📝 Line integrals over piecewise smooth curves (14 MCQs)

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

←Previous Topic:Working a Line Integral📝 Line integrals over piecewise smooth curvesIndependence of Path: Conservative Vector Fields:Next Topic→

What is Line integrals over piecewise smooth curves?

Line integrals over piecewise smooth curves:
Piecewise smooth curves consist of multiple smooth segments; the line integral is the sum of integrals over each segment.

Example:
Integrate $\mathbf{F} = \langle x, y \rangle$ over a triangle with vertices $(0,0), (1,0), (0,1)$ , parameterize each edge, and sum results.

Reason:
This extends line integrals to arbitrary paths, common in engineering and physics where curves are not always smooth.

Easy

Medium

Hard

📝 All Line integrals over piecewise smooth curves MCQs

Q1. A curve $C$ consists of two smooth pieces $C_1$ and $C_2$ joined at a common endpoint. Which expression correctly represents the line integral of a continuous scalar field $f$ over $C$ ?

A.The integral over

C_1

minus the integral over

C_2

B.The integral over

C_1

plus the integral over

C_2

✅

C.The integral over only the final piece

C_2

D.The integral over

C_1C_2

without separating the pieces

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: A piecewise smooth curve is handled by dividing it into its smooth components and adding the contributions from each component. Because the scalar line integral accumulates the quantity along the entire path, the total is $\int_{C_1}f\,ds+\int_{C_2}f\,ds$ . The pieces are not subtracted merely because the curve changes direction.

Q2. Suppose $C=C_1\cup C_2$ , where $C_1$ ends exactly where $C_2$ begins. Why can the line integral be evaluated piece by piece even though $C$ is not smooth at their joining point?

A.The joining point has zero arc length and does not affect the accumulated integral ✅

B.The derivative must be identical on both sides of the joining point

C.Only vector fields can be integrated over piecewise smooth curves

D.The function must be zero at the joining point

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: A finite collection of joining points contributes zero arc length, so isolated corners do not create a separate contribution to the integral. Each smooth segment can therefore be parameterized and integrated independently. The total integral is obtained by adding the contributions from all pieces, provided the relevant field is integrable along the curve.

Q3. A path travels from $A$ to $B$ along $C_1$ , then from $B$ to $C$ along $C_2$ . If a student reverses the parameterization of both pieces but preserves the geometric path, what happens to a scalar line integral $\int_C f\,ds$ ?

A.Its sign changes

B.Its value becomes zero

C.Its value remains unchanged ✅

D.Only the contribution from

C_2

changes sign

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For a scalar line integral with respect to arc length, the differential $ds$ is nonnegative and measures geometric distance. Reversing the direction of traversal does not change the arc length element. Therefore each piece contributes the same value after reversal, and the total scalar line integral remains unchanged.

Q4. A particle moves along two smooth path segments with a sharp corner at their junction. Which statement best explains why the total work done by a vector field can still be computed from the two segments separately?

A.Work depends only on the endpoints, regardless of the path

B.The tangent vector needs to be continuous at every point

C.The work integral is additive over consecutive curve segments ✅

D.A corner automatically makes the work integral undefined

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Work along a curve is represented by a line integral of the vector field against the displacement vector. If a path is divided into consecutive pieces, the displacement contributions from the pieces add. A corner does not invalidate the integral because differentiability is required on each smooth piece rather than necessarily at the isolated joining point.

Q5. Consider a path made from a straight segment followed by a circular arc. A student argues that the entire path must be reparameterized as one differentiable formula before a line integral can be evaluated. What is the best assessment?

A.The argument is correct because corners always prevent integration

B.The argument is incorrect because each smooth piece may be parameterized separately ✅

C.The argument is correct only for scalar line integrals

D.The argument is incorrect because line integrals never require parameterizations

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A piecewise smooth curve is specifically designed to allow separate smooth parameterizations for its components. A straight segment and a circular arc can be parameterized independently, their line integrals evaluated separately, and the results added. Requiring one globally differentiable parameterization is unnecessary and would confuse smoothness of pieces with smoothness of the whole curve.

Q6. A hiking trail consists of three smooth sections $C_1,C_2,C_3$ , with corners at two junctions. The terrain density is modeled by $f(x,y)$ . Which strategy most directly models the total accumulated mass of a thin trail of constant cross-sectional properties?

A.Evaluate only the longest section

B.Evaluate

\int_{C_1}f\,ds+\int_{C_2}f\,ds+\int_{C_3}f\,ds

✅

C.Evaluate

f

only at the two corners

D.Subtract the middle section because it is traversed after the first

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: If $f$ represents density per unit length, the accumulated mass along a path is obtained by integrating $f\,ds$ . Since the trail consists of three smooth sections, the contribution from each section is calculated separately. Additivity then gives the total mass as the sum of all three integrals, including the middle section.

Q7. A robot moves from $(0,0)$ to $(2,0)$ , then to $(2,3)$ . For the vector field $\mathbf F(x,y)=\langle y,x\rangle$ , what is the total work along this piecewise linear path?

A.0

B.3

C.6 ✅

D.9

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: On the first segment, $y=0$ and $dx$ varies while $dy=0$ , so $\mathbf F\cdot d\mathbf r=y\,dx+x\,dy=0$ . On the second segment, $x=2$ , $dx=0$ , and $dy$ runs from $0$ to $3$ . Thus the work is $\int_0^3 2\,dy=6$ , giving total work $6$ .

Q8. A student calculates $\int_C f\,ds$ for a two-piece curve and obtains a negative answer because the second segment is traversed in the opposite direction. Which error most likely occurred?

A.The student changed

ds

-ds

when reversing the second segment ✅

B.The student forgot that

f

must be continuous everywhere

C.The student used two parameterizations instead of one

D.The student added the two segment integrals

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: For scalar line integrals, reversing orientation does not make $ds$ negative. If a parameter $t$ runs backward, the derivative changes sign, but the arc-length element is $ds=|\mathbf r'(t)|dt$ , which remains nonnegative. A negative total therefore indicates an incorrect treatment of arc length or another computational mistake.

Q9. A student computes the work along $C_1$ and $C_2$ correctly but writes $W=\left|\int_{C_1}\mathbf F\cdot d\mathbf r-\int_{C_2}\mathbf F\cdot d\mathbf r\right|$ . The path actually moves continuously from the endpoint of $C_1$ into $C_2$ . What is wrong?

A.The absolute value should always be removed from every work integral

B.The two work contributions must be added using their actual orientations ✅

C.The second segment must always be parameterized backward

D.Work along a curve cannot be split into pieces

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Work is additive over consecutive oriented segments. The sign of each segment is determined by the direction in which that segment is traversed relative to the vector field. Therefore the correct total is $W=\int_{C_1}\mathbf F\cdot d\mathbf r+\int_{C_2}\mathbf F\cdot d\mathbf r$ . Subtracting the second integral reverses its physical orientation and can produce an incorrect result.

Q10. A graph shows a path consisting of a horizontal segment from $A$ to $B$ , followed by a diagonal segment from $B$ to $C$ . Along the first segment, the vector field points entirely upward; along the second, it points partly in the direction of motion. Which conclusion about work is most justified?

A.Both segments necessarily contribute positive work

B.The first segment contributes zero work, while the second can contribute positive work ✅

C.The first segment contributes negative work, while the second must be zero

D.The total work must be zero because the path has a corner

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Work depends on the component of the vector field parallel to the displacement. On the horizontal segment, an entirely upward field is perpendicular to the horizontal displacement, so its dot product with the displacement is zero. On the diagonal segment, a component in the direction of motion gives a positive dot product and therefore can produce positive work.

Q11. A delivery route has two alternatives between the same locations. Route $C$ consists of three smooth pieces and route $D$ consists of one smooth curve. If a vector field is conservative on the entire region containing both routes, which comparison is valid?

A.The number of pieces determines which route has greater work

B.The work depends only on the endpoints, so both routes give the same work ✅

C.The piecewise route always gives less work

D.The smooth route always gives less work

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For a conservative vector field, the line integral between two fixed endpoints is path independent, provided the paths lie within the appropriate region. Whether a path is smooth in one piece or piecewise smooth in several pieces does not change that conclusion. Thus both routes produce the same work when their initial and final points agree.

Q12. A force field is $\mathbf F=\langle 2x,2y\rangle$ . A path goes from $A=(0,0)$ to $B=(1,0)$ , then from $B$ to $C=(1,2)$ . A student computes each segment separately and obtains a total work of $5$ . Which result should be expected instead?

A.3

B.4 ✅

C.5

D.6

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The field is the gradient of the potential $f(x,y)=x^2+y^2$ , so the work depends only on the endpoints. At $A$ , the potential is $0$ , while at $C=(1,2)$ it is $1+4=5$ . Therefore the total work is $5$ , not $4$ . The student's result of $5$ is correct because the piecewise calculation agrees with endpoint dependence.

Q13. A piecewise smooth path is parameterized by $\mathbf r_1(t)$ for $0\le t\le1$ and $\mathbf r_2(t)$ for $1\le t\le2$ . The two formulas meet at $t=1$ , but their derivatives are different there. Which interpretation is mathematically appropriate?

A.The path is invalid because its derivative is discontinuous at

t=1

B.The path is piecewise smooth because each portion is smooth and the junction is isolated ✅

C.The path is smooth because different derivatives are automatically equal

D.The path cannot support any line integral

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Piecewise smoothness permits a finite number of junctions where the tangent direction may change abruptly. What matters is that each individual portion is sufficiently smooth for the line integral calculation. A discontinuity in the derivative at the isolated joining point therefore does not prevent the curve from being a valid piecewise smooth path.

Q14. Two piecewise smooth paths connect the same endpoints. Path $C_1$ consists of two segments, while $C_2$ consists of four segments. Suppose $\mathbf F=\nabla\phi$ throughout a simply connected region containing both paths. A researcher claims $C_2$ must require more total work because it has twice as many pieces. Which response is strongest?

A.Correct, because each corner adds extra work

B.Correct, because more parameter intervals always increase work

C.Incorrect, because conservativeness makes the work depend only on the endpoints ✅

D.Incorrect, because work is always proportional to the number of curve pieces

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: When $\mathbf F=\nabla\phi$ , the work along any suitable path between fixed endpoints equals the change in potential, $\phi(B)-\phi(A)$ . Splitting a path into additional smooth pieces changes only the computational organization, not the physical line integral. Consequently, the number of corners or parameter intervals does not determine the total work.

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