📝 How to evaluate flux integrals (16 MCQs)

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

←Previous Topic:Flux through a surface📝 How to evaluate flux integralsOrientation of nonparametric surfaces:Next Topic→

What is How to evaluate flux integrals?

How to evaluate flux integrals:
Parameterize the surface, compute the normal vector $\mathbf{r}_u \times \mathbf{r}_v$ , dot with $\mathbf{F}$ , and integrate over the parameter domain.

Example:
For $\mathbf{F} = \langle x, y, z \rangle$ over the hemisphere $z = \sqrt{1 - x^2 - y^2}$ , use spherical coordinates, compute $\mathbf{r}_\theta \times \mathbf{r}_\phi$ , and integrate to get $2\pi$ .

Reason:
Systematic flux evaluation handles complex surfaces, forming a core skill in vector calculus applications.

Easy

Medium

Hard

📝 All How to evaluate flux integrals MCQs

Q1. A vector field is given by $\mathbf{F}=\langle x,y,z\rangle$ , and $S$ is the sphere $x^2+y^2+z^2=4$ oriented outward. Without directly parameterizing the sphere, which approach most efficiently evaluates the outward flux through $S$ ?

A.Use symmetry to conclude the flux is zero

B.Use the divergence theorem and integrate

\nabla\cdot\mathbf{F}

over the enclosed ball ✅

C.Integrate

\mathbf{F}\cdot\mathbf{r}

along the boundary circle

D.Compute the curl of

\mathbf{F}

and integrate it over the sphere

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The divergence is $\nabla\cdot\mathbf{F}=1+1+1=3$ . Because the sphere is closed and outward oriented, the divergence theorem converts the surface flux into $3$ times the volume of the radius- $2$ ball. This avoids a difficult spherical parameterization and directly exploits the structure of the field.

Q2. For a parametrized surface $\mathbf{r}(u,v)$ , a student computes $\mathbf{r}_u\times\mathbf{r}_v$ and obtains a normal pointing opposite to the required orientation. What is the most direct correction before evaluating the flux integral?

A.Replace the vector field by its negative

B.Replace

\mathbf{r}_u\times\mathbf{r}_v

with

\mathbf{r}_v\times\mathbf{r}_u

✅

C.Take the absolute value of every component

D.Multiply the surface integral by the surface area

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Reversing the order of the cross product reverses the normal direction because $\mathbf{r}_v\times\mathbf{r}_u=-(\mathbf{r}_u\times\mathbf{r}_v)$ . This is essential because flux depends on the oriented normal. Taking absolute values would destroy directional information, while changing the vector field would solve a different problem.

Q3. Suppose a vector field is everywhere tangent to a smooth surface $S$ . What should be expected for the flux through $S$ , assuming the tangency is exact at every point?

A.The flux must be positive

B.The flux must be negative

C.The flux must be zero ✅

D.The flux equals the surface area

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Flux measures the component of the vector field normal to the surface. If the field is tangent everywhere, its dot product with the unit normal is zero at every point. Therefore the integrand $\mathbf{F}\cdot\mathbf{n}$ vanishes throughout the surface, making the total flux zero regardless of the surface area.

Q4. A rectangular surface lies in the plane $z=3$ , with upward orientation, and $\mathbf{F}=\langle x^2,y,4z\rangle$ . The rectangle is $0\le x\le2$ , $0\le y\le1$ . Which integral correctly represents the flux?

\int_0^2\int_0^1 x^2\,dy\,dx

\int_0^2\int_0^1 y\,dy\,dx

\int_0^2\int_0^1 12\,dy\,dx

✅

\int_0^2\int_0^1 4\,dy\,dx

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For the plane $z=3$ with upward orientation, the unit normal is $\mathbf{k}$ . Therefore $\mathbf{F}\cdot\mathbf{k}=4z=12$ on the surface. The flux is consequently represented by $\int_0^2\int_0^1 12\,dy\,dx$ . The other choices incorrectly use tangential components instead of the normal component.

Q5. A hemispherical surface is oriented outward, but its circular base is not included. A student applies the divergence theorem directly to the hemisphere and obtains a flux value. What crucial issue must be addressed first?

A.The field must be constant

B.The open surface must be closed by adding the base disk ✅

C.The radius must be converted to spherical coordinates

D.The normal vector must always point upward

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The divergence theorem applies to closed surfaces. A hemisphere alone has a boundary circle, so it is not closed. The standard strategy is to add the flat base disk, calculate the total outward flux of the closed surface, and then subtract the flux through the disk. This converts an open-surface problem into a closed one.

Q6. A fluid has velocity field $\mathbf{v}=\langle x,-y,2z\rangle$ . A rectangular sensor is placed in the plane $x=1$ , with its normal pointing in the positive $x$ -direction. If the sensor covers $0\le y\le2$ and $0\le z\le3$ , what is the flux through the sensor?

A.6 ✅

B.12

C.18

D.24

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The sensor lies in $x=1$ , so its positive $x$ -normal is $\mathbf{i}$ . Thus $\mathbf{v}\cdot\mathbf{n}=x=1$ . The area is $(2)(3)=6$ , giving flux $1\cdot6=6$ . This tests whether the student correctly identifies the normal component rather than integrating all three velocity components.

Q7. A student evaluates the flux of $\mathbf{F}=\langle x,y,z\rangle$ through a closed sphere and argues: 'The field points outward everywhere, so the flux must equal the sphere's surface area.' What is the flaw?

A.The field is not defined on the sphere

B.Flux depends only on the field magnitude

C.The normal component is not uniformly equal to

1

, so flux is not simply surface area ✅

D.Closed surfaces always have zero flux

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Although the field points outward, its magnitude on the sphere is not $1$ . On a sphere of radius $R$ , $\mathbf{F}$ has magnitude $R$ , and its normal component equals $R$ . Therefore the flux is $R$ times the surface area, not merely the surface area. The student's reasoning confuses direction with magnitude.

Q8. A surface is given as $z=g(x,y)$ and oriented upward. A student writes the flux integrand as $\mathbf{F}(x,y,g(x,y))\cdot\langle-g_x,-g_y,-1\rangle$ . Which conclusion is correct?

A.The expression is correct for upward orientation

B.The expression is correct only when

g_x=g_y=0

C.The vector has downward orientation and should be negated ✅

D.The vector must be divided by its magnitude before integration

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For $z=g(x,y)$ , the upward-oriented normal vector associated with the projection is $\langle-g_x,-g_y,1\rangle$ . The student's vector $\langle-g_x,-g_y,-1\rangle$ points downward. Negating it gives the required upward orientation. Normalization is unnecessary when the vector already represents the appropriate vector-area element.

Q9. A closed surface consists of a cylinder and its two circular caps. A vector field has constant divergence $5$ throughout the enclosed volume. If the enclosed volume is $20$ cubic units, what is the total outward flux?

A.4

B.25

C.50

D.100 ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: For a closed surface, the total outward flux equals the volume integral of the divergence. Since the divergence is constant at $5$ and the enclosed volume is $20$ , the flux is $5(20)=100$ . This avoids separately evaluating flux through the curved cylinder and both caps.

Q10. A graph of a vector field near a planar surface shows arrows crossing the surface strongly in the positive normal direction on the left half, while arrows are tangent to the surface on the right half. Which qualitative conclusion about total flux is most justified?

A.The total flux must equal zero

B.Only the tangential arrows contribute

C.The left region contributes positive flux while the right region contributes approximately zero ✅

D.Both regions contribute equally because the surface areas are equal

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Flux depends on the normal component of the field. The arrows crossing the surface in the positive normal direction contribute positive flux, while tangent arrows have approximately zero normal component and therefore contribute little or nothing. Equal surface areas do not imply equal flux because the vector-field direction and magnitude matter.

Q11. A diagram shows a closed box with outward-pointing normals. The vector field is stronger on the right face than on the left face and approximately tangent to the top and bottom faces. If the right and left faces have equal areas, which observation best predicts the net flux?

A.It must be zero because the box is closed

B.The right face can produce a larger positive contribution than the left face produces negative contribution ✅

C.Only the top face determines the flux

D.The total flux must equal the surface area of the box

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Flux is a signed quantity determined by the normal component of the field. On opposite faces, outward normals point in opposite directions. If the field crosses the right face more strongly than it crosses the left face, their contributions need not cancel. Tangential flow on the top and bottom contributes approximately zero.

Q12. Two methods are proposed for finding outward flux through a closed surface: Method I directly parameterizes every piece, while Method II computes the divergence and integrates over the enclosed volume. The field has a simple divergence and the surface contains several curved pieces. Which method is generally more efficient, and why?

A.Method I, because parameterization always gives an exact answer

B.Method II, because it replaces multiple surface integrals with one volume integral ✅

C.Method I, because divergence is unrelated to flux

D.Both methods must always require the same number of integrations

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For a closed surface, the divergence theorem can transform the flux into a volume integral of $\nabla\cdot\mathbf{F}$ . When the divergence is simple and the surface has complicated curved pieces, this can dramatically reduce the computational work. Direct parameterization remains valid but may require several difficult surface integrals.

Q13. A student claims that reversing the orientation of a surface changes the flux from $8$ to $12$ because the surface itself is unchanged but the parameterization becomes more complicated. Which correction is mathematically valid?

A.Reversing orientation changes

8

-8

✅

B.Reversing orientation leaves flux unchanged

C.Reversing orientation changes

8

64

D.Orientation affects only the surface area, not flux

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Changing orientation reverses the normal vector while leaving the geometric surface unchanged. Since the flux integrand contains $\mathbf{F}\cdot\mathbf{n}$ , reversing $\mathbf{n}$ multiplies the entire flux by $-1$ . Therefore a flux of $8$ becomes $-8$ , not $12$ or another unrelated value.

Q14. A closed surface encloses a region where $\nabla\cdot\mathbf{F}=0$ everywhere. A student concludes that $\mathbf{F}$ must be zero throughout the region. What is the best evaluation of this conclusion?

A.Correct, because zero divergence means no vector field exists

B.Correct only on the boundary

C.Incorrect, because zero divergence means zero net outward flux through every closed surface, not necessarily a zero field ✅

D.Incorrect because divergence is always positive

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Zero divergence does not imply that the vector field itself vanishes. It means there is no net local source or sink, and for an appropriate closed surface the total outward flux is zero. A constant nonzero field is a simple example: its divergence is zero even though the field is nonzero everywhere.

Q15. Let $\mathbf{F}=\langle 2x,2y,2z\rangle$ , and let $S$ be any closed surface enclosing a volume of $9$ cubic units. Without knowing the shape of $S$ , what is the outward flux?

A.9

B.18

C.27

D.54 ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: The divergence is constant: $\nabla\cdot\mathbf{F}=2+2+2=6$ . By the divergence theorem, the outward flux through any closed surface equals the volume integral of $6$ . Since the enclosed volume is $9$ , the flux is $6(9)=54$ . The shape and complexity of the surface are irrelevant once its enclosed volume is known.

Q16. A closed surface is divided into two pieces, $S_1$ and $S_2$ , sharing a common boundary curve. A calculation gives outward flux $14$ through $S_1$ and outward flux $9$ through $S_2$ . If both pieces use the outward orientation inherited from the closed surface, what is the total outward flux?

A.5

B.9

C.14

D.23 ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Flux is additive over non-overlapping portions of an oriented surface. Since $S_1$ and $S_2$ together form the complete closed surface and both flux values use the same outward orientation, their contributions are added: $14+9=23$ . Subtracting them would incorrectly treat the two pieces as oppositely oriented.

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←Previous Topic:Flux through a surface📝 How to evaluate flux integralsOrientation of nonparametric surfaces:Next Topic→