Computing Integrals via Stokes' Theorem - Calculus 3

1. Let $\mathbf{F} = (y^3 - z, z^3 - x, x^3 - y)$ and let $C$ be the curve of intersection between the cylinder $x^2 + y^2 = 1$ and the plane $z = x + y$, oriented counterclockwise when viewed from the point $(0, 0, 10)$. A student computes $\oint_C \mathbf{F} \cdot d\mathbf{r}$ using Stokes' theorem with the disk $x^2 + y^2 \leq 1$ in the plane $z = x + y$ as the surface. What is the most likely source of error if the student gets $3\pi$ instead of the correct answer $-3\pi$?
   Answer: Choosing the wrong orientation for the normal vector to the surface The fact that the student got exactly the negative of the correct answer ($3\pi$ instead of $-3\pi$) strongly suggests an orientation error rather than a computational mistake. In Stokes' theorem, $\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS$, and the sign of the result depends critically on choosing the correct orientation for the normal vector $\mathbf{n}$ that is consistent with the given orientation of the boundary curve $C$. Since the curve is oriented counterclockwise when viewed from $(0,0,10)$, there is a specific normal direction required by the right-hand rule. If the student chose the opposite normal direction, the integral would have the correct magnitude but opposite sign. Choices A, B, and D would typically lead to different magnitudes or more complex errors, not simply a sign change.
2. Let $\mathbf{F}(x,y,z) = (e^{xy} + z^2, xe^{xy} - 2yz, 2xz - y^2 + 1)$ be a vector field. Consider the curve $C$ formed by the boundary of the region where the paraboloid $z = x^2 + y^2$ intersects the cylinder $x^2 + y^2 = 4$, oriented counterclockwise when viewed from above. A student claims that $\oint_C \mathbf{F} \cdot d\mathbf{r} = 8\pi$ by applying Stokes' theorem. However, there is an error in the reasoning. What is the most likely mistake?
   Answer: The student computed $\nabla \times \mathbf{F}$ incorrectly, getting $(2x - (-2y), -(2z - 2z), xe^{xy} - xe^{xy}) = (2x + 2y, 0, 0)$ instead of the correct curl Let's compute $\nabla \times \mathbf{F}$ correctly: $\nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ e^{xy} + z^2 & xe^{xy} - 2yz & 2xz - y^2 + 1 \end{vmatrix}$. The $\mathbf{i}$ component is $\frac{\partial}{\partial y}(2xz - y^2 + 1) - \frac{\partial}{\partial z}(xe^{xy} - 2yz) = -2y - (-2y) = 0$. The $\mathbf{j}$ component is $-(\frac{\partial}{\partial x}(2xz - y^2 + 1) - \frac{\partial}{\partial z}(e^{xy} + z^2)) = -(2z - 2z) = 0$. The $\mathbf{k}$ component is $\frac{\partial}{\partial x}(xe^{xy} - 2yz) - \frac{\partial}{\partial y}(e^{xy} + z^2) = e^{xy} + x \cdot ye^{xy} - xe^{xy} = 0$. So $\nabla \times \mathbf{F} = (0,0,0)$, making the correct answer $0$, not $8\pi$. Choice A shows the most common computational error in curl calculation.
3. Let $\mathbf{F}(x,y,z) = (z - y, x - z, y - x)$ and let $C$ be the boundary of the triangle with vertices $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$, oriented counterclockwise when viewed from the origin. Two students compute $\oint_C \mathbf{F} \cdot d\mathbf{r}$ using Stokes' theorem but with different surfaces: Student A uses the triangular surface itself, while Student B uses the portion of the sphere $x^2 + y^2 + z^2 = 1$ bounded by the same curve $C$. What can be concluded about their answers?
   Answer: Both students get the same answer because Stokes' theorem guarantees this independence of surface choice when the surfaces have the same boundary orientation Stokes' theorem states that $\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS$ for any orientable surface $S$ bounded by $C$, provided the orientation of $C$ is consistent with the orientation of $S$. Since both students use surfaces with the same boundary curve $C$ and the same orientation (counterclockwise when viewed from the origin), Stokes' theorem guarantees they will get identical results. The choice of surface (flat triangle vs. curved spherical cap) does not affect the final answer. Choice B is incorrect because both get the same numerical result. Choice C incorrectly suggests the computational complexity affects the answer. Choice D misunderstands that proper orientation consistency is maintained in both cases.
4. Let $\mathbf{F}(x,y,z) = (yz^2, xz^2 + 2y, 2xyz + z)$ and consider the curve $C$ given by the intersection of the surfaces $x^2 + y^2 + z^2 = 9$ and $x + y + z = 3$, oriented so that $z$ increases as we traverse the curve in the direction of increasing $x$ when $y > 0$. To apply Stokes' theorem efficiently, which approach would minimize the computational complexity while ensuring correct orientation?
   Answer: Use the portion of the plane $x + y + z = 3$ bounded by the curve $C$, with normal vector $(1,1,1)$ The planar surface is computationally simpler than the spherical cap, making choice B preferable over A. For orientation, we need the normal vector that makes the boundary curve $C$ have the specified orientation. The curve lies on both the sphere and plane $x + y + z = 3$. Given that $z$ increases with $x$ when $y > 0$, and using the right-hand rule, the normal vector $(1,1,1)$ gives the correct orientation for the boundary. Choice C has the wrong orientation (normal pointing in opposite direction). Choice D incorrectly suggests avoiding Stokes' theorem when it actually simplifies the computation significantly. The curl calculation on a plane is much simpler than direct parametrization of the space curve intersection.
5. Consider the vector field $\mathbf{F} = (2xy + z^3, x^2 - 3yz^2, 3xz^2 - y^3)$ and the closed curve $C$ parametrized by $\mathbf{r}(t) = (2\cos t, 2\sin t, 3)$ for $0 \leq t \leq 2\pi$. A student attempts to use Stokes' theorem to compute $\oint_C \mathbf{F} \cdot d\mathbf{r}$ by choosing the disk $x^2 + y^2 \leq 4$ in the plane $z = 3$ as the surface, with normal vector pointing in the positive $z$ direction. What is the result?
   Answer: $0$ By Stokes' theorem, $\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS$. Computing the curl: $\nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 2xy + z^3 & x^2 - 3yz^2 & 3xz^2 - y^3 \end{vmatrix} = (-3y^2 - (-6yz), -(6xz - 3z^2), 2x - 2x) = (-3y^2 + 6yz, -6xz + 3z^2, 0)$. On the surface $z = 3$, we have $\nabla \times \mathbf{F} = (-3y^2 + 18y, -18x + 27, 0)$. With $\mathbf{n} = (0,0,1)$, we get $(\nabla \times \mathbf{F}) \cdot \mathbf{n} = 0$. Therefore, the surface integral is $0$. The other choices result from computational errors in the curl calculation or incorrect application of the surface integral.

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