Multivariable Calculus Quiz: Choosing Theorems
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Choosing TheoremsQuestion 1 of 18

Let SS be the surface of the ellipsoid defined by x2/9+y2/4+z2=1x^2/9 + y^2/4 + z^2 = 1. Consider the vector field F⃗(x,y,z)=⟨x3+yz,y3+xz,z3+xy⟩\vec{F}(x,y,z) = \langle x^3 + yz, y^3 + xz, z^3 + xy \rangle. Which theorem provides the most direct and efficient method to evaluate the flux integral ∯SF⃗⋅dS⃗\oiint_S \vec{F} \cdot d\vec{S}?

Stokes' Theorem, by relating the flux to a line integral over a boundary curve of the ellipsoid.
The Divergence Theorem, by converting the surface integral to a triple integral over the solid ellipsoid.
Green's Theorem, by projecting the ellipsoid onto one of the coordinate planes.
Direct computation, by parameterizing the surface of the ellipsoid and calculating the flux integral explicitly.
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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Choosing Theorems

Practice Choosing Theorems 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 Choosing Theorems, 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 ellipsoid defined by x2/9+y2/4+z2=1x^2/9 + y^2/4 + z^2 = 1. Consider the vector field F⃗(x,y,z)=⟨x3+yz,y3+xz,z3+xy⟩\vec{F}(x,y,z) = \langle x^3 + yz, y^3 + xz, z^3 + xy \rangle. Which theorem provides the most direct and efficient method to evaluate the flux integral ∯SF⃗⋅dS⃗\oiint_S \vec{F} \cdot d\vec{S}?

  1. Stokes' Theorem, by relating the flux to a line integral over a boundary curve of the ellipsoid.
  2. The Divergence Theorem, by converting the surface integral to a triple integral over the solid ellipsoid. (correct answer)
  3. Green's Theorem, by projecting the ellipsoid onto one of the coordinate planes.
  4. Direct computation, by parameterizing the surface of the ellipsoid and calculating the flux integral explicitly.
Explanation: The problem asks for the flux of a vector field over a closed surface SS. The Divergence Theorem is specifically designed for this, relating the flux integral to the triple integral of the divergence over the enclosed solid volume EE. Here, ∇⋅F⃗=3x2+3y2+3z2\nabla \cdot \vec{F} = 3x^2 + 3y^2 + 3z^2, which is a relatively simple function to integrate over the ellipsoid, likely using a change of variables. Direct computation would be extremely difficult. Stokes' Theorem applies to open surfaces with a boundary curve, not closed surfaces. Green's Theorem applies only to curves and regions in a 2D plane.

Question 2

Let CC be the curve of intersection of the cylinder x2+y2=4x^2 + y^2 = 4 and the plane z=y+3z = y + 3, oriented counterclockwise when viewed from above. To evaluate the line integral ∮C⟨y2,2x,z2⟩⋅dr⃗\oint_C \langle y^2, 2x, z^2 \rangle \cdot d\vec{r}, which theorem offers the most significant simplification?

  1. The Fundamental Theorem of Line Integrals, by finding a potential function for the vector field.
  2. The Divergence Theorem, by considering the volume enclosed by the cylinder and plane.
  3. Stokes' Theorem, by transforming the line integral into a surface integral over the planar ellipse bounded by CC. (correct answer)
  4. Green's Theorem, because the curve lies on the surface of a cylinder which can be unrolled into a plane.
Explanation: The integral is a line integral over a closed loop CC in R3\mathbb{R}^3. Stokes' Theorem is the ideal choice, as it converts this line integral into a surface integral of the curl of the vector field over any surface SS bounded by CC. The simplest such surface is the flat elliptical region of the plane z=y+3z=y+3 inside the cylinder. The curl of F⃗=⟨y2,2x,z2⟩\vec{F} = \langle y^2, 2x, z^2 \rangle is ∇×F⃗=⟨0,0,2−2y⟩\nabla \times \vec{F} = \langle 0, 0, 2-2y \rangle, which is a simple vector field to integrate over the planar surface. The field is not conservative, so the Fundamental Theorem does not apply. The Divergence Theorem relates flux over a closed surface to a triple integral, which is not relevant here. Green's Theorem applies only to curves in the xyxy-plane.

Question 3

Let F⃗=⟨2xy+z2,x2,2xz⟩\vec{F} = \langle 2xy+z^2, x^2, 2xz \rangle and let CC be the curve parameterized by r⃗(t)=⟨cos⁡(πt),sin⁡(πt),t2⟩\vec{r}(t) = \langle \cos(\pi t), \sin(\pi t), t^2 \rangle for t∈[0,1]t \in [0, 1]. Which of the following is the most efficient method for computing the work done by F⃗\vec{F} along CC, i.e., ∫CF⃗⋅dr⃗\int_C \vec{F} \cdot d\vec{r}?

  1. Apply the Fundamental Theorem of Line Integrals after verifying the vector field is conservative. (correct answer)
  2. Apply Stokes' Theorem by closing the path with a straight line and evaluating a surface integral.
  3. Apply the Divergence Theorem over a region whose boundary contains the curve CC.
  4. Directly compute the integral by substituting the parameterization of CC into ∫CF⃗⋅dr⃗\int_C \vec{F} \cdot d\vec{r}.
Explanation: The most efficient method for a line integral is often to check if the vector field is conservative. The curl of F⃗\vec{F} is ∇×F⃗=⟨(0−0),(2z−2z),(2x−2x)⟩=0⃗\nabla \times \vec{F} = \langle (0-0), (2z-2z), (2x-2x) \rangle = \vec{0}. Since the field is conservative, we can use the Fundamental Theorem of Line Integrals. A potential function is f(x,y,z)=x2y+xz2f(x,y,z) = x^2y + xz^2. The integral is f(r⃗(1))−f(r⃗(0))=f(−1,0,1)−f(1,0,0)=(−1)−(0)=−1f(\vec{r}(1)) - f(\vec{r}(0)) = f(-1,0,1) - f(1,0,0) = (-1) - (0) = -1. This is far simpler than direct computation. Stokes' and Green's theorems require a closed path, and the Divergence Theorem is for flux integrals.

Question 4

A vector field is given by F⃗=−yx2+y2i^+xx2+y2j^\vec{F} = \frac{-y}{x^2+y^2} \hat{i} + \frac{x}{x^2+y^2} \hat{j}. To evaluate ∮CF⃗⋅dr⃗\oint_C \vec{F} \cdot d\vec{r} where CC is the boundary of the annulus 1≤x2+y2≤41 \le x^2+y^2 \le 4, which mathematical tool is most essential for a correct analysis?

  1. The standard version of Green's Theorem applied to the annular region.
  2. The Fundamental Theorem for Line Integrals, since the curl of F⃗\vec{F} is zero.
  3. Green's Theorem for a multiply connected region, accounting for the inner boundary. (correct answer)
  4. Stokes' Theorem, treating the annulus as a surface in R3\mathbb{R}^3.
Explanation: The vector field F⃗\vec{F} is not defined at the origin (0,0)(0,0), which is in the hole of the annulus. For any simply connected region not containing the origin, ∂Q∂x−∂P∂y=0\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0. However, because the region of integration is not simply connected, we cannot apply the standard Green's Theorem or the Fundamental Theorem for Line Integrals across the entire region. The correct approach is to use the extension of Green's Theorem for multiply connected regions, which states that the integral over the outer boundary minus the integral over the inner boundary equals the double integral of (∂Q∂x−∂P∂y)(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) over the region. Since this is 0, the integral over the outer boundary equals the integral over the inner boundary (with appropriate orientations).

Question 5

A vector field F⃗\vec{F} is defined on all of R3\mathbb{R}^3 and has continuous partial derivatives. If it is known that the field is solenoidal (i.e., ∇⋅F⃗=0\nabla \cdot \vec{F} = 0 everywhere), which conclusion is guaranteed for any closed, piecewise smooth, oriented surface SS?

  1. The flux ∯SF⃗⋅dS⃗\oiint_S \vec{F} \cdot d\vec{S} is zero, by direct application of the Divergence Theorem. (correct answer)
  2. The flux ∯SF⃗⋅dS⃗\oiint_S \vec{F} \cdot d\vec{S} is zero, by direct application of Stokes' Theorem.
  3. The vector field F⃗\vec{F} must be conservative, and thus the flux is path-independent.
  4. The value of the flux ∯SF⃗⋅dS⃗\oiint_S \vec{F} \cdot d\vec{S} depends on the specific geometry of the surface SS.
Explanation: The question asks for the flux of F⃗\vec{F} over a closed surface SS. The Divergence Theorem states that ∯SF⃗⋅dS⃗=∭E(∇⋅F⃗) dV\oiint_S \vec{F} \cdot d\vec{S} = \iiint_E (\nabla \cdot \vec{F}) \, dV, where EE is the volume enclosed by SS. Since we are given that ∇⋅F⃗=0\nabla \cdot \vec{F} = 0 everywhere, the triple integral becomes ∭E0 dV=0\iiint_E 0 \, dV = 0. Therefore, the flux is always zero, regardless of the geometry of SS. Stokes' Theorem relates line integrals and curls, so it's not the correct tool. A solenoidal field is not necessarily conservative (e.g., F⃗=⟨y,−x,0⟩\vec{F} = \langle y, -x, 0 \rangle).

Question 6

You are asked to compute the line integral of F⃗=⟨yzexyz,xzexyz,xyexyz+2z⟩\vec{F} = \langle yz e^{xyz}, xz e^{xyz}, xy e^{xyz} + 2z \rangle along the helix r⃗(t)=⟨cos⁡t,sin⁡t,t⟩\vec{r}(t) = \langle \cos t, \sin t, t \rangle from t=0t=0 to t=2πt=2\pi. What is the most crucial initial step to determine the most efficient solution strategy?

  1. Begin by parameterizing the line integral using the given path r⃗(t)\vec{r}(t).
  2. Compute the divergence of F⃗\vec{F} to see if the Divergence Theorem can be applied.
  3. Compute the curl of F⃗\vec{F} to determine if the field is conservative. (correct answer)
  4. Check if the path is closed to decide if Stokes' Theorem is applicable.
Explanation: Before attempting a potentially very complicated line integral by direct parameterization, it is always best to check if the vector field is conservative. This is done by computing its curl. If ∇×F⃗=0⃗\nabla \times \vec{F} = \vec{0}, the Fundamental Theorem of Line Integrals can be used, which simplifies the problem to finding a potential function and evaluating it at the endpoints. In this case, the curl is indeed zero, and the potential function is f(x,y,z)=exyz+z2f(x,y,z) = e^{xyz} + z^2. Computing the divergence is irrelevant for a line integral. Checking if the path is closed is useful, but checking for conservativeness is more fundamental as it applies to both open and closed paths.

Question 7

Consider the vector field F⃗=⟨2y,−z,x2⟩\vec{F} = \langle 2y, -z, x^2 \rangle and the surface SS, which is the portion of the cone z=x2+y2z = \sqrt{x^2+y^2} below the plane z=2z=2, oriented downward. To simplify the computation of the flux ∬SF⃗⋅dS⃗\iint_S \vec{F} \cdot d\vec{S}, which would be the most effective strategy?

  1. Apply Stokes' Theorem by finding a vector potential for F⃗\vec{F} and integrating over the boundary circle.
  2. Parameterize the cone surface SS and compute the flux integral directly.
  3. Close the surface with a disk at z=2z=2 and apply the Divergence Theorem to the enclosed solid. (correct answer)
  4. Apply Green's Theorem to the projection of the surface onto the xyxy-plane.
Explanation: The surface SS is open. A powerful strategy for flux integrals over open surfaces is to close the surface and apply the Divergence Theorem. Let SdiskS_{disk} be the disk x2+y2≤4x^2+y^2 \le 4 at z=2z=2. The total flux through the closed surface S∪SdiskS \cup S_{disk} is given by ∭E(∇⋅F⃗) dV\iiint_E (\nabla \cdot \vec{F}) \, dV. Here, ∇⋅F⃗=∂∂x(2y)+∂∂y(−z)+∂∂z(x2)=0\nabla \cdot \vec{F} = \frac{\partial}{\partial x}(2y) + \frac{\partial}{\partial y}(-z) + \frac{\partial}{\partial z}(x^2) = 0. Thus, the total flux is zero. This means the flux through SS is the negative of the flux through SdiskS_{disk} (with appropriate orientation). Calculating the flux through the flat disk is much simpler than calculating the flux through the cone directly. Stokes' theorem is not for this type of flux integral, and Green's theorem is for 2D line integrals.

Question 8

Let SS be the part of the paraboloid z=9−x2−y2z = 9 - x^2 - y^2 that lies above the plane z=5z=5, oriented upward. Let G⃗=⟨yz,−xz,xy⟩\vec{G} = \langle yz, -xz, xy \rangle. What is the most effective theorem to simplify the evaluation of the surface integral ∬S(∇×G⃗)⋅dS⃗\iint_S (\nabla \times \vec{G}) \cdot d\vec{S}?

  1. The Divergence Theorem, because the divergence of a curl is always zero.
  2. Stokes' Theorem, which converts the integral to a line integral over the boundary of SS. (correct answer)
  3. The Fundamental Theorem for Line Integrals, as the integrand is a curl.
  4. Direct computation, by first calculating the curl of G⃗\vec{G} and then parameterizing the surface SS.
Explanation: The integral is of the form ∬S(curl)⋅dS⃗\iint_S (\text{curl}) \cdot d\vec{S}. Stokes' Theorem directly applies, converting this surface integral into a line integral of the original vector field G⃗\vec{G} over the boundary curve ∂S\partial S. The boundary is the circle x2+y2=4x^2+y^2=4 in the plane z=5z=5. This line integral is much simpler to compute than the original surface integral. The Divergence Theorem cannot be applied because SS is not a closed surface. Even if we closed it, the fact that ∇⋅(∇×G⃗)=0\nabla \cdot (\nabla \times \vec{G}) = 0 would only tell us the total flux through the closed surface is zero, not the flux through the open surface SS. The Fundamental Theorem is for line integrals of conservative fields.

Question 9

Let SS be the surface consisting of the four vertical sides of the cube with vertices at (±1,±1,±1)(\pm 1, \pm 1, \pm 1), oriented outward from the z-axis. Let F⃗=⟨x2,y2,z2⟩\vec{F} = \langle x^2, y^2, z^2 \rangle. What is the most efficient strategy to compute the flux ∬SF⃗⋅dS⃗\iint_S \vec{F} \cdot d\vec{S}?

  1. Apply Stokes' Theorem to relate the flux through SS to line integrals around the top and bottom boundary squares.
  2. Parameterize each of the four side faces of the cube and sum the resulting four flux integrals.
  3. Apply the Divergence Theorem directly to the open surface SS to find the total flux.
  4. Use the Divergence Theorem on the entire closed cube and subtract the flux through the top and bottom faces. (correct answer)
Explanation: Direct computation (B) requires four separate surface integrals, which is tedious. Stokes' Theorem (A) is not applicable as this is a flux integral of F⃗\vec{F}, not ∇×F⃗\nabla \times \vec{F}. The Divergence Theorem (C) requires a closed surface, and SS is open. The most clever strategy (D) is to close the surface by adding the top (z=1z=1) and bottom (z=−1z=-1) faces to form a closed cube, let's call it StotalS_{total}. By the Divergence Theorem, ∯StotalF⃗⋅dS⃗=∭E∇⋅F⃗ dV\oiint_{S_{total}} \vec{F} \cdot d\vec{S} = \iiint_E \nabla \cdot \vec{F} \, dV. The desired flux is ∬SF⃗⋅dS⃗=∯StotalF⃗⋅dS⃗−∬topF⃗⋅dS⃗−∬bottomF⃗⋅dS⃗\iint_S \vec{F} \cdot d\vec{S} = \oiint_{S_{total}} \vec{F} \cdot d\vec{S} - \iint_{top} \vec{F} \cdot d\vec{S} - \iint_{bottom} \vec{F} \cdot d\vec{S}. The triple integral and the two simple flux integrals over the flat top and bottom are much easier to compute than the four side integrals.

Question 10

A particle moves along the boundary of the triangle with vertices (0,0)(0,0), (2,0)(2,0), and (2,1)(2,1), oriented counterclockwise. The force field is given by F⃗(x,y)=⟨ex2−y,sin⁡(y3)+x⟩\vec{F}(x,y) = \langle e^{x^2} - y, \sin(y^3) + x \rangle. To compute the work done, which theorem is most appropriate and effective?

  1. The Fundamental Theorem of Line Integrals, since the path is a simple polygon.
  2. Stokes' Theorem, treating the triangle as a surface in three-dimensional space.
  3. The Divergence Theorem, by converting to an integral over the area of the triangle.
  4. Green's Theorem, by converting the line integral to a double integral over the triangular region. (correct answer)
Explanation: The problem asks for the work done, which is a line integral ∮CF⃗⋅dr⃗\oint_C \vec{F} \cdot d\vec{r}, over a simple closed curve CC in the xyxy-plane. Green's Theorem is specifically designed for this scenario. Let P(x,y)=ex2−yP(x,y) = e^{x^2} - y and Q(x,y)=sin⁡(y3)+xQ(x,y) = \sin(y^3) + x. Then ∂Q∂x−∂P∂y=1−(−1)=2\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 - (-1) = 2. By Green's Theorem, the integral equals ∬D2 dA\iint_D 2 \, dA, where DD is the triangular region. This is simply 2 times the area of the triangle, which is 2×(12×2×1)=22 \times (\frac{1}{2} \times 2 \times 1) = 2. This is much easier than parameterizing three line segments. Stokes' Theorem is the 3D generalization of Green's, so Green's is more specific and appropriate. The field is not conservative, so FTLI does not apply.

Question 11

Let F⃗\vec{F} be a vector field with ∇⋅F⃗=0\nabla \cdot \vec{F} = 0 on R3\mathbb{R}^3. Let S1S_1 be the upper hemisphere x2+y2+z2=1,z≥0x^2+y^2+z^2=1, z \ge 0 and S2S_2 be the disk x2+y2≤1,z=0x^2+y^2 \le 1, z=0. If both surfaces are oriented with an upward-pointing normal vector, which theorem is the primary tool to prove that the flux of F⃗\vec{F} through S1S_1 is equal to the flux of F⃗\vec{F} through S2S_2?

  1. The Divergence Theorem, applied to the region enclosed by S1S_1 and S2S_2. (correct answer)
  2. Stokes' Theorem, because S1S_1 and S2S_2 share the same boundary curve.
  3. Green's Theorem, by projecting the surfaces onto the xyxy-plane.
  4. The Fundamental Theorem of Line Integrals, applied to the boundary of the surfaces.
Explanation: Let SclosedS_{closed} be the closed surface formed by S1S_1 and −S2-S_2 (the disk with a downward normal). This surface encloses the upper semi-ball. The Divergence Theorem states ∯SclosedF⃗⋅dS⃗=∭E(∇⋅F⃗) dV\oiint_{S_{closed}} \vec{F} \cdot d\vec{S} = \iiint_E (\nabla \cdot \vec{F}) \, dV. Since ∇⋅F⃗=0\nabla \cdot \vec{F}=0, the total flux is 0. So, ∬S1F⃗⋅dS⃗+∬−S2F⃗⋅dS⃗=0\iint_{S_1} \vec{F} \cdot d\vec{S} + \iint_{-S_2} \vec{F} \cdot d\vec{S} = 0. This means ∬S1F⃗⋅dS⃗−∬S2F⃗⋅dS⃗=0\iint_{S_1} \vec{F} \cdot d\vec{S} - \iint_{S_2} \vec{F} \cdot d\vec{S} = 0, which implies the fluxes are equal. Stokes' Theorem would be relevant if we were integrating the curl of a field, but here we are integrating F⃗\vec{F} itself.

Question 12

Let F⃗\vec{F} be a continuously differentiable vector field on a simply-connected domain in R3\mathbb{R}^3 such that ∇×F⃗=0⃗\nabla \times \vec{F} = \vec{0}. For any simple, closed, piecewise smooth curve CC in the domain, which theorem most directly uses the given information to prove that ∮CF⃗⋅dr⃗=0\oint_C \vec{F} \cdot d\vec{r} = 0?

  1. The Divergence Theorem, which relates the divergence of F⃗\vec{F} to a surface integral.
  2. Green's Theorem, which applies to all closed curves in three-dimensional space.
  3. The Fundamental Theorem of Line Integrals, because the condition ∇×F⃗=0⃗\nabla \times \vec{F} = \vec{0} implies F⃗\vec{F} is a gradient field.
  4. Stokes' Theorem, which equates the line integral of F⃗\vec{F} to the surface integral of ∇×F⃗\nabla \times \vec{F}. (correct answer)
Explanation: Stokes' Theorem states ∮CF⃗⋅dr⃗=∬S(∇×F⃗)⋅dS⃗\oint_C \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot d\vec{S}. Given that ∇×F⃗=0⃗\nabla \times \vec{F} = \vec{0}, the right-hand side of the equation becomes ∬S0⃗⋅dS⃗=0\iint_S \vec{0} \cdot d\vec{S} = 0. This directly shows the line integral is zero using the given fact. While it is true that ∇×F⃗=0⃗\nabla \times \vec{F} = \vec{0} implies F⃗\vec{F} is conservative (the basis for choice C), this is a consequence of the condition, not a direct application of it in a theorem. Stokes' theorem uses the condition ∇×F⃗=0⃗\nabla \times \vec{F} = \vec{0} itself as the integrand. Therefore, it is the most direct justification. The Divergence Theorem and Green's Theorem are not the most appropriate choices for a general 3D curve.

Question 13

A vector field F=(P,Q,R)\mathbf{F} = (P, Q, R) satisfies ∇×F=(2z,x−y,3)\nabla \times \mathbf{F} = (2z, x-y, 3) on a region containing a surface SS bounded by curve CC. To find ∬S(∇×F)⋅n dS\iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS, which approach is most direct?

  1. Apply Stokes' theorem: ∬S(∇×F)⋅n dS=∮CF⋅dr\iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS = \oint_C \mathbf{F} \cdot d\mathbf{r}, requiring determination of the original vector field F\mathbf{F}
  2. Use the divergence theorem after extending to a closed surface, since ∇⋅(∇×F)=0\nabla \cdot (\nabla \times \mathbf{F}) = 0 provides immediate simplification
  3. Compute directly using ∬S(2z,x−y,3)⋅n dS\iint_S (2z, x-y, 3) \cdot \mathbf{n} \, dS with appropriate surface parameterization, since the curl is explicitly given (correct answer)
  4. Apply Green's theorem by projecting onto coordinate planes and decomposing the surface integral into three separate two-dimensional integrals
Explanation: Since ∇×F=(2z,x−y,3)\nabla \times \mathbf{F} = (2z, x-y, 3) is explicitly given, the most direct approach is to compute ∬S(2z,x−y,3)⋅n dS\iint_S (2z, x-y, 3) \cdot \mathbf{n} \, dS directly using surface parameterization. Choice A would require finding F\mathbf{F} from its curl, which is more complex than needed. Choice B misapplies the divergence theorem (we don't have a closed surface and the zero divergence doesn't help here). Choice D incorrectly suggests using Green's theorem for a 3D surface integral.

Question 14

For the surface integral ∬SF⋅n dS\iint_S \mathbf{F} \cdot \mathbf{n} \, dS where F=(yz,xz,xy)\mathbf{F} = (yz, xz, xy) and SS is the portion of the cylinder x2+y2=1x^2 + y^2 = 1 between z=0z = 0 and z=3z = 3, with outward normal, which computational strategy is most efficient?

  1. Use the divergence theorem by closing the surface with disks at z=0z = 0 and z=3z = 3, since ∇⋅F=0+0+0=0\nabla \cdot \mathbf{F} = 0 + 0 + 0 = 0 makes the volume integral zero (correct answer)
  2. Apply Stokes' theorem to convert to line integrals around the boundary circles at z=0z = 0 and z=3z = 3, avoiding the surface parameterization
  3. Parameterize directly using r(θ,z)=(cos⁡θ,sin⁡θ,z)\mathbf{r}(\theta, z) = (\cos \theta, \sin \theta, z) and compute ∬SF⋅(rθ×rz) dθ dz\iint_S \mathbf{F} \cdot (\mathbf{r}_\theta \times \mathbf{r}_z) \, d\theta \, dz
  4. Use Green's theorem by projecting the surface onto the xzxz-plane and yzyz-plane, then combining the results appropriately
Explanation: Since ∇⋅F=∂∂x(yz)+∂∂y(xz)+∂∂z(xy)=0+0+0=0\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(yz) + \frac{\partial}{\partial y}(xz) + \frac{\partial}{\partial z}(xy) = 0 + 0 + 0 = 0, we can close the cylindrical surface with disks at z=0z = 0 and z=3z = 3 to form a closed surface. The divergence theorem gives us that the total flux through the closed surface is zero, so the flux through the cylindrical surface equals the negative of the flux through the two disks. Choice B misapplies Stokes' theorem (which relates curl to circulation). Choice C works but is more computational work. Choice D misapplies Green's theorem.

Question 15

Consider the line integral ∫CF⋅dr\int_C \mathbf{F} \cdot d\mathbf{r} where F=(excos⁡y−y,−exsin⁡y−x+2)\mathbf{F} = (e^x \cos y - y, -e^x \sin y - x + 2) and CC is any path from (0,0)(0,0) to (1,π)(1,\pi). To determine the most efficient evaluation method, what should be checked first?

  1. Whether ∇×F=0\nabla \times \mathbf{F} = 0, and if so, apply Stokes' theorem to convert to a surface integral over any surface bounded by CC
  2. Whether F\mathbf{F} satisfies the conditions for Green's theorem by checking if CC bounds a simply connected region in the plane
  3. Whether the curve CC can be parameterized simply, since the transcendental components suggest direct computation may be necessary
  4. Whether ∂P∂y=∂Q∂x\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} where P=excos⁡y−yP = e^x \cos y - y and Q=−exsin⁡y−x+2Q = -e^x \sin y - x + 2, indicating path independence (correct answer)
Explanation: When evaluating line integrals, your first instinct should be to check for path independence, which can dramatically simplify calculations. A vector field F=(P,Q)\mathbf{F} = (P, Q) is conservative (path-independent) if and only if ∂P∂y=∂Q∂x\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} in a simply connected domain. Let's check this condition. For P=excos⁡y−yP = e^x \cos y - y and Q=−exsin⁡y−x+2Q = -e^x \sin y - x + 2:
  • ∂P∂y=−exsin⁡y−1\frac{\partial P}{\partial y} = -e^x \sin y - 1
  • ∂Q∂x=−exsin⁡y−1\frac{\partial Q}{\partial x} = -e^x \sin y - 1
Since these partial derivatives are equal, the field is conservative, meaning the integral depends only on the endpoints, not the path. This makes option D the most efficient first check. Option A mentions curl and Stokes' theorem, but this is a 2D problem where we should use the 2D test for conservativeness first. Option B discusses Green's theorem, which applies to closed curves, but our path goes from (0,0)(0,0) to (1,π)(1,\pi) and isn't closed. Option C suggests direct parameterization, but this ignores the possibility of path independence, which would make any complex parameterization unnecessary. Study tip: Always check for conservative vector fields first when evaluating line integrals between two points. If ∂P∂y=∂Q∂x\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}, find a potential function and use the Fundamental Theorem for Line Integrals—it's almost always faster than direct computation.

Question 16

A vector field G=(f(x,y,z),g(x,y,z),h(x,y,z))\mathbf{G} = (f(x,y,z), g(x,y,z), h(x,y,z)) satisfies ∇⋅G=xyz\nabla \cdot \mathbf{G} = xyz throughout a region containing a closed surface SS that encloses volume VV. If ∭Vxyz dV=12\iiint_V xyz \, dV = 12, which theorem directly provides ∬SG⋅n dS\iint_S \mathbf{G} \cdot \mathbf{n} \, dS?

  1. Stokes' theorem, since the curl of G\mathbf{G} can be related to the line integral around the boundary of any surface patch on SS
  2. Green's theorem applied to each coordinate plane projection, allowing reconstruction of the full surface integral from planar components
  3. Divergence theorem, directly giving ∬SG⋅n dS=∭V∇⋅G dV=∭Vxyz dV=12\iint_S \mathbf{G} \cdot \mathbf{n} \, dS = \iiint_V \nabla \cdot \mathbf{G} \, dV = \iiint_V xyz \, dV = 12 (correct answer)
  4. Fundamental theorem of line integrals, after verifying that G\mathbf{G} is conservative based on the given divergence condition
Explanation: The divergence theorem states that for a vector field G\mathbf{G} and closed surface SS bounding region VV: ∬SG⋅n dS=∭V∇⋅G dV\iint_S \mathbf{G} \cdot \mathbf{n} \, dS = \iiint_V \nabla \cdot \mathbf{G} \, dV. Since we're given that ∇⋅G=xyz\nabla \cdot \mathbf{G} = xyz and ∭Vxyz dV=12\iiint_V xyz \, dV = 12, we immediately get ∬SG⋅n dS=12\iint_S \mathbf{G} \cdot \mathbf{n} \, dS = 12. Choice A relates to circulation, not flux. Choice B unnecessarily complicates the problem. Choice D is about line integrals and conservativeness, not flux through closed surfaces.

Question 17

A closed surface SS bounds a region EE where a vector field F(x,y,z)=(x3,y3−z,z3+y)\mathbf{F}(x,y,z) = (x^3, y^3 - z, z^3 + y) is defined except at the origin, which lies inside EE. To evaluate ∬SF⋅n dS\iint_S \mathbf{F} \cdot \mathbf{n} \, dS, what is the most reliable approach?

  1. Apply the divergence theorem directly since ∇⋅F=3x2+3y2+3z2=3(x2+y2+z2)\nabla \cdot \mathbf{F} = 3x^2 + 3y^2 + 3z^2 = 3(x^2 + y^2 + z^2) and the singularity at the origin doesn't affect the volume integral
  2. Apply Stokes' theorem by decomposing the closed surface into patches with boundaries and evaluating line integrals around each boundary curve
  3. Parameterize the surface SS directly since the divergence theorem cannot be applied when the vector field has singularities inside the region
  4. Use the divergence theorem with a small sphere SϵS_\epsilon around the origin removed, computing ∭E∖Bϵ∇⋅F dV+∬SϵF⋅n dS\iiint_{E \setminus B_\epsilon} \nabla \cdot \mathbf{F} \, dV + \iint_{S_\epsilon} \mathbf{F} \cdot \mathbf{n} \, dS and taking the limit as ϵ→0\epsilon \to 0 (correct answer)
Explanation: When evaluating a flux integral where the vector field has a singularity inside the region, you need to carefully handle the discontinuity to apply the divergence theorem properly. The divergence theorem states that ∬SF⋅n dS=∭E∇⋅F dV\iint_S \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_E \nabla \cdot \mathbf{F} \, dV, but this requires F\mathbf{F} to be continuously differentiable throughout the region EE. Since F\mathbf{F} is undefined at the origin inside EE, we cannot apply the theorem directly to the entire region. The correct approach (D) uses a clever modification: remove a small sphere SϵS_\epsilon of radius ϵ\epsilon around the origin, creating a region where F\mathbf{F} is well-defined. Apply the divergence theorem to this modified region, then add the flux through the small sphere. As ϵ→0\epsilon \to 0, this gives the desired result. Option A incorrectly assumes the singularity doesn't matter. While the divergence formula is correct, the divergence theorem fails when F\mathbf{F} isn't defined everywhere in EE. Option B misapplies Stokes' theorem, which relates line integrals to surface integrals, not flux through closed surfaces. Plus, closed surfaces have no boundary curves. Option C gives up too quickly on the divergence theorem. While direct parameterization could work, it's unnecessarily complex when the modified divergence theorem approach handles the singularity elegantly. Remember: when vector fields have singularities inside your region, use the "punctured domain" technique—remove small neighborhoods around singularities, apply standard theorems, then take limits.

Question 18

A surface SS is defined by z=x2+y2z = x^2 + y^2 for x2+y2≤4x^2 + y^2 \leq 4, oriented upward. To evaluate ∬S(∇×F)⋅n dS\iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS where F=(z,−x,y)\mathbf{F} = (z, -x, y), which theorem provides the most efficient approach?

  1. Divergence theorem, converting to ∭E∇⋅(∇×F) dV\iiint_E \nabla \cdot (\nabla \times \mathbf{F}) \, dV where EE is bounded by SS and the disk z=0z = 0
  2. Stokes' theorem, converting to ∮CF⋅dr\oint_C \mathbf{F} \cdot d\mathbf{r} where CC is the boundary circle x2+y2=4,z=4x^2 + y^2 = 4, z = 4 oriented counterclockwise when viewed from above (correct answer)
  3. Green's theorem in the xyxy-plane, since the surface projects onto the disk x2+y2≤4x^2 + y^2 \leq 4 and allows direct computation
  4. Direct computation of the surface integral using parameterization r(u,v)=(ucos⁡v,usin⁡v,u2)\mathbf{r}(u,v) = (u\cos v, u\sin v, u^2) for 0≤u≤2,0≤v≤2π0 \leq u \leq 2, 0 \leq v \leq 2\pi
Explanation: For a surface integral of curl, Stokes' theorem is the natural choice: ∬S(∇×F)⋅n dS=∮CF⋅dr\iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS = \oint_C \mathbf{F} \cdot d\mathbf{r}. The boundary of surface SS is the circle where x2+y2=4x^2 + y^2 = 4 and z=4z = 4. Choice A is incorrect because ∇⋅(∇×F)=0\nabla \cdot (\nabla \times \mathbf{F}) = 0 always, and we don't have a closed surface. Choice C misapplies Green's theorem, which doesn't directly handle surface integrals of curl. Choice D ignores the available simplification from Stokes' theorem.