Multivariable Calculus Quiz: Line Integrals Vector Fields
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Line Integrals Vector FieldsQuestion 1 of 19

Let F⃗=⟨2xy3+cos⁡(x),3x2y2+ey⟩\vec{F} = \langle 2xy^3 + \cos(x), 3x^2y^2 + e^y \rangle. Calculate the work done by F⃗\vec{F} on a particle that moves from (0,0)(0,0) to (π,1)(\pi, 1) along the curve y=sin⁡2(x/2)y = \sin^2(x/2).

e−1e - 1
π2+e\pi^2 + e
1−e−π21 - e - \pi^2
π2+e−1\pi^2 + e - 1
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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Line Integrals Vector Fields

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

What this quiz covers

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

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

Let F⃗=⟨2xy3+cos⁡(x),3x2y2+ey⟩\vec{F} = \langle 2xy^3 + \cos(x), 3x^2y^2 + e^y \rangle. Calculate the work done by F⃗\vec{F} on a particle that moves from (0,0)(0,0) to (π,1)(\pi, 1) along the curve y=sin⁡2(x/2)y = \sin^2(x/2).

  1. e−1e - 1
  2. π2+e\pi^2 + e
  3. 1−e−π21 - e - \pi^2
  4. π2+e−1\pi^2 + e - 1 (correct answer)
Explanation: When you encounter a vector field and need to calculate work along a specific path, you have two main approaches: compute the line integral directly, or check if the field is conservative and use the fundamental theorem for line integrals. Let's check if F⃗=⟨2xy3+cos⁡(x),3x2y2+ey⟩\vec{F} = \langle 2xy^3 + \cos(x), 3x^2y^2 + e^y \rangle is conservative by testing if ∂P∂y=∂Q∂x\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}, where P=2xy3+cos⁡(x)P = 2xy^3 + \cos(x) and Q=3x2y2+eyQ = 3x^2y^2 + e^y. We have ∂P∂y=6xy2\frac{\partial P}{\partial y} = 6xy^2 and ∂Q∂x=6xy2\frac{\partial Q}{\partial x} = 6xy^2. Since these are equal, the field is conservative! This means the work depends only on the endpoints, not the specific path. For conservative fields, we need to find a potential function ff where ∇f=F⃗\nabla f = \vec{F}. Integrating ∂f∂x=2xy3+cos⁡(x)\frac{\partial f}{\partial x} = 2xy^3 + \cos(x) gives f=x2y3+sin⁡(x)+g(y)f = x^2y^3 + \sin(x) + g(y). Using ∂f∂y=3x2y2+ey\frac{\partial f}{\partial y} = 3x^2y^2 + e^y, we find g′(y)=eyg'(y) = e^y, so g(y)=ey+Cg(y) = e^y + C. Therefore, f(x,y)=x2y3+sin⁡(x)+eyf(x,y) = x^2y^3 + \sin(x) + e^y, and the work is: f(π,1)−f(0,0)=[π2(1)+sin⁡(π)+e1]−[0+sin⁡(0)+e0]=π2+0+e−0−0−1=π2+e−1f(\pi,1) - f(0,0) = [\pi^2(1) + \sin(\pi) + e^1] - [0 + \sin(0) + e^0] = \pi^2 + 0 + e - 0 - 0 - 1 = \pi^2 + e - 1 This matches answer D. Choices A, B, and C represent common arithmetic errors or incorrectly applying the fundamental theorem (like forgetting to subtract the initial point value or sign errors). Strategy tip: Always check for conservative vector fields first—it saves significant computation time compared to parametric line integrals.

Question 2

Consider the vector field F⃗(x,y)=(−yx2+y2,xx2+y2)\vec{F}(x,y) = \left(\frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2}\right) and two curves: C1C_1 is the upper semicircle of radius 2 centered at the origin from (2,0)(2,0) to (−2,0)(-2,0), and C2C_2 is the line segment from (2,0)(2,0) to (−2,0)(-2,0). Which statement about the line integrals ∫C1F⃗⋅dr⃗\int_{C_1} \vec{F} \cdot d\vec{r} and ∫C2F⃗⋅dr⃗\int_{C_2} \vec{F} \cdot d\vec{r} is correct?

  1. Both integrals equal π\pi since the vector field is conservative everywhere
  2. ∫C1F⃗⋅dr⃗=π\int_{C_1} \vec{F} \cdot d\vec{r} = \pi and ∫C2F⃗⋅dr⃗=0\int_{C_2} \vec{F} \cdot d\vec{r} = 0 (correct answer)
  3. Both integrals equal 0 since the curves have the same endpoints
  4. ∫C1F⃗⋅dr⃗=−π\int_{C_1} \vec{F} \cdot d\vec{r} = -\pi and ∫C2F⃗⋅dr⃗=0\int_{C_2} \vec{F} \cdot d\vec{r} = 0
Explanation: This vector field has the form F⃗=1r2(−sin⁡θ,cos⁡θ)\vec{F} = \frac{1}{r^2}(-\sin\theta, \cos\theta) in polar coordinates, which is not conservative everywhere because it has a singularity at the origin. For C1C_1 (upper semicircle), we can parameterize as r⃗(t)=(2cos⁡t,2sin⁡t)\vec{r}(t) = (2\cos t, 2\sin t) for t∈[0,π]t \in [0,\pi]. The calculation yields ∫C1F⃗⋅dr⃗=π\int_{C_1} \vec{F} \cdot d\vec{r} = \pi. For C2C_2 (line segment along x-axis), y=0y = 0, so F⃗=(−0/x2,x/x2)=(0,1/x)\vec{F} = (-0/x^2, x/x^2) = (0, 1/x). Since dr⃗=(dx,0)d\vec{r} = (dx, 0) along this path, F⃗⋅dr⃗=0\vec{F} \cdot d\vec{r} = 0, giving ∫C2F⃗⋅dr⃗=0\int_{C_2} \vec{F} \cdot d\vec{r} = 0. Choice A incorrectly assumes conservativity. Choice C ignores the singularity at origin. Choice D has the wrong sign for the semicircle integral.

Question 3

Let F⃗(x,y,z)=⟨y,−x,z2⟩\vec{F}(x,y,z) = \langle y, -x, z^2 \rangle. Calculate the work done by the vector field F⃗\vec{F} on a particle that moves along the helical path CC parameterized by r⃗(t)=⟨cos⁡(t),sin⁡(t),t⟩\vec{r}(t) = \langle \cos(t), \sin(t), t \rangle for 0≤t≤2π0 \le t \le 2\pi.

  1. 8π33−2π\frac{8\pi^3}{3} - 2\pi (correct answer)
  2. 2π−8π332\pi - \frac{8\pi^3}{3}
  3. −2π-2\pi
  4. 8π33\frac{8\pi^3}{3}
Explanation: The work done is given by the line integral W=∫CF⃗⋅dr⃗W = \int_C \vec{F} \cdot d\vec{r}. First, find the derivative of the parameterization: r⃗′(t)=⟨−sin⁡(t),cos⁡(t),1⟩\vec{r}'(t) = \langle -\sin(t), \cos(t), 1 \rangle. Next, evaluate the vector field along the path: F⃗(r⃗(t))=⟨sin⁡(t),−cos⁡(t),t2⟩\vec{F}(\vec{r}(t)) = \langle \sin(t), -\cos(t), t^2 \rangle. The dot product is F⃗(r⃗(t))⋅r⃗′(t)=(sin⁡t)(−sin⁡t)+(−cos⁡t)(cos⁡t)+(t2)(1)=−sin⁡2(t)−cos⁡2(t)+t2=−1+t2\vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) = (\sin t)(-\sin t) + (-\cos t)(\cos t) + (t^2)(1) = -\sin^2(t) - \cos^2(t) + t^2 = -1 + t^2. The integral for the work is then ∫02π(t2−1) dt=[t33−t]02π=(2π)33−2π=8π33−2π\int_0^{2\pi} (t^2 - 1) \, dt = \left[ \frac{t^3}{3} - t \right]_0^{2\pi} = \frac{(2\pi)^3}{3} - 2\pi = \frac{8\pi^3}{3} - 2\pi.

Question 4

Let F⃗(x,y,z)=⟨yzexyz+2x,xzexyz−3y2,xyexyz+1⟩\vec{F}(x,y,z) = \langle yz e^{xyz} + 2x, xz e^{xyz} - 3y^2, xy e^{xyz} + 1 \rangle. What is the work done by F⃗\vec{F} along the curve CC parameterized by r⃗(t)=⟨t,1−t2,t+1⟩\vec{r}(t) = \langle t, 1-t^2, t+1 \rangle for tt from 00 to 11?

  1. 3 (correct answer)
  2. -3
  3. 5
  4. 0
Explanation: Directly integrating ∫CF⃗⋅dr⃗\int_C \vec{F} \cdot d\vec{r} would be extremely complicated. The vector field F⃗\vec{F} is conservative because its domain is all of R3\mathbb{R}^3 and its curl is 0⃗\vec{0}. We can find a potential function ff such that ∇f=F⃗\nabla f = \vec{F}. By integration, we find f(x,y,z)=exyz+x2−y3+zf(x,y,z) = e^{xyz} + x^2 - y^3 + z. According to the Fundamental Theorem of Line Integrals, the work done is f(B)−f(A)f(B) - f(A), where AA and BB are the endpoints of the path. Here, A=r⃗(0)=⟨0,1,1⟩A = \vec{r}(0) = \langle 0, 1, 1 \rangle and B=r⃗(1)=⟨1,0,2⟩B = \vec{r}(1) = \langle 1, 0, 2 \rangle. Thus, the work is f(1,0,2)−f(0,1,1)=(e0+12−03+2)−(e0+02−13+1)=(1+1+2)−(1−1+1)=4−1=3f(1,0,2) - f(0,1,1) = (e^0 + 1^2 - 0^3 + 2) - (e^0 + 0^2 - 1^3 + 1) = (1+1+2) - (1-1+1) = 4 - 1 = 3.

Question 5

Evaluate the line integral ∮CF⃗⋅dr⃗\oint_C \vec{F} \cdot d\vec{r} where F⃗(x,y)=⟨−yx2+y2,xx2+y2⟩\vec{F}(x,y) = \left\langle \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} \right\rangle and CC is the boundary of the square with vertices (2,2)(2,2), (−2,2)(-2,2), (−2,−2)(-2,-2), and (2,−2)(2,-2), oriented counter-clockwise.

  1. 4π4\pi
  2. 00
  3. −2π-2\pi
  4. 2π2\pi (correct answer)
Explanation: When evaluating line integrals of vector fields, you should first check if the field is conservative or if Green's theorem applies. This particular vector field F⃗(x,y)=⟨−yx2+y2,xx2+y2⟩\vec{F}(x,y) = \left\langle \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} \right\rangle is special—it's the classic "circulation field" that's not defined at the origin. Since the origin (0,0)(0,0) lies inside the square, we can't directly apply Green's theorem (which requires the field to be defined everywhere inside the region). However, we can recognize this field geometrically: it represents pure counterclockwise rotation around the origin with magnitude 1x2+y2\frac{1}{x^2+y^2}. For any simple closed curve enclosing the origin once, this field produces a circulation of 2π2\pi. You can verify this by parameterizing any circle around the origin—the integral always yields 2π2\pi regardless of the radius. Since our square encloses the origin once and is oriented counterclockwise, the answer is 2π2\pi. Choice (A) 4π4\pi would result from incorrectly thinking the square somehow "wraps twice" around the origin. Choice (B) 00 represents the misconception that you can apply Green's theorem directly—if the field were conservative, this would be correct, but this field has non-zero curl everywhere except the origin. Choice (C) −2π-2\pi would occur if you mistakenly used clockwise orientation instead of the given counterclockwise orientation. Study tip: When you see the field ⟨−yx2+y2,xx2+y2⟩\left\langle \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} \right\rangle, immediately check if the origin is inside your curve. If yes, the answer is ±2π\pm 2\pi depending on orientation.

Question 6

Let DD be the region in the plane between the circles x2+y2=1x^2+y^2=1 and x2+y2=9x^2+y^2=9. Let F⃗(x,y)=⟨−y+sin⁡(x2),x+e−y2⟩\vec{F}(x,y) = \langle -y + \sin(x^2), x + e^{-y^2} \rangle. Calculate ∮CF⃗⋅dr⃗\oint_C \vec{F} \cdot d\vec{r}, where CC is the boundary of DD oriented such that the region DD is always on the left.

  1. 18π18\pi
  2. 8π8\pi
  3. 16π16\pi (correct answer)
  4. 20π20\pi
Explanation: When you encounter a line integral around the boundary of a region, immediately consider Green's theorem, which converts the line integral into a double integral over the region's interior. The boundary of region DD consists of two circles: the outer circle x2+y2=9x^2+y^2=9 (traversed counterclockwise) and the inner circle x2+y2=1x^2+y^2=1 (traversed clockwise to keep DD on the left). For F⃗(x,y)=⟨−y+sin⁡(x2),x+e−y2⟩\vec{F}(x,y) = \langle -y + \sin(x^2), x + e^{-y^2} \rangle, Green's theorem gives us: ∮CF⃗⋅dr⃗=∬D(∂Q∂x−∂P∂y)dA\oint_C \vec{F} \cdot d\vec{r} = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA where P=−y+sin⁡(x2)P = -y + \sin(x^2) and Q=x+e−y2Q = x + e^{-y^2}. Computing the partial derivatives: ∂Q∂x=1\frac{\partial Q}{\partial x} = 1 and ∂P∂y=−1\frac{\partial P}{\partial y} = -1. Therefore: ∂Q∂x−∂P∂y=1−(−1)=2\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 - (-1) = 2 The integral becomes: ∬D2 dA=2⋅Area(D)\iint_D 2 \, dA = 2 \cdot \text{Area}(D) The area of the annular region is: π(32)−π(12)=9π−π=8π\pi(3^2) - \pi(1^2) = 9\pi - \pi = 8\pi So the line integral equals 2⋅8π=16π2 \cdot 8\pi = 16\pi. Choice A (18π18\pi) likely comes from miscalculating the area as 9π9\pi. Choice B (8π8\pi) forgets the factor of 2 from the curl calculation. Choice D (20π20\pi) might result from using radius values incorrectly in the area formula. Study tip: Always check if Green's theorem applies before computing line integrals directly. When the curl is constant (as here), the problem reduces to finding the area of the region.

Question 7

Calculate the work done by the force field F⃗=⟨y2,2xy+1⟩\vec{F} = \langle y^2, 2xy+1 \rangle along the path CC, which is a segment of the parabola y=x2y=x^2 from (0,0)(0,0) to (2,4)(2,4). The problem notes that the path can be parameterized by either r⃗1(t)=⟨t,t2⟩\vec{r}_1(t) = \langle t, t^2 \rangle for t∈[0,2]t \in [0,2] or r⃗2(u)=⟨u,u⟩\vec{r}_2(u) = \langle \sqrt{u}, u \rangle for u∈[0,4]u \in [0,4].

  1. 34
  2. 32
  3. 36 (correct answer)
  4. 4
Explanation: When you encounter a work problem with a vector force field, you're calculating the line integral ∫CF⃗⋅dr⃗\int_C \vec{F} \cdot d\vec{r}. This measures how much the force contributes along your path of motion. Let's use the first parameterization: r⃗1(t)=⟨t,t2⟩\vec{r}_1(t) = \langle t, t^2 \rangle for t∈[0,2]t \in [0,2]. First, find dr⃗=r⃗′(t)dt=⟨1,2t⟩dtd\vec{r} = \vec{r}'(t)dt = \langle 1, 2t \rangle dt. Along this path, x=tx = t and y=t2y = t^2, so F⃗=⟨(t2)2,2t(t2)+1⟩=⟨t4,2t3+1⟩\vec{F} = \langle (t^2)^2, 2t(t^2)+1 \rangle = \langle t^4, 2t^3+1 \rangle. The dot product gives us: F⃗⋅dr⃗=⟨t4,2t3+1⟩⋅⟨1,2t⟩=t4+(2t3+1)(2t)=t4+4t4+2t=5t4+2t\vec{F} \cdot d\vec{r} = \langle t^4, 2t^3+1 \rangle \cdot \langle 1, 2t \rangle = t^4 + (2t^3+1)(2t) = t^4 + 4t^4 + 2t = 5t^4 + 2t Now integrate: ∫02(5t4+2t)dt=[t5+t2]02=32+4=36\int_0^2 (5t^4 + 2t) dt = \left[t^5 + t^2\right]_0^2 = 32 + 4 = 36 Answer (A) 34 likely comes from arithmetic errors in the integration. Answer (B) 32 results from forgetting the 2t2t term and only computing ∫025t4dt=32\int_0^2 5t^4 dt = 32. Answer (D) 4 might come from only evaluating t2t^2 at the bounds, missing the dominant t5t^5 term entirely. Study tip: Always double-check your parameterization and dot product calculation—these are the most common error sources. You can verify your answer using the alternate parameterization; both should give the same result since work depends only on the path, not how you parameterize it.

Question 8

Find the work done by the force field F⃗=⟨y2cos⁡(x),2ysin⁡(x)+ez,yez+3z2⟩\vec{F} = \langle y^2\cos(x), 2y\sin(x)+e^z, ye^z+3z^2 \rangle on a particle that moves along the path r⃗(t)=⟨πt2,sin⁡(πt/2),t⟩\vec{r}(t) = \langle \pi t^2, \sin(\pi t/2), t \rangle for t∈[1,2]t \in [1,2].

  1. e−7e - 7
  2. 7−e7 - e (correct answer)
  3. 88
  4. e+1e+1
Explanation: When you encounter a work problem with a vector field and parametric path, you're calculating W=∫CF⃗⋅dr⃗=∫abF⃗(r⃗(t))⋅r⃗′(t) dtW = \int_C \vec{F} \cdot d\vec{r} = \int_a^b \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) \, dt. This requires evaluating the force field along the path and taking the dot product with the velocity vector. First, find r⃗′(t)=⟨2πt,π2cos⁡(πt/2),1⟩\vec{r}'(t) = \langle 2\pi t, \frac{\pi}{2}\cos(\pi t/2), 1 \rangle. Next, substitute r⃗(t)\vec{r}(t) into F⃗\vec{F}: F⃗(r⃗(t))=⟨sin⁡2(πt/2)cos⁡(πt2),2sin⁡(πt/2)sin⁡(πt2)+et,sin⁡(πt/2)et+3t2⟩\vec{F}(\vec{r}(t)) = \langle \sin^2(\pi t/2)\cos(\pi t^2), 2\sin(\pi t/2)\sin(\pi t^2) + e^t, \sin(\pi t/2)e^t + 3t^2 \rangle. The dot product F⃗(r⃗(t))⋅r⃗′(t)\vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) gives: 2πtsin⁡2(πt/2)cos⁡(πt2)+πcos⁡(πt/2)sin⁡(πt/2)sin⁡(πt2)+πcos⁡(πt/2)et+sin⁡(πt/2)et+3t22\pi t \sin^2(\pi t/2)\cos(\pi t^2) + \pi\cos(\pi t/2)\sin(\pi t/2)\sin(\pi t^2) + \pi\cos(\pi t/2)e^t + \sin(\pi t/2)e^t + 3t^2 Using the identity sin⁡(πt/2)cos⁡(πt/2)=12sin⁡(πt)\sin(\pi t/2)\cos(\pi t/2) = \frac{1}{2}\sin(\pi t), this simplifies to: π2sin⁡(πt)sin⁡(πt2)+et(πcos⁡(πt/2)+sin⁡(πt/2))+3t2+2πtsin⁡2(πt/2)cos⁡(πt2)\frac{\pi}{2}\sin(\pi t)\sin(\pi t^2) + e^t(\pi\cos(\pi t/2) + \sin(\pi t/2)) + 3t^2 + 2\pi t\sin^2(\pi t/2)\cos(\pi t^2) Integrating from 1 to 2 yields 7−e7 - e. Choice A (e−7e - 7) has the wrong sign. Choice C (8) ignores the exponential terms entirely. Choice D (e+1e + 1) incorrectly handles the trigonometric integrations. Remember: work calculations require careful substitution and often involve complex trigonometric identities. Always double-check your parametric substitutions and use trigonometric identities to simplify before integrating.

Question 9

Let F⃗=⟨5y+x2,11x−sin⁡(y)⟩\vec{F} = \langle 5y + x^2, 11x - \sin(y) \rangle. Let CC be the boundary of a region DD in the xy-plane that has an area of 10. If CC is oriented counter-clockwise, what is the work done by F⃗\vec{F} around CC?

  1. 160
  2. 60 (correct answer)
  3. -60
  4. Cannot be determined from the given information.
Explanation: When you encounter a vector field and need to find work done around a closed curve, think about whether Green's theorem applies. This connects line integrals around closed curves to double integrals over the enclosed region. For work done by F⃗=⟨P,Q⟩\vec{F} = \langle P, Q \rangle around a counter-clockwise oriented curve CC, Green's theorem states: ∮CF⃗⋅dr⃗=∬D(∂Q∂x−∂P∂y)dA\oint_C \vec{F} \cdot d\vec{r} = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA Here, P=5y+x2P = 5y + x^2 and Q=11x−sin⁡(y)Q = 11x - \sin(y). Computing the partial derivatives:
  • ∂Q∂x=11\frac{\partial Q}{\partial x} = 11
  • ∂P∂y=5\frac{\partial P}{\partial y} = 5
So ∂Q∂x−∂P∂y=11−5=6\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 11 - 5 = 6 Since this difference is constant, the double integral becomes: ∬D6 dA=6∬DdA=6×Area(D)=6×10=60\iint_D 6 \, dA = 6 \iint_D dA = 6 \times \text{Area}(D) = 6 \times 10 = 60 The answer is B) 60. Choice A) 160 likely comes from miscalculating the partial derivatives or incorrectly using 11+5=1611 + 5 = 16 instead of the difference. Choice C) -60 represents the same calculation but with the wrong sign, perhaps from confusing clockwise versus counter-clockwise orientation. Choice D) Cannot be determined suggests missing that Green's theorem provides exactly the tools needed with the given information. Study tip: When you see work problems with closed curves and area given, immediately check if Green's theorem applies. The key insight is that constant differences in partial derivatives make the calculation straightforward regardless of the curve's specific shape.

Question 10

A conservative vector field F⃗\vec{F} exists throughout R3\mathbb{R}^3. The work done by F⃗\vec{F} along a path from point AA to point BB is 10. The work done by F⃗\vec{F} along a path from point CC to point BB is −3-3. What is the work done by F⃗\vec{F} along a path from point AA to point CC?

  1. 7
  2. 13 (correct answer)
  3. -7
  4. Cannot be determined from the given information.
Explanation: When you encounter a conservative vector field problem, remember that the fundamental property is path independence—the work done depends only on the starting and ending points, not the specific path taken. Since F⃗\vec{F} is conservative, you can use the additive property of work. Think of this like connecting destinations: if you know the "cost" to travel from A to B and from C to B, you can find the cost from A to C by considering how these paths relate. The work from A to C can be found by going from A to B, then from B to C. Since work from A to B is 10, and work from C to B is -3, the work from B to C must be +3 (reversing direction changes the sign). Therefore, work from A to C = work from A to B + work from B to C = 10 + 3 = 13. Choice A (7) represents the error of subtracting the magnitudes: 10 - 3 = 7. This ignores the proper vector addition needed when combining path segments. Choice C (-7) comes from incorrectly calculating 10 + (-3) - 10 = -7, showing confusion about how to combine the given information. Choice D suggests the answer cannot be determined, which would be true for non-conservative fields where work depends on the specific path, but conservative fields eliminate this path dependence. Remember: for conservative vector fields, work is path-independent, so you can always "connect the dots" between any three points using vector addition of work values, paying careful attention to direction.

Question 11

Let CC be the upper semi-circle of radius 1 centered at the origin, traversed from (1,0)(1,0) to (−1,0)(-1,0). Let W1W_1 be the work done by the vector field F⃗1=⟨−y,x⟩\vec{F}_1 = \langle -y, x \rangle along CC, and W2W_2 be the work done by F⃗2=⟨x,y⟩\vec{F}_2 = \langle x, y \rangle along CC. Which of the following statements is true?

  1. W1=W2W_1 = W_2
  2. W1<W2W_1 < W_2
  3. W1>W2W_1 > W_2 (correct answer)
  4. W1=−W2W_1 = -W_2 and W1≠0W_1 \ne 0
Explanation: When you encounter work problems with vector fields, the key insight is recognizing whether the field is conservative or has circulation properties. Work is calculated as W=∫CF⃗⋅dr⃗W = \int_C \vec{F} \cdot d\vec{r}, which measures how much the field "helps" the motion along the curve. For F⃗1=⟨−y,x⟩\vec{F}_1 = \langle -y, x \rangle, this field rotates counterclockwise around the origin. Since our upper semicircle also travels counterclockwise from (1,0)(1,0) to (−1,0)(-1,0), the field consistently points in the same general direction as our path. Using the standard parameterization r⃗(t)=⟨cos⁡t,sin⁡t⟩\vec{r}(t) = \langle \cos t, \sin t \rangle for t∈[0,π]t \in [0,\pi], we get W1=∫0ππdt=π>0W_1 = \int_0^{\pi} \pi dt = \pi > 0. For F⃗2=⟨x,y⟩\vec{F}_2 = \langle x, y \rangle, this is a radial field pointing outward from the origin. Since our semicircular path moves perpendicular to the radial direction at each point, the dot product F⃗2⋅dr⃗=0\vec{F}_2 \cdot d\vec{r} = 0 everywhere along the curve. Therefore, W2=0W_2 = 0. Since W1=π>0=W2W_1 = \pi > 0 = W_2, we have W1>W2W_1 > W_2, making C correct. Looking at the wrong answers: A suggests they're equal, but we showed W1=πW_1 = \pi and W2=0W_2 = 0. B claims W1<W2W_1 < W_2, which contradicts π>0\pi > 0. D suggests W1=−W2W_1 = -W_2, but this would require W2=−πW_2 = -\pi, not zero. Study tip: Always visualize vector fields geometrically. Rotational fields create circulation (nonzero work around closed or nearly closed paths), while radial fields typically produce zero work along circular paths due to perpendicularity.

Question 12

A particle moves in the force field F⃗(x,y)=⟨xy,x−y⟩\vec{F}(x,y) = \langle xy, x-y \rangle. Calculate the work done as the particle moves from (0,0)(0,0) to (2,4)(2,4) along the parabola y=x2y=x^2, and then from (2,4)(2,4) to (0,4)(0,4) along a horizontal line segment.

  1. −8-8
  2. 283\frac{28}{3}
  3. 43\frac{4}{3}
  4. −203-\frac{20}{3} (correct answer)
Explanation: When you encounter a work problem in a force field, you're calculating the line integral W=∫CF⃗⋅dr⃗W = \int_C \vec{F} \cdot d\vec{r}. This requires parameterizing each path segment and integrating along the entire journey. For the first segment from (0,0)(0,0) to (2,4)(2,4) along y=x2y = x^2, parameterize as r⃗1(t)=⟨t,t2⟩\vec{r}_1(t) = \langle t, t^2 \rangle where tt goes from 0 to 2. Then dr⃗1=⟨1,2t⟩dtd\vec{r}_1 = \langle 1, 2t \rangle dt. Along this path, F⃗=⟨t⋅t2,t−t2⟩=⟨t3,t−t2⟩\vec{F} = \langle t \cdot t^2, t - t^2 \rangle = \langle t^3, t - t^2 \rangle. The dot product gives: F⃗⋅dr⃗1=t3(1)+(t−t2)(2t)=t3+2t2−2t3=2t2−t3\vec{F} \cdot d\vec{r}_1 = t^3(1) + (t - t^2)(2t) = t^3 + 2t^2 - 2t^3 = 2t^2 - t^3. Integrating: ∫02(2t2−t3)dt=[2t33−t44]02=163−4=43\int_0^2 (2t^2 - t^3) dt = \left[\frac{2t^3}{3} - \frac{t^4}{4}\right]_0^2 = \frac{16}{3} - 4 = \frac{4}{3} For the second segment from (2,4)(2,4) to (0,4)(0,4) along the horizontal line, parameterize as r⃗2(t)=⟨2−t,4⟩\vec{r}_2(t) = \langle 2-t, 4 \rangle where tt goes from 0 to 2. Then dr⃗2=⟨−1,0⟩dtd\vec{r}_2 = \langle -1, 0 \rangle dt. Here F⃗=⟨(2−t)(4),(2−t)−4⟩=⟨8−4t,−2−t⟩\vec{F} = \langle (2-t)(4), (2-t) - 4 \rangle = \langle 8-4t, -2-t \rangle. The work is: ∫02(8−4t)(−1)dt=∫02(−8+4t)dt=[−8t+2t2]02=−16+8=−8\int_0^2 (8-4t)(-1) dt = \int_0^2 (-8+4t) dt = [-8t + 2t^2]_0^2 = -16 + 8 = -8 Total work: 43+(−8)=43−243=−203\frac{4}{3} + (-8) = \frac{4}{3} - \frac{24}{3} = -\frac{20}{3}, confirming answer D. Answer A gives only the second segment's work. Answer B represents the first segment multiplied by 7. Answer C gives only the first segment's work. Always break complex paths into manageable segments and carefully track signs in your parameterization—the direction matters for line integrals.

Question 13

A vector field F⃗(x,y)=(P(x,y),Q(x,y))\vec{F}(x,y) = (P(x,y), Q(x,y)) is given where ∂P∂y=∂Q∂x\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} throughout a simply connected domain. If the line integral ∫C1F⃗⋅dr⃗=5\int_{C_1} \vec{F} \cdot d\vec{r} = 5 for a curve C1C_1 from point AA to point BB, and ∫C2F⃗⋅dr⃗=−3\int_{C_2} \vec{F} \cdot d\vec{r} = -3 for a curve C2C_2 from point BB to point CC, what is ∫C3F⃗⋅dr⃗\int_{C_3} \vec{F} \cdot d\vec{r} where C3C_3 is any curve from point CC to point AA?

  1. −2-2 (correct answer)
  2. 22
  3. 88
  4. −8-8
Explanation: Since ∂P∂y=∂Q∂x\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}, the vector field is conservative in the simply connected domain. For conservative fields, line integrals are path-independent and ∮CF⃗⋅dr⃗=0\oint_C \vec{F} \cdot d\vec{r} = 0 for any closed curve. The path from AA to BB to CC and back to AA forms a closed curve, so ∫C1F⃗⋅dr⃗+∫C2F⃗⋅dr⃗+∫C3F⃗⋅dr⃗=0\int_{C_1} \vec{F} \cdot d\vec{r} + \int_{C_2} \vec{F} \cdot d\vec{r} + \int_{C_3} \vec{F} \cdot d\vec{r} = 0. Therefore, 5+(−3)+∫C3F⃗⋅dr⃗=05 + (-3) + \int_{C_3} \vec{F} \cdot d\vec{r} = 0, which gives ∫C3F⃗⋅dr⃗=−2\int_{C_3} \vec{F} \cdot d\vec{r} = -2. Choice B assumes the wrong sign. Choice C adds the values instead of using the closed curve property. Choice D incorrectly computes −(5+3)-(5+3).

Question 14

Consider the vector field F⃗(x,y,z)=(yz,xz,xy)\vec{F}(x,y,z) = (yz, xz, xy) and the curve CC which is the intersection of the cylinder x2+y2=1x^2 + y^2 = 1 and the plane z=x+yz = x + y, traversed once counterclockwise when viewed from above. To evaluate ∫CF⃗⋅dr⃗\int_C \vec{F} \cdot d\vec{r}, which parameterization and setup is most appropriate?

  1. r⃗(t)=(cos⁡t,sin⁡t,cos⁡t+sin⁡t)\vec{r}(t) = (\cos t, \sin t, \cos t + \sin t) for t∈[0,2π]t \in [0, 2\pi] with F⃗⋅dr⃗=(sin⁡t(cos⁡t+sin⁡t))(−sin⁡t)+(cos⁡t(cos⁡t+sin⁡t))(cos⁡t)+(cos⁡tsin⁡t)(cos⁡t−sin⁡t)\vec{F} \cdot d\vec{r} = (\sin t(\cos t + \sin t))(-\sin t) + (\cos t(\cos t + \sin t))(\cos t) + (\cos t \sin t)(\cos t - \sin t) (correct answer)
  2. r⃗(t)=(cos⁡t,sin⁡t,cos⁡t+sin⁡t)\vec{r}(t) = (\cos t, \sin t, \cos t + \sin t) for t∈[0,2π]t \in [0, 2\pi] with F⃗⋅dr⃗=(sin⁡t(cos⁡t+sin⁡t))(−sin⁡t)+(cos⁡t(cos⁡t+sin⁡t))(cos⁡t)+(cos⁡tsin⁡t)(cos⁡t+sin⁡t)\vec{F} \cdot d\vec{r} = (\sin t(\cos t + \sin t))(-\sin t) + (\cos t(\cos t + \sin t))(\cos t) + (\cos t \sin t)(\cos t + \sin t)
  3. r⃗(t)=(cos⁡t,sin⁡t,cos⁡t+sin⁡t)\vec{r}(t) = (\cos t, \sin t, \cos t + \sin t) for t∈[0,2π]t \in [0, 2\pi] with F⃗⋅dr⃗=(sin⁡t(cos⁡t+sin⁡t))(cos⁡t)+(cos⁡t(cos⁡t+sin⁡t))(sin⁡t)+(cos⁡tsin⁡t)(cos⁡t−sin⁡t)\vec{F} \cdot d\vec{r} = (\sin t(\cos t + \sin t))(\cos t) + (\cos t(\cos t + \sin t))(\sin t) + (\cos t \sin t)(\cos t - \sin t)
  4. r⃗(t)=(cos⁡t,sin⁡t,cos⁡t+sin⁡t)\vec{r}(t) = (\cos t, \sin t, \cos t + \sin t) for t∈[0,π]t \in [0, \pi] with F⃗⋅dr⃗=(sin⁡t(cos⁡t+sin⁡t))(−sin⁡t)+(cos⁡t(cos⁡t+sin⁡t))(cos⁡t)+(cos⁡tsin⁡t)(cos⁡t−sin⁡t)\vec{F} \cdot d\vec{r} = (\sin t(\cos t + \sin t))(-\sin t) + (\cos t(\cos t + \sin t))(\cos t) + (\cos t \sin t)(\cos t - \sin t)
Explanation: The curve is the intersection of cylinder x2+y2=1x^2 + y^2 = 1 and plane z=x+yz = x + y, so the parameterization r⃗(t)=(cos⁡t,sin⁡t,cos⁡t+sin⁡t)\vec{r}(t) = (\cos t, \sin t, \cos t + \sin t) for t∈[0,2π]t \in [0, 2\pi] is correct (one full revolution). We have r⃗′(t)=(−sin⁡t,cos⁡t,cos⁡t−sin⁡t)\vec{r}'(t) = (-\sin t, \cos t, \cos t - \sin t). The vector field evaluated on the curve is F⃗(r⃗(t))=(sin⁡t(cos⁡t+sin⁡t),cos⁡t(cos⁡t+sin⁡t),cos⁡tsin⁡t)\vec{F}(\vec{r}(t)) = (\sin t(\cos t + \sin t), \cos t(\cos t + \sin t), \cos t \sin t). The dot product is F⃗⋅r⃗′=(sin⁡t(cos⁡t+sin⁡t))(−sin⁡t)+(cos⁡t(cos⁡t+sin⁡t))(cos⁡t)+(cos⁡tsin⁡t)(cos⁡t−sin⁡t)\vec{F} \cdot \vec{r}' = (\sin t(\cos t + \sin t))(-\sin t) + (\cos t(\cos t + \sin t))(\cos t) + (\cos t \sin t)(\cos t - \sin t). Choice B incorrectly uses (cos⁡t+sin⁡t)(\cos t + \sin t) instead of (cos⁡t−sin⁡t)(\cos t - \sin t) in the third component. Choice C has incorrect signs in the first two components. Choice D uses the wrong interval [0,π][0,\pi] instead of [0,2π][0,2\pi].

Question 15

A vector field F⃗\vec{F} satisfies ∇×F⃗=0⃗\nabla \times \vec{F} = \vec{0} in a region RR, and ∫C1F⃗⋅dr⃗=8\int_{C_1} \vec{F} \cdot d\vec{r} = 8 where C1C_1 is a curve from point PP to point QQ lying entirely in RR. If C2C_2 is another curve from PP to QQ that passes through a small hole in region RR (where F⃗\vec{F} is undefined), which statement is necessarily true?

  1. ∫C2F⃗⋅dr⃗=8\int_{C_2} \vec{F} \cdot d\vec{r} = 8 because line integrals depend only on endpoints for curl-free fields
  2. ∫C2F⃗⋅dr⃗=8\int_{C_2} \vec{F} \cdot d\vec{r} = 8 because the hole doesn't affect the path independence property
  3. ∫C2F⃗⋅dr⃗\int_{C_2} \vec{F} \cdot d\vec{r} may not equal 8 because path independence requires the field to be defined on a simply connected domain (correct answer)
  4. ∫C2F⃗⋅dr⃗\int_{C_2} \vec{F} \cdot d\vec{r} is undefined because the curve passes through a region where F⃗\vec{F} is not defined
Explanation: While ∇×F⃗=0⃗\nabla \times \vec{F} = \vec{0} is a necessary condition for a conservative field, it is not sufficient unless the domain is simply connected. The presence of a hole makes the domain non-simply connected, which can break path independence even when the curl is zero. This is analogous to the vector field F⃗=(−y,x)x2+y2\vec{F} = \frac{(-y, x)}{x^2 + y^2} around the origin, which has zero curl everywhere it's defined but is not conservative due to the singularity. Choice A incorrectly assumes curl-free implies conservative everywhere. Choice B ignores the topological requirement for path independence. Choice D is wrong because C2C_2 passes through the hole but F⃗\vec{F} can still be evaluated along the parts of C2C_2 where it is defined.

Question 16

A vector field F⃗(x,y)=(yx2+y2+P(x,y),−xx2+y2+Q(x,y))\vec{F}(x,y) = \left(\frac{y}{x^2 + y^2} + P(x,y), \frac{-x}{x^2 + y^2} + Q(x,y)\right) is given, where P(x,y)P(x,y) and Q(x,y)Q(x,y) are smooth functions. If the line integral ∫CF⃗⋅dr⃗\int_C \vec{F} \cdot d\vec{r} around any simple closed curve CC not enclosing the origin equals zero, but ∫C0F⃗⋅dr⃗=2π\int_{C_0} \vec{F} \cdot d\vec{r} = 2\pi where C0C_0 is the unit circle centered at the origin oriented counterclockwise, what can be concluded about PP and QQ?

  1. ∂Q∂x−∂P∂y=0\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0 everywhere, and PP and QQ must be chosen to make F⃗\vec{F} conservative away from the origin
  2. ∂Q∂x−∂P∂y=0\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0 everywhere, but the field cannot be conservative due to the singularity at the origin (correct answer)
  3. PP and QQ must satisfy ∂Q∂x−∂P∂y=1x2+y2\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \frac{1}{x^2 + y^2} to compensate for the singular part
  4. PP and QQ must satisfy ∂Q∂x−∂P∂y=−2x2+y2\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = -\frac{2}{x^2 + y^2} to make the total curl zero except at the origin
Explanation: The vector field (yx2+y2,−xx2+y2)\left(\frac{y}{x^2 + y^2}, \frac{-x}{x^2 + y^2}\right) has the property that its line integral around the unit circle is −2π-2\pi (this is the standard angular field). Since the total field gives +2π+2\pi around C0C_0, the contribution from (P,Q)(P,Q) around C0C_0 must be 4π4\pi. However, the condition that integrals around curves not enclosing the origin are zero, combined with the fact that such curves can be continuously deformed to points without crossing the origin, implies that ∂Q∂x−∂P∂y=0\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0 in any simply connected region not containing the origin. But the field cannot be globally conservative because the line integral around C0C_0 is non-zero. Choice A incorrectly suggests the field can be conservative. Choices C and D propose incorrect relationships for the curl of PP and QQ.

Question 17

Consider the vector field F⃗(x,y)=(ex+y,xex+y+cos⁡y)\vec{F}(x,y) = (e^{x+y}, xe^{x+y} + \cos y) and the curve CC parameterized by r⃗(t)=(t,sin⁡t)\vec{r}(t) = (t, \sin t) for 0≤t≤π0 \leq t \leq \pi. To evaluate ∫CF⃗⋅dr⃗\int_C \vec{F} \cdot d\vec{r}, which expression correctly represents the integrand after substitution?

  1. et+sin⁡t+(tet+sin⁡t+cos⁡(sin⁡t))cos⁡te^{t + \sin t} + (te^{t + \sin t} + \cos(\sin t)) \cos t
  2. et+sin⁡t⋅1+(tet+sin⁡t+cos⁡(sin⁡t))⋅cos⁡te^{t + \sin t} \cdot 1 + (te^{t + \sin t} + \cos(\sin t)) \cdot \cos t
  3. et+sin⁡tcos⁡t+(tet+sin⁡t+cos⁡(sin⁡t))e^{t + \sin t} \cos t + (te^{t + \sin t} + \cos(\sin t))
  4. et+sin⁡t+tet+sin⁡tcos⁡t+cos⁡(sin⁡t)cos⁡te^{t + \sin t} + te^{t + \sin t} \cos t + \cos(\sin t) \cos t (correct answer)
Explanation: We have r⃗(t)=(t,sin⁡t)\vec{r}(t) = (t, \sin t), so r⃗′(t)=(1,cos⁡t)\vec{r}'(t) = (1, \cos t). The vector field along the curve is F⃗(r⃗(t))=F⃗(t,sin⁡t)=(et+sin⁡t,tet+sin⁡t+cos⁡(sin⁡t))\vec{F}(\vec{r}(t)) = \vec{F}(t, \sin t) = (e^{t + \sin t}, te^{t + \sin t} + \cos(\sin t)). The dot product is F⃗⋅r⃗′=(et+sin⁡t,tet+sin⁡t+cos⁡(sin⁡t))⋅(1,cos⁡t)=et+sin⁡t⋅1+(tet+sin⁡t+cos⁡(sin⁡t))⋅cos⁡t=et+sin⁡t+tet+sin⁡tcos⁡t+cos⁡(sin⁡t)cos⁡t\vec{F} \cdot \vec{r}' = (e^{t + \sin t}, te^{t + \sin t} + \cos(\sin t)) \cdot (1, \cos t) = e^{t + \sin t} \cdot 1 + (te^{t + \sin t} + \cos(\sin t)) \cdot \cos t = e^{t + \sin t} + te^{t + \sin t} \cos t + \cos(\sin t) \cos t. Choice A is missing the multiplication by the components of r⃗′(t)\vec{r}'(t). Choice B shows the structure but doesn't distribute the terms. Choice C has the components of the dot product in the wrong order.

Question 18

Let F⃗(x,y)=(2xy+y2,x2+2xy−1)\vec{F}(x,y) = (2xy + y^2, x^2 + 2xy - 1) be a vector field. If CC is the boundary of the region bounded by y=x2y = x^2 and y=2xy = 2x, oriented counterclockwise, and we want to evaluate ∫CF⃗⋅dr⃗\int_C \vec{F} \cdot d\vec{r} using Green's theorem, what is the value of ∂Q∂x−∂P∂y\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} where P=2xy+y2P = 2xy + y^2 and Q=x2+2xy−1Q = x^2 + 2xy - 1?

  1. 00 (the vector field is conservative) (correct answer)
  2. −2y-2y (indicating the field is not conservative)
  3. 2x−2y−2x=−2y2x - 2y - 2x = -2y
  4. 2x−4y−2x=−4y2x - 4y - 2x = -4y
Explanation: To apply Green's theorem, we need ∂Q∂x−∂P∂y\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}. We have P(x,y)=2xy+y2P(x,y) = 2xy + y^2 and Q(x,y)=x2+2xy−1Q(x,y) = x^2 + 2xy - 1. Computing the partial derivatives: ∂P∂y=2x+2y\frac{\partial P}{\partial y} = 2x + 2y and ∂Q∂x=2x+2y\frac{\partial Q}{\partial x} = 2x + 2y. Therefore, ∂Q∂x−∂P∂y=(2x+2y)−(2x+2y)=0\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = (2x + 2y) - (2x + 2y) = 0. This means the vector field is conservative, so the line integral around any closed curve is zero. Choices B, C, and D represent common computational errors in taking partial derivatives.

Question 19

Calculate the work done by the force field F⃗(x,y)=⟨y3,−x3⟩\vec{F}(x,y) = \langle y^3, -x^3 \rangle on a particle that moves along the circle x2+y2=4x^2 + y^2 = 4, oriented clockwise.

  1. 24π24\pi (correct answer)
  2. −24π-24\pi
  3. 16π16\pi
  4. 00
Explanation: We use Green's Theorem, ∮CP dx+Q dy=∬D(∂Q∂x−∂P∂y) dA\oint_C P\,dx + Q\,dy = \iint_D (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) \, dA. Here P=y3P=y^3 and Q=−x3Q=-x^3, so ∂Q∂x−∂P∂y=−3x2−3y2=−3(x2+y2)\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = -3x^2 - 3y^2 = -3(x^2+y^2). The region DD is the disk x2+y2≤4x^2+y^2 \le 4. Green's theorem assumes counter-clockwise orientation. Since the path is oriented clockwise, we must negate the result. Work W=−∬D−3(x2+y2) dA=3∬D(x2+y2) dAW = -\iint_D -3(x^2+y^2) \, dA = 3 \iint_D (x^2+y^2) \, dA. Converting to polar coordinates, x2+y2=r2x^2+y^2 = r^2 and dA=r dr dθdA = r\,dr\,d\theta. W=3∫02π∫02(r2)r dr dθ=3∫02π[r44]02 dθ=3∫02π4 dθ=12(2π)=24πW = 3 \int_0^{2\pi} \int_0^2 (r^2) r \, dr \, d\theta = 3 \int_0^{2\pi} \left[ \frac{r^4}{4} \right]_0^2 \, d\theta = 3 \int_0^{2\pi} 4 \, d\theta = 12(2\pi) = 24\pi.