Multivariable Calculus Quiz: Line Integrals Scalar Fields
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Line Integrals Scalar FieldsQuestion 1 of 13

Let CC be the arc of the parabola y=x2y = x^2 from (0,0)(0,0) to (2,4)(2,4). Which of the following integrals represents the value of ∫Cexy ds\int_C e^{xy} \, ds?

∫02et31+4t2 dt\int_0^2 e^{t^3} \sqrt{1+4t^2} \, dt
∫02et3 dt\int_0^2 e^{t^3} \, dt
∫02et21+4t2 dt\int_0^2 e^{t^2} \sqrt{1+4t^2} \, dt
∫04eyy1+12y dy\int_0^4 e^{y\sqrt{y}} \sqrt{1+\frac{1}{2\sqrt{y}}} \, dy
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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Line Integrals Scalar Fields

Practice Line Integrals Scalar Fields in Multivariable Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Line Integrals Scalar Fields, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

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Question 1

Let CC be the arc of the parabola y=x2y = x^2 from (0,0)(0,0) to (2,4)(2,4). Which of the following integrals represents the value of ∫Cexy ds\int_C e^{xy} \, ds?

  1. ∫02et31+4t2 dt\int_0^2 e^{t^3} \sqrt{1+4t^2} \, dt (correct answer)
  2. ∫02et3 dt\int_0^2 e^{t^3} \, dt
  3. ∫02et21+4t2 dt\int_0^2 e^{t^2} \sqrt{1+4t^2} \, dt
  4. ∫04eyy1+12y dy\int_0^4 e^{y\sqrt{y}} \sqrt{1+\frac{1}{2\sqrt{y}}} \, dy
Explanation: To set up the line integral, we first need to parameterize the curve CC. Since y=x2y=x^2, we can use xx as the parameter, say x=tx=t. Then y=t2y=t^2. The path starts at (0,0)(0,0) (so t=0t=0) and ends at (2,4)(2,4) (so t=2t=2). The parameterization is r(t)=⟨t,t2⟩\mathbf{r}(t) = \langle t, t^2 \rangle for t∈[0,2]t \in [0,2].\nNext, we find the arc length element dsds. The derivative is r′(t)=⟨1,2t⟩\mathbf{r}'(t) = \langle 1, 2t \rangle. Its magnitude is ∥r′(t)∥=12+(2t)2=1+4t2\|\mathbf{r}'(t)\| = \sqrt{1^2 + (2t)^2} = \sqrt{1+4t^2}. So, ds=1+4t2 dtds = \sqrt{1+4t^2} \, dt.\nFinally, we express the function f(x,y)=exyf(x,y) = e^{xy} in terms of the parameter tt: f(r(t))=e(t)(t2)=et3f(\mathbf{r}(t)) = e^{(t)(t^2)} = e^{t^3}.\nCombining these pieces, the line integral is ∫Cf ds=∫02f(r(t))∥r′(t)∥ dt=∫02et31+4t2 dt\int_C f \, ds = \int_0^2 f(\mathbf{r}(t)) \|\mathbf{r}'(t)\| \, dt = \int_0^2 e^{t^3} \sqrt{1+4t^2} \, dt.\n\nDistractor B incorrectly assumes ds=dtds = dt, omitting the arc length factor ∥r′(t)∥\|\mathbf{r}'(t)\|. Distractor C incorrectly substitutes y=ty=t into the function, evaluating exte^{xt} as et2e^{t^2} instead of ex(t)y(t)=et⋅t2=et3e^{x(t)y(t)} = e^{t \cdot t^2} = e^{t^3}. Distractor D results from an incorrect attempt to parameterize with respect to yy, with errors in both the function evaluation and the arc length element calculation.

Question 2

Let C1C_1 be the line segment from (0,0)(0,0) to (1,1)(1,1) and C2C_2 be the arc of the parabola y=x2y=x^2 from (0,0)(0,0) to (1,1)(1,1). Let I1=∫C1x dsI_1 = \int_{C_1} x \, ds and I2=∫C2x dsI_2 = \int_{C_2} x \, ds. Which of the following statements is true?

  1. I1<I2I_1 < I_2 (correct answer)
  2. I1>I2I_1 > I_2
  3. I1=I2I_1 = I_2
  4. The relationship depends on the parameterization.
Explanation: We need to compute both integrals and compare them.\nFor I1I_1: Parameterize C1C_1 by r(t)=⟨t,t⟩\mathbf{r}(t) = \langle t, t \rangle for 0≤t≤10 \le t \le 1. Then r′(t)=⟨1,1⟩\mathbf{r}'(t) = \langle 1, 1 \rangle, so ds=∥r′(t)∥dt=12+12 dt=2 dtds = \|\mathbf{r}'(t)\| dt = \sqrt{1^2+1^2} \, dt = \sqrt{2} \, dt. The function is f(x,y)=xf(x,y)=x, so f(r(t))=tf(\mathbf{r}(t)) = t. \nI1=∫01t2 dt=2[t22]01=22I_1 = \int_0^1 t \sqrt{2} \, dt = \sqrt{2} [\frac{t^2}{2}]_0^1 = \frac{\sqrt{2}}{2}.\n\nFor I2I_2: Parameterize C2C_2 by r(t)=⟨t,t2⟩\mathbf{r}(t) = \langle t, t^2 \rangle for 0≤t≤10 \le t \le 1. Then r′(t)=⟨1,2t⟩\mathbf{r}'(t) = \langle 1, 2t \rangle, so ds=∥r′(t)∥dt=12+(2t)2 dt=1+4t2 dtds = \|\mathbf{r}'(t)\| dt = \sqrt{1^2+(2t)^2} \, dt = \sqrt{1+4t^2} \, dt. The function is f(x,y)=xf(x,y)=x, so f(r(t))=tf(\mathbf{r}(t)) = t. \nI2=∫01t1+4t2 dtI_2 = \int_0^1 t \sqrt{1+4t^2} \, dt. We use u-substitution with u=1+4t2u = 1+4t^2, so du=8t dtdu = 8t \, dt. The limits change from t=0,1t=0,1 to u=1,5u=1,5. \nI2=∫15udu8=18[23u3/2]15=112(53/2−13/2)=55−112I_2 = \int_1^5 \sqrt{u} \frac{du}{8} = \frac{1}{8} [\frac{2}{3}u^{3/2}]_1^5 = \frac{1}{12} (5^{3/2} - 1^{3/2}) = \frac{5\sqrt{5}-1}{12}.\n\nNow we compare I1=22I_1 = \frac{\sqrt{2}}{2} and I2=55−112I_2 = \frac{5\sqrt{5}-1}{12}. We can approximate: I1≈1.4142=0.707I_1 \approx \frac{1.414}{2} = 0.707. For I2I_2, 5≈2.236\sqrt{5} \approx 2.236, so 55≈11.185\sqrt{5} \approx 11.18. Then I2≈11.18−112=10.1812≈0.848I_2 \approx \frac{11.18-1}{12} = \frac{10.18}{12} \approx 0.848. Since 0.707<0.8480.707 < 0.848, we have I1<I2I_1 < I_2.\n\nDistractor C is incorrect; it implies path independence, which is a property of line integrals of conservative vector fields, not scalar fields. Distractor D is incorrect as scalar line integrals are independent of parameterization.

Question 3

Let CC be the unit circle x2+y2=1x^2+y^2=1. It is given that ∫C(x2+y2) ds=2π\int_C (x^2+y^2) \, ds = 2\pi. Let CRC_R be the circle x2+y2=R2x^2+y^2=R^2 with R>0R>0. What is the value of the line integral ∫CR(x2+y2) ds\int_{C_R} (x^2+y^2) \, ds?

  1. 2πR2\pi R
  2. 2πR22\pi R^2
  3. 2πR32\pi R^3 (correct answer)
  4. 2πR42\pi R^4
Explanation: We can solve this by direct computation. Parameterize the circle CRC_R by r(t)=⟨Rcos⁡(t),Rsin⁡(t)⟩\mathbf{r}(t) = \langle R\cos(t), R\sin(t) \rangle for 0≤t≤2π0 \le t \le 2\pi. The derivative is r′(t)=⟨−Rsin⁡(t),Rcos⁡(t)⟩\mathbf{r}'(t) = \langle -R\sin(t), R\cos(t) \rangle. The arc length element is ds=∥r′(t)∥dt=(−Rsin⁡t)2+(Rcos⁡t)2 dt=R2(sin⁡2t+cos⁡2t) dt=R dtds = \|\mathbf{r}'(t)\| dt = \sqrt{(-R\sin t)^2 + (R\cos t)^2} \, dt = \sqrt{R^2(\sin^2 t + \cos^2 t)} \, dt = R \, dt. The function is f(x,y)=x2+y2f(x,y) = x^2+y^2. On the curve CRC_R, the value of the function is constant: f(Rcos⁡t,Rsin⁡t)=(Rcos⁡t)2+(Rsin⁡t)2=R2f(R\cos t, R\sin t) = (R\cos t)^2 + (R\sin t)^2 = R^2. The integral is:\n∫CR(x2+y2) ds=∫02π(R2)⋅(R dt)=R3∫02πdt=R3[t]02π=2πR3.\int_{C_R} (x^2+y^2) \, ds = \int_0^{2\pi} (R^2) \cdot (R \, dt) = R^3 \int_0^{2\pi} dt = R^3 [t]_0^{2\pi} = 2\pi R^3.\nAlternatively, one can use a scaling argument. Let f(x,y)=x2+y2f(x,y)=x^2+y^2 be a homogeneous function of degree k=2k=2. If a curve CC is scaled by a factor RR to become CRC_R, the line integral scales as ∫CRf ds=Rk+1∫Cf ds\int_{C_R} f \, ds = R^{k+1} \int_C f \, ds. Here k=2k=2, so the integral scales by R2+1=R3R^{2+1}=R^3. Given ∫C(x2+y2) ds=2π\int_C (x^2+y^2) \, ds = 2\pi, the new integral is R3(2π)R^3(2\pi).\n\nDistractor A, 2πR2\pi R, is the length of the circle CRC_R, which equals ∫CR1 ds\int_{C_R} 1 \, ds. Distractor B, 2πR22\pi R^2, results from correctly evaluating the function as R2R^2 on the circle but forgetting that dsds also scales with RR.

Question 4

A thin wire is bent in the shape of a quarter-circle of radius RR in the first quadrant, with endpoints at (R,0)(R,0) and (0,R)(0,R). The wire has a constant linear mass density. What is the xx-coordinate of the center of mass of the wire?

  1. R2\frac{R}{2}
  2. 2Rπ\frac{2R}{\pi} (correct answer)
  3. 4R3π\frac{4R}{3\pi}
  4. πR4\frac{\pi R}{4}
Explanation: The xx-coordinate of the center of mass, xˉ\bar{x}, of a wire with constant density is given by xˉ=∫Cx ds∫C1 ds\bar{x} = \frac{\int_C x \, ds}{\int_C 1 \, ds}. The denominator is the total length of the wire, LL. The curve is a quarter-circle of radius RR, so its length is L=14(2πR)=πR2L = \frac{1}{4}(2\pi R) = \frac{\pi R}{2}.\nTo compute the numerator, we parameterize the quarter-circle CC as r(t)=⟨Rcos⁡(t),Rsin⁡(t)⟩\mathbf{r}(t) = \langle R\cos(t), R\sin(t) \rangle for t∈[0,π/2]t \in [0, \pi/2]. The arc length element is ds=∥r′(t)∥dt=∥⟨−Rsin⁡(t),Rcos⁡(t)⟩∥dt=R dtds = \|\mathbf{r}'(t)\| dt = \|\langle -R\sin(t), R\cos(t) \rangle\| dt = R \, dt. The integral in the numerator is:\n∫Cx ds=∫0π/2(Rcos⁡(t))(R dt)=R2∫0π/2cos⁡(t) dt=R2[sin⁡(t)]0π/2=R2(sin⁡(π/2)−sin⁡(0))=R2(1−0)=R2.\int_C x \, ds = \int_0^{\pi/2} (R\cos(t)) (R \, dt) = R^2 \int_0^{\pi/2} \cos(t) \, dt = R^2 [\sin(t)]_0^{\pi/2} = R^2 (\sin(\pi/2) - \sin(0)) = R^2(1-0) = R^2.\nNow, we can find the xx-coordinate of the center of mass:\nxˉ=R2L=R2πR/2=2Rπ.\bar{x} = \frac{R^2}{L} = \frac{R^2}{\pi R / 2} = \frac{2R}{\pi}.\nDistractor A, R/2R/2, is an intuitive but incorrect guess based on the range of x-values. Distractor C, 4R3π\frac{4R}{3\pi}, is the xx-coordinate of the center of mass of a solid quarter-circular plate (a lamina), which is a common confusion. Distractor D, πR4\frac{\pi R}{4}, is half the arc length of the wire.

Question 5

A wire is shaped like a helix parameterized by r(t)=⟨cos⁡(t),sin⁡(t),t⟩\mathbf{r}(t) = \langle \cos(t), \sin(t), t \rangle for 0≤t≤2π0 \le t \le 2\pi. The density of the wire at any point (x,y,z)(x,y,z) is given by ρ(x,y,z)=z\rho(x,y,z) = z. What is the total mass of the wire?

  1. 2π22\pi^2
  2. 2π22\pi\sqrt{2}
  3. 2π222\pi^2\sqrt{2} (correct answer)
  4. 00
Explanation: The mass of the wire is given by the line integral of its density function over the curve CC, which is M=∫Cρ(x,y,z) dsM = \int_C \rho(x,y,z) \, ds. First, we find the arc length element dsds. The parameterization is r(t)=⟨cos⁡(t),sin⁡(t),t⟩\mathbf{r}(t) = \langle \cos(t), \sin(t), t \rangle. The derivative is r′(t)=⟨−sin⁡(t),cos⁡(t),1⟩\mathbf{r}'(t) = \langle -\sin(t), \cos(t), 1 \rangle. The magnitude of the derivative is ∥r′(t)∥=(−sin⁡(t))2+(cos⁡(t))2+12=sin⁡2(t)+cos⁡2(t)+1=1+1=2\|\mathbf{r}'(t)\| = \sqrt{(-\sin(t))^2 + (\cos(t))^2 + 1^2} = \sqrt{\sin^2(t) + \cos^2(t) + 1} = \sqrt{1+1} = \sqrt{2}. So, ds=2 dtds = \sqrt{2} \, dt. The density function along the curve is ρ(r(t))=z(t)=t\rho(\mathbf{r}(t)) = z(t) = t. Now we set up and evaluate the integral for the mass: M=∫02πt⋅2 dt=2∫02πt dt=2[t22]02π=2((2π)22−0)=24π22=2π22M = \int_0^{2\pi} t \cdot \sqrt{2} \, dt = \sqrt{2} \int_0^{2\pi} t \, dt = \sqrt{2} \left[ \frac{t^2}{2} \right]_0^{2\pi} = \sqrt{2} \left( \frac{(2\pi)^2}{2} - 0 \right) = \sqrt{2} \frac{4\pi^2}{2} = 2\pi^2\sqrt{2}.\n\nDistractor A, 2π22\pi^2, is the result of forgetting the 2\sqrt{2} factor from the arc length element dsds, effectively calculating ∫Cρ dt\int_C \rho \, dt. Distractor B, 2π22\pi\sqrt{2}, is the result of integrating a density of 1 instead of z=tz=t, which would calculate the length of the wire multiplied by 2\sqrt{2} for some reason. The length is 2π22\pi\sqrt{2}. Distractor D, 00, would be the result if the integral of an odd function were taken over a symmetric interval like [−π,π][-\pi, \pi], but the given interval is [0,2π][0, 2\pi].

Question 6

Let CC be the upper semicircle of radius 2 centered at the origin, traversed from (2,0)(2,0) to (−2,0)(-2,0). The value of the line integral ∫C(x2+y) ds\int_C (x^2+y) \, ds is KK. What is the value of the line integral ∫C′(x2+y) ds\int_{C'} (x^2+y) \, ds, where C′C' is the lower semicircle of radius 2, also traversed from (2,0)(2,0) to (−2,0)(-2,0)?

  1. KK
  2. −K-K
  3. K−8K-8
  4. K−16K-16 (correct answer)
Explanation: Let's calculate KK. The upper semicircle CC can be parameterized by r(t)=⟨2cos⁡(t),2sin⁡(t)⟩\mathbf{r}(t) = \langle 2\cos(t), 2\sin(t) \rangle for tt from 00 to π\pi. Then r′(t)=⟨−2sin⁡(t),2cos⁡(t)⟩\mathbf{r}'(t) = \langle -2\sin(t), 2\cos(t) \rangle, so ds=∥r′(t)∥dt=4sin⁡2(t)+4cos⁡2(t) dt=2 dtds = \|\mathbf{r}'(t)\| dt = \sqrt{4\sin^2(t) + 4\cos^2(t)} \, dt = 2 \, dt. The integrand is f(x,y)=x2+y=(2cos⁡t)2+2sin⁡t=4cos⁡2t+2sin⁡tf(x,y) = x^2+y = (2\cos t)^2 + 2\sin t = 4\cos^2 t + 2\sin t. So,\nK=∫0π(4cos⁡2t+2sin⁡t)(2 dt)=8∫0πcos⁡2t dt+4∫0πsin⁡t dt.K = \int_0^{\pi} (4\cos^2 t + 2\sin t) (2 \, dt) = 8\int_0^{\pi} \cos^2 t \, dt + 4\int_0^{\pi} \sin t \, dt.\nUsing cos⁡2t=1+cos⁡(2t)2\cos^2 t = \frac{1+\cos(2t)}{2}, the first integral is 8[t2+sin⁡(2t)4]0π=8(π2)=4π8[\frac{t}{2} + \frac{\sin(2t)}{4}]_0^{\pi} = 8(\frac{\pi}{2}) = 4\pi. The second integral is 4[−cos⁡t]0π=4(1−(−1))=84[-\cos t]_0^{\pi} = 4(1 - (-1)) = 8. Thus, K=4π+8K = 4\pi + 8.\n\nNow, let's calculate the integral over the lower semicircle C′C'. We can parameterize C′C' as g(t)=⟨2cos⁡(t),−2sin⁡(t)⟩\mathbf{g}(t) = \langle 2\cos(t), -2\sin(t) \rangle for tt from 00 to π\pi. Then g′(t)=⟨−2sin⁡(t),−2cos⁡(t)⟩\mathbf{g}'(t) = \langle -2\sin(t), -2\cos(t) \rangle, and ds=∥g′(t)∥dt=2 dtds = \|\mathbf{g}'(t)\| dt = 2 \, dt. The integrand is f(x,y)=x2+y=(2cos⁡t)2+(−2sin⁡t)=4cos⁡2t−2sin⁡tf(x,y) = x^2+y = (2\cos t)^2 + (-2\sin t) = 4\cos^2 t - 2\sin t. So,\n∫C′(x2+y) ds=∫0π(4cos⁡2t−2sin⁡t)(2 dt)=8∫0πcos⁡2t dt−4∫0πsin⁡t dt.\int_{C'} (x^2+y)\,ds = \int_0^{\pi} (4\cos^2 t - 2\sin t) (2 \, dt) = 8\int_0^{\pi} \cos^2 t \, dt - 4\int_0^{\pi} \sin t \, dt.\nThis evaluates to 4π−84\pi - 8. \nComparing this to K=4π+8K = 4\pi+8, we see that the new value is (4π+8)−16=K−16(4\pi+8) - 16 = K-16.\n\nDistractor A, KK, would be correct if the integrand were symmetric with respect to the x-axis (e.g., if it were just x2x^2). Distractor B, −K-K, is a common mistake confusing scalar line integrals with vector line integrals, where reversing path direction can negate the value. Distractor C, K−8K-8, is a possible calculation error.

Question 7

Let CC be a smooth curve of length L>0L>0 in the plane, and let f(x,y)f(x,y) be a continuous function defined on CC. Which of the following statements about the scalar line integral ∫Cf ds\int_C f \, ds is NOT always true?

  1. If m≤f(x,y)≤Mm \le f(x,y) \le M for all (x,y)(x,y) on CC, then mL≤∫Cf ds≤MLm L \le \int_C f \, ds \le M L.
  2. If −C-C is the curve CC with the opposite orientation, then ∫−Cf ds=∫Cf ds\int_{-C} f \, ds = \int_C f \, ds.
  3. If CC is a closed curve, then ∫Cf ds=0\int_C f \, ds = 0. (correct answer)
  4. ∫C(f1+f2) ds=∫Cf1 ds+∫Cf2 ds\int_C (f_1 + f_2) \, ds = \int_C f_1 \, ds + \int_C f_2 \, ds.
Explanation: Let's analyze each statement:\nA) This is a standard property of integrals. The integral is bounded by the minimum and maximum values of the integrand multiplied by the measure of the domain (in this case, arc length LL). This statement is always true.\nB) Scalar line integrals are defined in terms of arc length dsds, which is inherently positive regardless of the direction of traversal. Therefore, the integral is independent of the orientation of the curve. This statement is always true.\nC) This is not always true. A simple counterexample is integrating a positive constant function over a closed curve. For example, let CC be the unit circle and f(x,y)=1f(x,y)=1. Then ∫C1 ds\int_C 1 \, ds is the arc length of the unit circle, which is 2π2\pi, not 0. This property (integral over a closed path is zero) is characteristic of conservative vector fields, not scalar line integrals in general.\nD) This is the linearity property of integrals, which holds for scalar line integrals. This statement is always true.\nTherefore, the statement that is not always true is C.

Question 8

The average value of a function ff over a curve CC is given by the formula 1L∫Cf ds\frac{1}{L} \int_C f \, ds, where LL is the arc length of CC. Which expression represents the average value of f(x,y)=xf(x,y) = x over the curve CC given by the parabola y=x2y=x^2 from (0,0)(0,0) to (1,1)(1,1)?

  1. ∫01t1+4t2 dt∫011+4t2 dt\frac{\int_0^1 t \sqrt{1+4t^2} \, dt}{\int_0^1 \sqrt{1+4t^2} \, dt} (correct answer)
  2. ∫01t dt∫01 dt\frac{\int_0^1 t \, dt}{\int_0^1 \, dt}
  3. ∫01t1+4t2 dt\int_0^1 t \sqrt{1+4t^2} \, dt
  4. ∫01t21+4t2 dt∫011+4t2 dt\frac{\int_0^1 t^2 \sqrt{1+4t^2} \, dt}{\int_0^1 \sqrt{1+4t^2} \, dt}
Explanation: The average value is 1L∫Cf ds\frac{1}{L} \int_C f \, ds. The arc length LL is itself a line integral: L=∫C1 dsL = \int_C 1 \, ds. So the average value is ∫Cf ds∫C1 ds\frac{\int_C f \, ds}{\int_C 1 \, ds}.\nWe parameterize the curve CC by r(t)=⟨t,t2⟩\mathbf{r}(t) = \langle t, t^2 \rangle for 0≤t≤10 \le t \le 1. The derivative is r′(t)=⟨1,2t⟩\mathbf{r}'(t) = \langle 1, 2t \rangle, so the arc length element is ds=∥r′(t)∥dt=12+(2t)2 dt=1+4t2 dtds = \|\mathbf{r}'(t)\| dt = \sqrt{1^2+(2t)^2} \, dt = \sqrt{1+4t^2} \, dt.\nFor the numerator, we integrate f(x,y)=xf(x,y)=x. In terms of the parameter, f(r(t))=tf(\mathbf{r}(t)) = t. So, ∫Cx ds=∫01t1+4t2 dt\int_C x \, ds = \int_0^1 t \sqrt{1+4t^2} \, dt.\nFor the denominator, we integrate 11 to find the arc length LL. So, L=∫C1 ds=∫011⋅1+4t2 dtL = \int_C 1 \, ds = \int_0^1 1 \cdot \sqrt{1+4t^2} \, dt.\nCombining these, the average value is ∫01t1+4t2 dt∫011+4t2 dt\frac{\int_0^1 t \sqrt{1+4t^2} \, dt}{\int_0^1 \sqrt{1+4t^2} \, dt}.\n\nDistractor B is the expression for the average value if one incorrectly assumes ds=dtds=dt, ignoring the geometry of the curve. Distractor C is just the numerator, the integral of the function, not the average value. Distractor D is the average value of the function g(x,y)=yg(x,y)=y over the curve, not f(x,y)=xf(x,y)=x.

Question 9

Consider the curve CC that consists of the semicircle x2+y2=1x^2 + y^2 = 1 with y≥0y \geq 0, traversed counterclockwise, followed by the line segment from (−1,0)(-1, 0) to (1,0)(1, 0). If f(x,y)=xy+1f(x,y) = xy + 1, what is ∫Cf(x,y) ds\int_C f(x,y) \, ds?

  1. π+2\pi + 2 (correct answer)
  2. π+4\pi + 4
  3. 2π+22\pi + 2
  4. π2+2\frac{\pi}{2} + 2
Explanation: We split the integral into two parts. For C1C_1 (semicircle): parameterize as r⃗1(t)=(cos⁡t,sin⁡t)\vec{r}_1(t) = (\cos t, \sin t) for 0≤t≤π0 \leq t \leq \pi. Then f(cos⁡t,sin⁡t)=cos⁡tsin⁡t+1=sin⁡2t2+1f(\cos t, \sin t) = \cos t \sin t + 1 = \frac{\sin 2t}{2} + 1 and ∥r⃗1′(t)∥=sin⁡2t+cos⁡2t=1\|\vec{r}_1'(t)\| = \sqrt{\sin^2 t + \cos^2 t} = 1. So ∫C1f(x,y) ds=∫0π(sin⁡2t2+1) dt=[−cos⁡2t4+t]0π=[−14+π]−[−14+0]=π\int_{C_1} f(x,y) \, ds = \int_0^\pi (\frac{\sin 2t}{2} + 1) \, dt = [-\frac{\cos 2t}{4} + t]_0^\pi = [-\frac{1}{4} + \pi] - [-\frac{1}{4} + 0] = \pi. For C2C_2 (line segment from (−1,0)(-1,0) to (1,0)(1,0)): parameterize as r⃗2(t)=(−1+2t,0)\vec{r}_2(t) = (-1 + 2t, 0) for 0≤t≤10 \leq t \leq 1. Then f(−1+2t,0)=(−1+2t)⋅0+1=1f(-1 + 2t, 0) = (-1 + 2t) \cdot 0 + 1 = 1 and ∥r⃗2′(t)∥=2\|\vec{r}_2'(t)\| = 2. So ∫C2f(x,y) ds=∫011⋅2 dt=2\int_{C_2} f(x,y) \, ds = \int_0^1 1 \cdot 2 \, dt = 2. Therefore, ∫Cf(x,y) ds=π+2\int_C f(x,y) \, ds = \pi + 2.

Question 10

Let CC be the path consisting of two line segments: from (1,0)(1,0) to (0,1)(0,1) and then from (0,1)(0,1) to (−1,0)(-1,0). Compute the line integral ∫Cx2 ds\int_C x^2 \, ds.

  1. 00
  2. 23\frac{2}{3}
  3. 23\frac{\sqrt{2}}{3}
  4. 223\frac{2\sqrt{2}}{3} (correct answer)
Explanation: The path CC is composed of two segments, C1C_1 from (1,0)(1,0) to (0,1)(0,1) and C2C_2 from (0,1)(0,1) to (−1,0)(-1,0). We compute the integral over each segment and add the results.\n\nFor C1C_1, we can parameterize the path as r1(t)=⟨1−t,t⟩\mathbf{r}_1(t) = \langle 1-t, t \rangle for t∈[0,1]t \in [0,1]. Then r1′(t)=⟨−1,1⟩\mathbf{r}_1'(t) = \langle -1, 1 \rangle, and the arc length element is ds=∥r1′(t)∥dt=(−1)2+12 dt=2 dtds = \|\mathbf{r}_1'(t)\| dt = \sqrt{(-1)^2 + 1^2} \, dt = \sqrt{2} \, dt. The function is f(x,y)=x2f(x,y)=x^2, so on the path f(r1(t))=(1−t)2f(\mathbf{r}_1(t)) = (1-t)^2. The integral over C1C_1 is:\n∫C1x2 ds=∫01(1−t)22 dt=2[−(1−t)33]01=2(0−(−13))=23.\int_{C_1} x^2 \, ds = \int_0^1 (1-t)^2 \sqrt{2} \, dt = \sqrt{2} \left[ -\frac{(1-t)^3}{3} \right]_0^1 = \sqrt{2} \left(0 - \left(-\frac{1}{3}\right)\right) = \frac{\sqrt{2}}{3}.\n\nFor C2C_2, we can parameterize the path as r2(t)=⟨−t,1−t⟩\mathbf{r}_2(t) = \langle -t, 1-t \rangle for t∈[0,1]t \in [0,1]. Then r2′(t)=⟨−1,−1⟩\mathbf{r}_2'(t) = \langle -1, -1 \rangle, and ds=∥r2′(t)∥dt=(−1)2+(−1)2 dt=2 dtds = \|\mathbf{r}_2'(t)\| dt = \sqrt{(-1)^2 + (-1)^2} \, dt = \sqrt{2} \, dt. The function on the path is f(r2(t))=(−t)2=t2f(\mathbf{r}_2(t)) = (-t)^2 = t^2. The integral over C2C_2 is:\n∫C2x2 ds=∫01t22 dt=2[t33]01=23.\int_{C_2} x^2 \, ds = \int_0^1 t^2 \sqrt{2} \, dt = \sqrt{2} \left[ \frac{t^3}{3} \right]_0^1 = \frac{\sqrt{2}}{3}.\n\nThe total integral is the sum: ∫Cx2 ds=23+23=223\int_C x^2 \, ds = \frac{\sqrt{2}}{3} + \frac{\sqrt{2}}{3} = \frac{2\sqrt{2}}{3}.\n\nDistractor A is incorrect; the integrand is always non-negative. Distractor B is the result of forgetting the 2\sqrt{2} factor from dsds in both parts. Distractor C is the result of computing the integral over only one of the two segments.

Question 11

Let CC be the boundary of the rectangle with vertices (0,0)(0,0), (a,0)(a,0), (a,b)(a,b), and (0,b)(0,b), where a>0a > 0 and b>0b > 0. What is the value of the line integral ∫C(x+y) ds\int_C (x+y) \, ds?

  1. 00
  2. a(a+b)a(a+b)
  3. b(a+b)b(a+b)
  4. (a+b)2(a+b)^2 (correct answer)
Explanation: We can split the integral into four parts, one for each side of the rectangle.\nC1C_1: from (0,0)(0,0) to (a,0)(a,0). Parameterize as r(t)=⟨t,0⟩\mathbf{r}(t)=\langle t,0 \rangle for 0≤t≤a0 \le t \le a. ds=dtds=dt. ∫C1(x+y)ds=∫0a(t+0)dt=a22\int_{C_1} (x+y)ds = \int_0^a (t+0)dt = \frac{a^2}{2}.\nC2C_2: from (a,0)(a,0) to (a,b)(a,b). Parameterize as r(t)=⟨a,t⟩\mathbf{r}(t)=\langle a,t \rangle for 0≤t≤b0 \le t \le b. ds=dtds=dt. ∫C2(x+y)ds=∫0b(a+t)dt=ab+b22\int_{C_2} (x+y)ds = \int_0^b (a+t)dt = ab + \frac{b^2}{2}.\nC3C_3: from (a,b)(a,b) to (0,b)(0,b). Parameterize as r(t)=⟨a−t,b⟩\mathbf{r}(t)=\langle a-t,b \rangle for 0≤t≤a0 \le t \le a. ds=dtds=dt. ∫C3(x+y)ds=∫0a(a−t+b)dt=[at−t22+bt]0a=a2−a22+ab=a22+ab\int_{C_3} (x+y)ds = \int_0^a (a-t+b)dt = [at-\frac{t^2}{2}+bt]_0^a = a^2 - \frac{a^2}{2} + ab = \frac{a^2}{2} + ab.\nC4C_4: from (0,b)(0,b) to (0,0)(0,0). Parameterize as r(t)=⟨0,b−t⟩\mathbf{r}(t)=\langle 0,b-t \rangle for 0≤t≤b0 \le t \le b. ds=dtds=dt. ∫C4(x+y)ds=∫0b(0+b−t)dt=[bt−t22]0b=b2−b22=b22\int_{C_4} (x+y)ds = \int_0^b (0+b-t)dt = [bt-\frac{t^2}{2}]_0^b = b^2 - \frac{b^2}{2} = \frac{b^2}{2}.\nSumming the four parts: a22+(ab+b22)+(a22+ab)+b22=a2+b2+2ab=(a+b)2\frac{a^2}{2} + (ab + \frac{b^2}{2}) + (\frac{a^2}{2} + ab) + \frac{b^2}{2} = a^2 + b^2 + 2ab = (a+b)^2.\n\nAlternatively, we can compute ∫Cx ds\int_C x \, ds and ∫Cy ds\int_C y \, ds separately. ∫Cx ds=a22+∫0ba dt+∫0a(a−t) dt+0=a22+ab+a22=a2+ab=a(a+b)\int_C x \, ds = \frac{a^2}{2} + \int_0^b a \, dt + \int_0^a (a-t) \, dt + 0 = \frac{a^2}{2} + ab + \frac{a^2}{2} = a^2+ab = a(a+b). By symmetry (swapping roles of a,ba,b and x,yx,y), ∫Cy ds=b(a+b)\int_C y \, ds = b(a+b). The sum is a(a+b)+b(a+b)=(a+b)2a(a+b)+b(a+b) = (a+b)^2. Distractors B and C represent these partial results. Distractor A is the result of evaluating the vector line integral ∮C⟨x,y⟩⋅dr\oint_C \langle x,y \rangle \cdot d\mathbf{r}, which is 0 by Green's Theorem, a common point of confusion.

Question 12

Let CC be the cardioid given in polar coordinates by r=1+cos⁡(θ)r = 1 + \cos(\theta) for 0≤θ≤2π0 \le \theta \le 2\pi. Evaluate the line integral ∫Cx2+y2 ds\int_C \sqrt{x^2+y^2} \, ds.

  1. 00
  2. 163\frac{16}{3}
  3. 323\frac{32}{3} (correct answer)
  4. 2π2\pi
Explanation: The integrand is x2+y2\sqrt{x^2+y^2}, which in polar coordinates is simply rr. So we need to compute ∫Cr ds\int_C r \, ds. The arc length element in polar coordinates is ds=r2+(dr/dθ)2 dθds = \sqrt{r^2 + (dr/d\theta)^2} \, d\theta. Here, r=1+cos⁡(θ)r = 1+\cos(\theta), so dr/dθ=−sin⁡(θ)dr/d\theta = -\sin(\theta).\nds=(1+cos⁡θ)2+(−sin⁡θ)2 dθ=1+2cos⁡θ+cos⁡2θ+sin⁡2θ dθ=2+2cos⁡θ dθds = \sqrt{(1+\cos\theta)^2 + (-\sin\theta)^2} \, d\theta = \sqrt{1+2\cos\theta+\cos^2\theta+\sin^2\theta} \, d\theta = \sqrt{2+2\cos\theta} \, d\theta.\nUsing the half-angle identity 1+cos⁡θ=2cos⁡2(θ/2)1+\cos\theta = 2\cos^2(\theta/2), we get ds=4cos⁡2(θ/2) dθ=2∣cos⁡(θ/2)∣ dθds = \sqrt{4\cos^2(\theta/2)} \, d\theta = 2|\cos(\theta/2)| \, d\theta.\nThe integral becomes ∫02π(1+cos⁡θ)⋅2∣cos⁡(θ/2)∣ dθ\int_0^{2\pi} (1+\cos\theta) \cdot 2|\cos(\theta/2)| \, d\theta. We must split the integral because of the absolute value. For 0≤θ≤π0 \le \theta \le \pi, cos⁡(θ/2)≥0\cos(\theta/2) \ge 0. For π<θ≤2π\pi < \theta \le 2\pi, cos⁡(θ/2)<0\cos(\theta/2) < 0.\nAgain using 1+cos⁡θ=2cos⁡2(θ/2)1+\cos\theta = 2\cos^2(\theta/2), the integrand is (2cos⁡2(θ/2))⋅2∣cos⁡(θ/2)∣=4∣cos⁡3(θ/2)∣(2\cos^2(\theta/2)) \cdot 2|\cos(\theta/2)| = 4|\cos^3(\theta/2)|.\nIntegral =∫0π4cos⁡3(θ/2) dθ+∫π2π−4cos⁡3(θ/2) dθ= \int_0^{\pi} 4\cos^3(\theta/2) \, d\theta + \int_{\pi}^{2\pi} -4\cos^3(\theta/2) \, d\theta. \nLet's evaluate the first part: 4∫0πcos⁡3(θ/2) dθ=4∫0π(1−sin⁡2(θ/2))cos⁡(θ/2) dθ4\int_0^{\pi} \cos^3(\theta/2) \, d\theta = 4\int_0^{\pi} (1-\sin^2(\theta/2))\cos(\theta/2) \, d\theta. Let u=sin⁡(θ/2)u=\sin(\theta/2), du=12cos⁡(θ/2)dθdu = \frac{1}{2}\cos(\theta/2)d\theta. Limits become sin⁡(0)=0\sin(0)=0 to sin⁡(π/2)=1\sin(\pi/2)=1. The integral is 4∫01(1−u2)2du=8[u−u3/3]01=8(1−1/3)=16/34 \int_0^1 (1-u^2) 2du = 8[u-u^3/3]_0^1 = 8(1-1/3) = 16/3. By symmetry of the cardioid, the integral from π\pi to 2π2\pi will also be 16/316/3. So the total is 16/3+16/3=32/316/3+16/3 = 32/3.\n\nDistractor A, 00, arises from ignoring the absolute value and integrating cos⁡3(θ/2)\cos^3(\theta/2) over a full period, which gives zero. Distractor B, 16/316/3, is the result of integrating over only half the curve. Distractor D, 2π2\pi, is the result of computing ∫02π(1+cos⁡θ)dθ\int_0^{2\pi} (1+\cos\theta)d\theta, which ignores the dsds factor.

Question 13

Let CC be the square with vertices (1,0)(1,0), (0,1)(0,1), (−1,0)(-1,0), and (0,−1)(0,-1). What is the value of ∫C(x3+sin⁡y) ds\int_C (x^3 + \sin y) \, ds?

  1. 00 (correct answer)
  2. 11
  3. 424\sqrt{2}
  4. Cannot be determined without parameterization.
Explanation: The integral can be split by linearity: ∫C(x3+sin⁡y) ds=∫Cx3 ds+∫Csin⁡y ds\int_C (x^3 + \sin y) \, ds = \int_C x^3 \, ds + \int_C \sin y \, ds.\nWe can evaluate this integral using symmetry arguments. The curve CC (a square centered at the origin) is symmetric with respect to both the x-axis and the y-axis.\nConsider the integral ∫Cx3 ds\int_C x^3 \, ds. The function f(x,y)=x3f(x,y)=x^3 is an odd function of xx. For any point (x,y)(x,y) on the square, the point (−x,y)(-x,y) is also on the square (if y>0y>0, reflection across y-axis; if y<0y<0, reflection across y-axis). The value of dsds is the same for corresponding symmetric segments. Since f(−x,y)=(−x)3=−x3=−f(x,y)f(-x,y) = (-x)^3 = -x^3 = -f(x,y), the contributions to the integral from the right half of the square (where x>0x>0) and the left half (where x<0x<0) will cancel each other out. Thus, ∫Cx3 ds=0\int_C x^3 \, ds = 0.\nSimilarly, consider ∫Csin⁡y ds\int_C \sin y \, ds. The function g(x,y)=sin⁡yg(x,y)=\sin y is an odd function of yy. The square is symmetric with respect to the x-axis. For any point (x,y)(x,y), the point (x,−y)(x,-y) is also on the curve. Since g(x,−y)=sin⁡(−y)=−sin⁡y=−g(x,y)g(x,-y) = \sin(-y) = -\sin y = -g(x,y), the contributions from the upper half (y>0y>0) and lower half (y<0y<0) cancel out. Thus, ∫Csin⁡y ds=0\int_C \sin y \, ds = 0.\nThe total integral is 0+0=00+0=0.\n\nDistractor C, 424\sqrt{2}, is the length of the square (each side has length 2\sqrt{2}). This would be the answer for ∫C1 ds\int_C 1 \, ds. Distractors B and D tempt students who do not see the symmetry argument into thinking a complex calculation is required.