📝 Line Integrals in calculus (16 MCQs)

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

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What is Line Integrals in calculus?

Line Integrals in calculus:
A line integral integrates a function along a curve $C$ , written as $\int_C f \, ds$ for scalar fields or $\int_C \mathbf{F} \cdot d\mathbf{r}$ for vector fields, with $d\mathbf{r} = \langle dx, dy, dz \rangle$ .

Example:
$\int_C x^2 \, ds$ over a line segment from $(0,0)$ to $(1,1)$ using $ds = \sqrt{2} \, dt$ .

Reason:
Line integrals compute total quantities (mass, work) along paths, extending single-variable integration to curves in space.

Easy

Medium

Hard

📝 All Line Integrals in calculus MCQs

Q1. A particle moves along the curve $C$ from $A$ to $B$ under a force field $\mathbf F$ . Which expression correctly represents the work done by the field along the path?

\int_C \mathbf F\,ds

\int_C \mathbf F\cdot d\mathbf r

✅

\int_C |\mathbf F|\,d\mathbf r

\int_C \nabla\mathbf F\cdot d\mathbf r

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Work done by a vector field depends on the component of the force in the direction of displacement. Therefore it is represented by the line integral $\int_C \mathbf F\cdot d\mathbf r$ . The dot product accounts for both the magnitude of the force and how strongly it acts along the direction of motion.

Q2. For a scalar field $f(x,y)$ , two curves connect the same points $A$ and $B$ . Which statement best distinguishes $\int_C f\,ds$ from $\int_C \mathbf F\cdot d\mathbf r$ ?

A.Both are always independent of the path

B.The first accumulates scalar values with arc length, while the second measures directional field contribution ✅

C.The first requires a vector field, while the second requires a scalar field

D.Both always depend only on the endpoints

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: A scalar line integral $\int_C f\,ds$ accumulates a scalar quantity along physical distance and is generally path dependent. A vector line integral $\int_C \mathbf F\cdot d\mathbf r$ instead measures the component of a vector field along the displacement. Their mathematical roles and path dependence are therefore different.

Q3. A wire follows the curve $y=x^2$ from $x=0$ to $x=2$ , and its linear density is $\rho(x,y)=x+y$ . Which integral correctly gives the total mass of the wire?

\int_0^2 (x+x^2)\,dx

\int_0^2 (x+x^2)\sqrt{1+4x^2}\,dx

✅

\int_0^2 (x+x^2)(1+4x^2)\,dx

\int_0^2 (x+2x^2)\sqrt{1+4x^2}\,dx

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For a scalar line integral representing mass, the density must be multiplied by the differential arc length. Along $y=x^2$ , the density becomes $x+x^2$ , while $ds=\sqrt{1+(dy/dx)^2}\,dx=\sqrt{1+4x^2}\,dx$ . Combining these factors gives the stated integral.

Q4. A vector field is $\mathbf F=(2x,2y)$ , and a particle travels from $(1,0)$ to $(3,0)$ . Why can the work be evaluated without knowing the detailed shape of the path if the field is recognized as a gradient field?

A.Because every vector field has zero curl

B.Because

\mathbf F=\nabla(x^2+y^2)

, so the work depends only on the endpoints ✅

C.Because the speed of the particle is constant

D.Because the magnitude of

\mathbf F

is constant

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The field satisfies $\mathbf F=\nabla\phi$ for $\phi=x^2+y^2$ . For a conservative field, the line integral between two points equals the change in potential, $\phi(B)-\phi(A)$ . Thus the detailed route does not affect the work, provided the path remains in the relevant domain.

Q5. Consider a closed curve $C$ traversed once counterclockwise. If a vector field satisfies $\mathbf F=\nabla\phi$ throughout a region containing the curve and its interior, what should $\oint_C\mathbf F\cdot d\mathbf r$ equal?

A.A positive value equal to the perimeter

B.A negative value determined by orientation

C.Zero ✅

D.The maximum value of

\phi

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: A gradient field is conservative, so its line integral depends only on the initial and final points. For a closed curve, those points coincide, making the net change in potential zero. Consequently, $\oint_C\mathbf F\cdot d\mathbf r=0$ , regardless of the curve's shape or orientation.

Q6. A student argues that because a curve starts at $(0,0)$ and ends at $(2,2)$ , the value of $\int_C (y\,dx+x\,dy)$ must depend on the path. What is the best evaluation of this reasoning?

A.Correct, because both

x

and

y

vary

B.Incorrect, because

y\,dx+x\,dy=d(xy)

, so the integral depends only on the endpoints ✅

C.Correct, because the curve is not necessarily straight

D.Incorrect, because all line integrals are zero

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The differential expression $y\,dx+x\,dy$ is exactly $d(xy)$ . Therefore the integral is $xy$ evaluated at the endpoints, giving $4-0=4$ . The student's mistake is assuming that every line integral is path dependent without checking whether the integrand represents an exact differential.

Q7. A hiking trail is modeled by $\mathbf r(t)=(t,t^2)$ , $0\le t\le1$ , and the environmental field is $\mathbf F=(y,-x)$ . Which integral represents the accumulated directional effect of the field along the trail?

\int_0^1 (t^2- t^2)\,dt

\int_0^1 (t^2,-t)\cdot(1,2t)\,dt

✅

\int_0^1 (t^2,-t)\cdot(t,t^2)\,dt

\int_0^1 |(t^2,-t)|\,dt

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For a vector line integral, substitute the parameterization into the field and take its dot product with the velocity vector. Here $\mathbf F(\mathbf r(t))=(t^2,-t)$ and $\mathbf r'(t)=(1,2t)$ . Therefore the correct work-type integral is $\int_0^1\mathbf F(\mathbf r(t))\cdot\mathbf r'(t)\,dt$ .

Q8. A graph of a vector field shows arrows everywhere tangent to concentric circles centered at the origin, with arrow length increasing as the distance from the origin increases. A particle follows one complete circle counterclockwise. What can most reasonably be inferred about the work?

A.It must be zero because the path is closed

B.It is positive if the arrows point counterclockwise ✅

C.It is negative regardless of arrow direction

D.It depends only on the radius and cannot have a sign

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The arrows are tangent to the circular path, so the field has a nonzero component in the direction of motion. If the arrows point counterclockwise and the particle also moves counterclockwise, the dot product is positive throughout the path. Thus the work is positive, illustrating that a closed path does not automatically imply zero work.

Q9. A contour map of a scalar field shows level curves packed closely together near one part of a path and widely separated elsewhere. If the path represents a wire with uniform mass density determined by the scalar field value, which region is most likely to contribute more mass per unit length?

A.The region with widely spaced contours only

B.The region with closely spaced contours only, regardless of field value

C.It cannot be inferred from contour spacing alone ✅

D.Every region contributes exactly the same amount

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Contour spacing primarily indicates how rapidly a scalar field changes, not its actual value. A mass line integral depends on the density value multiplied by arc length. Therefore closely spaced contours alone do not determine which region contributes more mass; the field values and the amount of curve lying in each region must also be considered.

Q10. A student evaluates $\int_C (y\,dx-x\,dy)$ along two different paths having the same endpoints and obtains different values. They conclude that one calculation must be wrong because line integrals should depend only on endpoints. What is the strongest response?

A.The conclusion is correct for every vector field

B.The conclusion is false because general vector line integrals may depend on the path ✅

C.The conclusion is correct whenever the path is smooth

D.The conclusion is false only for scalar line integrals

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Path independence is a special property associated with conservative fields under appropriate domain conditions, not a universal property of line integrals. The field corresponding to $y\,dx-x\,dy$ generally has nonzero circulation, so different paths can produce different values even when their endpoints are identical.

Q11. Suppose a graph of a vector field shows arrows pointing directly outward from the origin, with magnitudes increasing proportionally to distance from the origin. Two paths connect the same points: a straight segment and a curved arc. Which strategy is most efficient for determining the work if the field can be recognized as $\mathbf F=(x,y)$ ?

A.Parameterize both paths and calculate both integrals

B.Use the potential function

\phi=(x^2+y^2)/2

and evaluate it at the endpoints ✅

C.Integrate only the magnitude of the field along the curved path

D.Assume the work is zero because the field is radial

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The field $\mathbf F=(x,y)$ is the gradient of $\phi=(x^2+y^2)/2$ . Therefore it is conservative and the work between fixed endpoints is simply the difference in potential values. This avoids unnecessary parameterization of either path and also demonstrates why the curved route and straight route give the same work.

Q12. A delivery robot travels along a closed rectangular route in a warehouse. The force field is $\mathbf F=(3y,3x)$ . A technician claims that the total work must be nonzero because the force changes direction around the rectangle. Which conclusion is correct?

A.The technician is correct because changing direction always produces work

B.The total work is zero because

\mathbf F=\nabla(3xy)

✅

C.The total work is positive because the rectangle is closed

D.The total work depends only on the rectangle's perimeter

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The field can be written as $\mathbf F=\nabla(3xy)$ , since the partial derivatives of $3xy$ are $3y$ and $3x$ . Hence the field is conservative. The robot returns to its starting point, so the net change in potential and therefore the total work around the closed route is zero.

Q13. For a parameterized curve $\mathbf r(t)$ , a student computes a vector line integral using $\int_a^b\mathbf F(\mathbf r(t))\,dt$ instead of $\int_a^b\mathbf F(\mathbf r(t))\cdot\mathbf r'(t)\,dt$ . What essential factor has been omitted?

A.The curvature of the path

B.The velocity or tangent vector that converts parameter change into displacement ✅

C.The magnitude of the force

D.The endpoint coordinates

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The differential displacement along a parameterized curve is $d\mathbf r=\mathbf r'(t)\,dt$ . Therefore a work-type line integral must use the dot product of the field with $\mathbf r'(t)$ . Omitting this factor ignores both the direction and rate of displacement, so the resulting expression does not correctly represent accumulated work.

Q14. A force field has magnitude $5$ everywhere along a path of length $8$ , but its direction makes a constant angle of $60^\circ$ with the direction of motion. What is the work done along the path?

A.20 ✅

B.40

C.10

D.5

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Work is the line integral of the tangential component of the force. Since the force makes a $60^\circ$ angle with the direction of motion, its tangential component is $5\cos60^\circ=2.5$ . Multiplying by the path length $8$ gives $W=2.5(8)=20$ .

Q15. A closed planar curve encloses a region $R$ . Suppose a vector field satisfies $\partial Q/\partial x-\partial P/\partial y=4$ , where $\mathbf F=(P,Q)$ , and the area enclosed by the curve is $7$ . For counterclockwise orientation, what is the circulation $\oint_C\mathbf F\cdot d\mathbf r$ ?

A.-11

B.-28 ✅

4/7

7/4

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: For a positively oriented simple closed curve, the circulation can be related to the area integral of $\partial Q/\partial x-\partial P/\partial y$ . Since this quantity is constantly $4$ , integrating it over a region of area $7$ gives $4\times7=28$ . This connects local rotational behavior with total circulation.

Q16. Let $C_1$ and $C_2$ be two paths from $A$ to $B$ . For a vector field $\mathbf F$ , suppose $\int_{C_1}\mathbf F\cdot d\mathbf r=12$ and $\int_{C_2}\mathbf F\cdot d\mathbf r=7$ . What can be concluded about the closed-loop circulation obtained by traversing $C_1$ from $A$ to $B$ and then $C_2$ backward from $B$ to $A$ ?

A.It is

19

B.It is

5

✅

C.It is

-5

D.It is always zero

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Reversing the direction of a path changes the sign of its line integral. Thus the combined closed loop has circulation $12+(-7)=5$ . This result also provides a practical test for path dependence: if two paths joining the same endpoints give different integrals, their difference equals the circulation around the closed loop formed by traversing one forward and the other backward.

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