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📝 Triple Integrals in Cylindrical and Spherical Coordinates (16 MCQs)

📖 From Calculus • 15. Multiple Integrals Calculus • 16 questions available

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What is Triple Integrals in Cylindrical and Spherical Coordinates?

Definition:
Cylindrical coordinates $(r, \theta, z)$ and spherical coordinates $(\rho, \theta, \phi)$ provide alternative systems for integrating over regions with cylindrical or spherical symmetry.

Example:
A cylinder is simple in cylindrical coords; a sphere is simple in spherical coords.

Reason:
These coordinate systems align with the symmetry of many physical objects, simplifying limits and integrands compared to Cartesian coordinates.

Easy

Medium

Hard

📝 All Triple Integrals in Cylindrical and Spherical Coordinates MCQs

Q1. A solid is rotationally symmetric about the $z$ -axis and its boundary is naturally described by $x^2+y^2=4$ . Which coordinate system is most likely to simplify the triple integral?

A.Cartesian coordinates

B.Cylindrical coordinates ✅

C.Spherical coordinates

D.A two-dimensional coordinate system

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Cylindrical coordinates are well suited because $x^2+y^2$ becomes $r^2$ , so the circular boundary becomes simply $r=2$ . This reduces the geometric complexity of the limits and avoids repeatedly expressing the same circular relationship in Cartesian variables.

Q2. When converting a volume integral from Cartesian coordinates to cylindrical coordinates, why does the differential volume become $dV=r\,dr\,d\theta\,dz$ ?

A.Because

z

changes with

r

B.Because the angular direction represents a changing arc length ✅

C.Because cylindrical coordinates contain three independent angles

D.Because the solid must be symmetric about the

z

-axis

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: In cylindrical coordinates, a small displacement in the angular direction corresponds to an arc length of approximately $r\,d\theta$ . Therefore, a small volume element has dimensions $dr$ , $r\,d\theta$ , and $dz$ , producing the factor $r$ .

Q3. A student argues that the cylindrical-coordinate limits $0\le r\le2$ , $0\le\theta\le2\pi$ , and $0\le z\le4-r^2$ describe only half of the solid because $\theta$ does not appear in the upper boundary. What is the best response?

A.The student is correct because

\theta

must always appear in every boundary

B.The limits describe the full solid because the boundary is rotationally symmetric ✅

C.The limits describe only a quarter of the solid

D.The limits are invalid because

r

cannot depend on

z

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The absence of $\theta$ from the boundary does not mean that the solid is incomplete. It indicates rotational symmetry about the $z$ -axis. Allowing $\theta$ to range from $0$ to $2\pi$ sweeps the entire circular region, while $z$ ranges between the specified surfaces.

Q4. Consider the region inside the sphere $x^2+y^2+z^2=9$ . Which spherical-coordinate description correctly represents the entire sphere?

0\le\rho\le3,\ 0\le\phi\le\pi,\ 0\le\theta\le2\pi

✅

0\le\rho\le9,\ 0\le\phi\le\pi,\ 0\le\theta\le2\pi

0\le\rho\le3,\ 0\le\phi\le\pi/2,\ 0\le\theta\le\pi

0\le\rho\le9,\ 0\le\phi\le\pi/2,\ 0\le\theta\le\pi

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: In spherical coordinates, $x^2+y^2+z^2=\rho^2$ , so the sphere of radius $3$ becomes $0\le\rho\le3$ . Covering the entire sphere requires the polar angle $\phi$ to range from $0$ to $\pi$ and the azimuthal angle $\theta$ to range through a full $2\pi$ .

Q5. A solid lies inside $x^2+y^2+z^2\le16$ and above the plane $z=2$ . Which feature makes spherical coordinates particularly useful for setting up its volume integral?

A.The plane becomes a constant value of

z

B.The spherical boundary becomes

\rho=4

, while the plane can be expressed using

\rho

and

\phi

✅

C.The

x

- and

y

-coordinates disappear completely

D.The Jacobian becomes

1

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The spherical boundary is especially simple because $x^2+y^2+z^2=\rho^2$ , giving $\rho=4$ . The plane $z=2$ can be written as $\rho\cos\phi=2$ , which leads to a manageable angular relationship and often makes the geometry easier to analyze.

Q6. A region is bounded by $z=0$ , $z=9-x^2-y^2$ , and $x^2+y^2\le9$ . A student writes the cylindrical integral for volume without the factor $r$ . What is the most likely consequence?

A.The result will still be correct because

r

is already contained in

x^2+y^2

B.The integral will generally underestimate or overestimate the volume because the cylindrical Jacobian is missing ✅

C.Only the angular limits will be affected

D.The integral will calculate surface area instead of volume

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The factor $r$ is essential because cylindrical coordinates stretch the volume element in the angular direction. Omitting it means the integral does not represent the actual three-dimensional volume element. Even if all geometric limits are correct, the resulting numerical value will generally be incorrect.

Q7. A cylindrical tank occupies $0\le r\le3$ , $0\le\theta\le2\pi$ , and $0\le z\le5$ . If the density depends only on distance from the axis according to $\delta(r)=2+r$ , which integral correctly models the total mass?

\int_0^5\int_0^{2\pi}\int_0^3(2+r)\,dr\,d\theta\,dz

\int_0^5\int_0^{2\pi}\int_0^3(2+r)r\,dr\,d\theta\,dz

✅

\int_0^3\int_0^{2\pi}\int_0^5(2+r)r\,dz\,dr\,d\theta

only if the tank has zero volume

\int_0^3\int_0^5\int_0^{2\pi}(2+r)\,d\theta\,dz\,dr

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Mass is obtained by integrating density over volume. In cylindrical coordinates, $dV=r\,dr\,d\theta\,dz$ , so the density $2+r$ must be multiplied by $r$ . The complete limits cover the entire circular tank and its full height, giving the correct physical model.

Q8. A spherical-coordinate setup for the region inside $\rho=5$ uses $0\le\phi\le\pi/3$ . A student claims this describes the entire sphere because $\theta$ still ranges from $0$ to $2\pi$ . What region is actually represented?

A.The entire sphere

B.A cone-shaped portion near the positive

z

-axis ✅

C.A hemisphere below the

xy

-plane

D.A cylindrical region

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The polar angle $\phi$ is measured from the positive $z$ -axis. Restricting it to $0\le\phi\le\pi/3$ selects only points within a cone around that axis. Although $\theta$ completes a full revolution, it cannot compensate for the restricted polar angle.

Q9. A region is enclosed between the paraboloid $z=x^2+y^2$ and the plane $z=6$ . Which cylindrical-coordinate limits most directly describe the solid?

0\le r\le6,\ r^2\le z\le6

0\le r\le\sqrt6,\ r^2\le z\le6

✅

0\le r\le\sqrt6,\ 0\le z\le r^2

0\le r\le6,\ 0\le z\le r^2

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The paraboloid becomes $z=r^2$ in cylindrical coordinates. Its intersection with $z=6$ occurs when $r^2=6$ , so $r=\sqrt6$ . For each point in the circular projection, $z$ extends from the paraboloid $r^2$ upward to the plane $6$ .

Q10. A graph shows a solid that is symmetric about the $z$ -axis, has a circular projection in the $xy$ -plane, and is bounded above and below by surfaces depending only on $x^2+y^2$ . Which approach would most naturally exploit the geometry?

A.Use cylindrical coordinates so radial symmetry is represented by

r

✅

B.Use spherical coordinates regardless of the vertical boundaries

C.Use Cartesian coordinates because symmetry prevents simplification

D.Use only polar coordinates because the region has volume

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The described symmetry is specifically rotational symmetry around the $z$ -axis. Cylindrical coordinates convert $x^2+y^2$ into $r^2$ , making both the circular projection and radial boundary simpler. Spherical coordinates are more advantageous when spherical surfaces or cones dominate the geometry.

Q11. A student evaluates the volume of a sphere of radius $a$ using spherical coordinates but obtains $\frac{4\pi a^3}{3}$ after integrating $\rho^2\sin\phi$ . Another student obtains $\frac{4\pi a^2}{3}$ . Which conclusion is justified?

A.The second answer is correct because volume has units of area

B.The first answer is dimensionally and mathematically consistent; the second is missing one power of length ✅

C.Both are correct for different coordinate conventions

D.Neither can be correct because spherical coordinates cannot represent a sphere

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The volume of a three-dimensional sphere must have units proportional to length cubed. In spherical coordinates, the Jacobian contributes $\rho^2\sin\phi$ , and integrating $\rho^2$ from $0$ to $a$ produces a factor proportional to $a^3$ . Thus the first result has the correct dimensional form.

Q12. A solid occupies the portion of the sphere $\rho\le6$ lying above the cone $\phi=\pi/4$ . Which description best captures the geometry needed before setting up the integral?

A.All points with

0\le\phi\le\pi/4

are included for every

0\le\rho\le6

B.The radial limit is fixed at

6

, while the angular limit selects points closer to the positive

z

-axis ✅

C.The cone determines

\rho

, so

\rho

must vary from

0

\pi/4

D.The azimuthal angle must be restricted because the cone removes half the sphere

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The sphere supplies the radial boundary $\rho=6$ , while the cone controls the polar angle. Points above the cone lie closer to the positive $z$ -axis, corresponding to $0\le\phi\le\pi/4$ . The full rotation around the axis remains, so $\theta$ ranges through $2\pi$ .

Q13. For the integral $\iiint_E (x^2+y^2+z^2)\,dV$ over a sphere of radius $3$ , which setup is most efficient in spherical coordinates?

\int_0^{2\pi}\int_0^\pi\int_0^3 \rho^4\sin\phi\,d\rho\,d\phi\,d\theta

✅

\int_0^{2\pi}\int_0^\pi\int_0^3 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta

\int_0^{2\pi}\int_0^\pi\int_0^3 \rho^3\sin\phi\,d\rho\,d\phi\,d\theta

\int_0^{\pi}\int_0^{2\pi}\int_0^3 \rho^4\,d\rho\,d\phi\,d\theta

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: In spherical coordinates, $x^2+y^2+z^2=\rho^2$ . The volume element contributes another factor of $\rho^2\sin\phi$ , so the integrand becomes $\rho^4\sin\phi$ . The radial limit is $0\le\rho\le3$ , while the full sphere requires $0\le\phi\le\pi$ and $0\le\theta\le2\pi$ .

Q14. A region is inside the sphere $\rho\le4$ and above the plane $z=\sqrt3\,r$ . Which polar-angle condition follows from the plane when spherical coordinates are used?

\phi\le\pi/6

✅

\phi\le\pi/3

\phi\ge\pi/3

\phi\ge2\pi/3

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Using $z=\rho\cos\phi$ and $r=\rho\sin\phi$ , the plane becomes $\rho\cos\phi=\sqrt3\rho\sin\phi$ . For nonzero $\rho$ , this gives $\tan\phi=1/\sqrt3$ , so $\phi=\pi/6$ . Being above the plane means points are closer to the positive $z$ -axis, hence $\phi\le\pi/6$ .

Q15. A student sets up the volume of the unit sphere using spherical coordinates as $\int_0^1\int_0^\pi\int_0^{2\pi}\rho^2\,d\theta\,d\phi\,d\rho$ . Which single correction is most important?

A.Replace

\rho^2

with

\rho

B.Insert the factor

\sin\phi

✅

C.Change the radial limit from

1

2

D.Remove the

\theta

-integration

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The spherical-coordinate Jacobian is $\rho^2\sin\phi$ , not merely $\rho^2$ . The radial and angular limits already cover the unit sphere correctly: $0\le\rho\le1$ , $0\le\phi\le\pi$ , and $0\le\theta\le2\pi$ . Therefore, the missing $\sin\phi$ is the essential correction.

Q16. A solid is simultaneously described by $x^2+y^2+z^2\le25$ and $x^2+y^2\le z^2$ , with $z\ge0$ . Which strategy gives the cleanest setup for its volume?

A.Use spherical coordinates and translate the second inequality into an angular restriction ✅

B.Use cylindrical coordinates and make both surfaces linear

C.Use Cartesian coordinates and ignore the symmetry

D.Use spherical coordinates but let

\rho

depend on

\theta

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The first inequality is a sphere, while the second describes a cone. Spherical coordinates make the sphere simply $\rho\le5$ , and the cone becomes an angular condition involving $\phi$ . Because the region is rotationally symmetric, $\theta$ can cover a full $2\pi$ , making this approach substantially simpler.

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