📝 How to use Stokes Theorem to Calculate Work (16 MCQs)

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

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What is How to use Stokes Theorem to Calculate Work?

How to use Stokes Theorem to Calculate Work:
Work $\oint_C \mathbf{F} \cdot d\mathbf{r}$ can be computed as $\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$ , choosing a convenient surface spanning $C$ .

Example:
For $\mathbf{F} = \langle yz, xz, xy \rangle$ around a circle in the $xy$ -plane, $\nabla \times \mathbf{F} = \mathbf{0}$ , so work = 0.

Reason:
Stokes' theorem simplifies work calculations by allowing surface choice, especially when the curl is easy to integrate.

Easy

Medium

Hard

📝 All How to use Stokes Theorem to Calculate Work MCQs

Q1. A vector field is $\mathbf F=\langle -y,x,0\rangle$ , and a closed curve $C$ bounds a planar region in the $xy$ -plane. Which surface-based expression correctly replaces the line integral for the work around $C$ , assuming the positive orientation is counterclockwise when viewed from above?

\iint_S(\nabla\times\mathbf F)\cdot\mathbf n\,dS

✅

\iint_S(\nabla\cdot\mathbf F)\,dS

\iint_S\mathbf F\cdot\mathbf n\,dS

\iint_S|\nabla\times\mathbf F|\,dS

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The work around a closed curve is represented by a circulation line integral. Stokes' theorem converts that circulation into the surface integral of the curl dotted with the consistently oriented unit normal. The divergence and flux expressions measure different properties and therefore cannot replace the circulation integral.

Q2. A vector field is $\mathbf{F}$ , and a closed curve $C$ bounds an oriented surface $S$ . Which expression correctly converts the work integral around $C$ into a surface integral?

\oint_C \mathbf{F}\cdot d\mathbf{r}=\iint_S \mathbf{F}\cdot dS

\oint_C \mathbf{F}\cdot d\mathbf{r}=\iint_S (\nabla\times\mathbf{F})\cdot d\mathbf{S}

✅

\oint_C \mathbf{F}\cdot d\mathbf{r}=\iint_S (\nabla\cdot\mathbf{F})\,dS

\oint_C \mathbf{F}\cdot d\mathbf{r}=\iint_S \nabla\mathbf{F}\cdot d\mathbf{S}

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The key conversion is from circulation along a closed boundary to the flux of the curl through a spanning surface. Thus the work integral $\oint_C\mathbf{F}\cdot d\mathbf{r}$ equals $\iint_S(\nabla\times\mathbf{F})\cdot d\mathbf{S}$ . The other choices confuse curl with divergence or omit the required vector operation.

Q3. A student chooses a complicated curved surface spanning a closed curve even though a flat disk also has the same boundary. Why can the disk be preferable when applying the theorem?

A.The theorem requires the surface to be flat

B.The curl is always zero on curved surfaces

C.A simpler surface can make the normal vector and surface integral much easier to evaluate ✅

D.Only planar surfaces satisfy the theorem

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The surface used in the surface integral need not be the original physical surface, provided it has the same oriented boundary and the field is sufficiently well behaved. Choosing a planar disk often simplifies the normal vector, parameterization, and limits, reducing a difficult calculation to a manageable one.

Q4. For a closed curve $C$ , reversing the direction in which $C$ is traversed changes the sign of the work integral. What corresponding change is required in the surface formulation?

A.The surface must be replaced by a different geometric surface

B.The curl must be replaced by the divergence

C.The orientation of the surface normal must also be reversed ✅

D.The surface integral becomes a volume integral

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Orientation is essential. If the traversal direction of $C$ is reversed, the compatible normal direction on $S$ must also reverse according to the orientation convention. Since the integrand contains a dot product with $d\mathbf{S}$ , reversing the normal changes the sign of the surface integral and therefore preserves the theorem's consistency.

Q5. A field has $\nabla\times\mathbf{F}=\langle 0,0,6\rangle$ . The boundary curve is the circle $x^2+y^2=4$ in the plane $z=3$ , oriented counterclockwise when viewed from above. What is the work around the curve?

12\pi

24\pi

✅

48\pi

6\pi

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The compatible normal for a counterclockwise traversal viewed from above is $\mathbf{k}$ . The curl therefore contributes $6$ per unit area. The circle has radius $2$ , so its area is $4\pi$ . Hence the work is $6(4\pi)=24\pi$ . The result follows directly from replacing the line integral with the curl flux.

Q6. Suppose $\mathbf{F}=\langle -y,x,0\rangle$ and $C$ is the boundary of the disk $x^2+y^2\leq9$ in the $xy$ -plane, traversed counterclockwise. Which strategy gives the work most efficiently?

A.Directly parameterize the boundary and integrate

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

B.Compute

\nabla\times\mathbf{F}

, then integrate its normal component over the disk ✅

C.Compute

\nabla\cdot\mathbf{F}

over the disk

D.Find the gradient of

\mathbf{F}

and integrate it along the boundary

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The curl is $\nabla\times\mathbf{F}=\langle0,0,2\rangle$ . For counterclockwise orientation viewed from above, the normal is $\mathbf{k}$ . The surface integral therefore becomes $\iint_S2\,dA$ , immediately giving $2(9\pi)=18\pi$ . Direct parameterization also works, but the curl method is substantially more efficient.

Q7. A circular boundary has radius $5$ , and a field's curl has constant component $8$ in the direction of the chosen unit normal. If the boundary orientation is compatible with that normal, what is the work?

40\pi

100\pi

200\pi

✅

8\pi

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: With constant curl component $8$ , the surface integral is $\iint_S8\,dS$ . A radius- $5$ disk has area $25\pi$ . Therefore the work is $8(25\pi)=200\pi$ . The important modelling step is recognizing that only the component of the curl along the oriented normal contributes.

Q8. A student computes $\nabla\times\mathbf{F}=\langle 0,0,-4\rangle$ and obtains a positive work value for a counterclockwise circle viewed from above. The student used the upward normal. What is the most likely error?

A.The circle's radius was necessarily incorrect

B.The student should use divergence instead of curl

C.The orientation sign is inconsistent with the computed sign of the curl ✅

D.Stokes' theorem applies only to clockwise curves

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For a counterclockwise traversal viewed from above, the compatible normal is upward. With curl $\langle0,0,-4\rangle$ , its dot product with the upward normal is negative. Thus the work must be negative for a positively oriented disk. A positive result indicates an algebraic, orientation, or sign-handling error.

Q9. A student says: 'Because two different surfaces span the same closed curve, Stokes' theorem gives different work values depending on which surface I choose.' Which response best evaluates this claim?

A.Correct, because surface area alone determines the work

B.Correct, because the curl changes whenever the surface changes

C.Incorrect under the theorem's regularity conditions; both valid oriented surfaces give the same boundary circulation ✅

D.Incorrect because only the smallest surface may be used

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The theorem relates the same boundary circulation to the curl flux through any suitable oriented surface spanning that boundary. Under the necessary smoothness and field conditions, different spanning surfaces produce the same result. If calculations disagree, the likely causes are orientation, parameterization, limits, or differentiation errors rather than a failure of the theorem.

Q10. A diagram shows a horizontal circular boundary viewed from above. The arrows around the circle point clockwise, while the selected surface normal points upward. A student obtains a positive surface integral. Which correction is required before interpreting the result as the work around the drawn curve?

A.Reverse the normal or reverse the curve orientation so the pair is compatible ✅

B.Replace curl with divergence

C.Multiply the result by the circle's circumference

D.Change the horizontal surface into a vertical one

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The boundary orientation and surface normal must satisfy the orientation convention. An upward normal corresponds to counterclockwise traversal when viewed from above, but the diagram specifies clockwise traversal. Therefore the pair is incompatible. Reversing either the curve direction or the normal restores consistency, changing the sign appropriately.

Q11. A field is $\mathbf{F}=\langle yz,xz,xy\rangle$ , and $C$ is the boundary of any closed surface patch lying in a region where the field is smooth. What observation can simplify the work calculation?

A.The curl is zero, so the work around

C

is zero ✅

B.The divergence is zero, so the work is zero

C.The field has three components, so direct integration is required

D.The work must equal the volume enclosed by

C

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Computing the curl gives $\nabla\times\mathbf{F}=\langle x-x, y-y, z-z\rangle=\mathbf{0}$ . Therefore the surface integral of the curl is zero for any suitable spanning surface. Stokes' theorem then implies that the circulation, and hence the work around the closed boundary, is zero.

Q12. A rectangular loop in the plane $z=0$ has side lengths $4$ and $7$ . For the chosen upward orientation, the curl of the field has constant value $\langle0,0,3\rangle$ . If the loop is traversed compatibly, what is the work?

A.-21

B.-42 ✅

C.-84

D.-10

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The upward normal is $\mathbf{k}$ , so the curl's normal component is $3$ . The rectangle's area is $4\times7=28$ . Stokes' theorem gives work $=\iint_S3\,dA=3(28)=84$ . Therefore the correct answer is $84$ , while $42$ results from an arithmetic or area-factor mistake.

Q13. Consider a vector field whose curl is $\langle 2x,2y,0\rangle$ . A circular boundary is the circle $x^2+y^2=R^2$ in the $xy$ -plane. Which reasoning best determines the work for counterclockwise traversal?

A.The work is always zero because the curl has no

z

-component

B.The work is

2\pi R^3

because the curl magnitude is constant on the boundary

C.The work is zero because

\langle2x,2y,0\rangle

has no upward component through the disk ✅

D.The work depends only on the circumference, not the enclosed surface

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: For the horizontal disk, the upward normal is $\mathbf{k}$ . The curl $\langle2x,2y,0\rangle$ has zero dot product with $\mathbf{k}$ everywhere on the disk. Hence the curl flux is zero, so Stokes' theorem gives zero circulation. This illustrates why curl magnitude alone is insufficient; its normal component matters.

Q14. A field is $\mathbf{F}=\langle -y/2,x/2,0\rangle$ , and a closed boundary encloses an arbitrary planar region $D$ in the $xy$ -plane. Without parameterizing the boundary, how can the work be expressed?

A.As the area of

D

✅

B.As twice the area of

D

C.As the negative of the area of

D

D.As the perimeter of

D

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The curl is $\nabla\times\mathbf{F}=\langle0,0,1\rangle$ . For positive counterclockwise orientation, the upward normal is $\mathbf{k}$ , so the curl flux equals $\iint_D1\,dA$ , which is simply the area of $D$ . This is a useful modelling situation because the exact boundary shape does not need to be parameterized.

Q15. A closed curve is the common boundary of two surfaces. On the first surface, the curl flux is computed as $15$ . A second valid surface gives a result of $-15$ . Which conclusion is most appropriate?

A.Both values are possible because surface choice changes circulation

B.The second surface must automatically be discarded

C.The orientations of the two surfaces are opposite relative to the same boundary orientation ✅

D.Stokes' theorem is invalid for nonplanar surfaces

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: For two valid spanning surfaces associated with the same directed boundary, Stokes' theorem requires compatible orientations. If one surface produces $15$ and the other produces $-15$ , the most natural explanation is that their normals induce opposite boundary orientations. Reorienting one surface makes the two flux calculations agree.

Q16. Let $\mathbf{F}=\langle y^2,x^2,0\rangle$ , and let $C$ be the positively oriented boundary of a unit disk in the $xy$ -plane. A student claims the work is $2\pi$ because the curl magnitude is $2$ everywhere. Which evaluation is correct?

A.The claim is correct because the disk has area

\pi

B.The work is zero because the curl has no

z

-component

C.The work is zero because the curl is

\langle0,0,2x-2y\rangle

, whose integral over the symmetric disk vanishes ✅

D.The work is

4\pi

because both derivatives contribute

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Here $\nabla\times\mathbf{F}=\langle0,0,2x-2y\rangle$ . With upward normal, the surface integral is $\iint_D(2x-2y)\,dA$ . The unit disk is symmetric about both coordinate axes, so the integrals of $x$ and $y$ separately vanish. Thus the work is zero, exposing the error of treating curl magnitude as constant.

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