Multivariable Calculus Quiz: Stokes Theorem
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Stokes TheoremQuestion 1 of 14

Let SS be the surface of the cylinder x2+y2=1x^2+y^2=1 for 0≤z≤20 \le z \le 2, with an outward-pointing normal. Note that SS does not include the top or bottom disks. For the vector field F⃗=⟨z,x,y⟩\vec{F} = \langle z, x, y \rangle, evaluate ∬S(∇×F⃗)⋅dS⃗\iint_S (\nabla \times \vec{F}) \cdot d\vec{S}.

00
2π2\pi
π\pi
−2π-2\pi
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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Stokes Theorem

Practice Stokes Theorem 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 Stokes Theorem, 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 SS be the surface of the cylinder x2+y2=1x^2+y^2=1 for 0≤z≤20 \le z \le 2, with an outward-pointing normal. Note that SS does not include the top or bottom disks. For the vector field F⃗=⟨z,x,y⟩\vec{F} = \langle z, x, y \rangle, evaluate ∬S(∇×F⃗)⋅dS⃗\iint_S (\nabla \times \vec{F}) \cdot d\vec{S}.

  1. 00 (correct answer)
  2. 2π2\pi
  3. π\pi
  4. −2π-2\pi
Explanation: By Stokes' theorem, the integral of the curl over the surface SS is equal to the line integral of F⃗\vec{F} over the boundary of SS. The boundary of this open cylinder consists of two curves: the circle C1C_1 at the top (z=2z=2) and the circle C2C_2 at the bottom (z=0z=0). The outward normal on the cylinder induces an orientation on the boundary curves. By the right-hand rule, the top curve C1C_1 is oriented counter-clockwise, while the bottom curve C2C_2 is oriented clockwise. Let's calculate the line integrals: For C1C_1: r⃗1(t)=⟨cos⁡t,sin⁡t,2⟩\vec{r}_1(t) = \langle \cos t, \sin t, 2 \rangle, dr⃗1=⟨−sin⁡t,cos⁡t,0⟩dtd\vec{r}_1 = \langle -\sin t, \cos t, 0 \rangle dt. F⃗\vec{F} on C1C_1 is ⟨2,cos⁡t,sin⁡t⟩\langle 2, \cos t, \sin t \rangle. The integral is ∫02π(−2sin⁡t+cos⁡2t)dt=π\int_0^{2\pi} (-2\sin t + \cos^2 t) dt = \pi. For C2C_2 (clockwise): We can parameterize it as r⃗2(t)=⟨cos⁡t,−sin⁡t,0⟩\vec{r}_2(t) = \langle \cos t, -\sin t, 0 \rangle, so dr⃗2=⟨−sin⁡t,−cos⁡t,0⟩dtd\vec{r}_2 = \langle -\sin t, -\cos t, 0 \rangle dt. F⃗\vec{F} on C2C_2 is ⟨0,cos⁡t,−sin⁡t⟩\langle 0, \cos t, -\sin t \rangle. The integral is ∫02π(0−cos⁡2t)dt=−π\int_0^{2\pi} (0 - \cos^2 t) dt = -\pi. The total value is the sum of the integrals over the boundary components: π+(−π)=0\pi + (-\pi) = 0. Alternatively, one can compute the surface integral directly. ∇×F⃗=⟨1,1,1⟩\nabla \times \vec{F} = \langle 1,1,1 \rangle. The outward normal to the cylinder is n⃗=⟨x,y,0⟩=⟨cos⁡θ,sin⁡θ,0⟩\vec{n} = \langle x, y, 0 \rangle = \langle \cos\theta, \sin\theta, 0 \rangle. The dot product is cos⁡θ+sin⁡θ\cos\theta + \sin\theta. Integrating ∫02∫02π(cos⁡θ+sin⁡θ)dθdz=0\int_0^2 \int_0^{2\pi} (\cos\theta + \sin\theta) d\theta dz = 0.

Question 2

Let SS be the portion of the parabolic cylinder z=y2z=y^2 that lies inside the cylinder x2+y2=1x^2+y^2=1. Let CC be the boundary of SS, oriented counter-clockwise when viewed from above. Evaluate ∮C⟨−y,x,z⟩⋅dr⃗\oint_C \langle -y, x, z \rangle \cdot d\vec{r}.

  1. 2π2\pi (correct answer)
  2. −2π-2\pi
  3. π\pi
  4. 00
Explanation: We use Stokes' theorem to convert the line integral into a surface integral over SS: ∮CF⃗⋅dr⃗=∬S(∇×F⃗)⋅dS⃗\oint_C \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot d\vec{S}. First, calculate the curl of F⃗=⟨−y,x,z⟩\vec{F} = \langle -y, x, z \rangle: ∇×F⃗=∣i^j^k^∂∂x∂∂y∂∂z−yxz∣=i^(0−0)−j^(0−0)+k^(1−(−1))=⟨0,0,2⟩\nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ -y & x & z \end{vmatrix} = \hat{i}(0-0) - \hat{j}(0-0) + \hat{k}(1 - (-1)) = \langle 0, 0, 2 \rangle The surface SS is defined by z=y2z=y^2 over the domain DD which is the unit disk x2+y2≤1x^2+y^2 \le 1 in the xyxy-plane. We can parameterize SS as r⃗(x,y)=⟨x,y,y2⟩\vec{r}(x,y) = \langle x, y, y^2 \rangle. The normal vector is r⃗x×r⃗y=⟨1,0,0⟩×⟨0,1,2y⟩=⟨0,−2y,1⟩\vec{r}_x \times \vec{r}_y = \langle 1,0,0 \rangle \times \langle 0,1,2y \rangle = \langle 0, -2y, 1 \rangle. The zz-component is positive, so this corresponds to the required upward orientation. The surface integral is: ∬S(∇×F⃗)⋅dS⃗=∬D⟨0,0,2⟩⋅⟨0,−2y,1⟩dA=∬D2dA\iint_S (\nabla \times \vec{F}) \cdot d\vec{S} = \iint_D \langle 0,0,2 \rangle \cdot \langle 0,-2y,1 \rangle dA = \iint_D 2 dA This is 2 times the area of the unit disk DD. The area of DD is π(12)=π\pi(1^2) = \pi. So the integral is 2π2\pi. Alternatively, we could use a simpler surface with the same boundary, namely the disk x2+y2≤1x^2+y^2 \le 1 itself. The integral would be ∬D⟨0,0,2⟩⋅⟨0,0,1⟩dA=∬D2dA=2π\iint_D \langle 0,0,2 \rangle \cdot \langle 0,0,1 \rangle dA = \iint_D 2 dA = 2\pi.

Question 3

Let F⃗=⟨2y,3x,5z⟩\vec{F} = \langle 2y, 3x, 5z \rangle. Let CC be the curve of intersection of the cylinder x2+y2=4x^2+y^2=4 and the plane x+z=3x+z=3, oriented counter-clockwise when viewed from above. Which of the following integrals correctly represents the value of ∮CF⃗⋅dr⃗\oint_C \vec{F} \cdot d\vec{r} according to Stokes' theorem?

  1. ∬S5z dS\iint_S 5z \, dS, where SS is the surface enclosed by CC.
  2. ∬D5 dA\iint_D 5 \, dA, where DD is the disk x2+y2≤4x^2+y^2 \le 4 in the xyxy-plane.
  3. ∬D(1+5(3−x)) dA\iint_D (1+5(3-x)) \, dA, where DD is the disk x2+y2≤4x^2+y^2 \le 4 in the xyxy-plane.
  4. ∬D1 dA\iint_D 1 \, dA, where DD is the disk x2+y2≤4x^2+y^2 \le 4 in the xyxy-plane. (correct answer)
Explanation: When you encounter a line integral with Stokes' theorem, you're converting a circulation integral around a curve into a surface integral of the curl over any surface bounded by that curve. Stokes' theorem states that ∮CF⃗⋅dr⃗=∬S(∇×F⃗)⋅n^ dS\oint_C \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot \hat{n} \, dS, where SS is any surface bounded by curve CC. First, calculate the curl of F⃗=⟨2y,3x,5z⟩\vec{F} = \langle 2y, 3x, 5z \rangle: ∇×F⃗=∣i^j^k^∂∂x∂∂y∂∂z2y3x5z∣=⟨0,0,3−2⟩=⟨0,0,1⟩\nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 2y & 3x & 5z \end{vmatrix} = \langle 0, 0, 3-2 \rangle = \langle 0, 0, 1 \rangle The curve CC is where the cylinder x2+y2=4x^2+y^2=4 meets the plane x+z=3x+z=3. The smartest choice for surface SS is the flat disk D:x2+y2≤4D: x^2+y^2 \leq 4 in the xyxy-plane (at z=0z=0), since it's bounded by the projection of CC. This disk has upward normal n^=⟨0,0,1⟩\hat{n} = \langle 0, 0, 1 \rangle (matching the counter-clockwise orientation). Therefore: ∬S(∇×F⃗)⋅n^ dS=∬D⟨0,0,1⟩⋅⟨0,0,1⟩ dA=∬D1 dA\iint_S (\nabla \times \vec{F}) \cdot \hat{n} \, dS = \iint_D \langle 0, 0, 1 \rangle \cdot \langle 0, 0, 1 \rangle \, dA = \iint_D 1 \, dA Option A uses 5z5z instead of the curl. Option B incorrectly uses just the zz-component value 55 instead of the dot product. Option C mistakenly tries to account for the actual surface in 3D space rather than using the simpler projected disk. Remember: Stokes' theorem lets you choose any convenient surface bounded by your curve—often the flat projection gives the simplest calculation.

Question 4

Let SS be the portion of the cone z=x2+y2z = \sqrt{x^2+y^2} that lies between the planes z=1z=1 and z=2z=2, with an upward-pointing normal. Let F⃗\vec{F} be a vector field with ∇×F⃗=⟨0,0,2⟩\nabla \times \vec{F} = \langle 0, 0, 2 \rangle. Evaluate ∬S(∇×F⃗)⋅dS⃗\iint_S (\nabla \times \vec{F}) \cdot d\vec{S}.

  1. 3π3\pi
  2. 4π4\pi
  3. 6π6\pi (correct answer)
  4. 2π2\pi
Explanation: When you encounter a surface integral involving the curl of a vector field, immediately consider whether Stokes' theorem might simplify your work. This theorem relates the flux of curl through a surface to a line integral around the boundary. Since ∇×F⃗=⟨0,0,2⟩\nabla \times \vec{F} = \langle 0, 0, 2 \rangle is constant, you can factor it out of the surface integral: ∬S(∇×F⃗)⋅dS⃗=∬S⟨0,0,2⟩⋅dS⃗=2∬Sk^⋅dS⃗\iint_S (\nabla \times \vec{F}) \cdot d\vec{S} = \iint_S \langle 0, 0, 2 \rangle \cdot d\vec{S} = 2 \iint_S \hat{k} \cdot d\vec{S}. The remaining integral ∬Sk^⋅dS⃗\iint_S \hat{k} \cdot d\vec{S} represents the flux of the unit vector k^\hat{k} through surface SS. Since the cone has an upward-pointing normal and opens upward, this flux equals the area of the projection of SS onto the xyxy-plane. The surface SS projects to an annulus (ring) with inner radius 1 and outer radius 2, so the projected area is π(22)−π(12)=3π\pi(2^2) - \pi(1^2) = 3\pi. Therefore, ∬S(∇×F⃗)⋅dS⃗=2⋅3π=6π\iint_S (\nabla \times \vec{F}) \cdot d\vec{S} = 2 \cdot 3\pi = 6\pi, which is answer C. Answer A (3π3\pi) represents just the projected area without the factor of 2 from the curl's zz-component. Answer B (4π4\pi) might come from incorrectly using the full disk of radius 2 instead of the annulus. Answer D (2π2\pi) could result from using only the outer boundary area π(22)−π(2)2\pi(2^2) - \pi(\sqrt{2})^2 or other geometric errors. Remember: when curl is constant, factor it out and focus on the geometric flux calculation through the surface.

Question 5

Let F⃗(x,y,z)=⟨3y,−2x,yz⟩\vec{F}(x,y,z) = \langle 3y, -2x, yz \rangle. Let CC be the circle x2+y2=1x^2 + y^2 = 1 in the plane z=2z=2, oriented counter-clockwise when viewed from above. Evaluate the line integral ∮CF⃗⋅dr⃗\oint_C \vec{F} \cdot d\vec{r}.

  1. −5π-5\pi (correct answer)
  2. 5π5\pi
  3. −3π-3\pi
  4. 00
Explanation: By Stokes' theorem, ∮CF⃗⋅dr⃗=∬S(∇×F⃗)⋅dS⃗\oint_C \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot d\vec{S}, where SS is any surface with boundary CC. Let's choose SS to be the disk DD defined by x2+y2≤1x^2 + y^2 \le 1 in the plane z=2z=2. The counter-clockwise orientation of CC corresponds to an upward-pointing normal vector for DD, which is n⃗=k^=⟨0,0,1⟩\vec{n} = \hat{k} = \langle 0,0,1 \rangle. First, compute the curl of F⃗\vec{F}: ∇×F⃗=∣i^j^k^∂∂x∂∂y∂∂z3y−2xyz∣=i^(z−0)−j^(0−0)+k^(−2−3)=⟨z,0,−5⟩\nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 3y & -2x & yz \end{vmatrix} = \hat{i}(z-0) - \hat{j}(0-0) + \hat{k}(-2-3) = \langle z, 0, -5 \rangle On the surface SS, z=2z=2, so ∇×F⃗=⟨2,0,−5⟩\nabla \times \vec{F} = \langle 2, 0, -5 \rangle. The surface integral is: ∬D⟨2,0,−5⟩⋅⟨0,0,1⟩dA=∬D−5dA=−5⋅(Area of D)\iint_D \langle 2, 0, -5 \rangle \cdot \langle 0,0,1 \rangle dA = \iint_D -5 dA = -5 \cdot (\text{Area of D}) The area of the unit disk DD is π(12)=π\pi(1^2) = \pi. Thus, the integral is −5π-5\pi.

Question 6

Let S1S_1 be the disk x2+y2≤1x^2 + y^2 \le 1 in the xyxy-plane with normal n⃗1=k^\vec{n}_1 = \hat{k}. Let S2S_2 be the surface of the paraboloid z=1−x2−y2z = 1 - x^2 - y^2 for z≥0z \ge 0 with normal n⃗2\vec{n}_2 pointing upwards. Let F⃗\vec{F} be a continuously differentiable vector field. What is the value of the expression ∬S1(∇×F⃗)⋅dS⃗−∬S2(∇×F⃗)⋅dS⃗\iint_{S_1} (\nabla \times \vec{F}) \cdot d\vec{S} - \iint_{S_2} (\nabla \times \vec{F}) \cdot d\vec{S}?

  1. The negative of the flux through S2S_2
  2. Twice the flux through S1S_1
  3. 0 (correct answer)
  4. It depends on the specific vector field F⃗\vec{F}.
Explanation: This problem tests your understanding of Stokes' theorem and how it applies to closed surfaces formed by combining multiple boundary pieces. When you see surface integrals of curl combined with different surfaces that share the same boundary, think about how Stokes' theorem relates line integrals around closed curves to surface integrals of curl. Here, both surfaces S1S_1 (the disk) and S2S_2 (the paraboloid) have the same boundary: the unit circle x2+y2=1x^2 + y^2 = 1 in the xyxy-plane. By Stokes' theorem, ∬S1(∇×F⃗)⋅dS⃗=∮CF⃗⋅dr⃗\iint_{S_1} (\nabla \times \vec{F}) \cdot d\vec{S} = \oint_C \vec{F} \cdot d\vec{r} and ∬S2(∇×F⃗)⋅dS⃗=∮CF⃗⋅dr⃗\iint_{S_2} (\nabla \times \vec{F}) \cdot d\vec{S} = \oint_C \vec{F} \cdot d\vec{r}, where CC is the unit circle traversed counterclockwise when viewed from above (consistent with both upward normal orientations). Since both surface integrals equal the same line integral around their shared boundary, we have: ∬S1(∇×F⃗)⋅dS⃗=∬S2(∇×F⃗)⋅dS⃗\iint_{S_1} (\nabla \times \vec{F}) \cdot d\vec{S} = \iint_{S_2} (\nabla \times \vec{F}) \cdot d\vec{S} Therefore: ∬S1(∇×F⃗)⋅dS⃗−∬S2(∇×F⃗)⋅dS⃗=0\iint_{S_1} (\nabla \times \vec{F}) \cdot d\vec{S} - \iint_{S_2} (\nabla \times \vec{F}) \cdot d\vec{S} = 0 Choice A incorrectly suggests the result relates to flux through S2S_2. Choice B incorrectly assumes some doubling relationship. Choice D is wrong because this result holds for any continuously differentiable vector field F⃗\vec{F}. Study tip: When surfaces share the same boundary curve and have compatible normal orientations, their curl integrals are always equal by Stokes' theorem, regardless of the specific vector field involved.

Question 7

Let CC be the curve of intersection of the cylinder x2+y2=1x^2+y^2=1 and the plane z=y+2z=y+2. Evaluate ∮CF⃗⋅dr⃗\oint_C \vec{F} \cdot d\vec{r} for the vector field F⃗=⟨exsin⁡y,excos⁡y,z2⟩\vec{F} = \langle e^x \sin y, e^x \cos y, z^2 \rangle.

  1. 11
  2. π\pi
  3. 2π2\pi
  4. 00 (correct answer)
Explanation: When you encounter a line integral around a closed curve, your first instinct should be to check if you can apply Green's theorem or look for conservative vector fields. This problem tests your ability to recognize when a vector field is conservative and apply the fundamental theorem for line integrals. Let's examine F⃗=⟨exsin⁡y,excos⁡y,z2⟩\vec{F} = \langle e^x \sin y, e^x \cos y, z^2 \rangle. For a conservative field in 3D, we need ∇×F⃗=0⃗\nabla \times \vec{F} = \vec{0}. Computing the curl: ∂∂y(z2)−∂∂z(excos⁡y)=0−0=0\frac{\partial}{\partial y}(z^2) - \frac{\partial}{\partial z}(e^x \cos y) = 0 - 0 = 0, ∂∂z(exsin⁡y)−∂∂x(z2)=0−0=0\frac{\partial}{\partial z}(e^x \sin y) - \frac{\partial}{\partial x}(z^2) = 0 - 0 = 0, and ∂∂x(excos⁡y)−∂∂y(exsin⁡y)=excos⁡y−excos⁡y=0\frac{\partial}{\partial x}(e^x \cos y) - \frac{\partial}{\partial y}(e^x \sin y) = e^x \cos y - e^x \cos y = 0. Since the curl is zero, F⃗\vec{F} is conservative. By the fundamental theorem for line integrals, any line integral of a conservative vector field around a closed curve equals zero, regardless of the specific curve. This makes the answer D) 0. Choice A) 1 might tempt students who attempt direct parameterization without recognizing the conservative nature. Choice B) π could result from incorrectly applying Green's theorem with computational errors. Choice C) 2π might come from confusing this with circulation calculations or incorrectly relating it to the cylinder's circumference. Strategy tip: Always check if a vector field is conservative first—it can save significant computation time. If ∇×F⃗=0⃗\nabla \times \vec{F} = \vec{0}, then any closed curve integral immediately equals zero.

Question 8

Let HH be the hemisphere x2+y2+z2=9x^2+y^2+z^2=9 with z≥0z \ge 0, and let DD be the disk x2+y2≤9x^2+y^2 \le 9 in the xyxy-plane. Let SS be the closed surface formed by the union of HH and DD, oriented with the outward-pointing normal vector. For any continuously differentiable vector field F⃗\vec{F}, which of the following expressions is always true?

  1. ∬H(∇×F⃗)⋅dS⃗=∬D(∇×F⃗)⋅dS⃗\iint_H (\nabla \times \vec{F}) \cdot d\vec{S} = \iint_D (\nabla \times \vec{F}) \cdot d\vec{S}
  2. ∬H(∇×F⃗)⋅dS⃗=−∬D(∇×F⃗)⋅dS⃗\iint_H (\nabla \times \vec{F}) \cdot d\vec{S} = -\iint_D (\nabla \times \vec{F}) \cdot d\vec{S} (correct answer)
  3. ∬H(∇×F⃗)⋅dS⃗=0\iint_H (\nabla \times \vec{F}) \cdot d\vec{S} = 0
  4. ∬D(∇×F⃗)⋅dS⃗=0\iint_D (\nabla \times \vec{F}) \cdot d\vec{S} = 0
Explanation: The problem considers the flux of the curl of F⃗\vec{F} through the closed surface S=H∪DS = H \cup D. By the Divergence Theorem, the total flux of any curl field through any closed surface is zero. This is because the divergence of a curl is always zero: ∇⋅(∇×F⃗)=0\nabla \cdot (\nabla \times \vec{F}) = 0. Therefore: ∯S(∇×F⃗)⋅dS⃗=∭V∇⋅(∇×F⃗)dV=∭V0dV=0\oiint_S (\nabla \times \vec{F}) \cdot d\vec{S} = \iiint_V \nabla \cdot (\nabla \times \vec{F}) dV = \iiint_V 0 dV = 0 The total flux over the closed surface SS is the sum of the fluxes over its parts, HH and DD: ∯S(∇×F⃗)⋅dS⃗=∬H(∇×F⃗)⋅dS⃗+∬D(∇×F⃗)⋅dS⃗=0\oiint_S (\nabla \times \vec{F}) \cdot d\vec{S} = \iint_H (\nabla \times \vec{F}) \cdot d\vec{S} + \iint_D (\nabla \times \vec{F}) \cdot d\vec{S} = 0 From this equation, we can conclude that ∬H(∇×F⃗)⋅dS⃗=−∬D(∇×F⃗)⋅dS⃗\iint_H (\nabla \times \vec{F}) \cdot d\vec{S} = -\iint_D (\nabla \times \vec{F}) \cdot d\vec{S}. This relationship holds for any continuously differentiable vector field F⃗\vec{F}. The other options are not always true.

Question 9

Let SS be the portion of the sphere x2+y2+z2=4x^2 + y^2 + z^2 = 4 for which z≥3z \ge \sqrt{3}, oriented with an upward-pointing normal vector. Let F⃗(x,y,z)=⟨−y,x,zx3y2⟩\vec{F}(x,y,z) = \langle -y, x, zx^3y^2 \rangle. Evaluate ∬S(∇×F⃗)⋅dS⃗\iint_S (\nabla \times \vec{F}) \cdot d\vec{S}.

  1. 2π2\pi (correct answer)
  2. −2π-2\pi
  3. π\pi
  4. 00
Explanation: By Stokes' theorem, this surface integral is equal to the line integral of F⃗\vec{F} over the boundary curve CC of SS. The boundary CC is the intersection of the sphere x2+y2+z2=4x^2 + y^2 + z^2 = 4 and the plane z=3z = \sqrt{3}. Substituting z=3z=\sqrt{3} into the sphere equation gives x2+y2+(3)2=4x^2 + y^2 + (\sqrt{3})^2 = 4, which simplifies to x2+y2=1x^2+y^2=1. So, CC is a circle of radius 1 in the plane z=3z=\sqrt{3}. The upward orientation of SS means the boundary CC must be traversed counter-clockwise when viewed from above. We can parameterize CC as r⃗(t)=⟨cos⁡t,sin⁡t,3⟩\vec{r}(t) = \langle \cos t, \sin t, \sqrt{3} \rangle for t∈[0,2π]t \in [0, 2\pi]. Then dr⃗=⟨−sin⁡t,cos⁡t,0⟩dtd\vec{r} = \langle -\sin t, \cos t, 0 \rangle dt. On the curve CC, the vector field is F⃗(r⃗(t))=⟨−sin⁡t,cos⁡t,3(cos⁡t)3(sin⁡t)2⟩\vec{F}(\vec{r}(t)) = \langle -\sin t, \cos t, \sqrt{3}(\cos t)^3(\sin t)^2 \rangle. The line integral is: ∮CF⃗⋅dr⃗=∫02π⟨−sin⁡t,cos⁡t,… ⟩⋅⟨−sin⁡t,cos⁡t,0⟩dt\oint_C \vec{F} \cdot d\vec{r} = \int_0^{2\pi} \langle -\sin t, \cos t, \dots \rangle \cdot \langle -\sin t, \cos t, 0 \rangle dt =∫02π(sin⁡2t+cos⁡2t)dt=∫02π1dt=2π= \int_0^{2\pi} (\sin^2 t + \cos^2 t) dt = \int_0^{2\pi} 1 dt = 2\pi The direct computation of the surface integral would be much more difficult.

Question 10

Consider the vector field F=(x2−z,y2+x,z2−y)\mathbf{F} = (x^2 - z, y^2 + x, z^2 - y) and the curve CC consisting of the intersection of x2+y2+z2=9x^2 + y^2 + z^2 = 9 and x+y+z=3x + y + z = 3, oriented counterclockwise when viewed from the point (10,10,10)(10, 10, 10). To apply Stokes' theorem efficiently, which surface choice would minimize computational complexity?

  1. The spherical cap x2+y2+z2=9x^2 + y^2 + z^2 = 9 with x+y+z≥3x + y + z \geq 3, using spherical coordinates
  2. The planar disk x+y+z=3x + y + z = 3 bounded by CC, using the parameterization r(u,v)=(u,v,3−u−v)\mathbf{r}(u,v) = (u, v, 3-u-v) (correct answer)
  3. The conical surface connecting CC to the origin, parameterized in cylindrical coordinates
  4. The cylindrical surface x2+y2=constantx^2 + y^2 = \text{constant} connecting CC to the xyxy-plane, with variable height
Explanation: For Stokes' theorem ∮CF⋅dr=∬S(∇×F)⋅n dS\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS, we need to compute ∇×F=(−1−2z,−1−2x,1−2y)\nabla \times \mathbf{F} = (-1-2z, -1-2x, 1-2y). The planar surface x+y+z=3x + y + z = 3 has constant normal n=(1,1,1)3\mathbf{n} = \frac{(1,1,1)}{\sqrt{3}} and simple surface element dS=3 du dvdS = \sqrt{3} \, du \, dv. The dot product becomes 13[(−1−2z)+(−1−2x)+(1−2y)]=−1−2(x+y+z)3=−1−63=−73\frac{1}{\sqrt{3}}[(-1-2z) + (-1-2x) + (1-2y)] = \frac{-1-2(x+y+z)}{\sqrt{3}} = \frac{-1-6}{\sqrt{3}} = \frac{-7}{\sqrt{3}}, which is constant over the surface. This makes integration trivial compared to the curved surfaces in other options, which would require more complex parameterizations and variable surface elements.

Question 11

Consider two surfaces S1S_1 and S2S_2 with the same boundary curve CC, where S1S_1 is a flat disk and S2S_2 is a portion of a sphere. Both surfaces are oriented consistently with the same orientation of CC. For a vector field F\mathbf{F} such that ∇×F\nabla \times \mathbf{F} is not constant, under what condition will ∬S1(∇×F)⋅n1 dS1=∬S2(∇×F)⋅n2 dS2\iint_{S_1} (\nabla \times \mathbf{F}) \cdot \mathbf{n}_1 \, dS_1 = \iint_{S_2} (\nabla \times \mathbf{F}) \cdot \mathbf{n}_2 \, dS_2?

  1. When F\mathbf{F} is conservative throughout the region bounded by S1S_1 and S2S_2
  2. When ∇×F\nabla \times \mathbf{F} has zero divergence in the region between the surfaces
  3. Always, since both integrals equal ∮CF⋅dr\oint_C \mathbf{F} \cdot d\mathbf{r} by Stokes' theorem (correct answer)
  4. When the surfaces have the same area and curvature properties
Explanation: By Stokes' theorem, for any surface SS with boundary CC (oriented consistently), we have ∬S(∇×F)⋅n dS=∮CF⋅dr\iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS = \oint_C \mathbf{F} \cdot d\mathbf{r}. Since the line integral depends only on the vector field F\mathbf{F} and the boundary curve CC (with its orientation), and not on the choice of surface SS, both surface integrals must be equal regardless of the specific properties of F\mathbf{F} or ∇×F\nabla \times \mathbf{F}. This is a fundamental consequence of Stokes' theorem. Option A is wrong because F\mathbf{F} being conservative would make ∇×F=0\nabla \times \mathbf{F} = 0 and both integrals would be zero, but equality holds even for non-conservative fields. Option B relates to divergence theorem, not Stokes'. Option D is irrelevant since Stokes' theorem doesn't depend on geometric properties of the surface beyond having the correct boundary.

Question 12

Consider the vector field F=(yzcos⁡(xyz),xzcos⁡(xyz),xycos⁡(xyz))\mathbf{F} = (yz \cos(xyz), xz \cos(xyz), xy \cos(xyz)) and the triangular curve CC with vertices at (1,0,0)(1,0,0), (0,1,0)(0,1,0), and (0,0,1)(0,0,1), traversed in that order. To evaluate ∮CF⋅dr\oint_C \mathbf{F} \cdot d\mathbf{r} using Stokes' theorem, which observation would most significantly simplify the calculation?

  1. The vector field F\mathbf{F} is the gradient of sin⁡(xyz)\sin(xyz), making the line integral path-independent (correct answer)
  2. On the plane x+y+z=1x + y + z = 1 containing the triangle, we have xyz≤127xyz \leq \frac{1}{27}, so cos⁡(xyz)≈1\cos(xyz) \approx 1
  3. The curl of F\mathbf{F} vanishes identically, making the surface integral zero by Stokes' theorem
  4. The triangle lies in a coordinate plane where one component of F\mathbf{F} is always zero
Explanation: We can verify that F=∇(sin⁡(xyz))\mathbf{F} = \nabla(\sin(xyz)) by computing: ∂∂x[sin⁡(xyz)]=yzcos⁡(xyz)\frac{\partial}{\partial x}[\sin(xyz)] = yz \cos(xyz), ∂∂y[sin⁡(xyz)]=xzcos⁡(xyz)\frac{\partial}{\partial y}[\sin(xyz)] = xz \cos(xyz), and ∂∂z[sin⁡(xyz)]=xycos⁡(xyz)\frac{\partial}{\partial z}[\sin(xyz)] = xy \cos(xyz). Since F\mathbf{F} is conservative (being the gradient of a scalar function), we have ∇×F=0\nabla \times \mathbf{F} = 0. By Stokes' theorem, ∮CF⋅dr=∬S(∇×F)⋅n dS=0\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS = 0. Alternatively, since F\mathbf{F} is conservative, the line integral around any closed curve is zero. Option B is a reasonable approximation but doesn't lead to exact simplification. Option C restates the conclusion without the key insight. Option D is incorrect - the triangle doesn't lie in a coordinate plane.

Question 13

A student claims that for any closed curve CC and vector field F\mathbf{F}, Stokes' theorem guarantees that ∮CF⋅dr=0\oint_C \mathbf{F} \cdot d\mathbf{r} = 0 because a closed curve bounds no surface. Which of the following best identifies the error in this reasoning?

  1. Stokes' theorem only applies to simple closed curves, not general closed curves with possible self-intersections
  2. The student confuses 'closed curve' with 'boundary of a surface'; every simple closed curve in R3\mathbb{R}^3 is the boundary of some surface (correct answer)
  3. Stokes' theorem requires the vector field to be conservative, which is not given in the general statement
  4. The theorem only applies when the curl of F\mathbf{F} is continuous and bounded in the region
Explanation: The fundamental error is conceptual: the student incorrectly believes that a closed curve cannot be the boundary of a surface. In fact, every simple closed curve in R3\mathbb{R}^3 bounds infinitely many surfaces (think of a rubber sheet stretched across a wire loop - you can deform it into many different shapes while keeping the same boundary). Stokes' theorem states that ∮CF⋅dr=∬S(∇×F)⋅n dS\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS for any surface SS with boundary CC. The line integral is generally nonzero unless ∇×F=0\nabla \times \mathbf{F} = 0 (i.e., F\mathbf{F} is conservative). Option A is wrong because Stokes' theorem does apply to simple closed curves. Option C is backwards - conservative fields have zero curl, making the line integral zero. Option D states technical conditions that are usually assumed to hold.

Question 14

Let CC be the triangular path from (1,0,0)(1,0,0) to (0,1,0)(0,1,0) to (0,0,1)(0,0,1) and back to (1,0,0)(1,0,0). Let F⃗(x,y,z)=⟨z2,x2,y2⟩\vec{F}(x,y,z) = \langle z^2, x^2, y^2 \rangle. Evaluate ∮CF⃗⋅dr⃗\oint_C \vec{F} \cdot d\vec{r}.

  1. −1-1
  2. 11 (correct answer)
  3. 1/31/3
  4. 2/32/3
Explanation: When you encounter a line integral around a closed path, your first instinct should be to check if the vector field is conservative or if you can apply Green's theorem (for planar curves) or Stokes' theorem (for 3D curves). Since this is a closed triangular path in 3D space, let's use Stokes' theorem: ∮CF⃗⋅dr⃗=∬S(∇×F⃗)⋅n⃗ dS\oint_C \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot \vec{n} \, dS, where SS is any surface bounded by CC. The curl of F⃗(x,y,z)=⟨z2,x2,y2⟩\vec{F}(x,y,z) = \langle z^2, x^2, y^2 \rangle is: ∇×F⃗=⟨2y−0,0−2z,2x−0⟩=⟨2y,−2z,2x⟩\nabla \times \vec{F} = \langle 2y - 0, 0 - 2z, 2x - 0 \rangle = \langle 2y, -2z, 2x \rangle The triangular path lies on the plane x+y+z=1x + y + z = 1, so we can use this as our surface SS. The unit normal vector is n⃗=13⟨1,1,1⟩\vec{n} = \frac{1}{\sqrt{3}}\langle 1, 1, 1 \rangle. Computing the dot product: (∇×F⃗)⋅n⃗=23(y−z+x)=23(x+y−z)(\nabla \times \vec{F}) \cdot \vec{n} = \frac{2}{\sqrt{3}}(y - z + x) = \frac{2}{\sqrt{3}}(x + y - z). On the plane x+y+z=1x + y + z = 1, we have x+y=1−zx + y = 1 - z, so (x+y−z)=1−2z(x + y - z) = 1 - 2z. The surface integral becomes 23∬S(1−2z) dS\frac{2}{\sqrt{3}} \iint_S (1 - 2z) \, dS. Since z=1−x−yz = 1 - x - y on our triangular region, and the area of the triangle is 32\frac{\sqrt{3}}{2}, this evaluates to 11. Answer choice A (−1-1) would result from using the wrong orientation of the normal vector. Answer C (1/31/3) and D (2/32/3) likely come from computational errors in the surface integral or incorrectly calculating the triangle's area. Key strategy: For closed path integrals in 3D, Stokes' theorem often simplifies calculations significantly compared to parameterizing each segment of the path.