Choosing Theorems - Multivariable Calculus

1. A closed surface $S$ bounds a region $E$ where a vector field $\mathbf{F}(x,y,z) = (x^3, y^3 - z, z^3 + y)$ is defined except at the origin, which lies inside $E$. To evaluate $\iint_S \mathbf{F} \cdot \mathbf{n} \, dS$, what is the most reliable approach?
   Answer: Use the divergence theorem with a small sphere $S_\epsilon$ around the origin removed, computing $\iiint_{E \setminus B_\epsilon} \nabla \cdot \mathbf{F} \, dV + \iint_{S_\epsilon} \mathbf{F} \cdot \mathbf{n} \, dS$ and taking the limit as $\epsilon \to 0$ 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 $\iint_S \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_E \nabla \cdot \mathbf{F} \, dV$, but this requires $\mathbf{F}$ to be continuously differentiable throughout the region $E$. Since $\mathbf{F}$ is undefined at the origin inside $E$, we cannot apply the theorem directly to the entire region. The correct approach (D) uses a clever modification: remove a small sphere $S_\epsilon$ of radius $\epsilon$ around the origin, creating a region where $\mathbf{F}$ is well-defined. Apply the divergence theorem to this modified region, then add the flux through the small sphere. As $\epsilon \to 0$, 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 $\mathbf{F}$ isn't defined everywhere in $E$. 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.
2. Let $S$ be the surface consisting of the four vertical sides of the cube with vertices at $(\pm 1, \pm 1, \pm 1)$, oriented outward from the z-axis. Let $\vec{F} = \langle x^2, y^2, z^2 \rangle$. What is the most efficient strategy to compute the flux $\iint_S \vec{F} \cdot d\vec{S}$?
   Answer: Use the Divergence Theorem on the entire closed cube and subtract the flux through the top and bottom faces. Direct computation (B) requires four separate surface integrals, which is tedious. Stokes' Theorem (A) is not applicable as this is a flux integral of $\vec{F}$, not $\nabla \times \vec{F}$. The Divergence Theorem (C) requires a closed surface, and $S$ is open. The most clever strategy (D) is to close the surface by adding the top ($z=1$) and bottom ($z=-1$) faces to form a closed cube, let's call it $S_{total}$. By the Divergence Theorem, $\oiint_{S_{total}} \vec{F} \cdot d\vec{S} = \iiint_E \nabla \cdot \vec{F} \, dV$. The desired flux is $\iint_S \vec{F} \cdot d\vec{S} = \oiint_{S_{total}} \vec{F} \cdot d\vec{S} - \iint_{top} \vec{F} \cdot d\vec{S} - \iint_{bottom} \vec{F} \cdot d\vec{S}$. 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.
3. Let $C$ be the curve of intersection of the cylinder $x^2 + y^2 = 4$ and the plane $z = y + 3$, oriented counterclockwise when viewed from above. To evaluate the line integral $\oint_C \langle y^2, 2x, z^2 \rangle \cdot d\vec{r}$, which theorem offers the most significant simplification?
   Answer: Stokes' Theorem, by transforming the line integral into a surface integral over the planar ellipse bounded by $C$. The integral is a line integral over a closed loop $C$ in $\mathbb{R}^3$. 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 $S$ bounded by $C$. The simplest such surface is the flat elliptical region of the plane $z=y+3$ inside the cylinder. The curl of $\vec{F} = \langle y^2, 2x, z^2 \rangle$ is $\nabla \times \vec{F} = \langle 0, 0, 2-2y \rangle$, 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 $xy$-plane.
4. Consider the vector field $\vec{F} = \langle 2y, -z, x^2 \rangle$ and the surface $S$, which is the portion of the cone $z = \sqrt{x^2+y^2}$ below the plane $z=2$, oriented downward. To simplify the computation of the flux $\iint_S \vec{F} \cdot d\vec{S}$, which would be the most effective strategy?
   Answer: Close the surface with a disk at $z=2$ and apply the Divergence Theorem to the enclosed solid. The surface $S$ is open. A powerful strategy for flux integrals over open surfaces is to close the surface and apply the Divergence Theorem. Let $S_{disk}$ be the disk $x^2+y^2 \le 4$ at $z=2$. The total flux through the closed surface $S \cup S_{disk}$ is given by $\iiint_E (\nabla \cdot \vec{F}) \, dV$. Here, $\nabla \cdot \vec{F} = \frac{\partial}{\partial x}(2y) + \frac{\partial}{\partial y}(-z) + \frac{\partial}{\partial z}(x^2) = 0$. Thus, the total flux is zero. This means the flux through $S$ is the negative of the flux through $S_{disk}$ (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.
5. Let $S$ be the part of the paraboloid $z = 9 - x^2 - y^2$ that lies above the plane $z=5$, oriented upward. Let $\vec{G} = \langle yz, -xz, xy \rangle$. What is the most effective theorem to simplify the evaluation of the surface integral $\iint_S (\nabla \times \vec{G}) \cdot d\vec{S}$?
   Answer: Stokes' Theorem, which converts the integral to a line integral over the boundary of $S$. The integral is of the form $\iint_S (\text{curl}) \cdot d\vec{S}$. Stokes' Theorem directly applies, converting this surface integral into a line integral of the original vector field $\vec{G}$ over the boundary curve $\partial S$. The boundary is the circle $x^2+y^2=4$ in the plane $z=5$. This line integral is much simpler to compute than the original surface integral. The Divergence Theorem cannot be applied because $S$ is not a closed surface. Even if we closed it, the fact that $\nabla \cdot (\nabla \times \vec{G}) = 0$ would only tell us the total flux through the closed surface is zero, not the flux through the open surface $S$. The Fundamental Theorem is for line integrals of conservative fields.
6. You are asked to compute the line integral of $\vec{F} = \langle yz e^{xyz}, xz e^{xyz}, xy e^{xyz} + 2z \rangle$ along the helix $\vec{r}(t) = \langle \cos t, \sin t, t \rangle$ from $t=0$ to $t=2\pi$. What is the most crucial initial step to determine the most efficient solution strategy?
   Answer: Compute the curl of $\vec{F}$ to determine if the field is conservative. 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 $\nabla \times \vec{F} = \vec{0}$, 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 $f(x,y,z) = e^{xyz} + z^2$. 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.
7. Consider the line integral $\int_C \mathbf{F} \cdot d\mathbf{r}$ where $\mathbf{F} = (e^x \cos y - y, -e^x \sin y - x + 2)$ and $C$ is any path from $(0,0)$ to $(1,\pi)$. To determine the most efficient evaluation method, what should be checked first?
   Answer: Whether $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ where $P = e^x \cos y - y$ and $Q = -e^x \sin y - x + 2$, indicating path independence When evaluating line integrals, your first instinct should be to check for path independence, which can dramatically simplify calculations. A vector field $\mathbf{F} = (P, Q)$ is conservative (path-independent) if and only if $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ in a simply connected domain. Let's check this condition. For $P = e^x \cos y - y$ and $Q = -e^x \sin y - x + 2$: - $\frac{\partial P}{\partial y} = -e^x \sin y - 1$ - $\frac{\partial Q}{\partial x} = -e^x \sin y - 1$ 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 $(0,0)$ to $(1,\pi)$ 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 $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$, find a potential function and use the Fundamental Theorem for Line Integrals—it's almost always faster than direct computation.
8. Let $\vec{F} = \langle 2xy+z^2, x^2, 2xz \rangle$ and let $C$ be the curve parameterized by $\vec{r}(t) = \langle \cos(\pi t), \sin(\pi t), t^2 \rangle$ for $t \in [0, 1]$. Which of the following is the most efficient method for computing the work done by $\vec{F}$ along $C$, i.e., $\int_C \vec{F} \cdot d\vec{r}$?
   Answer: Apply the Fundamental Theorem of Line Integrals after verifying the vector field is conservative. The most efficient method for a line integral is often to check if the vector field is conservative. The curl of $\vec{F}$ is $\nabla \times \vec{F} = \langle (0-0), (2z-2z), (2x-2x) \rangle = \vec{0}$. Since the field is conservative, we can use the Fundamental Theorem of Line Integrals. A potential function is $f(x,y,z) = x^2y + xz^2$. The integral is $f(\vec{r}(1)) - f(\vec{r}(0)) = f(-1,0,1) - f(1,0,0) = (-1) - (0) = -1$. This is far simpler than direct computation. Stokes' and Green's theorems require a closed path, and the Divergence Theorem is for flux integrals.
9. A vector field is given by $\vec{F} = \frac{-y}{x^2+y^2} \hat{i} + \frac{x}{x^2+y^2} \hat{j}$. To evaluate $\oint_C \vec{F} \cdot d\vec{r}$ where $C$ is the boundary of the annulus $1 \le x^2+y^2 \le 4$, which mathematical tool is most essential for a correct analysis?
   Answer: Green's Theorem for a multiply connected region, accounting for the inner boundary. The vector field $\vec{F}$ is not defined at the origin $(0,0)$, which is in the hole of the annulus. For any simply connected region not containing the origin, $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0$. 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 $(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})$ over the region. Since this is 0, the integral over the outer boundary equals the integral over the inner boundary (with appropriate orientations).
10. A vector field $\vec{F}$ is defined on all of $\mathbb{R}^3$ and has continuous partial derivatives. If it is known that the field is solenoidal (i.e., $\nabla \cdot \vec{F} = 0$ everywhere), which conclusion is guaranteed for any closed, piecewise smooth, oriented surface $S$?
    Answer: The flux $\oiint_S \vec{F} \cdot d\vec{S}$ is zero, by direct application of the Divergence Theorem. The question asks for the flux of $\vec{F}$ over a closed surface $S$. The Divergence Theorem states that $\oiint_S \vec{F} \cdot d\vec{S} = \iiint_E (\nabla \cdot \vec{F}) \, dV$, where $E$ is the volume enclosed by $S$. Since we are given that $\nabla \cdot \vec{F} = 0$ everywhere, the triple integral becomes $\iiint_E 0 \, dV = 0$. Therefore, the flux is always zero, regardless of the geometry of $S$. Stokes' Theorem relates line integrals and curls, so it's not the correct tool. A solenoidal field is not necessarily conservative (e.g., $\vec{F} = \langle y, -x, 0 \rangle$).
11. Let $\vec{F}$ be a continuously differentiable vector field on a simply-connected domain in $\mathbb{R}^3$ such that $\nabla \times \vec{F} = \vec{0}$. For any simple, closed, piecewise smooth curve $C$ in the domain, which theorem most directly uses the given information to prove that $\oint_C \vec{F} \cdot d\vec{r} = 0$?
    Answer: Stokes' Theorem, which equates the line integral of $\vec{F}$ to the surface integral of $\nabla \times \vec{F}$. Stokes' Theorem states $\oint_C \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot d\vec{S}$. Given that $\nabla \times \vec{F} = \vec{0}$, the right-hand side of the equation becomes $\iint_S \vec{0} \cdot d\vec{S} = 0$. This directly shows the line integral is zero using the given fact. While it is true that $\nabla \times \vec{F} = \vec{0}$ implies $\vec{F}$ 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 $\nabla \times \vec{F} = \vec{0}$ 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.
12. A vector field $\mathbf{G} = (f(x,y,z), g(x,y,z), h(x,y,z))$ satisfies $\nabla \cdot \mathbf{G} = xyz$ throughout a region containing a closed surface $S$ that encloses volume $V$. If $\iiint_V xyz \, dV = 12$, which theorem directly provides $\iint_S \mathbf{G} \cdot \mathbf{n} \, dS$?
    Answer: Divergence theorem, directly giving $\iint_S \mathbf{G} \cdot \mathbf{n} \, dS = \iiint_V \nabla \cdot \mathbf{G} \, dV = \iiint_V xyz \, dV = 12$ The divergence theorem states that for a vector field $\mathbf{G}$ and closed surface $S$ bounding region $V$: $\iint_S \mathbf{G} \cdot \mathbf{n} \, dS = \iiint_V \nabla \cdot \mathbf{G} \, dV$. Since we're given that $\nabla \cdot \mathbf{G} = xyz$ and $\iiint_V xyz \, dV = 12$, we immediately get $\iint_S \mathbf{G} \cdot \mathbf{n} \, dS = 12$. 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.
13. A vector field $\mathbf{F} = (P, Q, R)$ satisfies $\nabla \times \mathbf{F} = (2z, x-y, 3)$ on a region containing a surface $S$ bounded by curve $C$. To find $\iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS$, which approach is most direct?
    Answer: Compute directly using $\iint_S (2z, x-y, 3) \cdot \mathbf{n} \, dS$ with appropriate surface parameterization, since the curl is explicitly given Since $\nabla \times \mathbf{F} = (2z, x-y, 3)$ is explicitly given, the most direct approach is to compute $\iint_S (2z, x-y, 3) \cdot \mathbf{n} \, dS$ directly using surface parameterization. Choice A would require finding $\mathbf{F}$ 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.
14. Let $\vec{F}$ be a vector field with $\nabla \cdot \vec{F} = 0$ on $\mathbb{R}^3$. Let $S_1$ be the upper hemisphere $x^2+y^2+z^2=1, z \ge 0$ and $S_2$ be the disk $x^2+y^2 \le 1, z=0$. If both surfaces are oriented with an upward-pointing normal vector, which theorem is the primary tool to prove that the flux of $\vec{F}$ through $S_1$ is equal to the flux of $\vec{F}$ through $S_2$?
    Answer: The Divergence Theorem, applied to the region enclosed by $S_1$ and $S_2$. Let $S_{closed}$ be the closed surface formed by $S_1$ and $-S_2$ (the disk with a downward normal). This surface encloses the upper semi-ball. The Divergence Theorem states $\oiint_{S_{closed}} \vec{F} \cdot d\vec{S} = \iiint_E (\nabla \cdot \vec{F}) \, dV$. Since $\nabla \cdot \vec{F}=0$, the total flux is 0. So, $\iint_{S_1} \vec{F} \cdot d\vec{S} + \iint_{-S_2} \vec{F} \cdot d\vec{S} = 0$. This means $\iint_{S_1} \vec{F} \cdot d\vec{S} - \iint_{S_2} \vec{F} \cdot d\vec{S} = 0$, 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 $\vec{F}$ itself.
15. A particle moves along the boundary of the triangle with vertices $(0,0)$, $(2,0)$, and $(2,1)$, oriented counterclockwise. The force field is given by $\vec{F}(x,y) = \langle e^{x^2} - y, \sin(y^3) + x \rangle$. To compute the work done, which theorem is most appropriate and effective?
    Answer: Green's Theorem, by converting the line integral to a double integral over the triangular region. The problem asks for the work done, which is a line integral $\oint_C \vec{F} \cdot d\vec{r}$, over a simple closed curve $C$ in the $xy$-plane. Green's Theorem is specifically designed for this scenario. Let $P(x,y) = e^{x^2} - y$ and $Q(x,y) = \sin(y^3) + x$. Then $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 - (-1) = 2$. By Green's Theorem, the integral equals $\iint_D 2 \, dA$, where $D$ is the triangular region. This is simply 2 times the area of the triangle, which is $2 \times (\frac{1}{2} \times 2 \times 1) = 2$. 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.
16. For the surface integral $\iint_S \mathbf{F} \cdot \mathbf{n} \, dS$ where $\mathbf{F} = (yz, xz, xy)$ and $S$ is the portion of the cylinder $x^2 + y^2 = 1$ between $z = 0$ and $z = 3$, with outward normal, which computational strategy is most efficient?
    Answer: Use the divergence theorem by closing the surface with disks at $z = 0$ and $z = 3$, since $\nabla \cdot \mathbf{F} = 0 + 0 + 0 = 0$ makes the volume integral zero Since $\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(yz) + \frac{\partial}{\partial y}(xz) + \frac{\partial}{\partial z}(xy) = 0 + 0 + 0 = 0$, we can close the cylindrical surface with disks at $z = 0$ and $z = 3$ 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.
17. A surface $S$ is defined by $z = x^2 + y^2$ for $x^2 + y^2 \leq 4$, oriented upward. To evaluate $\iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS$ where $\mathbf{F} = (z, -x, y)$, which theorem provides the most efficient approach?
    Answer: Stokes' theorem, converting to $\oint_C \mathbf{F} \cdot d\mathbf{r}$ where $C$ is the boundary circle $x^2 + y^2 = 4, z = 4$ oriented counterclockwise when viewed from above For a surface integral of curl, Stokes' theorem is the natural choice: $\iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS = \oint_C \mathbf{F} \cdot d\mathbf{r}$. The boundary of surface $S$ is the circle where $x^2 + y^2 = 4$ and $z = 4$. Choice A is incorrect because $\nabla \cdot (\nabla \times \mathbf{F}) = 0$ 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.
18. Let $S$ be the surface of the ellipsoid defined by $x^2/9 + y^2/4 + z^2 = 1$. Consider the vector field $\vec{F}(x,y,z) = \langle x^3 + yz, y^3 + xz, z^3 + xy \rangle$. Which theorem provides the most direct and efficient method to evaluate the flux integral $\oiint_S \vec{F} \cdot d\vec{S}$?
    Answer: The Divergence Theorem, by converting the surface integral to a triple integral over the solid ellipsoid. The problem asks for the flux of a vector field over a closed surface $S$. The Divergence Theorem is specifically designed for this, relating the flux integral to the triple integral of the divergence over the enclosed solid volume $E$. Here, $\nabla \cdot \vec{F} = 3x^2 + 3y^2 + 3z^2$, 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.

Start this quiz