta$ in cylindrical coordinates if and only if it has rotational symmetry about the $z$-axis, meaning $x$ and $y$ appear only through $x^2 + y^2$. Test Option B: $x^2 + y^2 + z^2 = 16$ becomes $r^2 + z^2 = 16$, which depends only on $r$ and $z$. Why distractors fail: Option A: $x^2 - y^2 = r^2\cos^2\theta - r^2\sin^2\theta = r^2\cos(2\theta)$, which depends on $\theta$. Option C: $x^2 + 2xy + y^2 = (x+y)^2 = r^2(\cos\theta + \sin\theta)^2$, which depends on $\theta$. Option D: $xz = rz\cos\theta$, which depends on $\theta$.

Answer: The right-hand side is 0 rather than a nonzero constant, so the surface is a cone, not a hyperboloid

Distinguish cones from hyperboloids: The standard hyperboloid of one sheet has the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1$, while a cone has $\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 0$. The right-hand side is the critical distinction. Why Option B is correct: Setting the equation equal to 0 means the surface degenerates: it passes through the origin and has a vertex there. The cross-sections at constant $z$ grow linearly, forming a cone. Why distractors fail: Option A is false — the rearrangement is algebraically valid. Option C is wrong — a paraboloid has one variable appearing linearly, not squared on both sides. Option D imposes a nonexistent requirement.

Answer: The two squared terms have opposite signs, making it a hyperbolic paraboloid rather than an elliptic paraboloid

Identify the sign structure: In $y^2 - x^2 = z$, the two squared terms have opposite signs: $y^2$ is positive and $x^2$ is negative. Distinguish elliptic from hyperbolic paraboloid: An elliptic paraboloid has the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = z$ (same signs). A hyperbolic paraboloid has the form $\frac{y^2}{b^2} - \frac{x^2}{a^2} = z$ (opposite signs). The sign difference is the defining distinction. Why distractors fail: Option A is false — the number of terms is fine for a paraboloid. Option C is false — any variable can be the linear one, depending on orientation. Option D is wrong — a hyperboloid requires all three variables to be squared.

Answer: $r = 3$

Use the cylindrical coordinate identity: Since $x^2 + y^2 = r^2$ in cylindrical coordinates, the equation $x^2 + y^2 = 9$ becomes $r^2 = 9$, so $r = 3$ (taking the positive root). Why Option B is correct: $r = 3$ represents a cylinder of radius 3 centered on the $z$-axis, which is exactly the surface $x^2 + y^2 = 9$. Why distractors fail: Option A confuses $r^2 = 9$ with $r = 9$. Option C introduces $z$, which is not in the original equation. Option D describes a different surface (a cylinder offset from the $z$-axis).

Answer: $(2, \frac{\pi}{4}, \frac{\pi}{4})$

Compute ρ: $\rho = \sqrt{x^2 + y^2 + z^2} = \sqrt{1 + 1 + 2} = 2$. Compute φ: $\cos\phi = \frac{z}{\rho} = \frac{\sqrt{2}}{2}$, so $\phi = \frac{\pi}{4}$. Compute θ: Since $x = 1$ and $y = 1$, $\tan\theta = \frac{y}{x} = 1$, giving $\theta = \frac{\pi}{4}$. Why distractors fail: Option B uses $\theta = \frac{\pi}{2}$, which would require $x = 0$. Option C uses $\rho = \sqrt{2}$, omitting $z^2$. Option D uses $\phi = \frac{\pi}{3}$, which would give $\cos\phi = \frac{1}{2}$.

Answer: Spherical coordinates, because the cone becomes $\phi = \frac{\pi}{4}$ and the sphere becomes $\rho = 2\sqrt{2}$

Express the cone in spherical coordinates: The cone $z = \sqrt{x^2+y^2}$ means $\rho\cos\phi = \rho\sin\phi$, which simplifies to $\tan\phi = 1$, so $\phi = \frac{\pi}{4}$. Express the sphere in spherical coordinates: The sphere $x^2+y^2+z^2 = 8$ becomes $\rho^2 = 8$, so $\rho = 2\sqrt{2}$. Why spherical is most efficient: Both boundaries become single-coordinate constraints ($\phi = \frac{\pi}{4}$ and $\rho = 2\sqrt{2}$), making the region trivially described by constant bounds on individual coordinates. Why distractors fail: Option A keeps both surfaces as multi-variable expressions. Option B simplifies the cone to $z = r$ but the sphere remains two-variable. Option D is just cylindrical coordinates described differently.

Answer: They add a third coordinate $z$ to polar coordinates $(r, \theta)$, leaving the $xy$-plane parameterization unchanged

Relate polar and cylindrical coordinates: Polar coordinates $(r, \theta)$ describe points in the plane. Cylindrical coordinates $(r, \theta, z)$ simply append a vertical height $z$, using the same polar description for the horizontal plane. Why Option B is correct: The relationships $x = r\cos\theta$, $y = r\sin\theta$ remain identical; only $z$ is added as an independent coordinate. Why distractors fail: Option A describes something closer to spherical coordinates. Option C conflates cylindrical with spherical. Option D incorrectly restricts their applicability.