📝 Work using Green's theorem (15 MCQs)

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

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What is Work using Green's theorem?

Work using Green's theorem:
Work $\oint_C \mathbf{F} \cdot d\mathbf{r}$ can be computed as $\iint_D (Q_x - P_y) \, dA$ , avoiding path parameterization.

Example:
Work by $\mathbf{F} = \langle x^2, y^2 \rangle$ around unit circle: $Q_x - P_y = 0 - 0 = 0$ , so work = 0.

Reason:
This application dramatically reduces computational effort for closed paths, especially in complex regions.

Easy

Medium

Hard

📝 All Work using Green's theorem MCQs

Q1. A positively oriented closed curve $C$ encloses a region $R$ . For a vector field $\mathbf F=\langle P,Q\rangle$ , which expression directly converts the work integral $\oint_C \mathbf F\cdot d\mathbf r$ into a double integral over $R$ ?

\iint_R(P_x-Q_y)\,dA

\iint_R(Q_x-P_y)\,dA

✅

\iint_R(P_y-Q_x)\,dA

\iint_R(P_x+Q_y)\,dA

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: For a positively oriented simple closed curve, the work integral is $\oint_C P\,dx+Q\,dy$ . The corresponding double integral uses the curl-like quantity $Q_x-P_y$ , so the correct conversion is $\iint_R(Q_x-P_y)\,dA$ . The other choices either reverse the sign or represent a different differential combination.

Q2. Suppose $\mathbf F=\langle -y,x\rangle$ and $C$ is a positively oriented closed curve enclosing area $A$ . Without knowing the exact shape of $C$ , what can be concluded about the work done around $C$ ?

A.It is always zero because the curve is closed

B.It depends on the perimeter of

C

C.It equals

2A

✅

D.It equals

A/2

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Here $P=-y$ and $Q=x$ , so $Q_x=1$ and $P_y=-1$ . Therefore $Q_x-P_y=2$ , a constant. Green's Theorem converts the work into $\iint_R2,dA=2A$ . Thus the detailed boundary shape is irrelevant; only the enclosed area matters.

Q3. A student argues that because a path is closed, the work integral $\oint_C\mathbf F\cdot d\mathbf r$ must be zero. Which response best identifies the flaw in this reasoning?

A.Closed paths always have zero displacement, so the student is correct

B.A closed path can have nonzero circulation when

Q_x-P_y

is not identically zero inside the region ✅

C.Work can never be evaluated on closed paths

D.The work is zero only when the curve has positive orientation

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A closed curve returns to its starting point, but that does not imply zero work for a general vector field. Green's Theorem shows that circulation depends on the accumulated quantity $Q_x-P_y$ over the enclosed region. Unless that quantity integrates to zero, the closed-path work can be nonzero.

Q4. For $\mathbf F=\langle y^2,3x^2\rangle$ , a positively oriented curve encloses a region of area $5$ . What is the work around the curve?

A.0

B.15 ✅

C.30

D.45

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The integrand is $6x-2y$ , which is not constant. Knowing only that the region has area $5$ does not determine its integral because different regions with the same area can have different averages of $x$ and $y$ . Hence the correct mathematical conclusion should be that the work cannot be determined from the stated information alone.

Q5. A rectangular region is bounded by $0\le x\le2$ and $0\le y\le3$ , traversed counterclockwise. For $\mathbf F=\langle -2y,4x\rangle$ , what is the work around the boundary?

A.12

B.24 ✅

C.36

D.48

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Here $P=-2y$ and $Q=4x$ , giving $Q_x-P_y=4-(-2)=6$ . The rectangle has area $2\cdot3=6$ . Green's Theorem therefore gives the work as $6\cdot6=36$ . Thus option C is correct. The distractors arise from using perimeter, area, or an incorrect derivative combination.

Q6. A positively oriented triangular region has vertices $(0,0)$ , $(4,0)$ , and $(0,2)$ . If $\mathbf F=\langle -y,x\rangle$ , what is the work around its boundary?

A.2

B.4

C.8 ✅

D.16

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For $\mathbf F=\langle-y,x\rangle$ , the quantity $Q_x-P_y=1-(-1)=2$ . The triangle has area $\frac12(4)(2)=4$ . Therefore Green's Theorem gives work $=2(4)=8$ . The result depends only on the enclosed area because the relevant derivative combination is constant.

Q7. A rectangular loop is traversed clockwise instead of counterclockwise. If the counterclockwise work would be $18$ , what is the clockwise work for the same vector field and region?

A.18

B.9

C.-18 ✅

D.-36

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Reversing the direction of traversal reverses the sign of a line integral. Green's Theorem is normally stated for positive, counterclockwise orientation, so a clockwise traversal introduces a negative sign. Therefore a counterclockwise value of $18$ becomes $-18$ , not $18$ or a value based on a changed area.

Q8. A student computes $\iint_R(Q_x-P_y)\,dA$ for a clockwise boundary and obtains $12$ . They report the work as $12$ . What is the most appropriate correction?

A.Keep

12

, because orientation never matters in Green's Theorem

B.Use

-12

, because the given boundary orientation is opposite to the standard positive orientation ✅

C.Divide

12

by the area

D.Replace

Q_x-P_y

with

Q_x+P_y

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The standard Green's Theorem circulation form assumes counterclockwise orientation. If the actual boundary is clockwise, the line integral has the opposite sign from the double integral computed with positive orientation. Thus the student's magnitude is correct but the sign is wrong, so the corrected work is $-12$ .

Q9. A diagram shows a positively oriented circular boundary centered at the origin with radius $3$ . The vector field is $\mathbf F=\langle -y,x\rangle$ . A student estimates the work by multiplying the circumference $6\pi$ by an average field magnitude of $3$ . Why is this approach unreliable, and what is the exact work?

A.It is reliable and gives

18\pi

B.It ignores directional variation; the exact work is

18\pi

✅

C.It ignores directional variation; the exact work is

9\pi

D.The field is conservative, so the exact work is zero

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Along the circular path, the vector field is tangent to the circle and has magnitude $r=3$ , so the direct computation happens to produce $18\pi$ . However, treating the field magnitude as an average without analyzing direction is generally unsafe. Green's Theorem gives $Q_x-P_y=2$ , while the disk area is $9\pi$ , yielding exactly $18\pi$ .

Q10. Two different positively oriented simple closed curves enclose regions $R_1$ and $R_2$ . For a vector field satisfying $Q_x-P_y=5$ everywhere, curve $C_1$ encloses area $7$ and curve $C_2$ encloses area $11$ . Which comparison is correct?

A.The work is equal because both curves are closed

B.The work around

C_2

exceeds that around

C_1

20

✅

C.The work around

C_2

exceeds that around

C_1

4

D.The work depends only on the perimeter difference

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Because $Q_x-P_y=5$ is constant, Green's Theorem gives work equal to $5$ times the enclosed area. The first work is $35$ , while the second is $55$ . Their difference is $55-35=20$ . This illustrates why enclosed area, rather than perimeter or detailed shape, controls the result in this special situation.

Q11. A force field is modeled by $\mathbf F=\langle y+2x, x+3y\rangle$ . An engineer wants the net work around any positively oriented closed boundary enclosing a region. Which feature of the model determines whether Green's Theorem predicts zero circulation for every such region?

A.Whether

P+Q

is constant

B.Whether

Q_x-P_y

is identically zero ✅

C.Whether

P_x+Q_y

is identically zero

D.Whether the region has zero area

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: For $P=y+2x$ and $Q=x+3y$ , we have $Q_x=1$ and $P_y=1$ , so $Q_x-P_y=0$ . Green's Theorem then makes the circulation integral zero for every suitable positively oriented closed region. The divergence-like expression $P_x+Q_y$ is unrelated to circulation in this formula.

Q12. A student evaluates a complicated boundary integral directly and obtains $14$ . Another student uses Green's Theorem and obtains $-14$ . The boundary is actually oriented clockwise, while the region was entered into the double integral using positive orientation. Which conclusion is most reasonable?

A.The direct calculation must be wrong because Green's Theorem always gives the same sign

B.The Green's Theorem result should be

14

because clockwise orientation is positive

C.The two results are consistent because reversing orientation changes the sign of the line integral ✅

D.Both results must be zero because the boundary is closed

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Green's Theorem uses positive counterclockwise orientation. If the direct boundary integral follows the clockwise direction while the double integral is evaluated over the same region using the positive convention, the two calculations naturally differ by a sign. Thus $14$ and $-14$ can represent the same circulation under opposite orientations.

Q13. Let $\mathbf F=\langle P,Q\rangle$ satisfy $Q_x-P_y=2x$ . A positively oriented region is symmetric about the $y$ -axis. Without knowing its exact dimensions, what can be concluded about the work around its boundary?

A.It must be positive

B.It must be negative

C.It must be zero ✅

D.It must equal twice the region's area

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Green's Theorem converts the work into $\iint_R2x\,dA$ . Because the region is symmetric about the $y$ -axis, every point with positive $x$ is paired with a point having negative $x$ , producing cancellation. Therefore the integral of $2x$ over the region is zero, so the closed-path work is zero.

Q14. Suppose a positively oriented region is the union of two adjacent subregions, and $Q_x-P_y$ is positive on one while negative on the other. Which strategy is most appropriate for determining the total work?

A.Use only the largest subregion because it dominates the result

B.Assume the contributions cancel automatically because the signs differ

C.Compute the signed double integral over the entire region, or split it into subregions and add their contributions ✅

D.Use the total perimeter instead of area

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Green's Theorem requires the signed integral of $Q_x-P_y$ across the entire enclosed region. Positive and negative portions may partially cancel, but cancellation cannot be assumed without evaluating their magnitudes and areas. Splitting the region into manageable pieces and adding the resulting integrals is often the safest modelling strategy.

Q15. For a positively oriented family of closed curves, suppose $Q_x-P_y=k$ , where $k$ is a constant. One curve encloses area $A$ , while another encloses exactly $4A$ . What is the ratio of their work integrals, provided $k\ne0$ ?

A.0.04236111111111107

B.0.043055555555555625

C.0.04444444444444451

D.0.16736111111111107 ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: Green's Theorem gives the work as $\iint_Rk\,dA=kA$ for the first region. The second region has area $4A$ , so its work is $4kA$ . Therefore the ratio of the first work to the second is $1:4$ . This conclusion requires recognizing proportionality rather than performing a boundary parametrization.

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