📝 Line integral around closed path (16 MCQs)

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

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What is Line integral around closed path?

Line integral around closed path:
For a closed path $C$ , $\oint_C \mathbf{F} \cdot d\mathbf{r} = 0$ if $\mathbf{F}$ is conservative; otherwise, it measures circulation.

Example:
For $\mathbf{F} = \langle -y, x \rangle$ , $\oint_C \mathbf{F} \cdot d\mathbf{r} = 2\pi \cdot \text{area}$ for a circle, indicating nonzero circulation.

Reason:
Closed path integrals distinguish conservative from non-conservative fields and are fundamental to Stokes' and Green's theorems.

Easy

Medium

Hard

📝 All Line integral around closed path MCQs

Q1. A vector field $F$ is continuous on a region containing a closed curve $C$ . Which statement most directly describes the meaning of the closed-path line integral $\oint_C F\cdot dr$ ?

A.It measures only the length of

C

B.It measures the accumulated tangential component of

F

around the complete loop ✅

C.It always equals zero

D.It measures the area enclosed by

C

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: For a closed curve, the line integral $\oint_C F\cdot dr$ accumulates the component of the vector field tangent to the direction of travel over the entire loop. It is not automatically zero, nor is it generally equal to either the curve's length or enclosed area.

Q2. A student claims that because a curve begins and ends at the same point, its line integral must be zero. Which response best evaluates the claim?

A.The claim is always correct because the endpoints coincide

B.The claim is correct only for straight-line curves

C.The claim is incorrect because a closed curve can accumulate nonzero circulation ✅

D.The claim is correct whenever the vector field is continuous

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Coincident endpoints alone do not force a line integral to vanish. A field can consistently have a tangential component in the direction of traversal, producing nonzero circulation around a loop. Additional properties of the field and its domain are required to conclude that the integral is zero.

Q3. Suppose $\oint_C F\cdot dr=7$ for a counterclockwise traversal of a closed curve $C$ . What is the value when the same curve is traversed clockwise?

A.-7

B.-14

C.0

D.-7 ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: Reversing the orientation of a path reverses every differential displacement $dr$ . Therefore, the contribution from every small segment changes sign, so the entire line integral changes sign. Thus, a counterclockwise value of $7$ becomes $-7$ for clockwise traversal.

Q4. Two closed curves $C_1$ and $C_2$ enclose different regions, but both lie entirely in a region where a vector field has zero circulation around every closed loop. A student computes both integrals separately. What is the most efficient conclusion?

A.Both integrals are equal to the circumference of their curves

B.Both integrals must be zero ✅

C.Only the smaller curve has zero integral

D.The integrals must have opposite signs

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: If a vector field has zero circulation around every closed loop in the region under consideration, then the line integral over each closed curve is zero. The geometric size, shape, or orientation of the curves does not change that conclusion.

Q5. A force field acts on a particle moving once around a closed track. The particle returns to its starting position, but the force consistently has a component in the direction of motion. Which conclusion is most reasonable?

A.The total work must be zero because displacement is zero

B.The total work can be nonzero despite zero net displacement ✅

C.The total work must equal the track length

D.The total work depends only on the starting point

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Net displacement being zero does not imply zero work for a variable force field. Work is accumulated locally through $F\cdot dr$ , not determined solely by the final minus initial position. A tangential force component can therefore produce nonzero total work over a closed track.

Q6. A closed path consists of two curves joining the same points, with one traversed forward and the other backward. If the integrals along the two curves are $5$ and $2$ in their stated directions, what is the closed-path integral?

A.-7

B.-3 ✅

C.3

D.-3

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The second curve is traversed in the direction opposite to the direction used when its integral was stated. Its contribution therefore becomes $-2$ . Combining the two oriented contributions gives $5+(-2)=3$ , illustrating why orientation must be tracked carefully.

Q7. A rectangular loop is divided into four directed sides. A calculation gives contributions $6,-3,-8,$ and $5$ in traversal order. What does the resulting value imply?

A.The circulation is

16

B.The circulation is

0

✅

C.The circulation is

10

D.The circulation cannot be determined

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For a piecewise smooth closed path, the total line integral is the sum of the integrals over all individual segments. Here $6-3-8+5=0$ . The cancellation results from the vector field's contributions along the oriented sides, not merely from the rectangle being closed.

Q8. A computational model gives $\oint_C F\cdot dr=4$ for a circular path. Another program traces exactly the same circle in the opposite direction and reports $4$ . What is the strongest diagnosis?

A.Both results are necessarily correct

B.The second result should be

-4

, so orientation was likely mishandled ✅

C.The correct result must be

0

D.The radius must have changed

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: For the same geometric curve, reversing traversal reverses the sign of a line integral. Therefore, if the first oriented calculation gives $4$ , the opposite orientation should give $-4$ . Reporting $4$ again strongly suggests that the computational model failed to reverse the path parameterization or direction vector.

Q9. A drone follows a closed polygonal route through a velocity-dependent vector field. Measurements along successive segments produce $2.4,-1.1,3.0,-4.7,$ and $0.8$ . Which modelling interpretation is best?

A.The final sum estimates the circulation around the route ✅

B.The largest positive value alone determines the circulation

C.Only the segment with negative contribution should be included

D.The values must all be converted to positive quantities before summing

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: A closed-path line integral is obtained by adding the signed contributions from every oriented segment. The values $2.4,-1.1,3.0,-4.7,$ and $0.8$ must therefore be combined algebraically. Negative contributions represent portions where the field opposes the chosen direction of travel.

Q10. A student parameterizes a circle as $r(t)=(\cos t,\sin t)$ for $0\le t\le2\pi$ , but later uses $r(t)=(\cos t,-\sin t)$ without changing the integration limits. Which issue should be investigated first?

A.The curve has become a line

B.The second parameterization reverses the orientation, affecting the sign of the integral ✅

C.The circle now has twice the radius

D.The line integral becomes independent of the vector field

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The parameterization $(\cos t,-\sin t)$ traces the same unit circle but in the opposite orientation compared with $(\cos t,\sin t)$ . Since reversing orientation changes the sign of a line integral, the student must account for this change when comparing the two calculations.

Q11. A graph of a vector field shows arrows that are approximately tangent to a circular path and point counterclockwise along most of the circle. Which prediction is most defensible before performing any calculation?

A.The circulation is likely positive for counterclockwise traversal ✅

B.The circulation must be exactly zero

C.The circulation must be negative for counterclockwise traversal

D.The enclosed area must be zero

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: When the vector field arrows tend to align with the counterclockwise tangent direction along a closed curve, the dot product $F\cdot dr$ is expected to be predominantly positive. Therefore, the circulation is plausibly positive, although an exact value requires quantitative information.

Q12. A sketch shows a closed curve divided into two portions. On the first portion, arrows strongly oppose the direction of travel; on the second, arrows strongly align with it. The first portion is twice as long as the second. Which conclusion is safest?

A.The circulation is certainly positive

B.The circulation is certainly negative

C.The circulation cannot be determined from lengths and arrow directions alone ✅

D.The circulation must be zero

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Length and qualitative alignment provide useful intuition but are insufficient for a definite sign when field magnitudes vary. A shorter segment with a much stronger aligned field could outweigh a longer opposing segment. Quantitative information about $F\cdot dr$ is therefore necessary.

Q13. A graph suggests that a vector field is everywhere perpendicular to the tangent direction of a closed curve $C$ . What would this imply for the line integral along $C$ ?

A.It is positive

B.It is negative

C.It is zero ✅

D.It equals the curve's area

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: If the vector field is perpendicular to the tangent direction everywhere on the curve, then $F\cdot dr=0$ at every point because the dot product of perpendicular vectors is zero. Integrating these zero contributions around the entire closed path therefore gives zero.

Q14. For a closed path $C$ , one method evaluates the integral directly by parameterizing four segments, while another uses a known structural property of the vector field. The direct method gives $3.2$ , and the structural method gives $3.2$ . What is the best interpretation?

A.The agreement supports the consistency of both approaches ✅

B.The direct method must be wrong because it uses segments

C.The structural method is invalid for closed paths

D.The value must actually be zero because the path is closed

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Independent methods producing the same value provide evidence that the orientation, parameterization, and field properties have been handled consistently. A closed path does not itself force the integral to vanish, so $3.2$ is entirely possible when the field has nonzero circulation.

Q15. A closed triangular path $C$ is traversed counterclockwise. The field has the property that its tangential contribution increases proportionally with distance from the triangle's center. Which modelling strategy is most appropriate for determining the circulation?

A.Ignore orientation because the path is closed

B.Compute or approximate

F\cdot dr

along each directed side and combine the signed contributions ✅

C.Multiply the triangle's area by its perimeter

D.Use only the field value at the center

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: For a polygonal closed path, the safest direct modelling approach is to evaluate the field's tangential contribution along each directed segment. Because both the field and the direction can vary, the signed line integrals from all sides must be combined rather than replaced by a simple area-perimeter product.

Q16. Consider a closed curve $C$ surrounding a region that contains no singular points of $F(x,y)=\left(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2}\right)$ . A student argues that the integral must be zero because the components have continuous-looking formulas away from the origin. What is the key flaw?

A.The curve is not closed

B.The field is undefined at the origin, and whether the origin lies inside the loop matters ✅

C.The field has no

x

-component

D.Closed curves always have zero circulation

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The field is undefined at $(0,0)$ , so its behavior cannot be analyzed as though it were globally regular on the entire enclosed region. For closed-loop circulation, the presence or absence of such an excluded point inside the loop can fundamentally affect the result, making domain analysis essential.

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