📝 How to Find Flux Using Divergence Theorem (15 MCQs)

📖 From Calculus • 16. Topics in vector Calculus • 15 questions available

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What is How to Find Flux Using Divergence Theorem?

How to Find Flux Using Divergence Theorem:
Compute $\nabla \cdot \mathbf{F}$ , set up a triple integral over the volume enclosed by the surface, and evaluate, provided the surface is closed and smooth.

Example:
For $\mathbf{F} = \langle x^2, y^2, z^2 \rangle$ over a cube $[0,1]^3$ , $\nabla \cdot \mathbf{F} = 2x+2y+2z$ , integral = $\int_0^1 \int_0^1 \int_0^1 (2x+2y+2z) \, dx dy dz = 3$ .

Reason:
This method drastically simplifies flux calculations for complex closed surfaces, a staple in physics and engineering.

Easy

Medium

Hard

📝 All How to Find Flux Using Divergence Theorem MCQs

Q1. For a closed surface $S$ enclosing a volume $V$ , which quantity must be integrated over $V$ to determine the outward flux of a sufficiently smooth vector field $\mathbf{F}$ using the divergence theorem?

A.The magnitude of

\mathbf{F}

B.The divergence

\nabla\cdot\mathbf{F}

✅

C.The curl

\nabla\times\mathbf{F}

D.The gradient of

\mathbf{F}

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The divergence theorem converts the outward flux through a closed surface into a triple integral over the enclosed volume. Therefore, the required volume integrand is $\nabla\cdot\mathbf{F}$ . The magnitude, curl, and gradient describe different properties and do not generally determine the total outward flux.

Q2. A vector field has constant divergence $6$ throughout a closed region whose volume is $20$ . Without evaluating any surface integral directly, what is the outward flux through the boundary?

A.26

B.120 ✅

C.3.33

D.6

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Because the divergence is constant, the volume integral simplifies immediately to $\iiint_V 6\,dV=6(\text{Volume})$ . With volume $20$ , the flux is $120$ . The problem tests recognition that the theorem can replace a potentially complicated surface calculation with a much simpler volume calculation.

Q3. A closed surface encloses a region in which $\nabla\cdot\mathbf{F}$ is positive everywhere except on a small subregion where it is strongly negative. Which conclusion is logically justified?

A.The outward flux must be positive

B.The outward flux must be negative

C.The sign of the flux depends on the total volume integral of the divergence ✅

D.The outward flux must be zero

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The divergence theorem relates total outward flux to the integral of divergence over the entire enclosed volume. A small region with negative divergence can outweigh a larger region with positive divergence if its magnitude is sufficiently large. Therefore, the sign cannot be determined from local sign information alone.

Q4. A spherical closed surface is replaced by a cube that encloses exactly the same volume. A vector field has constant divergence throughout both regions. How do their total outward fluxes compare?

A.The sphere has greater flux because its area is larger

B.The cube has greater flux because it has more edges

C.Both surfaces have the same flux ✅

D.The flux depends only on the surface shape

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For a constant divergence, the divergence theorem gives total flux as divergence multiplied by enclosed volume. Since both surfaces enclose exactly the same volume, their total outward fluxes are identical. Surface area, curvature, and the number of edges do not affect the total flux in this situation.

Q5. A closed surface encloses the box $0\le x\le2$ , $0\le y\le3$ , $0\le z\le4$ . For $\mathbf{F}=\langle x^2,y^2,z^2\rangle$ , what is the outward flux?

A.72

B.96

C.144 ✅

D.192

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The divergence is $\nabla\cdot\mathbf{F}=2x+2y+2z$ . Integrating over the box gives $96+48+96=240$ , so none of the displayed numerical choices matches the actual result. This exposes a flawed item rather than a mathematical difficulty. A properly constructed assessment should replace the options with values including $240$ .

Q6. A student computes the flux through a closed surface by integrating $\mathbf{F}\cdot\mathbf{n}\,dS$ separately over six complicated faces. Another student calculates $\iiint_V\nabla\cdot\mathbf{F}\,dV$ . When is the second approach generally preferable?

A.Only when the surface is a sphere

B.When the surface is closed and the volume integral is substantially easier ✅

C.Only when divergence is zero

D.Only when the vector field is constant

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The divergence theorem is especially useful when a closed surface has complicated geometry or many pieces, while the divergence is simple and the enclosed volume is easier to describe. It does not require a spherical surface or constant field. The key structural requirement is that the surface be closed and appropriately oriented.

Q7. Consider a closed surface surrounding a region where $\nabla\cdot\mathbf{F}=4x-2$ . The region is symmetric about the $yz$ -plane. What can be concluded about the total flux?

A.It must equal

4

times the volume

B.It must be negative

C.It is determined entirely by the

-2

term

D.It is zero because the

4x

contribution cancels while the constant term remains ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: By symmetry, the integral of $x$ over a region symmetric about the $yz$ -plane is zero. Thus the $4x$ contribution vanishes, while the constant $-2$ contributes $-2$ times the volume. Therefore the flux is not generally zero; it equals $-2\operatorname{Vol}(V)$ .

Q8. A student claims that if $\nabla\cdot\mathbf{F}=0$ at the center of a closed region, then the total outward flux through its boundary must be zero. What is the best evaluation of this reasoning?

A.Correct, because the center controls the whole region

B.Correct, because divergence is always constant

C.Incorrect, because flux depends on the divergence throughout the entire volume ✅

D.Incorrect, because flux depends only on surface area

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The divergence theorem requires integrating divergence over the entire enclosed volume, not evaluating it at a single point. A zero value at the center provides almost no information about the total volume integral. Divergence may be positive or negative elsewhere, producing a nonzero total flux.

Q9. A graph of a closed three-dimensional region shows it is symmetric about the $xy$ -plane. The divergence is $\nabla\cdot\mathbf{F}=z+5$ . Which feature of the graph is most useful for evaluating the flux efficiently?

A.The region's symmetry causes the

z

-term to integrate to zero ✅

B.The region's symmetry makes the entire flux zero

C.The upper half contributes twice as much as the lower half

D.The

z

-term can be replaced by

5

everywhere

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Because the region is symmetric about the $xy$ -plane, positive and negative values of $z$ occur in matching pairs. Hence $\iiint_V z\,dV=0$ . The remaining constant contribution is $5\operatorname{Vol}(V)$ , so the graph's symmetry substantially simplifies the volume integral without making the total flux zero.

Q10. A diagram shows a closed region composed of a cylinder and two hemispherical caps. Directly integrating flux over every curved piece is difficult, but the divergence of the field is a simple function of $x,y,z$ . Which strategy is most efficient?

A.Parameterize every curved piece immediately

B.Compute curl first and then integrate it

C.Use the volume description and integrate the divergence over the entire enclosed region ✅

D.Ignore the hemispherical pieces because their curvature is difficult

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The surface is closed, so the divergence theorem allows the entire flux to be replaced by one volume integral. The geometry can often be described naturally using cylindrical or spherical coordinates. This avoids separate normal-vector calculations on the cylinder and curved caps.

Q11. A student evaluates $\iiint_V\nabla\cdot\mathbf{F}\,dV$ for a closed surface but obtains the negative of the expected answer. Inspection shows that the surface normal in the original problem points inward. What is the most appropriate correction?

A.Change the divergence to its negative

B.Reverse the final flux sign ✅

C.Square the computed flux

D.Multiply the flux by the surface area

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The divergence theorem normally gives outward flux when the boundary orientation is outward. If the problem specifies inward orientation, the inward flux is the negative of the outward flux. The divergence itself is not changed; only the orientation of the surface changes the sign of the resulting flux.

Q12. A computational model reports a total outward flux of $30$ for a closed region. A second model uses the divergence theorem and obtains $30$ . If the divergence is known to be nonnegative everywhere, what does this agreement most strongly indicate?

A.The surface must be planar

B.The field must have zero curl

C.The integrated divergence over the enclosed volume is consistent with the computed boundary flux ✅

D.The surface area must equal

30

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The divergence theorem states that total outward flux equals the volume integral of divergence for a suitable closed surface. Agreement between the two independent computational approaches therefore provides evidence that the surface flux and accumulated volumetric source strength are consistent. Surface planarity, area, and curl are not required for this conclusion.

Q13. Suppose a closed region is divided into two adjacent subregions $V_1$ and $V_2$ . Each is assigned its own outward boundary orientation. If the fluxes across their boundaries are added, what happens to the flux across their common internal interface?

A.It is counted twice with the same sign

B.It cancels because the two induced normals are opposite ✅

C.It becomes zero only when divergence is zero

D.It is converted into a volume integral automatically

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: On the common interface, the outward normal for $V_1$ points in the opposite direction from the outward normal for $V_2$ . Therefore the two flux contributions are equal in magnitude and opposite in sign, so they cancel when the subregion fluxes are added. Only the external boundary remains.

Q14. A closed region has volume $V$ , and its divergence is $3x^2+3y^2+3z^2$ . The region is centered at the origin and is contained within a sphere of radius $R$ . Which observation provides the strongest route toward comparing its flux with that of a larger concentric region?

A.Only the surface areas need to be compared

B.The flux can be compared through the volume integrals of the nonnegative divergence ✅

C.The flux must be identical because both regions share the same center

D.The flux is determined only by

R

and never by the actual region

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The divergence $3(x^2+y^2+z^2)$ is nonnegative and increases with distance from the origin. The divergence theorem converts flux into its volume integral, so comparing regions requires comparing the accumulated divergence over their actual volumes. Equal centers do not imply equal flux, and surface area alone is insufficient.

Q15. Let $V$ be the unit ball and let $\mathbf{F}=\langle x,y,z\rangle$ . A student argues that the outward flux must be $4\pi$ because the unit sphere has surface area $4\pi$ and the vector field has magnitude $1$ on the sphere. Is the conclusion correct?

A.Yes, because the magnitude is constant on the sphere

B.No, because the flux is

3

times the volume, which equals

4\pi

✅

C.No, because the divergence is

1

and the volume is

4\pi/3

, giving flux

4\pi/3

D.Yes, because the field is tangent to the sphere

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Here $\nabla\cdot\mathbf{F}=1+1+1=3$ . The unit ball has volume $4\pi/3$ , so the divergence theorem gives flux $3(4\pi/3)=4\pi$ . The student's numerical conclusion is correct, but the reasoning should be justified by the normal component of the field or the volume integral, not merely by surface area and magnitude.

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