What is Triple Integrals Conversion from Rectangular to Cylindrical Coordinates?
Definition:
Substitute x=rcosθ,y=rsinθ,z=z, replace dV with rdzdrdθ, and convert limits to describe the region in r,θ,z.
Example:
The region x2+y2≤1,0≤z≤2 becomes 0≤r≤1,0≤θ≤2π,0≤z≤2.
Reason:
Conversion simplifies integrals involving x2+y2 or circular boundaries, reducing algebraic complexity.
📝 All Triple Integrals Conversion from Rectangular to Cylindrical Coordinates MCQs
Q1. A solid is described in rectangular coordinates by x2+y2≤9 and 0≤z≤5. Which cylindrical-coordinate description represents the same solid?
A.0≤r≤3, 0≤θ≤2π, 0≤z≤5 ✅ B.0≤r≤9, 0≤θ≤π, 0≤z≤5 C.0≤r≤3, 0≤θ≤π, 0≤z≤5 D.0≤r≤9, 0≤θ≤2π, 0≤z≤5 💡 Difficulty: easy | ✅ Correct: A
📖 Explanation: The condition x2+y2≤9 becomes r2≤9, so 0≤r≤3. Because the entire circular region is included, the angle must cover a full revolution, 0≤θ≤2π. The vertical restriction remains unchanged, giving 0≤z≤5.
Q2. When converting a triple integral from rectangular coordinates to cylindrical coordinates, which combination correctly describes the transformations needed for both the integrand and the volume element?
A.Replace x2+y2 by r and use dxdydz=drdθdz B.Replace x by rcosθ, y by rsinθ, and use dV=rdrdθdz ✅ C.Replace x by rsinθ, y by rcosθ, and use dV=drdθdz D.Replace x2+y2 by r2 but keep dV=dxdydz 💡 Difficulty: easy | ✅ Correct: B
📖 Explanation: Cylindrical coordinates use x=rcosθ and y=rsinθ, so x2+y2=r2. The rectangular volume element must also be transformed because the coordinate grid stretches with radius. The correct volume element is dV=rdrdθdz, where the factor r is the Jacobian.
Q3. A region satisfies 4≤x2+y2≤16, 0≤z≤2, and all possible angles are included. A student writes 2≤r≤16. What is the correct radial interval?
B.2≤r≤4 ✅ C.4≤r≤16 💡 Difficulty: medium | ✅ Correct: B
📖 Explanation: Since x2+y2=r2, the inequality becomes 4≤r2≤16. Because r represents distance from the z-axis and is nonnegative, taking square roots gives 2≤r≤4. The student's upper bound incorrectly uses the value 16 without taking its square root.
Q4. A rectangular-coordinate region is restricted by x2+y2≤6x. Which cylindrical-coordinate description correctly captures the radial boundary?
A.0≤r≤6cosθ ✅ B.0≤r≤6sinθ C.0≤r2≤6cosθ D.0≤r≤6 for every θ 💡 Difficulty: medium | ✅ Correct: A
📖 Explanation: Substituting x=rcosθ and x2+y2=r2 gives r2≤6rcosθ. For r≥0, this becomes r≤6cosθ. The region is therefore not a full disk centered at the origin; its radial boundary depends on the angle.
Q5. Suppose a solid has horizontal projection bounded by x2+y2≤4, with 1≤z≤x2+y2+2. Which cylindrical-coordinate limits correctly describe the solid?
A.0≤r≤2, 0≤θ≤2π, 1≤z≤r2+2 ✅ B.0≤r≤4, 0≤θ≤2π, 1≤z≤r+2 C.0≤r≤2, 0≤θ≤π, 1≤z≤r2+2 D.0≤r≤4, 0≤θ≤π, 1≤z≤r2+2 💡 Difficulty: medium | ✅ Correct: A
📖 Explanation: The projection condition x2+y2≤4 becomes r2≤4, hence 0≤r≤2. The full projection requires 0≤θ≤2π. Finally, z lies between 1 and x2+y2+2=r2+2, producing the stated limits.
Q6. A cylindrical region is described by 1≤r≤3, 0≤θ≤π/2, and 0≤z≤4. If the original integrand is x2+y2, which transformed integral correctly represents its triple integral over the region?
A.∫04∫0π/2∫13r2drdθdz B.∫04∫0π/2∫13r3drdθdz ✅ C.∫04∫0π/2∫13rdrdθdz D.∫04∫0π/2∫13r2drdθ 💡 Difficulty: medium | ✅ Correct: B
📖 Explanation: The integrand x2+y2 becomes r2. However, changing coordinates also changes the volume element to rdrdθdz. Therefore the complete integrand-volume combination is r2⋅r=r3. The z-integration must also remain present because the region is three-dimensional.
Q7. A circular plate-shaped projection is centered at (2,0) and satisfies x2+y2≤4x. Which radial boundary results after conversion to cylindrical coordinates?
A.r≤2cosθ B.r≤4cosθ ✅ C.r≤4sinθ D.r≤2sinθ 💡 Difficulty: medium | ✅ Correct: B
📖 Explanation: Using x=rcosθ and x2+y2=r2, the boundary becomes r2≤4rcosθ. For r≥0, divide by r to obtain r≤4cosθ. The factor 4, rather than 2, comes directly from the original equation.
Q8. A student converts x2+y2≤4 into r≤4. The student then argues that the result represents the same region because r is the distance from the origin. What is the key error?
A.The angle should be restricted to 0≤θ≤π B.The correct radial bound is r≤2, because r2=x2+y2 ✅ C.The variable r can be negative D.The z-coordinate must always be replaced by r 💡 Difficulty: easy | ✅ Correct: B
📖 Explanation: The definition r2=x2+y2 is essential. Therefore x2+y2≤4 becomes r2≤4, and because r≥0, the correct result is 0≤r≤2. Using r≤4 would double the disk's radius and therefore describe a larger region.
Q9. A student changes x2+y2 to r2 correctly but writes dV=drdθdz. Another student includes the factor r. Which statement best evaluates their work?
A.The first is correct because only the integrand changes
B.The second is correct because the cylindrical volume element contains the Jacobian factor r ✅ C.Both are correct if the limits are adjusted
D.Neither is correct because cylindrical coordinates require r2 in the volume element 💡 Difficulty: medium | ✅ Correct: B
📖 Explanation: The coordinate transformation changes not only the algebraic form of the integrand but also the volume element. In cylindrical coordinates, dV=rdrdθdz. Omitting r changes the value of the integral even if every transformed function and boundary is otherwise correct.
Q10. A graph of a horizontal region shows a semicircular area above the x-axis with radius 3, centered at the origin. Which angular interval should be used when converting this region to cylindrical coordinates?
A.0≤θ≤2π B.−π/2≤θ≤π/2 C.0≤θ≤π ✅ D.π≤θ≤2π 💡 Difficulty: medium | ✅ Correct: C
📖 Explanation: The upper half-plane corresponds to angles beginning at the positive x-axis and moving counterclockwise to the negative x-axis. Thus the appropriate interval is 0≤θ≤π. A full 2π interval would include the lower semicircle and therefore describe a different region.
Q11. A model describes a solid whose projection is the region between the circles x2+y2=1 and x2+y2=9, while its height is z=10−(x2+y2) above the xy-plane. Which cylindrical description is most appropriate for setting up the integral of 1 over the solid?
A.0≤r≤3, 0≤θ≤2π, 0≤z≤10−r2 B.1≤r≤3, 0≤θ≤2π, 0≤z≤10−r2 ✅ C.1≤r≤9, 0≤θ≤2π, 0≤z≤10−r D.1≤r≤3, 0≤θ≤π, 0≤z≤10−r2 💡 Difficulty: medium | ✅ Correct: B
📖 Explanation: The projection lies between radii 1 and 3, since 1≤x2+y2≤9 becomes 1≤r≤3. The complete annular region requires 0≤θ≤2π. The upper surface becomes z=10−r2, so the vertical bounds are 0≤z≤10−r2.
Q12. Consider the rectangular-coordinate integral over a region where x2+y2≤9: ∭E(x2+y2+z)dV. Which transformed integrand should appear after changing to cylindrical coordinates?
💡 Difficulty: easy | ✅ Correct: A
📖 Explanation: The substitution x2+y2=r2 transforms the first part of the integrand directly into r2. The variable z is already a cylindrical coordinate and therefore remains z. Thus the transformed function is r2+z. The Jacobian factor r belongs to the volume element, not this function itself.
Q13. A region is defined by x2+y2≤2x and lies above the xy-plane. A student uses 0≤r≤2 for every 0≤θ≤2π. Which correction is necessary?
A.Use 0≤r≤2cosθ and restrict angles to where cosθ≥0 ✅ B.Use 0≤r≤2sinθ and let 0≤θ≤2π C.Use 0≤r≤1 for 0≤θ≤2π D.Use 1≤r≤2cosθ and restrict angles to 0≤θ≤π 💡 Difficulty: hard | ✅ Correct: A
📖 Explanation: Substitution gives r2≤2rcosθ, so the radial limit is 0≤r≤2cosθ. Since r cannot have a negative upper bound, only angles satisfying cosθ≥0 are permitted, which can be represented by −π/2≤θ≤π/2.
Q14. Two students convert the same integral over a disk centered at the origin. Student A uses 0≤θ≤2π, while Student B uses −π≤θ≤π. Both use the same radial and vertical limits. Which conclusion is mathematically justified?
A.Only Student A can represent the full disk
B.Only Student B can represent the full disk
C.Both can represent the full disk because each angular interval spans 2π without changing the region ✅ D.Neither can represent the full disk because θ must start at 0 💡 Difficulty: medium | ✅ Correct: C
📖 Explanation: Both intervals have angular width 2π, so each makes one complete revolution around the z-axis. Although the starting angles differ, the geometric region covered is identical. Therefore either interval can correctly describe a full disk, provided the radial and vertical limits are also correct.
Q15. A region in the xy-plane is bounded by x2+y2≤4x and x2+y2≤4y. Which strategy gives the most efficient cylindrical description before introducing a z-range?
A.Use 0≤r≤4 and determine the region only after integration B.Convert the two inequalities to r≤4cosθ and r≤4sinθ, then determine which bound is smaller over the relevant angular intervals ✅ C.Replace both inequalities directly by r2≤4 D.Use 0≤θ≤2π with r bounded by the average of 4cosθ and 4sinθ 💡 Difficulty: hard | ✅ Correct: B
📖 Explanation: The two boundaries become r≤4cosθ and r≤4sinθ. Because both restrictions must hold simultaneously, the radial limit is the smaller of the two expressions. The relevant region lies where both sine and cosine are nonnegative, and comparing them identifies where each boundary controls the radial extent.