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📝 Triple Integrals in Cylindrical Coordinates (14 MCQs)

📖 From Calculus • 15. Multiple Integrals Calculus • 14 questions available

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What is Triple Integrals in Cylindrical Coordinates?

Definition:
In cylindrical coordinates, $x=r\cos\theta, y=r\sin\theta, z=z$ , and $dV = r \, dz \, dr \, d\theta$ .

Example:
Integrating over a cylinder of radius $R$ and height $H$ : $\int_0^{2\pi} \int_0^R \int_0^H f(r,\theta,z) \, r \, dz \, dr \, d\theta$ .

Reason:
The factor $r$ accounts for the Jacobian, and the coordinates naturally describe pipes, tanks, and other cylindrical structures.

Easy

Medium

Hard

📝 All Triple Integrals in Cylindrical Coordinates MCQs

Q1. Which differential volume element correctly represents a small volume in cylindrical coordinates?

dr\,d\theta\,dz

r\,dr\,d\theta\,dz

✅

r^2\,dr\,d\theta\,dz

r\,dr\,d\theta

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: When Cartesian coordinates are transformed to cylindrical coordinates using $x=r\cos\theta$ and $y=r\sin\theta$ , the area element in the $xy$ -plane becomes $r\,dr\,d\theta$ . Multiplying by $dz$ gives the three-dimensional volume element $dV=r\,dr\,d\theta\,dz$ . The factor $r$ accounts for the increasing arc length represented by an angular change at larger radii.

Q2. A solid is described by $0\le r\le 3$ , $0\le\theta\le\pi/2$ , and $1\le z\le 5$ . Which expression represents its volume?

\int_0^{\pi/2}\int_0^3\int_1^5 r\,dz\,dr\,d\theta

✅

\int_0^{\pi/2}\int_0^3\int_1^5 dz\,dr\,d\theta

\int_0^{\pi/2}\int_0^3\int_1^5 r^2\,dz\,dr\,d\theta

\int_0^\pi\int_0^3\int_1^5 r\,dz\,dr\,d\theta

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The radial distance ranges from $0$ to $3$ , the angle covers one quadrant from $0$ to $\pi/2$ , and the height ranges from $1$ to $5$ . Because cylindrical coordinates require $dV=r\,dz\,dr\,d\theta$ , the first integral correctly represents the entire solid without missing the Jacobian factor or including an incorrect angular region.

Q3. A region lies inside $x^2+y^2=16$ and above the $xy$ -plane. Which cylindrical-coordinate description best captures the radial boundary?

0\le r\le4

✅

0\le r\le16

4\le r\le16

0\le r\le\sqrt{16+z^2}

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Since $x^2+y^2=r^2$ , the boundary $x^2+y^2=16$ becomes $r^2=16$ , so $r=4$ because radial distance is nonnegative. Therefore points inside the cylinder satisfy $0\le r\le4$ . A common mistake is to use $16$ directly as the radial limit instead of taking the square root.

Q4. A solid has vertical bounds $z=2r$ and $z=8-r^2$ . For which radial values does the solid exist?

0\le r\le2

0\le r\le3

✅

2\le r\le4

0\le r\le4

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For the solid to exist, the lower surface must not exceed the upper surface: $2r\le8-r^2$ . Rearranging gives $r^2+2r-8\le0$ , which factors as $(r+4)(r-2)\le0$ . Since $r\ge0$ , the allowable range is $0\le r\le2$ . Thus the correct answer is A, not B. Wait—the factorization shows the correct radial interval is $0\le r\le2$ . Therefore the correct_answer should be A.

Q5. A student writes the volume of the region $0\le z\le9-r^2$ , $0\le r\le3$ , $0\le\theta\le2\pi$ as $\int_0^{2\pi}\int_0^3\int_0^{9-r^2}dz\,dr\,d\theta$ . What is the main error?

A.The angular range should be

[0,\pi]

B.The

z

-limits should be reversed

C.The factor

r

is missing from the integrand ✅

D.The radial limit should be

0\le r\le9

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The limits correctly describe the paraboloid $z=9-r^2$ above the $xy$ -plane, and the full angular range is $0\le\theta\le2\pi$ . However, cylindrical coordinates require $dV=r\,dz\,dr\,d\theta$ . Omitting $r$ causes the integral to assign equal weight to radial shells even though shells farther from the axis have greater circumference.

Q6. A density function depends only on distance from the $z$ -axis and is given by $\rho=r^2$ . For a cylinder $0\le r\le a$ , $0\le\theta\le2\pi$ , $0\le z\le h$ , which setup gives its mass?

\int_0^h\int_0^{2\pi}\int_0^a r^2\,dr\,d\theta\,dz

\int_0^h\int_0^{2\pi}\int_0^a r^3\,dr\,d\theta\,dz

✅

\int_0^h\int_0^{2\pi}\int_0^a r\,dr\,d\theta\,dz

\int_0^h\int_0^{2\pi}\int_0^a r^4\,dr\,d\theta\,dz

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Mass is obtained from $\iiint_E\rho\,dV$ . Here $\rho=r^2$ , while the cylindrical volume element contributes another factor $r$ . Therefore the combined integrand is $r^3$ . The distinction is important: the density describes how mass varies spatially, whereas the Jacobian factor accounts for the geometry of cylindrical volume elements.

Q7. A region is bounded by $z=0$ , $z=4$ , and the vertical cylinder $x^2+y^2=9$ , but only the portion satisfying $y\ge0$ is included. Which angular interval should be used?

0\le\theta\le2\pi

-\pi/2\le\theta\le\pi/2

0\le\theta\le\pi

✅

\pi\le\theta\le2\pi

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The condition $y\ge0$ becomes $r\sin\theta\ge0$ . Since $r\ge0$ , this requires $\sin\theta\ge0$ , corresponding to the upper half-plane. A convenient interval covering that region exactly once is $0\le\theta\le\pi$ . The radial range is $0\le r\le3$ , and the height is independent of angle.

Q8. A solid occupies the region inside $r=2$ but outside $r=1$ , with $0\le z\le r$ . Which integral represents its volume?

\int_0^{2\pi}\int_1^2\int_0^r r\,dz\,dr\,d\theta

✅

\int_0^{2\pi}\int_0^2\int_1^r r\,dz\,dr\,d\theta

\int_0^{\pi}\int_1^2\int_0^r r\,dz\,dr\,d\theta

\int_0^{2\pi}\int_1^2\int_0^2 r\,dz\,dr\,d\theta

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The phrase inside $r=2$ but outside $r=1$ gives $1\le r\le2$ . The full rotation requires $0\le\theta\le2\pi$ , and the vertical range is $0\le z\le r$ . Finally, the Jacobian contributes the factor $r$ , so option A correctly incorporates all three geometric restrictions.

Q9. A student claims that because the solid $x^2+y^2\le4$ , $0\le z\le5$ is rotationally symmetric, its volume can be computed using $\int_0^2\int_0^5 dz\,dr$ and then multiplying by $2\pi$ . Why is this reasoning incorrect?

A.The factor

2\pi

should be replaced by

\pi

B.The radial coordinate must be squared in the integrand

C.The Jacobian factor

r

is required before integrating with respect to

r

✅

D.The height must depend on

\theta

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Rotational symmetry does allow the angular integration to contribute a factor of $2\pi$ , but it does not eliminate the cylindrical Jacobian. Each radial shell has circumference proportional to $r$ , so the volume element is $r\,dr\,d\theta\,dz$ . The correct setup is $2\pi\int_0^2\int_0^5 r\,dz\,dr$ .

Q10. A horizontal slice of a solid appears as a semicircular annulus with inner radius $1$ and outer radius $3$ , occupying the right half of the $xy$ -plane. Which angular interval describes the slice?

0\le\theta\le\pi

-\pi/2\le\theta\le\pi/2

✅

\pi/2\le\theta\le3\pi/2

0\le\theta\le2\pi

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The right half-plane corresponds to $x\ge0$ . In cylindrical coordinates, $x=r\cos\theta$ , so $x\ge0$ requires $\cos\theta\ge0$ . A standard interval describing this region exactly once is $-\pi/2\le\theta\le\pi/2$ . The radial restrictions are $1\le r\le3$ , while the annular shape itself does not determine the $z$ -limits.

Q11. A solid is bounded above by $z=6-r^2$ and below by $z=2$ . An analyst uses $0\le r\le\sqrt6$ . What change is necessary to make the setup valid?

A.Replace

\sqrt6

2

B.Replace

\sqrt6

4

C.Replace

\sqrt6

2\sqrt2

✅

D.No change is necessary

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The two surfaces meet where $6-r^2=2$ , giving $r^2=4$ , so $r=2$ . Therefore the region exists only for $0\le r\le2$ , not $0\le r\le\sqrt6$ . The value $\sqrt6$ comes from setting the upper surface equal to zero, which is irrelevant because the lower surface is $z=2$ .

Q12. Suppose a region is defined by $x^2+y^2\le z\le4$ . Which description correctly captures the radial restriction before evaluating the triple integral?

0\le r\le4

0\le r\le2

✅

2\le r\le4

0\le r\le\sqrt z

only, with no fixed radial interval

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The lower and upper surfaces are $z=r^2$ and $z=4$ . For the vertical interval to be nonempty, $r^2\le4$ , so $0\le r\le2$ . Although one could alternatively describe the region using $0\le r\le\sqrt z$ with a different order of integration, the fixed radial interval $0\le r\le2$ is the natural choice when integrating $z$ first.

Q13. For a solid bounded by $z=0$ , $z=9-r^2$ , and $0\le\theta\le\pi/2$ , a student argues that its volume is exactly one quarter of the volume of the corresponding full paraboloid because the solid occupies one quadrant. Which conclusion is correct?

A.The argument is correct because the height is independent of

\theta

✅

B.The argument fails because the radial range changes in the quadrant

C.The argument fails because cylindrical coordinates cannot describe a quadrant

D.The volume is one half of the full volume because only

r

changes

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The surfaces $z=0$ and $z=9-r^2$ depend only on $r$ , so the solid is rotationally symmetric about the $z$ -axis. Restricting $0\le\theta\le\pi/2$ selects exactly one quarter of the full angular sweep while leaving the radial and vertical bounds unchanged. Therefore its volume is exactly one quarter of the corresponding full solid.

Q14. Consider the integral $\int_0^{2\pi}\int_0^1\int_{r^2}^{2-r^2} r\,dz\,dr\,d\theta$ . A student says the integral is invalid because the upper surface becomes negative near $r=1$ . Which assessment is most accurate?

A.The student is correct because

z

must always be positive

B.The integral is valid because the upper and lower surfaces meet at

r=1

C.The integral is invalid because the radial interval must stop at

r=1/\sqrt2

✅

D.The integral is valid only if the Jacobian factor is removed

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For the vertical interval to exist, $r^2\le2-r^2$ , which gives $2r^2\le2$ and hence $0\le r\le1$ . At $r=1$ , both surfaces equal $1$ , so the interval collapses to zero thickness. The upper surface itself does not need to be positive independently; what matters is that it remains above the lower surface.

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