📝 Green's theorem for multiply connected regions (16 MCQs)

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

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What is Green's theorem for multiply connected regions?

Green's theorem for multiply connected regions:
For regions with holes, Green's theorem applies by summing integrals over the outer and inner boundaries, with appropriate orientation (outer counterclockwise, inner clockwise).

Example:
For an annulus $1 \le r \le 2$ , $\oint_{\text{outer}} \mathbf{F} \cdot d\mathbf{r} - \oint_{\text{inner}} \mathbf{F} \cdot d\mathbf{r} = \iint_D (Q_x - P_y) \, dA$ .

Reason:
This extends Green's theorem to non-simply connected domains, essential for real-world applications with obstacles.

Easy

Medium

Hard

📝 All Green's theorem for multiply connected regions MCQs

Q1. A positively oriented region has an outer boundary $C_1$ and one inner boundary $C_2$ . Which orientation is required when applying Green's Theorem directly to the entire multiply connected region?

A.Both boundaries must be counterclockwise.

B.The outer boundary must be counterclockwise and the inner boundary clockwise. ✅

C.The outer boundary must be clockwise and the inner boundary counterclockwise.

D.Both boundaries must be clockwise.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For a positively oriented multiply connected region, the outer boundary is traversed counterclockwise so the region remains on the left. Each inner boundary must be traversed clockwise because the region lies outside the hole. This orientation correctly accounts for the missing interior when evaluating the boundary integral.

Q2. Suppose a vector field has $P_y-Q_x=5$ throughout a region whose area is $12$ , including a circular hole. If the stated area already excludes the hole, what is the value of the positively oriented boundary integral?

A.17

B.60 ✅

12/5

5/12

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Green's Theorem relates the circulation integral around the complete positively oriented boundary to the double integral of $P_y-Q_x$ over the actual region. Since the given area is already the area after removing the hole, the integral is $5(12)=60$ . The hole requires no additional subtraction.

Q3. Why can an inner boundary not simply be traversed counterclockwise when using Green's Theorem on a region containing a hole?

A.Because counterclockwise curves always produce zero circulation.

B.Because the inner curve would place the region on its right rather than its left. ✅

C.Because Green's Theorem applies only to circular boundaries.

D.Because the vector field must vanish inside every hole.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The positive orientation convention requires the region itself to remain on the left while moving along the boundary. Around an inner hole, counterclockwise motion places the surrounding region on the right. Therefore, the inner boundary must be clockwise so that the region stays on the left.

Q4. A student claims that a hole can be ignored because Green's Theorem integrates only around the outside boundary. Which response best evaluates the claim?

A.Correct, because holes never affect circulation.

B.Correct, but only when the hole is circular.

C.Incorrect, because the complete boundary includes the inner boundary and the region excludes the hole. ✅

D.Incorrect, because Green's Theorem cannot be used for regions with holes.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The claim confuses the outer boundary with the complete boundary of the region. A hole creates an additional boundary component, and its contribution is essential. Green's Theorem applies when all boundary components are included with the proper orientation. Ignoring the inner boundary generally produces an incorrect circulation value.

Q5. Consider a field satisfying $P_y-Q_x=0$ everywhere on an annular region. What can be concluded about the total circulation around the positively oriented outer and inner boundaries together?

A.It must equal the area of the annulus.

B.It must be positive.

C.It must be zero. ✅

D.It depends only on the radius of the inner circle.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Green's Theorem gives the total circulation over all boundary components as the double integral of $P_y-Q_x$ across the region. Because this quantity is identically zero, the double integral is zero regardless of the annulus dimensions. Thus, the combined circulation of the correctly oriented boundaries is zero.

Q6. A region is bounded by an outer curve and two disjoint holes. The outer boundary contributes $18$ to a circulation integral, while the two inner boundaries contribute $-7$ and $-5$ . What is the total circulation for the complete positively oriented boundary?

A.4 ✅

B.16

C.20

D.30

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The boundary contributions must be added algebraically, including the signs associated with the inner curves. Here the calculation is $18+(-7)+(-5)=6$ . Therefore, none of the listed numerical options is mathematically correct. This illustrates why every boundary component and its orientation must be tracked carefully rather than relying on geometric intuition alone.

Q7. An annular region has outer radius $4$ and inner radius $2$ . If $P_y-Q_x=x^2+y^2$ , what is the circulation around the complete positively oriented boundary?

30\pi

60\pi

120\pi

✅

240\pi

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Using Green's Theorem, integrate $x^2+y^2=r^2$ over the annulus. In polar coordinates the integral becomes $\int_0^{2\pi}\int_2^4 r^3\,dr\,d\theta$ . Evaluating gives $2\pi(4^4-2^4)/4=120\pi$ . The hole is handled by the lower radial limit $r=2$ , so no separate boundary calculation is necessary.

Q8. A student evaluates an annular circulation by integrating $P_y-Q_x$ over the entire disk of radius $4$ , forgetting the hole. If $P_y-Q_x=1$ , what conceptual error has occurred?

A.The student used polar coordinates unnecessarily.

B.The student included points that are not part of the actual region. ✅

C.The student reversed the outer boundary orientation.

D.The student assumed the field was conservative.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Green's Theorem requires integration over the actual region enclosed by all boundary components. An annulus is not the full disk because its central hole is excluded. Integrating over the whole disk counts area where the theorem's region does not exist, producing an answer that is too large when the integrand is positive.

Q9. A diagram shows a large counterclockwise outer circle and a smaller clockwise inner circle. A field has constant curl $3$ . If the outer radius is $5$ and the hole radius is $3$ , which expression represents the circulation?

3\pi(5^2+3^2)

3\pi(5^2-3^2)

✅

3\pi(5-3)^2

3\pi(5^2)

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The diagram represents an annulus, so the relevant area is the outer disk area minus the inner disk area: $\pi(5^2-3^2)$ . Since the curl $P_y-Q_x$ is $3$ , Green's Theorem gives $3\pi(25-9)=48\pi$ . The clockwise inner boundary automatically accounts for subtracting the hole.

Q10. Two methods are proposed for an annular circulation. Method I integrates $P_y-Q_x$ over the annulus. Method II separately computes the line integrals over the outer and inner circles. Under which condition should the two methods agree?

A.Only when the inner radius is zero.

B.Only when both circles have the same orientation.

C.When all boundary components in Method II use the correct induced orientations. ✅

D.Only when the vector field is conservative.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Green's Theorem permits either a double-integral approach over the actual multiply connected region or a boundary-integral approach over every boundary component. The methods agree only when the outer boundary is counterclockwise and each inner boundary is clockwise. Incorrect orientation changes signs and destroys the equivalence.

Q11. A student computes the circulation around the outer boundary of an annulus and obtains $40$ . They then compute the inner circle counterclockwise and obtain $12$ . They report $52$ as the circulation of the annulus. What is the best correction?

A.The answer should be

28

, because the inner boundary must be reversed. ✅

B.The answer should remain

52

, because both curves are boundaries.

C.The answer should be

48

, because the inner circle contributes twice.

D.The calculation is impossible because Green's Theorem cannot handle annuli.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The positive orientation of the outer boundary is counterclockwise, but the inner boundary must be clockwise. If $12$ was obtained using counterclockwise orientation, reversing that traversal changes its contribution to $-12$ . Thus the complete boundary integral is $40-12=28$ , not $52$ .

Q12. A region has two holes. A graph indicates that $P_y-Q_x$ is positive in the region and larger near the outer edge than near either hole. Which qualitative conclusion is most justified?

A.The circulation must be zero.

B.The circulation is positive, with its magnitude determined by the accumulated curl over the region. ✅

C.The holes force the circulation to be negative.

D.Only the largest hole contributes to the circulation.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Green's Theorem converts the total positively oriented boundary circulation into the double integral of $P_y-Q_x$ over the region. If this quantity is positive throughout the region, the total must be positive. The holes affect the domain of integration by removing area, but they do not automatically reverse the overall sign.

Q13. An engineer models fluid circulation around a plate containing two holes. The measured circulation around the outer edge is $25$ , while clockwise measurements around the holes are $4$ and $6$ . What total circulation corresponds to the complete positively oriented boundary?

A.-15

B.-23

C.-35 ✅

D.-25

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: For a positively oriented multiply connected region, the outer boundary is counterclockwise and the holes are clockwise. Since the stated hole measurements already use clockwise orientation, their contributions are added to the outer contribution. Therefore, the total circulation is $25+4+6=35$ . Orientation must be interpreted before combining measurements.

Q14. A region has an outer boundary $C_0$ and one hole $C_1$ . A student writes $\oint_{C_0}P\,dx+Q\,dy=\iint_R(P_y-Q_x)\,dA$ , claiming the hole does not matter. Which statement most accurately diagnoses the equation?

A.It is always correct for any vector field.

B.It becomes correct only if

C_1

contributes zero.

C.It is incomplete because the left side must include the integral around the inner boundary with its proper orientation. ✅

D.It is invalid because double integrals cannot represent line integrals.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: For a multiply connected region, Green's Theorem relates the double integral over the region to the sum of line integrals over every boundary component. The outer boundary alone is insufficient in general. The missing inner-boundary term accounts for the excluded hole and can substantially change the final result.

Q15. An annular region has outer radius $R$ and inner radius $r$ , with $R>r>0$ . Suppose $P_y-Q_x=k$ , where $k$ is constant. Which expression correctly models the total circulation as the hole expands while $R$ stays fixed?

2\pi k(R+r)

\pi k(R-r)

\pi k(R^2-r^2)

✅

2\pi k(R^2+r^2)

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The circulation equals the double integral of the constant curl $k$ over the annular region. The annular area is $\pi R^2-\pi r^2=\pi(R^2-r^2)$ . Therefore the circulation is $k\pi(R^2-r^2)$ . As the hole expands, less area remains, so the circulation decreases when $k>0$ .

Q16. For a region with several holes, suppose the curl is zero everywhere in the region, but the vector field is not necessarily defined inside the holes. Which conclusion follows most directly from Green's Theorem?

A.Every individual boundary integral must be zero.

B.The sum of the correctly oriented integrals over all boundary components is zero. ✅

C.The outer boundary integral must equal the area of all holes.

D.The field must be conservative throughout the entire plane.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Green's Theorem applies on the specified region, not necessarily inside the holes. If $P_y-Q_x=0$ throughout that region, the double integral is zero, so the sum of all correctly oriented boundary integrals is zero. This does not require the field to be defined or conservative inside excluded holes.

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