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📝 Triple Integrals in Spherical Coordinates (14 MCQs)

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

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What is Triple Integrals in Spherical Coordinates?

Definition:
In spherical coordinates, $x=\rho\sin\phi\cos\theta, y=\rho\sin\phi\sin\theta, z=\rho\cos\phi$ , and $dV = \rho^2\sin\phi \, d\rho \, d\phi \, d\theta$ .

Example:
Integrating over a sphere of radius $R$ : $\int_0^{2\pi} \int_0^\pi \int_0^R f(\rho,\phi,\theta) \, \rho^2\sin\phi \, d\rho \, d\phi \, d\theta$ .

Reason:
The factor $\rho^2\sin\phi$ is the Jacobian, and these coordinates are ideal for spheres, cones, and other radially symmetric objects.

Easy

Medium

Hard

📝 All Triple Integrals in Spherical Coordinates MCQs

Q1. A solid is described by $0\le \rho\le 4$ , $0\le \phi\le \pi/3$ , and $0\le \theta\le 2\pi$ . Which geometric description best matches this region?

A.A full sphere of radius 4

B.A spherical cone of half-angle

\pi/3

together with all points inside it up to radius 4 ✅

C.A hemisphere of radius 4

D.A cylinder of radius 4 and height 4

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The variable $\rho$ limits the distance from the origin, while $\phi$ measures the angle from the positive $z$ -axis. Restricting $\phi$ to $\pi/3$ selects a cone-shaped portion around that axis. Since $\theta$ covers the full $2\pi$ , the region extends completely around the axis.

Q2. When evaluating a triple integral after converting to spherical coordinates, which differential-volume element should be used?

dV=d\rho\,d\phi\,d\theta

dV=\rho\,d\rho\,d\phi\,d\theta

dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta

✅

dV=\rho^2\cos\phi\,d\rho\,d\phi\,d\theta

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The spherical coordinate transformation changes a small rectangular coordinate box into a curved volume element. Its scaling factor is $\rho^2\sin\phi$ . Therefore, the correct volume element is $dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ . Omitting either factor produces an incorrect measure and therefore an incorrect integral.

Q3. A solid consists of all points inside the sphere $x^2+y^2+z^2\le 9$ and above the cone $z=\sqrt{3(x^2+y^2)}$ . Which spherical bounds describe the solid?

0\le\rho\le3,\ 0\le\phi\le\pi/6,\ 0\le\theta\le2\pi

✅

0\le\rho\le3,\ 0\le\phi\le\pi/3,\ 0\le\theta\le2\pi

0\le\rho\le3,\ \pi/3\le\phi\le\pi,\ 0\le\theta\le2\pi

0\le\rho\le3,\ 0\le\phi\le2\pi/3,\ 0\le\theta\le2\pi

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The sphere gives $\rho\le3$ . For the cone, $z=\sqrt{3}\,r$ . Substituting $z=\rho\cos\phi$ and $r=\rho\sin\phi$ yields $\cos\phi=\sqrt3\sin\phi$ , so $\tan\phi=1/\sqrt3$ and $\phi=\pi/6$ . Since the region is above the cone, $0\le\phi\le\pi/6$ , with $0\le\theta\le2\pi$ .

Q4. A solid is inside the sphere $x^2+y^2+z^2\le16$ and lies between the cones $\phi=\pi/6$ and $\phi=\pi/3$ . A student claims the spherical limits are $0\le\rho\le4,\ 0\le\phi\le\pi/3,\ 0\le\theta\le2\pi$ . What is the main error?

A.The radial bound should be

0\le\rho\le16

B.The angular region must exclude

0\le\phi<\pi/6

✅

C.The azimuthal angle must stop at

\pi

D.The sphere cannot be represented using

\rho

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The radial equation $x^2+y^2+z^2=16$ becomes $\rho=4$ , so the radial bound is correct. The important issue is the region between two cones. Therefore, $\phi$ must range from $\pi/6$ to $\pi/3$ , not from zero to $\pi/3$ . Starting at zero incorrectly includes the cone's interior.

Q5. A solid is inside a sphere of radius $a$ and above the $xy$ -plane. A physical model has density proportional to distance from the origin. Which integral correctly represents its mass, ignoring the constant of proportionality?

\int_0^{2\pi}\int_0^{\pi/2}\int_0^a \rho\cdot\rho^2\sin\phi\,d\rho\,d\phi\,d\theta

✅

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

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

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

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Density proportional to distance means the density function is $\rho$ . The upper hemisphere requires $0\le\phi\le\pi/2$ , while full rotation requires $0\le\theta\le2\pi$ . Multiplying the density by the spherical volume element $\rho^2\sin\phi$ produces the integrand $\rho^3\sin\phi$ , giving option A.

Q6. A region is defined by $1\le x^2+y^2+z^2\le9$ and $z\ge0$ . Which change of variables most efficiently expresses the radial restriction?

A.Use

1\le\rho\le9

B.Use

1\le\rho^2\le9

and then integrate directly

C.Use

1\le\rho\le3

✅

D.Use

0\le\rho\le3

and subtract the inner sphere later

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Since $x^2+y^2+z^2=\rho^2$ , the inequality becomes $1\le\rho^2\le9$ . Because $\rho$ represents a nonnegative distance, taking square roots gives $1\le\rho\le3$ . The condition $z\ge0$ separately gives $0\le\phi\le\pi/2$ , so no subtraction is necessary.

Q7. A solid is bounded by the sphere $\rho=5$ and the cone $\phi=\pi/4$ , occupying the region closer to the positive $z$ -axis. If the goal is to compute its volume, which setup is most appropriate?

\int_0^{2\pi}\int_0^{\pi/4}\int_0^5 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta

✅

\int_0^{\pi/4}\int_0^5\int_0^{2\pi} \rho\sin\phi\,d\theta\,d\rho\,d\phi

\int_0^{2\pi}\int_0^{\pi/2}\int_0^5 \rho^2\cos\phi\,d\rho\,d\phi\,d\theta

\int_0^{2\pi}\int_{\pi/4}^{\pi}\int_0^5 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The cone selects angles from the positive $z$ -axis up to $\pi/4$ , so $0\le\phi\le\pi/4$ . The sphere gives $0\le\rho\le5$ , and rotational symmetry gives $0\le\theta\le2\pi$ . Finally, the spherical volume element contributes $\rho^2\sin\phi$ , making option A correct.

Q8. A student converts $x^2+y^2+z^2\le25$ into $\rho\le25$ and argues that this is correct because $\rho$ represents the radius. Which reasoning best identifies the mistake?

A.The spherical radius is

\rho^2

, not

\rho

B.The original expression equals

\rho^2

, so the correct radial bound is

\rho\le5

✅

C.The inequality must become

\rho\ge5

D.Spherical coordinates cannot represent spheres centered at the origin

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The key distinction is between the squared distance and the distance itself. Since $x^2+y^2+z^2=\rho^2$ , the inequality becomes $\rho^2\le25$ . Because $\rho$ is nonnegative, this simplifies to $0\le\rho\le5$ . Using $\rho\le25$ describes a much larger sphere.

Q9. A graph shows a sphere centered at the origin with a shaded region forming a narrow cap around the positive $z$ -axis. The boundary of the shaded region is a cone making an angle $\alpha$ with the positive $z$ -axis. Which description of the angular bound matches the graph?

0\le\phi\le\alpha

✅

\alpha\le\phi\le\pi-\alpha

0\le\theta\le\alpha

\alpha\le\theta\le2\pi

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: In spherical coordinates, $\phi$ measures the angle measured downward from the positive $z$ -axis. A narrow cap around that axis therefore corresponds to small values of $\phi$ . If the cone boundary makes angle $\alpha$ with the positive $z$ -axis, the cap is represented by $0\le\phi\le\alpha$ .

Q10. Suppose a spherical-coordinate integral has integrand $f(\rho,\phi,\theta)\rho^2\sin\phi$ . Another method evaluates the same volume integral using cylindrical coordinates. Which observation is most useful when deciding between the two methods for a solid bounded by a sphere centered at the origin?

A.Spherical coordinates usually simplify the spherical boundary directly ✅

B.Cylindrical coordinates cannot represent spheres

C.Spherical coordinates always eliminate every variable from the integrand

D.Cylindrical coordinates always require more integrations

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A sphere centered at the origin has the especially simple spherical equation $\rho=R$ . This can make radial limits straightforward, particularly when the remaining boundaries are cones or planes through the origin. Cylindrical coordinates can also represent spheres, but their radial and vertical limits may require more complicated piecewise descriptions.

Q11. A solid is inside $\rho=2$ , restricted by $0\le\phi\le\pi/2$ , and has no restriction on $\theta$ . An integrand contains only $\rho$ . Which feature can be exploited to simplify the calculation before performing the full integration?

A.The

\theta

-integration contributes a factor of

2\pi

✅

B.The

\phi

-integration can be replaced by

2\pi

C.The

\rho

-integration can be replaced by

2

D.The factor

\sin\phi

can be removed because the integrand contains only

\rho

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Because the region covers a complete rotation around the $z$ -axis and the integrand has no dependence on $\theta$ , integrating with respect to $\theta$ simply gives $2\pi$ . The spherical Jacobian still contains $\sin\phi$ , so that factor cannot be discarded. This symmetry reduces the problem without changing its geometry.

Q12. Consider the integral over a ball of radius $R$ of the function $x^2+y^2+z^2$ . Which spherical-coordinate expression results after correctly transforming both the function and the volume element?

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

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

✅

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

\int_0^{2\pi}\int_0^\pi\int_0^R \rho^2\cos\phi\,d\rho\,d\phi\,d\theta

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The function $x^2+y^2+z^2$ becomes $\rho^2$ in spherical coordinates. The volume element contributes another factor $\rho^2\sin\phi$ . Their product is therefore $\rho^4\sin\phi$ . The full ball requires $0\le\rho\le R$ , $0\le\phi\le\pi$ , and $0\le\theta\le2\pi$ .

Q13. A researcher models heat generation inside a spherical region where the rate depends only on distance from the center and increases as the square of that distance. If the region is a full ball of radius $R$ , which reasoning correctly determines the power of $\rho$ in the radial integrand?

A.Only the model contributes, giving

\rho^2

B.Only the volume element contributes, giving

\rho^2

C.The model contributes

\rho^2

, while the volume element contributes another

\rho^2

, giving

\rho^4

✅

D.The model contributes

\rho

, while the volume element contributes

\rho^2

, giving

\rho^3

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The heat-generation rate is proportional to the square of the distance, so its spherical form contributes $\rho^2$ . The volume element contributes $\rho^2\sin\phi$ . Multiplying these factors gives $\rho^4\sin\phi$ . This distinction between the physical model and the coordinate Jacobian is essential when setting up the integral.

Q14. Let $I=\iiint_B (x^2+y^2+z^2)^{3/2}\,dV$ , where $B$ is the ball $x^2+y^2+z^2\le a^2$ . Without evaluating every elementary antiderivative, which radial dependence must appear in the spherical-coordinate integral?

\rho^3

\rho^4

\rho^5

✅

\rho^6

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Inside the ball, $(x^2+y^2+z^2)^{3/2}=(\rho^2)^{3/2}=\rho^3$ , since $\rho\ge0$ . The volume element adds a factor of $\rho^2\sin\phi$ . Therefore, the radial part of the integrand becomes $\rho^5$ , making option C the correct choice.

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