📝 Stokes Theorem in Calculus (15 MCQs)

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

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What is Stokes Theorem in Calculus?

Stokes Theorem in Calculus:
Stokes' theorem relates the surface integral of curl to the line integral around its boundary: $\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$ .

Example:
For $\mathbf{F} = \langle -y, x, 0 \rangle$ over a disk in the $xy$ -plane, $\nabla \times \mathbf{F} = \langle 0,0,2 \rangle$ , so LHS = $2 \cdot \text{area} = 2\pi a^2$ .

Reason:
Stokes' theorem generalizes Green's theorem to 3D, converting boundary circulation to surface curl, essential in electromagnetism.

Easy

Medium

Hard

📝 All Stokes Theorem in Calculus MCQs

Q1. For a smooth oriented surface $S$ with boundary curve $C$ , which expression correctly represents the relationship used in Stokes' Theorem?

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

✅

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

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

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

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Stokes' Theorem equates the circulation of a vector field around a closed boundary curve with the flux of its curl through any suitably oriented surface spanning that curve. The normal direction and boundary orientation must be compatible, so the correct relationship is the one given in option A.

Q2. A student says that Stokes' Theorem can be applied to any open curve $C$ because the surface integral only depends on the curve's endpoints. What is the most important correction?

A.The curve must be closed and serve as the oriented boundary of the surface. ✅

B.The curve may be open if the field is conservative.

C.The curve must always be a circle.

D.The curve must lie in the

xy

-plane.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Stokes' Theorem relates a surface integral over a surface to a line integral around its closed boundary. An open curve does not generally form the complete boundary of a surface, so the student's reasoning fails at the fundamental geometric requirement. The curve need not be circular or planar.

Q3. Two different smooth surfaces $S_1$ and $S_2$ span the same closed curve $C$ , and both use the compatible orientation. A vector field has a continuous curl throughout the region between the surfaces. What conclusion follows?

A.The two surface integrals of the curl are equal. ✅

B.The surface integral over

S_1

must be zero.

C.The surface integral over

S_2

must be larger because its area may differ.

D.The two integrals are equal only when the surfaces have the same area.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Stokes' Theorem states that each compatible surface integral of $(\nabla\times\mathbf F)\cdot\mathbf n$ equals the same circulation $\oint_C\mathbf F\cdot d\mathbf r$ . Therefore, changing the spanning surface can simplify the calculation without changing the final value, provided the required smoothness and field conditions hold.

Q4. Suppose $\nabla\times\mathbf F=\langle 0,0,6\rangle$ , and a closed boundary $C$ encloses a planar region of area $4$ with upward orientation. What is the circulation around $C$ ?

A.6

B.10

C.24 ✅

3/2

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The curl is constant and points in the positive $z$ -direction. With an upward normal, the dot product is $6$ . Therefore, the surface integral becomes $6$ times the enclosed area, giving $6(4)=24$ . Stokes' Theorem then identifies this flux of curl with the boundary circulation.

Q5. A circular boundary of radius $3$ lies in the $xy$ -plane and is oriented counterclockwise when viewed from above. If $\nabla\times\mathbf F=\langle 0,0,2\rangle$ , which strategy is most efficient for finding $\oint_C\mathbf F\cdot d\mathbf r$ ?

A.Use the disk bounded by

C

and evaluate the curl flux. ✅

B.Parametrize the circle and compute four separate line integrals.

C.Use the total surface area of a sphere of radius

3

D.Use the divergence of

\mathbf F

over the disk.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The boundary is a circle, so the flat disk it encloses is a natural spanning surface. Since the curl is constant and perpendicular to that disk, the surface integral is simply $2$ multiplied by the disk area $9\pi$ , giving $18\pi$ . Stokes' Theorem avoids unnecessary parametrization.

Q6. Let $\mathbf F=\langle -y,x,0\rangle$ , and let $C$ be the boundary of any smooth surface whose upward projection onto the $xy$ -plane has area $10$ . What is $\oint_C\mathbf F\cdot d\mathbf r$ , assuming the compatible positive orientation?

A.10

B.20 ✅

5\pi

D.0

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For this field, $\nabla\times\mathbf F=\langle0,0,2\rangle$ . Stokes' Theorem permits replacing the original surface with the simpler planar projection when the boundary orientation is preserved. The curl flux is therefore $2$ times the projected area $10$ , producing $20$ .

Q7. A hemispherical surface and its circular base have the same boundary circle. A student chooses the curved hemisphere because it appears more directly related to the vector field. Why might this be a poor computational choice?

A.The hemisphere cannot have an orientation.

B.Stokes' Theorem only works on planar surfaces.

C.A simpler spanning surface may produce a much easier curl flux integral. ✅

D.The boundary of a hemisphere is not a closed curve.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Stokes' Theorem allows different spanning surfaces with the same oriented boundary. The curved hemisphere is mathematically valid when properly oriented, but its geometry may create complicated normal vectors and surface elements. If a planar disk has the same boundary, it can often reduce the problem to a much simpler integral.

Q8. A calculation gives $\oint_C\mathbf F\cdot d\mathbf r=12$ using a surface normal pointing upward. Another student reverses the normal but keeps the same direction around $C$ . They still obtain $12$ . What is wrong?

A.Reversing the normal requires reversing the boundary orientation, so the result should change sign. ✅

B.The result remains

12

because normal direction never matters in Stokes' Theorem.

C.Only the magnitude of the normal matters, so the sign is unchanged.

D.The curve must be reparameterized with twice its original length.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The surface orientation and boundary orientation are linked by the right-hand rule. Reversing the normal changes the orientation of the boundary. If the student reverses only the normal while retaining the original traversal direction, the two sides no longer correspond consistently, causing a sign error. The circulation should become $-12$ under reversed boundary orientation.

Q9. A student computes $\nabla\times\mathbf F$ , integrates it over a convenient disk, and obtains a positive value. However, the original curve $C$ is traversed clockwise when viewed from above, while the disk normal used was upward. What should the student do?

A.Keep the positive result because the curl determines the sign independently.

B.Change the disk normal to downward or reverse the direction of traversal. ✅

C.Multiply the result by the disk radius.

D.Replace the curl with the divergence of the field.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: An upward normal corresponds, by the right-hand rule, to counterclockwise boundary traversal when viewed from above. Since the given curve is clockwise, it is incompatible with the upward normal. The student must either use a downward normal or reverse the curve orientation. Otherwise, the computed sign does not represent the requested circulation.

Q10. A graph of a scalar quantity $g(t)$ representing the accumulated circulation contribution along a boundary shows $g(t)$ increasing from $0$ to $8$ as $t$ moves once around the closed curve. Which interpretation is most consistent with Stokes' Theorem if the graph represents the cumulative line integral?

A.The final value

8

represents the circulation around the boundary. ✅

B.The maximum slope must equal the surface area.

C.The final value must always be zero because the curve is closed.

D.The graph directly gives the divergence of the field.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: For a cumulative line integral, the value after one complete traversal represents the total circulation $\oint_C\mathbf F\cdot d\mathbf r$ . Stokes' Theorem identifies that circulation with the flux of the curl through a compatible spanning surface. A closed path does not imply zero circulation unless additional conditions apply.

Q11. A graph shows a closed planar curve enclosing two regions: one where the normal component of $\nabla\times\mathbf F$ is strongly positive and another where it is weakly negative. The positive region has approximately twice the area and three times the magnitude of the negative region. What is the most reasonable conclusion?

A.The total circulation is certainly zero because the curve is closed.

B.The positive contribution dominates, so the circulation should be positive. ✅

C.The negative region must dominate because its sign is negative.

D.The circulation cannot be determined without calculating the divergence.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Stokes' Theorem converts the boundary circulation into the surface integral of the normal component of the curl. The positive region contributes approximately twice the area times three times the magnitude, while the negative contribution is smaller. Thus the net curl flux is positive, so the circulation should also be positive.

Q12. A field is $\mathbf F=\langle y,z,x\rangle$ , and $C$ is a closed curve bounding a surface oriented upward. Which feature of the field determines the circulation through Stokes' Theorem?

A.The divergence

\nabla\cdot\mathbf F

B.The magnitude of

\mathbf F

everywhere on the surface

C.The component of

\nabla\times\mathbf F

normal to the surface ✅

D.The value of

\mathbf F

only at the center of the surface

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Stokes' Theorem does not use the divergence or merely the magnitude of the original field. It uses the curl and specifically its component normal to the chosen surface. For $\mathbf F=\langle y,z,x\rangle$ , the curl is $\langle-1,-1,-1\rangle$ , which must then be dotted with the oriented normal.

Q13. A researcher needs the circulation of a field around a complicated closed boundary. Direct parametrization requires several piecewise expressions, while a simple planar surface has the same boundary and makes $\nabla\times\mathbf F$ easy to integrate. Which modelling decision is best?

A.Use the complicated parametrization because Stokes' Theorem requires the original surface.

B.Replace the boundary with a different curve that is easier to parameterize.

C.Choose the simple spanning surface and apply Stokes' Theorem. ✅

D.Ignore orientation because the two methods must give the same positive value.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The strength of Stokes' Theorem is that the line integral can be replaced by a surface integral over any suitable spanning surface with the same oriented boundary. A simpler surface can dramatically reduce computational complexity. The original boundary cannot simply be replaced, and orientation remains essential because reversing it changes the sign.

Q14. A vector field satisfies $\nabla\times\mathbf F=\mathbf 0$ throughout a simply connected region containing a closed curve $C$ . Without directly parameterizing $C$ , what can be concluded?

\oint_C\mathbf F\cdot d\mathbf r=0

✅

B.The circulation equals the perimeter of

C

C.The circulation equals the area enclosed by

C

D.The circulation is nonzero unless

C

is circular.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: By Stokes' Theorem, the circulation equals the surface integral of the curl over any suitable spanning surface. If the curl is identically zero throughout the relevant region, the surface integral is zero. The simply connected condition ensures that the closed curve can be treated within a region where the field has the required curl-free behavior.

Q15. Let $C$ be a closed curve bounding a surface $S$ , and suppose $\nabla\times\mathbf F=\langle 2x,2y,2z\rangle$ . The curve lies on a sphere centered at the origin, and its chosen spanning surface is perpendicular to the radial direction everywhere. What is the most important observation for simplifying the Stokes integral?

A.The curl is always tangent to the surface, so the integral is zero.

B.The curl is radial, so its normal component can be evaluated directly from the surface geometry. ✅

C.The curl has constant magnitude, so the integral equals surface area times

2

D.The curl must be replaced by the divergence before integration.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The curl field $\langle2x,2y,2z\rangle$ is radial because it is proportional to the position vector. If the surface normal is aligned with the radial direction, then the dot product $(\nabla\times\mathbf F)\cdot\mathbf n$ becomes especially simple. This geometric alignment can turn a difficult-looking surface integral into a direct scalar integration.

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