📝 Working a Line Integral (15 MCQs)

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

←Previous Topic:Integration a Vector Field Along a Curve📝 Working a Line IntegralLine integrals over piecewise smooth curves:Next Topic→

What is Working a Line Integral?

Working of a Line Integral:
To compute, parameterize $C$ , find $\mathbf{r}'(t)$ , substitute into $\mathbf{F}$ , compute the dot product, and integrate over $t$ .

Example:
For $\mathbf{F} = \langle x^2, yz \rangle$ and $C: \mathbf{r}(t) = \langle t, t^2, t^3 \rangle$ , $0 \le t \le 1$ , calculate $\int_0^1 \langle t^2, t^5 \rangle \cdot \langle 1, 2t, 3t^2 \rangle dt = \int_0^1 (t^2 + 2t^6) dt = 1/3 + 2/7 = 13/21$ .

Reason:
Step-by-step evaluation ensures accuracy and builds confidence in applying vector calculus to real problems.

Easy

Medium

Hard

📝 All Working a Line Integral MCQs

Q1. A particle moves from $A=(0,0)$ to $B=(2,1)$ under the force field $\mathbf F(x,y)=\langle y,x\rangle$ . Which expression correctly represents the work along a parametrized path $\mathbf r(t)=\langle x(t),y(t)\rangle$ , $a\le t\le b$ ?

W=\int_a^b [x(t)+y(t)]\,dt

W=\int_a^b [y(t)x&#039;(t)+x(t)y&#039;(t)]\,dt

✅

W=\int_a^b [y(t)+x(t)]x&#039;(t)y&#039;(t)\,dt

W=\int_a^b [x(t)y(t)]\,dt

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Work done by a vector field along a curve is obtained from $\int_C\mathbf F\cdot d\mathbf r$ . Since $d\mathbf r=\langle x'(t),y'(t)\rangle dt$ , the dot product becomes $yx'+xy'$ . The key issue is that both the force components and the direction of motion contribute to the work.

Q2. For a force field $\mathbf F=\langle P,Q\rangle$ , two students evaluate work along the same oriented curve. Student 1 computes $\int_C P\,dx+Q\,dy$ , while Student 2 computes $\int_C P\,dy+Q\,dx$ . Which statement best explains the difference?

A.Both expressions always give the same work

B.Student 1 uses the standard work integral, while Student 2 generally interchanges the force components with the coordinate differentials ✅

C.Student 2 is correct only when the curve is closed

D.Student 1 is correct only for straight-line paths

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: For $\mathbf F=\langle P,Q\rangle$ , the work integral is $\int_C\mathbf F\cdot d\mathbf r=\int_C P\,dx+Q\,dy$ . Swapping $dx$ and $dy$ changes the dot product and therefore generally changes the physical quantity being calculated. The error is conceptual, not merely algebraic.

Q3. A force field is $\mathbf F(x,y)=\langle 2x,2y\rangle$ . A particle travels from $(1,0)$ to $(0,2)$ along any smooth path. Without parametrizing a particular curve, what is the work?

A.-2

B.-3

C.-4 ✅

D.-5

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The field is conservative because $\mathbf F=\nabla(x^2+y^2)$ . Therefore the work depends only on the endpoints. The potential changes from $1^2+0^2=1$ to $0^2+2^2=4$ , so the work is $4-1=3$ . Thus the correct choice is B, not C.

Q4. A student argues: 'If a force field has positive components $P>0$ and $Q>0$ everywhere on a path, the work must be positive.' Which response most accurately evaluates the argument?

A.It is always correct because both force components add energy

B.It is correct only for closed curves

C.It can be false because

dx

dy

may be negative during the motion ✅

D.It is false only if the force field is conservative

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The work is $\int_C P\,dx+Q\,dy$ , so the signs of the coordinate changes matter. Even if $P$ and $Q$ are positive, motion in a negative $x$ - or $y$ -direction can produce negative contributions. Work depends on the force component parallel to the displacement, not merely on force-component signs.

Q5. A particle travels along the path $x=t,\ y=t^2$ , from $t=0$ to $t=2$ , under $\mathbf F=\langle y,2x\rangle$ . Which integral correctly models the work?

\int_0^2 [t^2+4t]\,dt

✅

\int_0^2 [t^2+4t^2]\,dt

\int_0^2 [2t+t^3]\,dt

\int_0^2 [t+2t^2]\,dt

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Along the path, $x=t$ , $y=t^2$ , $dx=dt$ , and $dy=2t\,dt$ . Therefore $P=y=t^2$ contributes $t^2dt$ , while $Q=2x=2t$ contributes $2t(2t\,dt)=4t^2dt$ . Hence the correct integral is $\int_0^2 5t^2dt$ , which means option B, not A.

Q6. A force field is $\mathbf F=\langle y, x+y\rangle$ . A particle moves from $(0,0)$ to $(1,1)$ along the straight segment $y=x$ . What is the work?

A.-1

3/2

C.-2 ✅

5/2

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Parameterize the segment by $x=t,\ y=t$ , $0\le t\le1$ . Then $dx=dt$ and $dy=dt$ . The field becomes $\langle t,2t\rangle$ , so $\mathbf F\cdot d\mathbf r=t\,dt+2t\,dt=3t\,dt$ . Integrating gives $3/2$ , so option B is correct.

Q7. A particle moves around a closed curve in a force field. One student reverses the direction of traversal but leaves every algebraic expression unchanged. What should happen to the calculated work?

A.It remains unchanged because the curve is the same

B.It becomes zero automatically

C.Its sign reverses because

d\mathbf r

reverses direction ✅

D.Its magnitude must double

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Reversing orientation changes $d\mathbf r$ to $-d\mathbf r$ . Since work is the line integral $\int_C\mathbf F\cdot d\mathbf r$ , every contribution changes sign. Therefore the work for the reversed traversal is the negative of the original work. The geometric path is unchanged, but the direction of motion is essential.

Q8. A student evaluates the work of $\mathbf F=\langle 3y,3x\rangle$ along a path from $(0,0)$ to $(2,2)$ and obtains different answers using two smooth paths. Which conclusion is most justified?

A.The student must have found two different physical forces

B.The field is conservative, so different answers indicate an error in at least one calculation ✅

C.Different paths always produce different work

D.The field cannot be tested for path independence

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Here $\mathbf F=\nabla(3xy)$ , because the partial derivative of $3xy$ with respect to $x$ is $3y$ , and with respect to $y$ is $3x$ . Therefore the field is conservative and work depends only on the endpoints. Different results from two paths signal an evaluation or parametrization error.

Q9. Consider the force field $\mathbf F=\langle y,-x\rangle$ . A particle moves counterclockwise around the unit circle. Which statement best predicts the sign of the work before doing the integral?

A.The work is positive because the motion is counterclockwise

B.The work is negative because the force opposes the counterclockwise tangent direction ✅

C.The work is zero because the path is closed

D.The sign cannot be determined from the field and orientation

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: On the unit circle, the counterclockwise tangent is $\langle -y,x\rangle$ , while the force is $\langle y,-x\rangle=-\langle -y,x\rangle$ . Thus the force points exactly opposite the direction of motion at every point. Their dot product is negative, so the total work is negative.

Q10. A graph of a planar path shows a particle moving rightward while a force vector at each point points upward and slightly rightward. At another portion, the particle moves leftward while the force still points upward and slightly rightward. Which interpretation is most accurate?

A.Both portions necessarily contribute positive work

B.The rightward motion contributes positive work from the horizontal force component, while the leftward motion contributes negative work from it ✅

C.Only the upward component determines work

D.The vertical force component always makes the work positive

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Work depends on the dot product between force and displacement. During rightward motion, a rightward force component contributes positively. During leftward motion, that same rightward component contributes negatively. The upward component contributes only when there is vertical displacement. Thus the graph must be interpreted using local alignment between force and motion.

Q11. A plotted trajectory starts at $(0,0)$ , moves to $(2,0)$ , then curves upward to $(2,2)$ . A force field has a horizontal component that is positive on the first segment and a vertical component that is negative on the second segment. Which qualitative conclusion follows?

A.Both segments must contribute positive work

B.The first segment contributes positive work, while the second contributes negative work ✅

C.The total work must be zero

D.Neither segment contributes work because the path changes direction

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: On the first horizontal segment, displacement is rightward and the positive horizontal force component aligns with it, producing positive work. On the second segment, displacement is upward while the vertical force component is negative, so their dot product is negative. The total work is therefore the sum of opposing contributions.

Q12. For $\mathbf F=\langle 2x+y,x\rangle$ , a student claims the work from $A=(0,0)$ to $B=(1,1)$ is path-independent because the field components are both linear functions. What is the best diagnosis?

A.The reasoning is correct because every linear field is conservative

B.The reasoning is incorrect; the relevant mixed partial derivatives are unequal ✅

C.The reasoning is correct only for closed curves

D.The reasoning is irrelevant because line integrals cannot depend on path

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Linearity of the components does not guarantee conservativeness. Here $P=2x+y$ gives $P_y=1$ , while $Q=x$ gives $Q_x=1$ , so in fact they are equal and the field is conservative. Therefore the student's conclusion is correct, but the stated justification is incomplete. The correct test compares cross-partial derivatives, making B factually incorrect.

Q13. A force field is $\mathbf F=\langle P,Q\rangle$ , and a particle follows a curve with $x$ increasing throughout but $y$ first increasing and then decreasing. Which modelling strategy is most reliable for calculating work?

A.Replace

dy

dx

everywhere

B.Split the curve into intervals where the behavior of

y

is simple, then evaluate

\int(P\,dx+Q\,dy)

on each part ✅

C.Ignore the

Q\,dy

term because

x

is increasing

D.Use only the initial and final values of

x

and

y

, regardless of the field

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Because $y$ changes direction, its derivative changes sign and the vertical contribution to work cannot be treated as having one fixed sign. Splitting the path into convenient pieces makes the orientation and derivatives explicit. The total work is the sum of the line integrals over those pieces.

Q14. Suppose $\mathbf F=\langle y^2,2xy\rangle$ . A particle travels from $(0,1)$ to $(2,3)$ along an arbitrary smooth path. Which method is most efficient for determining the work?

A.Parameterize every possible path and compare the results

B.Find a potential function and evaluate its change between the endpoints ✅

C.Compute only the magnitude of the force at each endpoint

D.Assume the work is zero because the endpoints are different

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The field is the gradient of $f(x,y)=xy^2$ , since $f_x=y^2$ and $f_y=2xy$ . Therefore the work is path-independent and equals $f(2,3)-f(0,1)=18$ . Recognizing the potential function avoids unnecessary parametrization and demonstrates how conservative structure simplifies a line-integral problem.

Q15. Let $\mathbf F=\langle y,-x\rangle$ . Among all smooth curves connecting $(1,0)$ to $(0,1)$ , one curve is the quarter-circle $x^2+y^2=1$ traversed counterclockwise. Another is the two-segment path from $(1,0)$ to $(0,0)$ to $(0,1)$ . Which statement is correct about their work?

A.Both works are equal because they share endpoints

B.The quarter-circle gives

-\pi/2

, while the two-segment path gives

1

✅

C.The quarter-circle gives

\pi/2

, while the two-segment path gives

-1

D.Both works are zero because the endpoints lie on the same circle

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: For the quarter-circle, use $x=\cos t,\ y=\sin t$ , $0\le t\le\pi/2$ . The force is $\langle\sin t,-\cos t\rangle$ , while $d\mathbf r=\langle-\sin t,\cos t\rangle dt$ , giving a constant integrand $-1$ , hence work $-\pi/2$ . Along the two segments, direct evaluation gives $-1$ . The different values demonstrate path dependence.

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←Previous Topic:Integration a Vector Field Along a Curve📝 Working a Line IntegralLine integrals over piecewise smooth curves:Next Topic→