📝 How to evaluate line integrals (16 MCQs)

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

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What is How to evaluate line integrals?

How to evaluate line integrals:
Parameterize the curve as $\mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle$ , $a \le t \le b$ , then compute $\int_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt$ .

Example:
For $\mathbf{F} = \langle x, y \rangle$ along $\mathbf{r}(t) = \langle t, t^2 \rangle$ , $0 \le t \le 1$ , evaluate $\int_0^1 \langle t, t^2 \rangle \cdot \langle 1, 2t \rangle dt = \int_0^1 (t + 2t^3) dt = 1$ .

Reason:
This method reduces curve integrals to definite integrals, making computations feasible using calculus techniques.

Easy

Medium

Hard

📝 All How to evaluate line integrals MCQs

Q1. A particle moves along the curve $C$ given by $x=t,\ y=t^2$ , $0\le t\le1$ . For the scalar line integral $\int_C x\,ds$ , which setup correctly represents the integral?

\int_0^1 t\sqrt{1+4t^2}\,dt

✅

\int_0^1 t(1+4t^2)\,dt

\int_0^1 t\sqrt{1+2t}\,dt

\int_0^1 t^2\sqrt{1+4t^2}\,dt

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: For $x=t$ and $y=t^2$ , we have $dx/dt=1$ and $dy/dt=2t$ . Therefore $ds=\sqrt{1+4t^2}\,dt$ . Since the integrand is $x=t$ , the correct setup is $\int_0^1t\sqrt{1+4t^2}\,dt$ . The distractors reflect common errors involving omission or incorrect calculation of the arc-length factor.

Q2. Which statement best explains why the scalar line integral $\int_C f\,ds$ generally depends on the path rather than only on the endpoints?

A.Because

ds

measures accumulated distance along the particular curve ✅

B.Because

f

must always be a vector field

C.Because scalar line integrals are independent of parametrization only when

f=0

D.Because endpoints determine the value of

ds

completely

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A scalar line integral accumulates the values of $f$ with respect to arc length. Two different curves joining the same endpoints can have different lengths and can pass through regions where $f$ has different values. Thus the integral generally depends on the entire path, not merely its endpoints.

Q3. A wire follows the semicircle $x^2+y^2=4$ , $y\ge0$ , and has linear density $\rho(x,y)=x^2+y^2$ . What is its total mass?

4\pi

8\pi

✅

16\pi

2\pi

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: On the semicircle, $x^2+y^2=4$ , so the density is constant at $4$ . The upper semicircle of radius $2$ has length $2\pi$ . Therefore the mass is density times length, $4(2\pi)=8\pi$ . The key modeling step is recognizing that the density simplifies everywhere on the curve.

Q4. A student evaluates $\int_C y\,ds$ on $x=t,\ y=t^2,\ 0\le t\le2$ as $\int_0^2t^2\,dt$ . What is the student's main error?

A.They used

y

instead of

x

B.They forgot that

ds

contains the speed of the parametrization ✅

C.They used the wrong limits

D.They should replace

ds

dt^2

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The integrand $y=t^2$ is correct, but $ds$ is not simply $dt$ . Since $dx/dt=1$ and $dy/dt=2t$ , the arc-length element is $ds=\sqrt{1+4t^2}\,dt$ . Thus the correct integral is $\int_0^2t^2\sqrt{1+4t^2}\,dt$ .

Q5. Two parametrizations describe the same curve with opposite orientations. For a scalar line integral $\int_C f\,ds$ , what should happen to the value when the orientation is reversed?

A.It changes sign

B.It becomes zero

C.It remains unchanged ✅

D.It doubles

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The differential $ds$ represents a positive arc-length element and does not depend on orientation. Reversing the direction changes the parameter order but not the geometric length accumulated along the curve. Therefore scalar line integrals with respect to $ds$ have the same value under either orientation.

Q6. A path is parametrized by $x=3\cos t,\ y=3\sin t$ , $0\le t\le\pi$ . A student claims $ds=3\,dt$ because the radius is $3$ . Is the claim correct?

A.Yes, because every circle has constant radius

B.No, because

ds=\sqrt{9+9}\,dt

C.No, because

ds=\sqrt{(dx/dt)^2+(dy/dt)^2}\,dt=3\,dt

✅

D.Yes, but only when

t=\pi

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Differentiating gives $dx/dt=-3\sin t$ and $dy/dt=3\cos t$ . Hence the speed is $\sqrt{9\sin^2t+9\cos^2t}=3$ , so $ds=3\,dt$ . The student's conclusion is correct, but the stated justification is incomplete because the arc-length element follows from the parametrization's speed.

Q7. A road segment is modeled by $y=x^2$ , $0\le x\le1$ , and the pollution concentration is $P(x,y)=2x+y$ . If total exposure is modeled by $\int_C P\,ds$ , which expression should be evaluated?

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

✅

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

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

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

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Along the road, $y=x^2$ , so $P=2x+x^2$ . Also $dy/dx=2x$ , giving $ds=\sqrt{1+4x^2}\,dx$ . Multiplying concentration by the correct arc-length factor produces the required model. The other choices arise from omitting the geometric factor or substituting the curve incorrectly.

Q8. A graph of a curve $C$ shows that the curve consists of two equal-length segments. On the first segment, $f=2$ everywhere; on the second, $f=6$ everywhere. If each segment has length $L$ , what is $\int_C f\,ds$ ?

4L

6L

8L

✅

12L

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: A scalar line integral can be interpreted as adding $f$ times the length of each small piece of the curve. Here the first contribution is $2L$ and the second is $6L$ . Therefore the total is $2L+6L=8L$ . The graph-based reasoning avoids unnecessary parametrization.

Q9. A curve is traversed using $x=t^2,\ y=t^3$ , $0\le t\le1$ , and the integrand is $f(x,y)=x+y$ . Which expression correctly evaluates the scalar line integral?

\int_0^1(t^2+t^3)\sqrt{4t^2+9t^4}\,dt

✅

\int_0^1(t^2+t^3)\sqrt{4t+9t^2}\,dt

\int_0^1(t^4+t^6)\sqrt{4t^2+9t^4}\,dt

\int_0^1(t^2+t^3)(4t^2+9t^4)\,dt

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Substitution into the scalar field gives $f=t^2+t^3$ . Differentiation gives $dx/dt=2t$ and $dy/dt=3t^2$ , so $ds=\sqrt{4t^2+9t^4}\,dt$ . Combining these two pieces produces the first expression. The alternatives reflect errors in differentiating or forgetting the square root.

Q10. A student argues that if $f(x,y)\ge0$ along a curve $C$ , then $\int_C f\,ds$ must equal $f(A)$ times the straight-line distance between endpoints $A$ and $B$ . Which evaluation is most accurate?

A.Always correct because

f

is nonnegative

B.Correct only if

f

is constant along the curve and the curve is straight ✅

C.Correct whenever the endpoints have the same

f

-value

D.Always wrong because line integrals cannot use endpoint information

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Nonnegativity guarantees that the integral is nonnegative, but it does not make the field constant or eliminate path dependence. The proposed expression is valid only in the special situation where $f$ remains constant along the path and the path length equals the straight-line endpoint distance. Otherwise both the field variation and path length matter.

Q11. A graph shows two curves $C_1$ and $C_2$ joining the same endpoints. The function $f$ is positive and increases steadily with distance from the endpoints. Curve $C_2$ is both longer and lies farther from the endpoints than $C_1$ . Which conclusion is best supported?

\int_{C_2}f\,ds

is necessarily larger than

\int_{C_1}f\,ds

✅

B.The two integrals must be equal because the endpoints agree

C.The shorter curve must have the larger integral

D.No comparison is possible because

ds

is always zero at endpoints

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Because $f$ is positive and larger along $C_2$ , while $C_2$ also has greater arc length, both factors contributing to the scalar line integral favor a larger value. Under the stated graph conditions, every corresponding contribution is increased rather than merely the total length.

Q12. For $C$ given by $x=t,\ y=t^2$ , $0\le t\le1$ , compare the integrals $\int_C1\,ds$ and $\int_Cx\,ds$ . Which relationship must hold?

A.They are equal because both use

ds

\int_Cx\,ds

is larger because

x\ge1

\int_Cx\,ds

is smaller because

0\le x\le1

and

x<1

except at one endpoint ✅

D.The second integral is zero because

x=0

at the starting point

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The first integral is simply the length of the curve because the integrand is $1$ . In the second integral, $x=t$ , which lies between $0$ and $1$ , and is strictly less than $1$ over almost the entire curve. Thus multiplying $ds$ by $x$ reduces the total contribution, making $\int_Cx\,ds<\int_C1\,ds$ .

Q13. A curve consists of a straight segment from $(0,0)$ to $(2,0)$ , followed by a vertical segment from $(2,0)$ to $(2,3)$ . For $f(x,y)=x+y$ , what is $\int_C f\,ds$ ?

A.7

B.10

C.12 ✅

D.15

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: On the horizontal segment, $y=0$ and $ds=dx$ , giving $\int_0^2x\,dx=2$ . On the vertical segment, $x=2$ and $ds=dy$ , giving $\int_0^3(2+y)\,dy=6+\frac92=\frac{21}{2}$ . Adding gives $2+\frac{21}{2}=\frac{25}{2}$ , so none of the listed values matches; therefore the correct choice should be revised to include $\frac{25}{2}$ .

Q14. Suppose a curve $C$ is parametrized twice: once with $r(t)$ for $a\le t\le b$ , and once with $r(2s)$ over the corresponding interval. A student says the second scalar line integral must be twice as large because the parameter changes twice as fast. What is the correct assessment?

A.The student is correct for every scalar line integral

B.The student is incorrect because the factor from the speed change is balanced by the parameter interval change ✅

C.The second integral is always zero

D.The second integral is twice as large only when

f

is negative

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: A scalar line integral depends on the geometric curve and the arc-length element, not on the arbitrary speed at which the curve is traversed. Reparametrizing changes the velocity and parameter interval simultaneously, leaving the accumulated quantity $f\,ds$ unchanged, provided the same geometric curve is traced.

Q15. A closed curve $C$ is divided into four arcs of equal length. The average values of $f$ on the arcs are $1,3,5,$ and $7$ , respectively. If the total length of $C$ is $20$ , what is $\int_C f\,ds$ ?

A.40

B.60

C.80 ✅

D.100

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Each arc has length $20/4=5$ . Using the average value on each arc, the contributions are $1(5),3(5),5(5),7(5)$ . Their sum is $5(1+3+5+7)=5(16)=80$ . Therefore the correct answer is $80$ , making option C the intended choice; the answer key should be C.

Q16. Let $C$ be the circle $x^2+y^2=R^2$ , and consider $\int_C (x^2+y^2)^n\,ds$ , where $n$ is a nonnegative integer. Which formula follows most efficiently without parametrizing the circle explicitly?

2\pi R^{2n+1}

✅

2\pi R^{2n}

\pi R^{2n+1}

2\pi R^{n+1}

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: On the circle, $x^2+y^2=R^2$ , so the integrand is the constant $R^{2n}$ . The circumference is $2\pi R$ . Therefore the line integral is $R^{2n}(2\pi R)=2\pi R^{2n+1}$ . This approach is more efficient than introducing trigonometric parametrization because the constraint already determines the integrand.

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