📝 The Divergence Theorem in calculus (14 MCQs)

📖 From Calculus • 16. Topics in vector Calculus • 14 questions available

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What is The Divergence Theorem in calculus?

The Divergence Theorem in calculus:
The Divergence Theorem states $\iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_E \nabla \cdot \mathbf{F} \, dV$ , relating flux through a closed surface to divergence inside volume $E$ .

Example:
For $\mathbf{F} = \langle x,y,z \rangle$ over a unit sphere, flux = $\iiint_E 3 \, dV = 3 \cdot 4\pi/3 = 4\pi$ .

Reason:
This theorem converts surface integrals to volume integrals, simplifying flux computations and providing physical insight into sources/sinks.

Easy

Medium

Hard

📝 All The Divergence Theorem in calculus MCQs

Q1. A vector field is $\mathbf{F}(x,y,z)=\langle x^2,y^2,z^2\rangle$ , and $V$ is the solid cube $0\le x,y,z\le1$ . Which expression correctly represents the total outward flux through the entire boundary of $V$ using a volume integral?

\iiint_V (x+y+z)\,dV

\iiint_V 2(x+y+z)\,dV

✅

\iiint_V (x^2+y^2+z^2)\,dV

\iiint_V 6xyz\,dV

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The divergence is obtained by adding the partial derivatives of the three components: $\nabla\cdot\mathbf F=2x+2y+2z=2(x+y+z)$ . For a closed surface with outward orientation, the divergence theorem converts the total flux into the triple integral of this divergence over the volume. Therefore option B is correct.

Q2. Which condition is essential before replacing a closed-surface flux integral by a volume integral involving divergence?

A.The vector field must be conservative everywhere.

B.The surface must be a planar surface.

C.The surface must be closed and the vector field sufficiently smooth on the enclosed region. ✅

D.The divergence must be constant throughout the region.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The divergence theorem relates the outward flux across a closed boundary surface to the volume integral of divergence over the enclosed solid. The field must have the required smoothness on the region. A planar surface or constant divergence is unnecessary, and the field does not need to be conservative.

Q3. A closed container occupies a region $V$ , and a fluid velocity field satisfies $\nabla\cdot\mathbf F=5$ everywhere in $V$ . If the volume of the container is $12$ cubic units, what is the net outward flux through its boundary?

A.-2.4

B.-17

C.-60 ✅

D.-125

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The divergence theorem states that the net outward flux equals the volume integral of divergence. Because the divergence is the constant $5$ , the integral becomes $5\operatorname{Vol}(V)=5(12)=60$ . Thus the total outward flux is $60$ square-unit-equivalent flux units.

Q4. A researcher computes the outward flux through a complicated closed surface surrounding a region of volume $20$ . The field has divergence $3x^2+2y^2+z^2$ . Which strategy is generally preferable when the surface itself is difficult to parametrize but the enclosed volume has simple bounds?

A.Directly parameterize every piece of the surface.

B.Replace the surface integral with a triple integral of

3x^2+2y^2+z^2

over the volume. ✅

C.Set the divergence equal to zero because the surface is closed.

D.Integrate the field components separately over the surface and ignore orientation.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The major advantage of the divergence theorem is that it replaces a potentially complicated closed-surface flux calculation with a volume integral of divergence. When the volume has simple bounds, integrating $3x^2+2y^2+z^2$ over that region is usually substantially easier than parameterizing all surface pieces.

Q5. A student claims that because $\nabla\cdot\mathbf F=0$ , the flux through every surface contained in the region must be zero. Which correction is most accurate?

A.Zero divergence guarantees zero flux through every open surface.

B.Zero divergence guarantees zero flux through every closed surface, assuming the required smoothness conditions hold. ✅

C.Zero divergence means the field itself is zero everywhere.

D.Zero divergence guarantees positive flux through closed surfaces.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Zero divergence implies zero net outward flux through closed surfaces under the divergence theorem. It does not imply that the flux through an individual open surface is zero. A divergence-free field can cross an open surface substantially while having its total flux balance to zero across a suitable closed boundary.

Q6. Consider $\mathbf F=\langle x^2,yz,z\rangle$ and a closed region whose projection onto the $xy$ -plane is a disk. Which calculation must be performed first if the divergence theorem is used?

A.Compute

x^2+yz+z

B.Compute

2x+y+1

C.Compute

2x+z+1

. ✅

D.Compute

x+y+z

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For $\mathbf F=\langle P,Q,R\rangle$ , divergence is $P_x+Q_y+R_z$ . Here $P_x=2x$ , $Q_y=z$ , and $R_z=1$ , giving $\nabla\cdot\mathbf F=2x+z+1$ . This divergence, rather than the original field magnitude or component sum, becomes the volume-integrand.

Q7. A student evaluates the flux through a closed surface and obtains $24$ . Another student reverses the normal orientation everywhere and obtains $-24$ . Which interpretation is correct?

A.Only the first value can be correct because closed surfaces always use outward normals.

B.Only the second value can be correct because divergence theorem requires inward normals.

C.Both calculations can be correct for their respective orientations; the standard outward flux is

24

. ✅

D.The sign must be zero because the surface is closed.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Reversing the normal vector changes the sign of a flux integral. The divergence theorem conventionally uses outward orientation, so $24$ is the standard outward flux. If every normal is reversed, the corresponding inward flux is $-24$ . The two results are therefore consistent rather than contradictory.

Q8. A closed surface is represented visually by a distorted sphere. Arrows of a vector field appear to leave the surface more strongly on the right side and enter more strongly on the left side. If the field is smooth and the divergence is positive everywhere inside, what must be true about the total outward flux?

A.It must be negative because some arrows enter.

B.It must be zero because entering and leaving arrows always cancel.

C.It must be positive because positive divergence indicates net local expansion throughout the enclosed volume. ✅

D.Its sign cannot be determined from divergence.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Positive divergence represents local net expansion or source behavior. If it is positive throughout the enclosed volume, its volume integral is positive. By the divergence theorem, that integral equals the total outward flux. Individual surface regions can have inward flux, but the overall outward flux must remain positive.

Q9. A graph of divergence over a solid shows positive values in one half and negative values in the other half. The magnitudes are unequal, with the positive region contributing a larger volume-weighted integral. What can be concluded about the total outward flux?

A.It must be zero because both signs occur.

B.It is positive because the integral of divergence over the entire volume is positive. ✅

C.It is negative because any negative divergence dominates a flux calculation.

D.It cannot be related to divergence because the field is not constant.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The divergence theorem depends on the integral of divergence over the entire volume, not merely on whether positive and negative values occur. If the positive region has a larger volume-weighted contribution than the negative region, the total integral is positive. Consequently, the net outward flux is positive.

Q10. A rectangular box has dimensions $2\times3\times4$ , and $\mathbf F=\langle x^2,3y,2z\rangle$ . An engineer wants the net outward flow through the box. Which method is most efficient and why?

A.Compute six surface integrals because divergence cannot handle boxes.

B.Use

\nabla\cdot\mathbf F=2x+5

, then integrate over the box; this avoids evaluating six separate face contributions. ✅

C.Use the magnitude of

\mathbf F

and multiply it by the box surface area.

D.Integrate only the

z

-component because the box has four units of height.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The divergence is $2x+3+2=2x+5$ . Since the box is closed and has simple rectangular bounds, the divergence theorem converts the total flux into $\int_0^2\int_0^3\int_0^4(2x+5)\,dz\,dy\,dx$ . This is more efficient than computing six separate oriented surface integrals.

Q11. Suppose a computational model produces a smooth vector field inside a closed region. Numerical evaluation gives a surface flux of $18.7$ , while integrating the divergence over the same region gives $18.6$ . Which diagnosis is most reasonable before declaring the theorem invalid?

A.The theorem fails for numerical models.

B.The discrepancy may result from numerical approximation, discretization, or inconsistent orientation and should be investigated before drawing a theoretical conclusion. ✅

C.Divergence theorem only works when the two values are exactly equal to integers.

D.The surface integral should always be twice the volume integral.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: For a sufficiently smooth field and correctly matched closed region, the two quantities should agree mathematically. Small numerical discrepancies can arise from mesh resolution, quadrature error, boundary approximation, or orientation inconsistencies. A difference such as $0.1$ should therefore prompt numerical verification rather than rejection of the theorem.

Q12. Let $\mathbf F=\langle x^3,y^3,z^3\rangle$ , and let $V$ be any closed solid symmetric under reflection across each coordinate plane. Without knowing the exact shape of $V$ , what can be concluded about the total outward flux?

A.It is necessarily zero.

B.It is necessarily positive. ✅

C.It is necessarily negative.

D.It depends only on the surface area, not the volume.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The divergence is $3x^2+3y^2+3z^2$ , which is nonnegative everywhere and positive except at the origin. For any three-dimensional solid with positive volume, its integral over the region is therefore positive. By the divergence theorem, the total outward flux is positive, regardless of the detailed symmetric shape.

Q13. A field $\mathbf F$ has divergence $4$ throughout a solid $V$ . The solid is divided into two subregions $V_1$ and $V_2$ by an internal interface. A student adds the outward fluxes of $V_1$ and $V_2$ and obtains the flux through the original outer boundary. Why does this work?

A.The internal interface contributes equal and opposite fluxes for the two subregions. ✅

B.The internal interface has zero area.

C.Divergence is always zero on internal interfaces.

D.Both subregions use the same outward normal on the interface.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: When the two subregions are treated separately, the outward normal for one region on the internal interface points opposite to the outward normal for the other. Their interface fluxes therefore cancel when the two total fluxes are added. Only the original external boundary remains, illustrating flux additivity.

Q14. Consider a smooth vector field whose divergence is $6x^2+6y^2+6z^2-12$ . A family of closed regions is centered at the origin, with increasing radius. For small regions around the origin the divergence is negative, while for sufficiently large regions positive values dominate. What behavior should the net outward flux exhibit as the regions expand?

A.It must remain negative for every radius.

B.It must remain zero for every radius.

C.It can change from negative to positive as the volume integral of divergence changes sign. ✅

D.It must always equal the surface area.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The net outward flux equals the volume integral of $6x^2+6y^2+6z^2-12$ . Near the origin the integrand is mostly negative, so small centered regions can have negative flux. As the radius grows, the quadratic positive terms increasingly dominate, allowing the total volume integral, and hence the flux, to change sign.

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