📝 Green's theorem vs Stokes' theorem (14 MCQs)

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

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What is Green's theorem vs Stokes' theorem?

Green's theorem vs Stokes' theorem:
Green's theorem is the 2D planar case of Stokes' theorem, where $S$ lies in the $xy$ -plane and $\nabla \times \mathbf{F} \cdot d\mathbf{S} = (Q_x - P_y) \, dA$ .

Example:
For $\mathbf{F} = \langle P, Q, 0 \rangle$ , Stokes' theorem reduces to $\oint_C P \, dx + Q \, dy = \iint_D (Q_x - P_y) \, dA$ , exactly Green's theorem.

Reason:
Stokes' theorem is a 3D generalization, showing Green's theorem as a special case, unifying plane and space vector calculus.

Easy

Medium

Hard

📝 All Green's theorem vs Stokes' theorem MCQs

Q1. Which statement best describes why Green's theorem can be viewed as a special case of Stokes' theorem?

A.Green's theorem applies only to conservative vector fields, while Stokes' theorem applies to all fields

B.Green's theorem uses a planar region and its boundary, which can be embedded in three-dimensional space so that the surface curl relation becomes the planar circulation relation ✅

C.Stokes' theorem is obtained from Green's theorem only when the boundary is a straight line

D.Green's theorem calculates flux, whereas Stokes' theorem calculates only divergence

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Green's theorem relates circulation around a positively oriented planar boundary to a double integral involving the scalar curl. Stokes' theorem relates circulation around a space curve to the surface integral of the curl. By placing the planar region in three-dimensional space with an appropriate normal, the two formulations become equivalent.

Q2. A student claims that Green's theorem and Stokes' theorem must produce different answers whenever the problem is written in three dimensions. Which observation most directly challenges the claim?

A.A planar surface can be regarded as a surface in three-dimensional space, allowing Stokes' theorem to reproduce the corresponding Green's theorem calculation ✅

B.Green's theorem always gives a larger value because it uses a double integral

C.Stokes' theorem ignores the orientation of the boundary

D.Three-dimensional vector fields cannot be restricted to planar regions

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: A planar region in the $xy$ -plane is also a legitimate surface in three-dimensional space. If its upward normal and positively oriented boundary are chosen consistently, Stokes' theorem gives the same circulation relation as Green's theorem. Thus the difference is mainly formulation and geometric setting, not the underlying result.

Q3. For a planar vector field $F=\langle P,Q\rangle$ , which quantity in Green's circulation form corresponds most directly to the normal component of the three-dimensional curl used in Stokes' theorem?

P_x+Q_y

Q_x-P_y

✅

P_y+Q_x

P_x-Q_y

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For $F=\langle P,Q\rangle$ , Green's circulation theorem uses $Q_x-P_y$ . If the field is embedded as $\mathbf F=\langle P,Q,0\rangle$ , then its curl is $\nabla\times\mathbf F=\langle -Q_z,P_z,Q_x-P_y\rangle$ . On a planar $xy$ -surface, the upward normal selects the third component, giving exactly $Q_x-P_y$ .

Q4. A rectangular region lies in the $xy$ -plane and its boundary is traversed counterclockwise when viewed from above. A student wants to use Stokes' theorem instead of Green's theorem. What is the most important geometric choice?

A.Use a downward normal because counterclockwise orientation always corresponds to a downward normal

B.Use an upward normal so that the right-hand rule produces the counterclockwise boundary orientation ✅

C.Choose either normal because Stokes' theorem is independent of orientation

D.Use a normal parallel to the

x

-axis because the region is rectangular

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The orientation of the boundary and surface normal must satisfy the right-hand rule. For a surface in the $xy$ -plane, an upward normal points in the positive $z$ -direction. Viewed from above, that normal induces counterclockwise boundary orientation. Reversing the normal would reverse the boundary direction and change the sign.

Q5. A circulation integral around a closed planar curve appears difficult to evaluate directly because the curve is piecewise and contains several arcs. Which reasoning best justifies replacing the line integral with a surface integral?

A.The complicated boundary can be ignored because all closed curves have zero circulation

B.The curl may be integrated over the enclosed region, often turning a complicated boundary calculation into a simpler double integral ✅

C.The field must first be proven constant along the boundary

D.A surface integral is always numerically smaller than a line integral

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Both theorems provide a strategic alternative to direct boundary integration. Instead of parameterizing every segment or arc, one can calculate the appropriate curl over the enclosed region. If the resulting double integral is easier, the theorem converts geometric complexity in the boundary into potentially simpler area integration.

Q6. Suppose a planar region is rotated rigidly in space while the vector field is interpreted as a three-dimensional field. Which principle determines whether the Stokes formulation still corresponds to the original planar Green formulation?

A.Only the numerical area of the region matters

B.The transformed surface, its normal orientation, and the induced boundary orientation must remain geometrically consistent ✅

C.The boundary must remain parallel to the

x

-axis

D.The vector field must have zero curl everywhere

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Green's theorem is tied to a planar orientation, while Stokes' theorem is coordinate-independent and works on oriented surfaces. After rotating the surface, the relevant normal changes, so the curl must be dotted with that new normal. The boundary orientation must also follow the right-hand rule associated with the transformed normal.

Q7. A model of fluid motion has planar velocity field $F=\langle -y,x\rangle$ . A circular boundary is difficult to parameterize accurately in a numerical simulation. Which approach is most efficient for finding circulation around the boundary?

A.Compute

P_x+Q_y

over the region

B.Compute

Q_x-P_y

over the region and integrate it over the enclosed area ✅

C.Integrate

P+Q

along the boundary without parameterization

D.Compute the divergence and multiply by the circumference

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Here $P=-y$ and $Q=x$ , so $Q_x-P_y=1-(-1)=2$ . Green's circulation form therefore replaces the boundary calculation by integrating $2$ over the enclosed disk. This is especially useful computationally because the area is easy to represent even when boundary parameterization is inconvenient.

Q8. A student uses Green's theorem on a planar curve but obtains the negative of the expected answer. Inspection shows that the curve was traversed clockwise while the problem's orientation was counterclockwise. What is the most likely correction?

A.Change the curl to its absolute value

B.Reverse the sign of the computed circulation ✅

C.Replace the double integral by a triple integral

D.Multiply the result by the perimeter

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Green's theorem requires positive orientation, normally counterclockwise for a region in the $xy$ -plane with the upward normal. Traversing the same boundary clockwise reverses the direction of the line integral. Therefore the circulation changes sign. No change to the vector field or the magnitude of the curl is required.

Q9. A graph shows a closed curve in the $xy$ -plane traversed counterclockwise. The vector field has $Q_x-P_y>0$ throughout the enclosed region. Without performing any integration, what can be concluded about the circulation?

A.It must be negative

B.It must be zero

C.It must be positive ✅

D.Its sign cannot be determined from the given information

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For counterclockwise orientation in the $xy$ -plane, Green's circulation form uses the integral of $Q_x-P_y$ over the region. If this quantity is positive everywhere and the region has positive area, its double integral is positive. Therefore the circulation must also be positive.

Q10. A student computes a circulation using Stokes' theorem and obtains $12$ . Another student applies Green's theorem to the same planar region but obtains $-12$ . Both use the same vector field and boundary. Which explanation is most plausible?

A.The theorems disagree in three dimensions

B.One student likely used opposite orientations for the boundary and surface normal ✅

C.Green's theorem cannot handle polynomial vector fields

D.Stokes' theorem always doubles the answer from Green's theorem

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: For the same planar region, Green's theorem and Stokes' theorem must agree when their orientations are consistent. A sign difference strongly indicates that one calculation used a reversed boundary direction or an inconsistent normal. Reversing orientation changes the circulation's sign, explaining $12$ versus $-12$ without contradicting either theorem.

Q11. Consider a planar region whose boundary consists of several line segments and a circular arc. A direct line integral requires separate parameterizations for each piece, while the corresponding curl is a constant $5$ . Which method is likely to minimize computational effort, and why?

A.Direct integration is always preferable because Green's theorem adds another variable

B.Use the area integral because integrating the constant curl over the region reduces the problem to

5

times the region's area ✅

C.Use divergence because circulation depends only on sources and sinks

D.Use a triple integral because the boundary contains a circular arc

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Green's theorem transforms the circulation into a double integral of the scalar curl. Since the curl is constant at $5$ , the integral becomes $5A$ , where $A$ is the enclosed area. This avoids separate parameterizations of every boundary piece and illustrates how theorem selection can simplify modeling.

Q12. A researcher models circulation around a planar boundary and notices that the same vector field can be embedded as $\mathbf F=\langle P,Q,0\rangle$ . The researcher then computes $\nabla\times\mathbf F$ and takes its dot product with the upward normal. What additional fact is essential for concluding that the result reproduces the planar circulation theorem?

A.The

z

-component of curl must equal

Q_x-P_y

, matching the scalar integrand in the planar formulation ✅

B.The divergence must equal zero everywhere

C.The surface must have unit area

D.The boundary must consist only of straight lines

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Embedding $F=\langle P,Q\rangle$ as $\mathbf F=\langle P,Q,0\rangle$ gives a curl whose $z$ -component is $Q_x-P_y$ . The upward unit normal $\mathbf k$ selects that component in the surface integral. Thus the Stokes expression becomes the same double integral used for planar circulation.

Q13. A closed planar curve encloses a region containing a point where the vector field is undefined. A student immediately applies Green's theorem over the entire enclosed region and concludes that the circulation is determined by the curl integral. What should be examined first?

A.Whether the field satisfies the required smoothness conditions throughout the region and on an appropriate neighborhood ✅

B.Whether the curve has an even number of corners

C.Whether the area is measured in square units

D.Whether the field has positive divergence at the center

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The usual theorem conditions require sufficient smoothness of the vector field on the relevant region. If the field is undefined at an interior point, applying the theorem directly over the entire region may be invalid. The region may need to be modified, such as by removing a small neighborhood around the singularity and analyzing the resulting boundaries.

Q14. For a smooth planar field, suppose every positively oriented closed curve in a simply connected region has circulation zero. What stronger structural conclusion is most consistent with the relationship between the two theorems?

A.The planar curl must vanish throughout the region, so the corresponding three-dimensional curl has zero normal component on the plane ✅

B.The divergence must be positive everywhere

C.The field must have nonzero curl away from every curve

D.The region must have infinite area

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: If circulation is zero around every closed curve in a suitable simply connected region, Green's theorem implies that the integral of $Q_x-P_y$ over every such subregion is zero. Under appropriate smoothness, this forces $Q_x-P_y=0$ throughout the region. In the three-dimensional embedding, this is precisely the normal component of the curl, linking the planar and spatial viewpoints.

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