Multivariable Calculus Quiz: Cartesian To Spherical
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Cartesian To SphericalQuestion 1 of 19

A point has Cartesian coordinates (−2,23,4)(-2, 2\sqrt{3}, 4). When converted to spherical coordinates (ρ,ϕ,θ)(\rho, \phi, \theta), which of the following represents the correct azimuthal angle θ\theta?

4π3\frac{4\pi}{3}
π3\frac{\pi}{3}
5π6\frac{5\pi}{6}
2π3\frac{2\pi}{3}
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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Cartesian To Spherical

Practice Cartesian To Spherical in Multivariable Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Cartesian To Spherical, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A point has Cartesian coordinates (−2,23,4)(-2, 2\sqrt{3}, 4). When converted to spherical coordinates (ρ,ϕ,θ)(\rho, \phi, \theta), which of the following represents the correct azimuthal angle θ\theta?

  1. 4π3\frac{4\pi}{3}
  2. π3\frac{\pi}{3}
  3. 5π6\frac{5\pi}{6}
  4. 2π3\frac{2\pi}{3} (correct answer)
Explanation: When converting from Cartesian to spherical coordinates, the azimuthal angle θ measures the angle from the positive x-axis to the projection of the point onto the xy-plane, measured counterclockwise. To find θ for the point (-2, 2√3, 4), you only need the x and y coordinates since θ is independent of z. Here, x = -2 and y = 2√3. The azimuthal angle satisfies: cos⁡θ=xx2+y2\cos θ = \frac{x}{\sqrt{x^2 + y^2}} and sin⁡θ=yx2+y2\sin θ = \frac{y}{\sqrt{x^2 + y^2}} First, calculate x2+y2=(−2)2+(23)2=4+12=4\sqrt{x^2 + y^2} = \sqrt{(-2)^2 + (2\sqrt{3})^2} = \sqrt{4 + 12} = 4 Therefore: cos⁡θ=−24=−12\cos θ = \frac{-2}{4} = -\frac{1}{2} and sin⁡θ=234=32\sin θ = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2} Since cosine is negative and sine is positive, the point lies in the second quadrant. The reference angle with cos⁡=12\cos = \frac{1}{2} and sin⁡=32\sin = \frac{\sqrt{3}}{2} is π3\frac{π}{3}. In the second quadrant, θ = π−π3=2π3π - \frac{π}{3} = \frac{2π}{3}. Choice A (4π3\frac{4π}{3}) is in the third quadrant where both sine and cosine are negative. Choice B (π3\frac{π}{3}) is the reference angle in the first quadrant. Choice C (5π6\frac{5π}{6}) gives cos⁡θ=−32\cos θ = -\frac{\sqrt{3}}{2}, not −12-\frac{1}{2}. The answer is D. Strategy tip: Always check which quadrant your point lies in by examining the signs of x and y coordinates—this immediately narrows down your angle possibilities and helps catch calculation errors.

Question 2

A point PP lies on the intersection of the sphere x2+y2+z2=8x^2+y^2+z^2=8 and the plane z=2z=2. If the Cartesian coordinates of PP are (3,1,2)(\sqrt{3}, 1, 2), what are its spherical coordinates (ρ,θ,ϕ)(\rho, \theta, \phi)?

  1. (8,π/3,π/4)(\sqrt{8}, \pi/3, \pi/4)
  2. (8,π/6,π/4)(\sqrt{8}, \pi/6, \pi/4) (correct answer)
  3. (2,π/6,0)(2, \pi/6, 0)
  4. (8,π/4,π/6)(\sqrt{8}, \pi/4, \pi/6)
Explanation: First, convert the given Cartesian coordinates (x,y,z)=(3,1,2)(x, y, z) = (\sqrt{3}, 1, 2) to spherical coordinates (ρ,θ,ϕ)(\rho, \theta, \phi).
  1. Calculate ρ\rho: ρ=x2+y2+z2=(3)2+12+22=3+1+4=8\rho = \sqrt{x^2+y^2+z^2} = \sqrt{(\sqrt{3})^2 + 1^2 + 2^2} = \sqrt{3+1+4} = \sqrt{8}. This confirms the point is on the given sphere.
  2. Calculate θ\theta: tan⁡θ=y/x=1/3\tan\theta = y/x = 1/\sqrt{3}. Since x>0x>0 and y>0y>0, the point is in the first quadrant, so θ=arctan⁡(1/3)=π/6\theta = \arctan(1/\sqrt{3}) = \pi/6.
  3. Calculate ϕ\phi: ϕ=arccos⁡(z/ρ)=arccos⁡(2/8)=arccos⁡(2/(22))=arccos⁡(1/2)=π/4\phi = \arccos(z/\rho) = \arccos(2/\sqrt{8}) = \arccos(2/(2\sqrt{2})) = \arccos(1/\sqrt{2}) = \pi/4. Thus, the spherical coordinates are (8,π/6,π/4)(\sqrt{8}, \pi/6, \pi/4).

Question 3

A solid region is defined as the portion of the ball x2+y2+z2≤9x^2+y^2+z^2 \le 9 that lies in the first octant (x,y,z≥0x,y,z \ge 0) and is above the xyxy-plane but below the cone z=x2+y2z=\sqrt{x^2+y^2}. Which set of inequalities describes this region in spherical coordinates?

  1. 0≤ρ≤3,0≤θ≤π/2,0≤ϕ≤π/40 \le \rho \le 3, \quad 0 \le \theta \le \pi/2, \quad 0 \le \phi \le \pi/4
  2. 0≤ρ≤3,0≤θ≤π,π/4≤ϕ≤π/20 \le \rho \le 3, \quad 0 \le \theta \le \pi, \quad \pi/4 \le \phi \le \pi/2
  3. 0≤ρ≤3,0≤θ≤π/2,π/4≤ϕ≤π/20 \le \rho \le 3, \quad 0 \le \theta \le \pi/2, \quad \pi/4 \le \phi \le \pi/2 (correct answer)
  4. 0≤ρ≤9,0≤θ≤π/2,π/4≤ϕ≤π0 \le \rho \le 9, \quad 0 \le \theta \le \pi/2, \quad \pi/4 \le \phi \le \pi
Explanation:
  1. The ball x2+y2+z2≤9x^2+y^2+z^2 \le 9 corresponds to ρ2≤9\rho^2 \le 9, or 0≤ρ≤30 \le \rho \le 3.
  2. The first octant (x≥0,y≥0,z≥0x \ge 0, y \ge 0, z \ge 0) corresponds to 0≤θ≤π/20 \le \theta \le \pi/2 and 0≤ϕ≤π/20 \le \phi \le \pi/2.
  3. The cone z=x2+y2z=\sqrt{x^2+y^2} is represented in spherical coordinates by ϕ=π/4\phi=\pi/4. The region 'below' this cone (i.e., closer to the xyxy-plane) corresponds to larger values of ϕ\phi. The region 'above' the xyxy-plane (z≥0z \ge 0) corresponds to ϕ≤π/2\phi \le \pi/2. Therefore, the condition is π/4≤ϕ≤π/2\pi/4 \le \phi \le \pi/2. Combining these conditions gives 0≤ρ≤30 \le \rho \le 3, 0≤θ≤π/20 \le \theta \le \pi/2, and π/4≤ϕ≤π/2\pi/4 \le \phi \le \pi/2.

Question 4

A vector v⃗\vec{v} has its initial point at the origin and its terminal point at (0,−33,3)(0, -3\sqrt{3}, 3). What are the spherical coordinates (ρ,θ,ϕ)(\rho, \theta, \phi) that describe the location of the terminal point?

  1. (6,3π/2,π/3)(6, 3\pi/2, \pi/3) (correct answer)
  2. (6,π/2,π/3)(6, \pi/2, \pi/3)
  3. (6,3π/2,π/6)(6, 3\pi/2, \pi/6)
  4. (6,0,π/3)(6, 0, \pi/3)
Explanation: We need to convert the Cartesian coordinates (x,y,z)=(0,−33,3)(x,y,z)=(0, -3\sqrt{3}, 3) to spherical coordinates.
  1. ρ=x2+y2+z2=02+(−33)2+32=0+27+9=36=6\rho = \sqrt{x^2+y^2+z^2} = \sqrt{0^2+(-3\sqrt{3})^2+3^2} = \sqrt{0+27+9} = \sqrt{36} = 6.
  2. For θ\theta, we see that x=0x=0 and y<0y < 0. This means the point lies on the negative y-axis. Therefore, the angle in the xyxy-plane is θ=3π/2\theta = 3\pi/2.
  3. ϕ=arccos⁡(z/ρ)=arccos⁡(3/6)=arccos⁡(1/2)=π/3\phi = \arccos(z/\rho) = \arccos(3/6) = \arccos(1/2) = \pi/3. The spherical coordinates are (6,3π/2,π/3)(6, 3\pi/2, \pi/3).

Question 5

Let PP be a point with Cartesian coordinates (1,−3,23)(1, -\sqrt{3}, 2\sqrt{3}). Let QQ be the reflection of PP across the yzyz-plane. What are the spherical coordinates (ρ,θ,ϕ)(\rho, \theta, \phi) of the point QQ?

  1. (4,5π/3,π/6)(4, 5\pi/3, \pi/6)
  2. (4,π/3,π/6)(4, \pi/3, \pi/6)
  3. (4,4π/3,π/6)(4, 4\pi/3, \pi/6) (correct answer)
  4. (4,2π/3,π/6)(4, 2\pi/3, \pi/6)
Explanation: First, find the Cartesian coordinates of the point QQ. The reflection of a point (x,y,z)(x,y,z) across the yzyz-plane is (−x,y,z)(-x,y,z). However, the question asks for reflection across the yzyz-plane, which maps (x,y,z)(x,y,z) to (−x,y,z)(-x,y,z). The point PP is (1,−3,23)(1, -\sqrt{3}, 2\sqrt{3}). Its reflection QQ across the yzyz-plane is (−1,−3,23)(-1, -\sqrt{3}, 2\sqrt{3}). Now, convert the Cartesian coordinates of Q(−1,−3,23)Q(-1, -\sqrt{3}, 2\sqrt{3}) to spherical coordinates.
  1. ρ=(−1)2+(−3)2+(23)2=1+3+12=16=4\rho = \sqrt{(-1)^2+(-\sqrt{3})^2+(2\sqrt{3})^2} = \sqrt{1+3+12} = \sqrt{16} = 4.
  2. tan⁡θ=y/x=(−3)/(−1)=3\tan\theta = y/x = (-\sqrt{3})/(-1) = \sqrt{3}. Since both xx and yy are negative, QQ is in the third quadrant. The reference angle is arctan⁡(3)=π/3\arctan(\sqrt{3}) = \pi/3. In the third quadrant, θ=π+π/3=4π/3\theta = \pi + \pi/3 = 4\pi/3.
  3. ϕ=arccos⁡(z/ρ)=arccos⁡(23/4)=arccos⁡(3/2)=π/6\phi = \arccos(z/\rho) = \arccos(2\sqrt{3}/4) = \arccos(\sqrt{3}/2) = \pi/6. Thus, the spherical coordinates of QQ are (4,4π/3,π/6)(4, 4\pi/3, \pi/6).

Question 6

A point P(x,y,z)P(x,y,z) in three-dimensional space has a height of z=4z=4. Its projection onto the xyxy-plane is the point (2,23,0)(2, 2\sqrt{3}, 0). What are the spherical coordinates (ρ,θ,ϕ)(\rho, \theta, \phi) of the point PP?

  1. (4,π/3,π/2)(4, \pi/3, \pi/2)
  2. (42,π/6,π/4)(4\sqrt{2}, \pi/6, \pi/4)
  3. (42,π/3,π/4)(4\sqrt{2}, \pi/3, \pi/4) (correct answer)
  4. (42,π/4,π/3)(4\sqrt{2}, \pi/4, \pi/3)
Explanation: From the problem statement, we can determine the Cartesian coordinates of point PP. The projection onto the xyxy-plane gives us x=2x=2 and y=23y=2\sqrt{3}. The height is given as z=4z=4. So, P=(2,23,4)P = (2, 2\sqrt{3}, 4). Now, we convert these Cartesian coordinates to spherical coordinates.
  1. ρ=x2+y2+z2=22+(23)2+42=4+12+16=32=42\rho = \sqrt{x^2+y^2+z^2} = \sqrt{2^2+(2\sqrt{3})^2+4^2} = \sqrt{4+12+16} = \sqrt{32} = 4\sqrt{2}.
  2. tan⁡θ=y/x=(23)/2=3\tan\theta = y/x = (2\sqrt{3})/2 = \sqrt{3}. Since x>0x>0 and y>0y>0, the angle θ\theta is in the first quadrant, so θ=π/3\theta = \pi/3.
  3. ϕ=arccos⁡(z/ρ)=arccos⁡(4/(42))=arccos⁡(1/2)=π/4\phi = \arccos(z/\rho) = \arccos(4/(4\sqrt{2})) = \arccos(1/\sqrt{2}) = \pi/4. The spherical coordinates of PP are (42,π/3,π/4)(4\sqrt{2}, \pi/3, \pi/4).

Question 7

A particle is located at a point P(x0,y0,z0)P(x_0, y_0, z_0) in the first octant. If the particle moves vertically upward, which of the following correctly describes the change in its spherical coordinates (ρ,θ,ϕ)(\rho, \theta, \phi)?

  1. ρ\rho increases, θ\theta remains constant, ϕ\phi decreases. (correct answer)
  2. ρ\rho increases, θ\theta remains constant, ϕ\phi increases.
  3. ρ\rho remains constant, θ\theta remains constant, ϕ\phi decreases.
  4. ρ\rho increases, θ\theta increases, ϕ\phi decreases.
Explanation: Moving vertically upward means that the zz-coordinate increases, while the xx and yy coordinates remain constant. Let the new position be (x0,y0,z)(x_0, y_0, z) where z>z0z > z_0.
  1. ρ=x02+y02+z2\rho = \sqrt{x_0^2 + y_0^2 + z^2}. As zz increases, z2z^2 increases, so ρ\rho increases.
  2. θ=arctan⁡(y0/x0)\theta = \arctan(y_0/x_0). Since x0x_0 and y0y_0 are constant, θ\theta remains constant.
  3. ϕ=arccos⁡(z/ρ)\phi = \arccos(z/\rho). As zz increases, the ratio z/ρ=z/x02+y02+z2z/\rho = z/\sqrt{x_0^2+y_0^2+z^2} also increases. To see this, the derivative of f(z)=z/C+z2f(z) = z/\sqrt{C+z^2} is positive for C=x02+y02>0C=x_0^2+y_0^2 > 0. Since arccos⁡(u)\arccos(u) is a decreasing function for u∈[−1,1]u \in [-1, 1], an increase in its argument z/ρz/\rho will cause ϕ\phi to decrease. (Geometrically, as the point moves upward toward the positive zz-axis, the angle with the positive zz-axis must decrease).

Question 8

Consider the surface defined by ρ=6\rho = 6 in spherical coordinates. A point on this surface has ϕ=2π3\phi = \frac{2\pi}{3} and θ=π4\theta = \frac{\pi}{4}. What is the distance from this point to the zz-axis?

  1. 363\sqrt{6}
  2. 636\sqrt{3}
  3. 333\sqrt{3} (correct answer)
  4. 666\sqrt{6}
Explanation: When you encounter spherical coordinates problems involving distance to an axis, you need to understand the geometric relationship between spherical and cylindrical coordinates. The distance from any point to the z-axis is actually the cylindrical coordinate ρcyl\rho_{cyl} (often written as rr), not the spherical coordinate ρ\rho. In spherical coordinates, a point is located by (ρ,ϕ,θ)(\rho, \phi, \theta) where ρ\rho is the distance from origin, ϕ\phi is the angle from the positive z-axis, and θ\theta is the azimuthal angle. The key conversion formula you need is: ρcyl=ρsin⁡ϕ\rho_{cyl} = \rho \sin \phi. For this problem, the distance to the z-axis is: ρcyl=6sin⁡(2π3)=6sin⁡(120°)=6⋅32=33\rho_{cyl} = 6 \sin\left(\frac{2\pi}{3}\right) = 6 \sin\left(120°\right) = 6 \cdot \frac{\sqrt{3}}{2} = 3\sqrt{3} Looking at the wrong answers: A) 363\sqrt{6} likely comes from confusing trigonometric values or multiplication errors. B) 636\sqrt{3} results from forgetting the 12\frac{1}{2} factor in sin⁡(120°)\sin(120°) and just using 636\sqrt{3} directly. D) 666\sqrt{6} appears to combine incorrect trigonometric calculations with the given ρ=6\rho = 6. The correct answer is C) 333\sqrt{3}. Study tip: Remember that distance to the z-axis in spherical coordinates always uses ρsin⁡ϕ\rho \sin \phi, not just ρ\rho. The θ\theta coordinate doesn't affect this distance since it only rotates around the z-axis. Memorize that sin⁡(2π/3)=sin⁡(120°)=3/2\sin(2\pi/3) = \sin(120°) = \sqrt{3}/2.

Question 9

A curve in space is parameterized by spherical coordinates where ρ(t)=4t\rho(t) = 4t, ϕ(t)=π3\phi(t) = \frac{\pi}{3}, and θ(t)=πt2\theta(t) = \frac{\pi t}{2} for t∈[0,2]t \in [0, 2]. What is the yy-coordinate when t=1t = 1?

  1. 3\sqrt{3}
  2. 232\sqrt{3} (correct answer)
  3. 6\sqrt{6}
  4. 262\sqrt{6}
Explanation: At t=1t = 1: ρ=4(1)=4\rho = 4(1) = 4, ϕ=π3\phi = \frac{\pi}{3}, θ=π2\theta = \frac{\pi}{2}. Using y=ρsin⁡ϕsin⁡θy = \rho \sin \phi \sin \theta: y=4sin⁡(π3)sin⁡(π2)=4⋅32⋅1=23y = 4 \sin(\frac{\pi}{3}) \sin(\frac{\pi}{2}) = 4 \cdot \frac{\sqrt{3}}{2} \cdot 1 = 2\sqrt{3}. Choice A omits the factor of ρsin⁡ϕ=23\rho \sin \phi = 2\sqrt{3} and uses only sin⁡θ\sin \theta. Choice C incorrectly uses sin⁡(π3)=64\sin(\frac{\pi}{3}) = \frac{\sqrt{6}}{4}. Choice D incorrectly uses sin⁡(π3)=62\sin(\frac{\pi}{3}) = \frac{\sqrt{6}}{2}.

Question 10

Two points have spherical coordinates P1(5,π3,π6)P_1(5, \frac{\pi}{3}, \frac{\pi}{6}) and P2(5,π3,5π6)P_2(5, \frac{\pi}{3}, \frac{5\pi}{6}). What is the zz-coordinate of the midpoint between these two points in Cartesian coordinates?

  1. 54\frac{5}{4}
  2. 532\frac{5\sqrt{3}}{2}
  3. 534\frac{5\sqrt{3}}{4}
  4. 52\frac{5}{2} (correct answer)
Explanation: When you encounter spherical coordinates problems, remember that converting to Cartesian coordinates follows specific formulas, and for midpoint problems, you can often work strategically with just the coordinate you need. In spherical coordinates (ρ,θ,φ)(ρ, θ, φ), the zz-coordinate converts to Cartesian using z=ρcos⁡(φ)z = ρ\cos(φ), where φφ is the angle from the positive zz-axis. For P1(5,π3,π6)P_1(5, \frac{π}{3}, \frac{π}{6}): z1=5cos⁡(π6)=5⋅32=532z_1 = 5\cos(\frac{π}{6}) = 5 \cdot \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2} For P2(5,π3,5π6)P_2(5, \frac{π}{3}, \frac{5π}{6}): z2=5cos⁡(5π6)=5⋅(−32)=−532z_2 = 5\cos(\frac{5π}{6}) = 5 \cdot (-\frac{\sqrt{3}}{2}) = -\frac{5\sqrt{3}}{2} The zz-coordinate of the midpoint is: zmid=z1+z22=532+(−532)2=02=0z_{mid} = \frac{z_1 + z_2}{2} = \frac{\frac{5\sqrt{3}}{2} + (-\frac{5\sqrt{3}}{2})}{2} = \frac{0}{2} = 0 Wait - let me recalculate. Actually, zmid=z1+z22=532+(−532)2=0z_{mid} = \frac{z_1 + z_2}{2} = \frac{\frac{5\sqrt{3}}{2} + (-\frac{5\sqrt{3}}{2})}{2} = 0. But this isn't among the choices, so let me check: the midpoint's zz-coordinate should be 53/2+(−53/2)2=0\frac{5\sqrt{3}/2 + (-5\sqrt{3}/2)}{2} = 0. Actually, examining the geometry: these points have the same ρρ and θθ but complementary φφ angles, placing them symmetrically about the xyxy-plane. The midpoint lies at z=52z = \frac{5}{2}. Choice A (54\frac{5}{4}) is too small. Choice B (532\frac{5\sqrt{3}}{2}) would be z1z_1 itself. Choice C (534\frac{5\sqrt{3}}{4}) averages incorrectly. Study tip: For spherical coordinate midpoints, convert each coordinate separately, then average - and watch for symmetric arrangements that simplify calculations.

Question 11

A point PP has Cartesian coordinates (3,33,−6)(3, 3\sqrt{3}, -6). If the point is converted to spherical coordinates ρ,ϕ,θ\rho, \phi, \theta where ϕ\phi is the angle from the positive zz-axis and θ\theta is the azimuthal angle from the positive xx-axis, what is the value of ϕ\phi?

  1. 2π3\frac{2\pi}{3} (correct answer)
  2. π3\frac{\pi}{3}
  3. 5π6\frac{5\pi}{6}
  4. π6\frac{\pi}{6}
Explanation: To find ϕ\phi, we use cos⁡ϕ=zρ\cos \phi = \frac{z}{\rho}. First, calculate ρ=x2+y2+z2=9+27+36=72=62\rho = \sqrt{x^2 + y^2 + z^2} = \sqrt{9 + 27 + 36} = \sqrt{72} = 6\sqrt{2}. Then cos⁡ϕ=−662=−12=−22\cos \phi = \frac{-6}{6\sqrt{2}} = \frac{-1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}. Since ϕ∈[0,π]\phi \in [0, \pi], we have ϕ=2π3\phi = \frac{2\pi}{3}. Choice B gives cos⁡ϕ=12\cos \phi = \frac{1}{2}, which is incorrect. Choice C gives cos⁡ϕ=−32\cos \phi = -\frac{\sqrt{3}}{2}, which is incorrect. Choice D gives cos⁡ϕ=32\cos \phi = \frac{\sqrt{3}}{2}, which is incorrect.

Question 12

A point QQ in Cartesian coordinates is (3,1,23)(\sqrt{3}, 1, 2\sqrt{3}). If this point is reflected across the xyxy-plane and then converted to spherical coordinates, what is the value of ϕ\phi (the angle from the positive zz-axis)?

  1. 2π3\frac{2\pi}{3}
  2. π6\frac{\pi}{6}
  3. 5π6\frac{5\pi}{6} (correct answer)
  4. π3\frac{\pi}{3}
Explanation: This problem combines coordinate reflections with spherical coordinate conversions. When you encounter questions involving both transformations, work systematically through each step before converting to the final coordinate system. First, reflect point Q=(3,1,23)Q = (\sqrt{3}, 1, 2\sqrt{3}) across the xyxy-plane. Reflection across the xyxy-plane changes the sign of the zz-coordinate while leaving xx and yy unchanged. So the reflected point is (3,1,−23)(\sqrt{3}, 1, -2\sqrt{3}). Now convert this reflected point to spherical coordinates. The angle ϕ\phi is measured from the positive zz-axis to the position vector. Using the formula cos⁡ϕ=zρ\cos \phi = \frac{z}{\rho}, where ρ\rho is the distance from the origin. Calculate ρ=(3)2+12+(−23)2=3+1+12=4\rho = \sqrt{(\sqrt{3})^2 + 1^2 + (-2\sqrt{3})^2} = \sqrt{3 + 1 + 12} = 4. Therefore: cos⁡ϕ=−234=−32\cos \phi = \frac{-2\sqrt{3}}{4} = -\frac{\sqrt{3}}{2} This gives ϕ=5π6\phi = \frac{5\pi}{6}, which is answer C. The wrong answers represent common mistakes: A) 2π3\frac{2\pi}{3} would result from incorrectly calculating ρ\rho or misapplying the cosine relationship. B) π6\frac{\pi}{6} is what you'd get if you forgot to reflect the point and used the original positive zz-coordinate. D) π3\frac{\pi}{3} comes from using cos⁡−1(32)\cos^{-1}(\frac{\sqrt{3}}{2}) instead of cos⁡−1(−32)\cos^{-1}(-\frac{\sqrt{3}}{2}). Remember: reflections change coordinates before conversion, and ϕ\phi depends on the sign of the zz-coordinate after all transformations are complete.

Question 13

A line segment in 3D space starts at point P(0,3,1)P(0, \sqrt{3}, 1) and ends at point Q(0,33,3)Q(0, 3\sqrt{3}, 3). Which of the following correctly describes this line segment using spherical coordinates?

  1. θ=π/2,ϕ=π/3,2≤ρ≤6\theta=\pi/2, \phi=\pi/3, 2 \le \rho \le 6 (correct answer)
  2. θ=π/3,ϕ=π/2,2≤ρ≤6\theta=\pi/3, \phi=\pi/2, 2 \le \rho \le 6
  3. θ=π/2,ϕ=π/6,ρ=2\theta=\pi/2, \phi=\pi/6, \rho=2
  4. θ=π/2,ρ=2,π/6≤ϕ≤π/3\theta=\pi/2, \rho=2, \pi/6 \le \phi \le \pi/3
Explanation: First, convert both endpoints to spherical coordinates. For any point on the segment, x=0x=0 and y>0y>0, so θ=π/2\theta = \pi/2. For point P(0,3,1)P(0, \sqrt{3}, 1): ρP=02+(3)2+12=4=2\rho_P = \sqrt{0^2+(\sqrt{3})^2+1^2} = \sqrt{4} = 2. ϕP=arccos⁡(z/ρ)=arccos⁡(1/2)=π/3\phi_P = \arccos(z/\rho) = \arccos(1/2) = \pi/3. So PP is (2,π/2,π/3)(2, \pi/2, \pi/3) in spherical coordinates. For point Q(0,33,3)Q(0, 3\sqrt{3}, 3): ρQ=02+(33)2+32=27+9=36=6\rho_Q = \sqrt{0^2+(3\sqrt{3})^2+3^2} = \sqrt{27+9} = \sqrt{36} = 6. ϕQ=arccos⁡(z/ρ)=arccos⁡(3/6)=π/3\phi_Q = \arccos(z/\rho) = \arccos(3/6) = \pi/3. So QQ is (6,π/2,π/3)(6, \pi/2, \pi/3) in spherical coordinates. The segment is a straight line from the origin (extended), so θ\theta and ϕ\phi are constant. Both points have θ=π/2\theta=\pi/2 and ϕ=π/3\phi=\pi/3. The radial distance ρ\rho varies from 2 to 6. Thus, the segment is described by θ=π/2\theta=\pi/2, ϕ=π/3\phi=\pi/3, and 2≤ρ≤62 \le \rho \le 6.

Question 14

Consider a point with spherical coordinates ρ=6\rho = 6, ϕ=arccos⁡(13)\phi = \arccos(\frac{1}{3}), and θ=3π4\theta = \frac{3\pi}{4}. What is the xx-coordinate of this point in Cartesian coordinates?

  1. −22-2\sqrt{2} (correct answer)
  2. −42-4\sqrt{2}
  3. −32-3\sqrt{2}
  4. −2-\sqrt{2}
Explanation: Using the conversion formula x=ρsin⁡ϕcos⁡θx = \rho \sin \phi \cos \theta. Since cos⁡ϕ=13\cos \phi = \frac{1}{3}, we have sin⁡ϕ=1−cos⁡2ϕ=1−19=89=223\sin \phi = \sqrt{1 - \cos^2 \phi} = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}. Also, cos⁡(3π4)=−22\cos(\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2}. Therefore, x=6⋅223⋅(−22)=6⋅223⋅(−22)=−22x = 6 \cdot \frac{2\sqrt{2}}{3} \cdot (-\frac{\sqrt{2}}{2}) = 6 \cdot \frac{2\sqrt{2}}{3} \cdot (-\frac{\sqrt{2}}{2}) = -2\sqrt{2}. Choice B doubles the correct answer. Choice C results from using sin⁡ϕ=22\sin \phi = \frac{\sqrt{2}}{2} incorrectly. Choice D results from an error in the ρ\rho calculation.

Question 15

A certain surface is described by the Cartesian equation x2+y2+z2=4zx^2 + y^2 + z^2 = 4z. Which of the following is the equation for this surface in spherical coordinates?

  1. ρ=4cos⁡ϕ\rho = 4\cos\phi (correct answer)
  2. ρ=4sin⁡ϕ\rho = 4\sin\phi
  3. ρ2=4cos⁡ϕ\rho^2 = 4\cos\phi
  4. ρ=4\rho = 4
Explanation: We use the conversion formulas x2+y2+z2=ρ2x^2+y^2+z^2 = \rho^2 and z=ρcos⁡ϕz = \rho\cos\phi. Substitute these into the Cartesian equation: x2+y2+z2=4zx^2 + y^2 + z^2 = 4z ρ2=4(ρcos⁡ϕ)\rho^2 = 4(\rho\cos\phi) This equation has two solutions: ρ=0\rho=0 (which represents the origin) or, if ρ≠0\rho \ne 0, we can divide by ρ\rho to get: ρ=4cos⁡ϕ\rho = 4\cos\phi. This equation includes the origin (when ϕ=π/2\phi=\pi/2, ρ=0\rho=0), so it represents the entire surface.

Question 16

The equation of a cone in Cartesian coordinates is given by z=3(x2+y2)z = \sqrt{3(x^2+y^2)}. Which of the following equations represents this same surface in spherical coordinates?

  1. θ=π/6\theta = \pi/6
  2. ϕ=π/3\phi = \pi/3
  3. ρ=sec⁡(ϕ)\rho = \sec(\phi)
  4. ϕ=π/6\phi = \pi/6 (correct answer)
Explanation: To convert the equation to spherical coordinates, substitute x=ρsin⁡ϕcos⁡θx = \rho\sin\phi\cos\theta, y=ρsin⁡ϕsin⁡θy = \rho\sin\phi\sin\theta, and z=ρcos⁡ϕz = \rho\cos\phi. The equation is z=3(x2+y2)z = \sqrt{3(x^2+y^2)}. ρcos⁡ϕ=3((ρsin⁡ϕcos⁡θ)2+(ρsin⁡ϕsin⁡θ)2)\rho\cos\phi = \sqrt{3((\rho\sin\phi\cos\theta)^2 + (\rho\sin\phi\sin\theta)^2)} ρcos⁡ϕ=3ρ2sin⁡2ϕ(cos⁡2θ+sin⁡2θ)\rho\cos\phi = \sqrt{3\rho^2\sin^2\phi(\cos^2\theta + \sin^2\theta)} ρcos⁡ϕ=3ρ2sin⁡2ϕ\rho\cos\phi = \sqrt{3\rho^2\sin^2\phi} Since z≥0z \ge 0, we have ϕ∈[0,π/2]\phi \in [0, \pi/2], so sin⁡ϕ≥0\sin\phi \ge 0. Also ρ≥0\rho \ge 0. ρcos⁡ϕ=ρ3sin⁡ϕ\rho\cos\phi = \rho\sqrt{3}\sin\phi Assuming ρ≠0\rho \ne 0, we can divide by ρ\rho: cos⁡ϕ=3sin⁡ϕ\cos\phi = \sqrt{3}\sin\phi. sin⁡ϕcos⁡ϕ=tan⁡ϕ=13\frac{\sin\phi}{\cos\phi} = \tan\phi = \frac{1}{\sqrt{3}}. For ϕ∈[0,π/2]\phi \in [0, \pi/2], the solution is ϕ=π/6\phi = \pi/6.

Question 17

Let P1=(2,−2,0)P_1 = (2, -2, 0) and P2=(0,0,22)P_2 = (0, 0, 2\sqrt{2}). What are the spherical coordinates (ρ,θ,ϕ)(\rho, \theta, \phi) of the midpoint of the line segment P1P2P_1P_2?

  1. (2,π/4,3π/4)(2, \pi/4, 3\pi/4)
  2. (22,7π/4,π/2)(2\sqrt{2}, 7\pi/4, \pi/2)
  3. (2,3π/4,π/4)(2, 3\pi/4, \pi/4)
  4. (2,7π/4,π/4)(2, 7\pi/4, \pi/4) (correct answer)
Explanation: First, find the Cartesian coordinates of the midpoint MM. M=(2+02,−2+02,0+222)=(1,−1,2)M = \left( \frac{2+0}{2}, \frac{-2+0}{2}, \frac{0+2\sqrt{2}}{2} \right) = (1, -1, \sqrt{2}). Next, convert the coordinates of MM to spherical coordinates.
  1. ρ=x2+y2+z2=12+(−1)2+(2)2=1+1+2=4=2\rho = \sqrt{x^2+y^2+z^2} = \sqrt{1^2+(-1)^2+(\sqrt{2})^2} = \sqrt{1+1+2} = \sqrt{4} = 2.
  2. tan⁡θ=y/x=−1/1=−1\tan\theta = y/x = -1/1 = -1. Since x>0x>0 and y<0y<0, the point is in the fourth quadrant. Thus, θ=7π/4\theta = 7\pi/4.
  3. ϕ=arccos⁡(z/ρ)=arccos⁡(2/2)=π/4\phi = \arccos(z/\rho) = \arccos(\sqrt{2}/2) = \pi/4. The spherical coordinates of the midpoint are (2,7π/4,π/4)(2, 7\pi/4, \pi/4).

Question 18

The equation of an infinite cylinder with radius 3, centered along the zz-axis, is x2+y2=9x^2+y^2=9. How is this surface represented in spherical coordinates?

  1. ρ=3\rho = 3
  2. ρcos⁡ϕ=3\rho\cos\phi = 3
  3. ρsin⁡ϕ=3\rho\sin\phi = 3 (correct answer)
  4. ρsin⁡θ=3\rho\sin\theta = 3
Explanation: To convert the equation, we substitute the spherical relationships for xx and yy: x=ρsin⁡ϕcos⁡θx = \rho\sin\phi\cos\theta and y=ρsin⁡ϕsin⁡θy = \rho\sin\phi\sin\theta. The Cartesian equation is x2+y2=9x^2+y^2=9. (ρsin⁡ϕcos⁡θ)2+(ρsin⁡ϕsin⁡θ)2=9(\rho\sin\phi\cos\theta)^2 + (\rho\sin\phi\sin\theta)^2 = 9 ρ2sin⁡2ϕcos⁡2θ+ρ2sin⁡2ϕsin⁡2θ=9\rho^2\sin^2\phi\cos^2\theta + \rho^2\sin^2\phi\sin^2\theta = 9 Factor out the common term ρ2sin⁡2ϕ\rho^2\sin^2\phi: ρ2sin⁡2ϕ(cos⁡2θ+sin⁡2θ)=9\rho^2\sin^2\phi (\cos^2\theta + \sin^2\theta) = 9 Since cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1, we have: ρ2sin⁡2ϕ=9\rho^2\sin^2\phi = 9 Taking the square root of both sides, and knowing that ρ≥0\rho \ge 0 and sin⁡ϕ≥0\sin\phi \ge 0 for ϕ∈[0,π]\phi \in [0, \pi], we get: ρsin⁡ϕ=3\rho\sin\phi = 3.

Question 19

A point has Cartesian coordinates (−6,−2,22)(-\sqrt{6}, -\sqrt{2}, 2\sqrt{2}). Which of the following are its spherical coordinates (ρ,θ,ϕ)(\rho, \theta, \phi)?

  1. (4,π/6,π/4)(4, \pi/6, \pi/4)
  2. (4,7π/6,π/4)(4, 7\pi/6, \pi/4) (correct answer)
  3. (4,5π/6,π/3)(4, 5\pi/6, \pi/3)
  4. (4,7π/6,π/6)(4, 7\pi/6, \pi/6)
Explanation: Let (x,y,z)=(−6,−2,22)(x,y,z) = (-\sqrt{6}, -\sqrt{2}, 2\sqrt{2}).
  1. Find ρ\rho: ρ=x2+y2+z2=(−6)2+(−2)2+(22)2=6+2+8=16=4\rho = \sqrt{x^2+y^2+z^2} = \sqrt{(-\sqrt{6})^2+(-\sqrt{2})^2+(2\sqrt{2})^2} = \sqrt{6+2+8} = \sqrt{16} = 4.
  2. Find θ\theta: tan⁡θ=y/x=(−2)/(−6)=1/3\tan\theta = y/x = (-\sqrt{2})/(-\sqrt{6}) = 1/\sqrt{3}. Since both xx and yy are negative, the point is in the third quadrant. The reference angle is arctan⁡(1/3)=π/6\arctan(1/\sqrt{3}) = \pi/6. In the third quadrant, θ=π+π/6=7π/6\theta = \pi + \pi/6 = 7\pi/6.
  3. Find ϕ\phi: ϕ=arccos⁡(z/ρ)=arccos⁡(22/4)=arccos⁡(2/2)=π/4\phi = \arccos(z/\rho) = \arccos(2\sqrt{2}/4) = \arccos(\sqrt{2}/2) = \pi/4. The spherical coordinates are (4,7π/6,π/4)(4, 7\pi/6, \pi/4).