📝 Line integrals with respect to x y z (17 MCQs)

📖 From Calculus • 16. Topics in vector Calculus • 17 questions available

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What is Line integrals with respect to x y z?

Line integrals with respect to x, y, z:
Line integrals can be expressed as $\int_C P \, dx + Q \, dy + R \, dz$ , evaluated by substituting parameterization and integrating each term separately.

Example:
$\int_C x \, dx + y \, dy$ along $\mathbf{r}(t) = \langle t, t \rangle$ , $0 \le t \le 1$ , becomes $\int_0^1 (t \cdot 1 + t \cdot 1) dt = 1$ .

Reason:
This differential form is useful in physics for work and in complex analysis, linking to exact differentials and path independence.

Easy

Medium

Hard

📝 All Line integrals with respect to x y z MCQs

Q1. A curve $C$ is parameterized by $x=t^2,\ y=t,\ z=2t$ , where $0\le t\le1$ . Which expression correctly represents the line integral $\int_C f(x,y,z)\,dx$ for a scalar field $f$ ?

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

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

✅

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

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

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For a parameterized curve, $dx$ must be replaced by $x'(t)\,dt$ . Since $x=t^2$ , we have $dx=2t\,dt$ . Substituting $x=t^2,\ y=t,\ z=2t$ into the field gives $f(t^2,t,2t)$ , so the complete integral is $\int_0^1 f(t^2,t,2t)(2t)\,dt$ .

Q2. For a parameterized curve $x=x(t),y=y(t),z=z(t)$ , which statement best distinguishes $\int_C f\,dx$ , $\int_C f\,dy$ , and $\int_C f\,dz$ ?

A.They are always equal because

dx,dy,dz

describe the same curve.

B.They differ only when the scalar field changes.

C.They weight the same field by the respective coordinate changes

x&#039;(t),y&#039;(t),z&#039;(t)

. ✅

D.They are all equivalent to

\int_C f\,ds

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The three integrals use the same scalar field but different differential components. Under parameterization, $dx=x'(t)dt$ , $dy=y'(t)dt$ , and $dz=z'(t)dt$ . Therefore, their values can differ substantially even on the same curve because each measures the field contribution according to a different coordinate direction.

Q3. Suppose $f(x,y,z)=x+y+z$ along the straight-line path from $(0,0,0)$ to $(1,2,3)$ . Which approach is most efficient for evaluating $\int_C f\,dx$ ?

A.Use

ds=\sqrt{14}\,dt

because every line integral requires arc length.

B.Parameterize the path, substitute into

f

, and multiply by

dx/dt

. ✅

C.Replace

dx

dy+dz

because the path is straight.

D.Integrate

f

with respect to

x

while treating

y

and

z

as zero.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: For a line integral with respect to $x$ , the differential $dx$ determines the required weighting. A straight-line parameterization such as $x=t,\ y=2t,\ z=3t$ makes the substitution direct. Then $f=6t$ and $dx=dt$ , so the integral becomes an ordinary single-variable integral.

Q4. A student claims that if a curve begins and ends at the same point, then $\int_C f\,dx=0$ for every scalar field $f$ . Which observation most directly disproves the claim?

A.The curve may have nonzero length.

B.The field

f

may be negative.

C.The integral depends on the entire path and on how

x

changes along that path, not merely on the endpoints. ✅

D.A closed curve cannot be parameterized.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: A closed path has zero net change in $x$ , but $\int_C f\,dx$ is generally a weighted accumulation of the field along the path. Unless special conditions cause cancellation, positive and negative contributions need not balance. Thus identical endpoints alone do not force a line integral with respect to $x$ to vanish.

Q5. A curve is given by $x=t,\ y=t^2,\ z=t^3$ , $0\le t\le2$ , and $f(x,y,z)=y$ . A student evaluates $\int_C f\,dy$ as $\int_0^2 t^2\,dt$ . What is the student's key error?

A.The field should be replaced by

z

B.The limits should be reversed.

C.The factor

dy/dt=2t

was omitted. ✅

D.The parameterization cannot be used for line integrals.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Since $y=t^2$ , its differential is $dy=2t\,dt$ . The field becomes $f=y=t^2$ , but the integral must include the coordinate differential. Therefore, the correct setup is $\int_0^2 t^2(2t)\,dt$ . Omitting $2t$ changes the weighting and produces an incorrect result.

Q6. A wire follows $x=t,\ y=1-t,\ z=t^2$ for $0\le t\le1$ . If the density is $f(x,y,z)=x+z$ , which integral models the total contribution measured with respect to $z$ ?

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

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

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

✅

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

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Substitution gives $f=x+z=t+t^2$ . Because the integral is taken with respect to $z$ , the differential must be included as $dz=z'(t)dt=2t\,dt$ . Hence the correct model is $\int_0^1(t+t^2)(2t)\,dt$ . This requires identifying both the field and the appropriate coordinate differential.

Q7. Consider the path $x=t^2,\ y=2t,\ z=1-t$ , $0\le t\le1$ . For $f(x,y,z)=xz$ , which expression correctly represents $\int_C f\,dy$ ?

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

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

✅

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

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

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Here $f=xz=t^2(1-t)$ . Since the integral is with respect to $y$ , we calculate $dy/dt=2$ , giving $dy=2dt$ . Therefore, the correct expression is $\int_0^1 t^2(1-t)(2)\,dt$ . The distractors correspond to confusing $dy$ with $dx$ , $dz$ , or omitting the differential factor.

Q8. Two paths connect the same endpoints. Along path $C_1$ , $x$ increases monotonically while along $C_2$ , $x$ first increases and then decreases. For a positive scalar field $f$ , which conclusion about $\int_C f\,dx$ is most justified?

A.Both integrals must be equal because their endpoints agree.

B.The integral along

C_2

can contain cancellation because

dx

changes sign. ✅

C.The integral along

C_1

must be zero because

x

changes.

D.Both integrals are positive because

f>0

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Although $f$ is positive, $dx$ carries the sign of the change in $x$ . On $C_2$ , the portion where $x$ decreases contributes negatively, potentially canceling positive contributions from the increasing portion. This demonstrates why a line integral with respect to a coordinate differential is sensitive to path behavior and orientation.

Q9. A graph of a projected path onto the $xy$ -plane shows a motion from $x=0$ to $x=4$ , followed by a return from $x=4$ to $x=1$ . Suppose $f$ is positive everywhere. Which statement about $\int_C f\,dx$ is necessarily true?

A.It must be positive because

f>0

B.It must be negative because the path eventually moves left.

C.Its sign cannot be determined without comparing the weighted contributions of the two portions. ✅

D.It must equal

3f

regardless of how

f

varies.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The first portion has $dx>0$ , while the return portion has $dx<0$ . Since the field is positive, these portions contribute with opposite signs. The final sign depends on the magnitude of the weighted contributions, which requires information about how $f$ behaves along the path. Positivity of $f$ alone is insufficient.

Q10. A plotted space curve has $z$ increasing steadily as the parameter increases, while $x$ oscillates and $y$ remains constant. If $f>0$ , which integral is least likely to exhibit sign cancellation caused by coordinate reversal?

\int_C f\,dx

\int_C f\,dy

\int_C f\,dz

✅

D.All three must have the same cancellation.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Because $z$ increases steadily, $dz$ retains the same sign throughout the path. Therefore, with $f>0$ , contributions to $\int_C f\,dz$ do not cancel because of reversals in the $z$ -direction. In contrast, the oscillation of $x$ can produce positive and negative $dx$ contributions.

Q11. A student parameterizes a curve correctly but reverses the parameter interval from $t=1$ to $t=0$ without changing the differential. What happens to $\int_C f\,dy$ under this reversal?

A.It remains unchanged because

f

is scalar.

B.It changes sign because

dy

changes orientation with the path. ✅

C.It becomes an arc-length integral.

D.It becomes zero for every field.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Reversing the parameter direction reverses the orientation of the curve. Since $dy=y'(t)dt$ , the differential changes sign under the reversed traversal. Consequently, a line integral with respect to $dy$ changes sign. This is different from scalar arc-length integrals involving $ds$ , which remain unchanged under orientation reversal.

Q12. Let $f(x,y,z)=2x-y+z$ , and let $C$ be parameterized by $x=t,\ y=t,\ z=2t$ , $0\le t\le1$ . Which pair of integrals has equal values, and why?

\int_C f\,dx

and

\int_C f\,dy

, because

dx=dy

along the path. ✅

\int_C f\,dx

and

\int_C f\,dz

, because

dx=dz

\int_C f\,dy

and

\int_C f\,dz

, because

dy=dz

D.None, because differentials can never be equal.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Along the given parameterization, $x=t$ and $y=t$ , so $dx=dt$ and $dy=dt$ . Therefore, the same field values are multiplied by identical differentials, making $\int_C f\,dx=\int_C f\,dy$ . Meanwhile, $dz=2dt$ , so the $z$ -integral generally has twice the weighting.

Q13. A modeling problem describes heat accumulation along a moving sensor path. The measured intensity is $f(x,y,z)$ , and the quantity of interest is accumulated only as the sensor moves vertically. Which mathematical model is most appropriate?

\int_C f\,ds

\int_C f\,dx

\int_C f\,dy

\int_C f\,dz

✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: If the modeled accumulation is specifically associated with vertical movement, the relevant coordinate differential is $dz$ . Thus $\int_C f\,dz$ directly weights the measured intensity by vertical displacement. An arc-length integral would instead measure accumulation per unit distance, while $dx$ and $dy$ describe horizontal coordinate changes.

Q14. Suppose $f(x,y,z)=x^2+y^2+z^2$ and $C$ is any closed curve lying entirely on the plane $x=2$ . Which conclusion about $\int_C f\,dx$ follows immediately?

A.It is positive because

f>0

B.It is negative because the curve is closed.

C.It is zero because

dx=0

everywhere on the curve. ✅

D.It equals the length of the curve.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Every point of the curve lies on the plane $x=2$ , so $x$ is constant along the entire path. Therefore $dx=0$ at every point, regardless of the behavior of $y$ , $z$ , or the value of $f$ . Hence the entire line integral $\int_C f\,dx$ is zero.

Q15. Consider a closed curve $C$ and the three integrals $\int_C x\,dx$ , $\int_C y\,dy$ , and $\int_C z\,dz$ . Which reasoning correctly determines all three values without explicitly parameterizing the curve?

A.Each equals zero because the curve is closed and every line integral over a closed curve vanishes.

B.Each is zero because

x\,dx

y\,dy

, and

z\,dz

are exact differentials of

\frac{x^2}{2},\frac{y^2}{2},\frac{z^2}{2}

, respectively. ✅

C.They are generally nonzero because the field changes along the curve.

D.Only

\int_C z\,dz

is zero because

z

is the third coordinate.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The key is to recognize exact differentials: $x\,dx=d(x^2/2)$ , $y\,dy=d(y^2/2)$ , and $z\,dz=d(z^2/2)$ . The integral of an exact differential around a closed curve is zero. This conclusion follows from structure rather than from the particular shape or parameterization of the curve.

Q16. A researcher compares two parameterizations of the same oriented curve: one uses $t$ , and the other uses a smooth increasing parameter $s$ . Which statement best explains why both should produce the same value for $\int_C f\,dz$ ?

A.The field becomes constant under reparameterization.

B.The coordinate

z

disappears from the integral.

C.The chain rule transforms

dz

consistently, preserving the oriented accumulation along the same geometric path. ✅

D.Both parameterizations must have identical derivatives.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: A smooth increasing reparameterization changes how quickly the curve is traversed but does not change its orientation or geometric path. By the chain rule, $dz$ transforms consistently with the new parameter, so the accumulated quantity represented by $\int_C f\,dz$ remains unchanged. The parameter itself is therefore not part of the intrinsic result.

Q17. Let $C$ be a closed curve for which $x\ge0$ , and suppose $f=x$ . A student argues that $\int_C x\,dx>0$ because both $x$ and the length of the curve are nonnegative. What is the strongest correction?

A.The argument is correct because

dx

is always positive.

B.The integral is zero because

x\,dx=d(x^2/2)

, so the closed-path integral vanishes. ✅

C.The integral is negative because the curve is closed.

D.The integral equals the circumference multiplied by the average value of

x

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The student's mistake is treating $dx$ like a nonnegative length element. Unlike $ds$ , $dx$ can be positive or negative depending on direction. More importantly, $x\,dx$ is the exact differential $d(x^2/2)$ . Therefore, its integral around any closed curve is zero, regardless of the nonnegative values of $x$ .

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