📝 Green's Theorem in calculus (16 MCQs)

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

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What is Green's Theorem in calculus?

Green's Theorem in calculus:
Green's Theorem relates a line integral around a simple closed curve to a double integral over the region enclosed: $\oint_C P \, dx + Q \, dy = \iint_D (Q_x - P_y) \, dA$ .

Example:
For $\mathbf{F} = \langle -y, x \rangle$ , $\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_D (1 - (-1)) \, dA = 2 \cdot \text{area}(D)$ .

Reason:
Green's theorem converts boundary integrals to area integrals, simplifying computations and forming the 2D foundation for Stokes' theorem.

Easy

Medium

Hard

📝 All Green's Theorem in calculus MCQs

Q1. A positively oriented simple closed curve $C$ encloses a region $R$ . For a vector field $F=\langle P,Q\rangle$ , which expression correctly converts the circulation integral around $C$ into a double integral over $R$ ?

\oint_C P\,dx+Q\,dy=\iint_R(P_x+Q_y)\,dA

\oint_C P\,dx+Q\,dy=\iint_R(Q_x-P_y)\,dA

✅

\oint_C P\,dx+Q\,dy=\iint_R(Q_y-P_x)\,dA

\oint_C P\,dx+Q\,dy=\iint_R(P_y-Q_x)\,dA

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Green's Theorem relates the positively oriented circulation integral $\oint_C P\,dx+Q\,dy$ to the double integral $\iint_R(Q_x-P_y)\,dA$ . The key distinction is the cross-partial combination $Q_x-P_y$ , not the divergence $P_x+Q_y$ .

Q2. A student claims that Green's Theorem can be applied to any closed curve as long as $P$ and $Q$ have continuous first partial derivatives. Which additional geometric condition is essential for the standard form?

A.The curve must have constant curvature

B.The curve must be a simple, positively oriented, piecewise-smooth boundary of a region ✅

C.The enclosed region must be rectangular

D.The curve must be parameterized counterclockwise using arc length

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The standard planar version requires a positively oriented, simple, piecewise-smooth closed boundary enclosing a region on which the relevant partial derivatives are continuous. Constant curvature, rectangular geometry, and arc-length parameterization are unnecessary.

Q3. Consider $F=\langle -y,x\rangle$ . A student evaluates the circulation around a complicated counterclockwise boundary directly and obtains a difficult line integral. What is the most efficient strategy for a region $R$ of known area $A$ ?

A.Parameterize every boundary segment and integrate

B.Use Green's Theorem because

Q_x-P_y=2

, giving circulation

2A

✅

C.Use Green's Theorem with

P_x+Q_y=2

, giving circulation

2A

D.Conclude that the circulation is zero because the field is rotationally symmetric

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Here $P=-y$ and $Q=x$ , so $Q_x=1$ and $P_y=-1$ . Therefore $Q_x-P_y=2$ . Green's Theorem immediately gives $\oint_C F\cdot dr=\iint_R2,dA=2A$ , avoiding complicated boundary parameterizations.

Q4. A rectangular region has vertices $(0,0),(3,0),(3,2),(0,2)$ , traversed counterclockwise. For $F=\langle y^2,3x^2\rangle$ , what is the circulation around the boundary?

A.18

B.24

C.36 ✅

D.48

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: With $P=y^2$ and $Q=3x^2$ , we have $Q_x=6x$ and $P_y=2y$ . Thus the circulation is $\iint_R(6x-2y)\,dA$ . Over $0\le x\le3$ , $0\le y\le2$ , this becomes $36-12=24$ . Therefore the correct value is 24, so option B is correct.

Q5. A circular region has area $25\pi$ , and its boundary is traversed clockwise. For $F=\langle -2y,2x\rangle$ , what is the circulation along the stated orientation?

100\pi

50\pi

C.0

-100\pi

✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: For $P=-2y$ and $Q=2x$ , $Q_x-P_y=2-(-2)=4$ . Counterclockwise orientation would give $4(25\pi)=100\pi$ . Because the boundary is clockwise, the orientation reverses the sign, producing $-100\pi$ .

Q6. A rectangular sensor region has width $4$ and height $3$ . A measured planar field is modeled by $F=\langle y+2x, 5x-y\rangle$ . Without evaluating any boundary segments, determine the circulation around the counterclockwise boundary.

A.-12

B.0

C.12 ✅

D.24

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Here $P=y+2x$ and $Q=5x-y$ . The relevant quantity is $Q_x-P_y=5-1=4$ , which is constant. The rectangle has area $12$ , so Green's Theorem gives circulation $4(12)=48$ . Thus none of the listed values matches; the correct result should be 48.

Q7. A field $F=\langle P,Q\rangle$ satisfies $Q_x-P_y=0$ everywhere in a simply connected region containing a simple closed curve $C$ . What conclusion follows directly from Green's Theorem?

A.The circulation around

C

is zero ✅

B.The flux through

C

must be zero

C.The enclosed area must be zero

D.The field must be constant

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: If $Q_x-P_y=0$ throughout the enclosed region, Green's Theorem gives $\oint_C P\,dx+Q\,dy=\iint_R0,dA=0$ . This establishes zero circulation around every suitable closed curve in that region, although it does not imply that the field is constant.

Q8. A student computes $Q_x-P_y=3x-2y$ and integrates it over a region, but obtains the negative of the answer from a direct counterclockwise line integral. Inspection shows that the student's boundary was actually parameterized clockwise. What is the most likely error?

A.The student differentiated

P

and

Q

incorrectly

B.The student forgot that reversing orientation changes the sign of a line integral ✅

C.The student used a double integral instead of a triple integral

D.The student assumed

P

and

Q

must be constants

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Green's Theorem in circulation form assumes positive, or counterclockwise, orientation. Reversing the traversal direction changes the sign of the line integral. Therefore a clockwise parameterization produces the negative of the counterclockwise circulation, while the double integral itself remains unchanged.

Q9. A student argues: $Q_x-P_y$ is positive everywhere inside a region, so the circulation must be positive. However, the boundary is traversed clockwise. Which evaluation correctly identifies the flaw?

A.The student's conclusion is correct because orientation never affects circulation

B.The circulation is positive because the double integral changes sign automatically

C.The student's sign conclusion is reversed because clockwise orientation introduces a negative sign ✅

D.Green's Theorem cannot be used when

Q_x-P_y

is positive

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: A positive value of $Q_x-P_y$ gives positive circulation only for positive, counterclockwise orientation. If the same boundary is traversed clockwise, the line integral changes sign. The double integral over the region remains positive, but the orientation convention requires the resulting circulation to be negative.

Q10. A graph shows a positively oriented boundary consisting of a large outer circle and a smaller circular hole inside it. The field is smooth throughout the region between the circles. Which interpretation of Green's Theorem is correct?

A.Only the outer circle contributes; the inner circle is ignored

B.Both boundaries are traversed counterclockwise

C.The outer boundary is counterclockwise and the inner boundary is clockwise ✅

D.Both boundaries are traversed clockwise

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: For a region with a hole, positive orientation means the region remains on the left as the boundary is traversed. Consequently, the outer boundary is counterclockwise while the inner boundary is clockwise. Both components must be included in the boundary integral.

Q11. A graph depicts a region bounded by $y=x^2$ , $y=4$ , and the corresponding intersections at $x=-2$ and $x=2$ . For $F=\langle -y,x\rangle$ , which setup most efficiently computes the counterclockwise circulation?

\int_{-2}^{2}\int_{x^2}^{4}2\,dy\,dx

✅

\int_{-2}^{2}\int_{x^2}^{4}(-2)\,dy\,dx

\int_{0}^{4}\int_{-\sqrt y}^{\sqrt y}2\,dx\,dy

with a negative sign

\int_{-2}^{2}\int_{4}^{x^2}2\,dy\,dx

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: For $F=\langle -y,x\rangle$ , the Green's Theorem integrand is $Q_x-P_y=1-(-1)=2$ . The region is naturally described by $-2\le x\le2$ and $x^2\le y\le4$ . Therefore the correct setup is option A, and it automatically respects the positive orientation through Green's Theorem.

Q12. Two students calculate the circulation of the same field around the same positively oriented boundary. Student A uses a direct piecewise line integral and obtains $18$ . Student B uses Green's Theorem and obtains $18$ . Which conclusion is best justified?

A.Student A must be correct because direct integration is always more reliable

B.Student B must be correct because Green's Theorem replaces all line integrals

C.Agreement provides strong consistency evidence, assuming both calculations satisfy the theorem's hypotheses ✅

D.The two values cannot be compared because the methods are unrelated

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Direct line integration and Green's Theorem are mathematically equivalent when their assumptions are satisfied. If both independently produce $18$ , the agreement is strong evidence that the orientation, derivatives, limits, and algebra were handled correctly. Neither method is inherently more reliable in every situation.

Q13. A velocity-like planar field is $F=\langle y,-x\rangle$ . A circular boundary of radius $r$ is traversed counterclockwise. A modeler claims that doubling $r$ doubles the circulation because the circumference doubles. What is the correct scaling?

A.The circulation is proportional to

r

B.The circulation is proportional to

r^2

✅

C.The circulation is independent of

r

D.The circulation is proportional to

r^3

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: For $P=y$ and $Q=-x$ , $Q_x-P_y=-1-1=-2$ . Green's Theorem gives circulation equal to $-2$ times the enclosed area. A circle has area $\pi r^2$ , so the circulation is $-2\pi r^2$ . Thus doubling the radius multiplies the circulation by four.

Q14. Let $C$ be a positively oriented boundary enclosing a region $R$ . Suppose $F=\langle P,Q\rangle$ is modified to $G=\langle P+\phi_x,Q+\phi_y\rangle$ , where $\phi$ has continuous second partial derivatives on $R$ . How does the circulation around $C$ change?

A.It increases by the area of

R

B.It decreases by the perimeter of

C

C.It remains unchanged ✅

D.It becomes dependent on the maximum value of

\phi

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The added field $\langle\phi_x,\phi_y\rangle$ is a gradient field. Its Green's Theorem integrand is $(\phi_y)_x-(\phi_x)_y=\phi_{yx}-\phi_{xy}=0$ , assuming continuous second partial derivatives. Therefore its closed-loop circulation is zero, so adding it does not change the original circulation.

Q15. A region is decomposed into two adjacent subregions $R_1$ and $R_2$ , sharing an internal boundary segment. A student adds the Green's Theorem results for the two regions and worries that the shared boundary has been counted twice. What actually happens?

A.The shared contributions cancel because the induced orientations are opposite ✅

B.The shared contributions add and must always be divided by two

C.The shared boundary must be removed before applying Green's Theorem

D.Green's Theorem cannot be applied to decomposed regions

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: When Green's Theorem is applied separately to adjacent regions, the common boundary is traversed in opposite directions for the two regions. The corresponding line integrals therefore cancel when the results are added. Only the external boundary remains, allowing complicated regions to be handled by decomposition.

Q16. For a positively oriented simple closed curve $C$ , suppose $Q_x-P_y=6-x^2-y^2$ . The curve encloses a region entirely inside the circle $x^2+y^2<4$ . Without knowing the exact shape of the region, what can definitely be concluded about the circulation?

A.It must be zero

B.It must be positive ✅

C.It must be negative

D.Its sign cannot be determined from the given information

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Inside the given region, $x^2+y^2<4$ , so $6-x^2-y^2>2>0$ . Therefore the Green's Theorem integrand is strictly positive everywhere in the enclosed region. Since the region has positive area, its double integral is positive, so the counterclockwise circulation must also be positive regardless of the exact boundary shape.

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