Stokes' Theorem - Multivariable Calculus

1. Let $C$ be the triangular path from $(1,0,0)$ to $(0,1,0)$ to $(0,0,1)$ and back to $(1,0,0)$. Let $\vec{F}(x,y,z) = \langle z^2, x^2, y^2 \rangle$. Evaluate $\oint_C \vec{F} \cdot d\vec{r}$.
   Answer: $1$ 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: $\oint_C \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot \vec{n} \, dS$, where $S$ is any surface bounded by $C$. The curl of $\vec{F}(x,y,z) = \langle z^2, x^2, y^2 \rangle$ is: $\nabla \times \vec{F} = \langle 2y - 0, 0 - 2z, 2x - 0 \rangle = \langle 2y, -2z, 2x \rangle$ The triangular path lies on the plane $x + y + z = 1$, so we can use this as our surface $S$. The unit normal vector is $\vec{n} = \frac{1}{\sqrt{3}}\langle 1, 1, 1 \rangle$. Computing the dot product: $(\nabla \times \vec{F}) \cdot \vec{n} = \frac{2}{\sqrt{3}}(y - z + x) = \frac{2}{\sqrt{3}}(x + y - z)$. On the plane $x + y + z = 1$, we have $x + y = 1 - z$, so $(x + y - z) = 1 - 2z$. The surface integral becomes $\frac{2}{\sqrt{3}} \iint_S (1 - 2z) \, dS$. Since $z = 1 - x - y$ on our triangular region, and the area of the triangle is $\frac{\sqrt{3}}{2}$, this evaluates to $1$. Answer choice A ($-1$) would result from using the wrong orientation of the normal vector. Answer C ($1/3$) and D ($2/3$) 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.
2. Let $\vec{F}(x,y,z) = \langle 3y, -2x, yz \rangle$. Let $C$ be the circle $x^2 + y^2 = 1$ in the plane $z=2$, oriented counter-clockwise when viewed from above. Evaluate the line integral $\oint_C \vec{F} \cdot d\vec{r}$.
   Answer: $-5\pi$ By Stokes' theorem, $\oint_C \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot d\vec{S}$, where $S$ is any surface with boundary $C$. Let's choose $S$ to be the disk $D$ defined by $x^2 + y^2 \le 1$ in the plane $z=2$. The counter-clockwise orientation of $C$ corresponds to an upward-pointing normal vector for $D$, which is $\vec{n} = \hat{k} = \langle 0,0,1 \rangle$. First, compute the curl of $\vec{F}$: $\nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 3y & -2x & yz \end{vmatrix} = \hat{i}(z-0) - \hat{j}(0-0) + \hat{k}(-2-3) = \langle z, 0, -5 \rangle$ On the surface $S$, $z=2$, so $\nabla \times \vec{F} = \langle 2, 0, -5 \rangle$. The surface integral is: $\iint_D \langle 2, 0, -5 \rangle \cdot \langle 0,0,1 \rangle dA = \iint_D -5 dA = -5 \cdot (\text{Area of D})$ The area of the unit disk $D$ is $\pi(1^2) = \pi$. Thus, the integral is $-5\pi$.
3. Let $H$ be the hemisphere $x^2+y^2+z^2=9$ with $z \ge 0$, and let $D$ be the disk $x^2+y^2 \le 9$ in the $xy$-plane. Let $S$ be the closed surface formed by the union of $H$ and $D$, oriented with the outward-pointing normal vector. For any continuously differentiable vector field $\vec{F}$, which of the following expressions is always true?
   Answer: $\iint_H (\nabla \times \vec{F}) \cdot d\vec{S} = -\iint_D (\nabla \times \vec{F}) \cdot d\vec{S}$ The problem considers the flux of the curl of $\vec{F}$ through the closed surface $S = H \cup D$. 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: $\nabla \cdot (\nabla \times \vec{F}) = 0$. Therefore: $\oiint_S (\nabla \times \vec{F}) \cdot d\vec{S} = \iiint_V \nabla \cdot (\nabla \times \vec{F}) dV = \iiint_V 0 dV = 0$ The total flux over the closed surface $S$ is the sum of the fluxes over its parts, $H$ and $D$: $\oiint_S (\nabla \times \vec{F}) \cdot d\vec{S} = \iint_H (\nabla \times \vec{F}) \cdot d\vec{S} + \iint_D (\nabla \times \vec{F}) \cdot d\vec{S} = 0$ From this equation, we can conclude that $\iint_H (\nabla \times \vec{F}) \cdot d\vec{S} = -\iint_D (\nabla \times \vec{F}) \cdot d\vec{S}$. This relationship holds for any continuously differentiable vector field $\vec{F}$. The other options are not always true.
4. Consider the vector field $\mathbf{F} = (yz \cos(xyz), xz \cos(xyz), xy \cos(xyz))$ and the triangular curve $C$ with vertices at $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$, traversed in that order. To evaluate $\oint_C \mathbf{F} \cdot d\mathbf{r}$ using Stokes' theorem, which observation would most significantly simplify the calculation?
   Answer: The vector field $\mathbf{F}$ is the gradient of $\sin(xyz)$, making the line integral path-independent We can verify that $\mathbf{F} = \nabla(\sin(xyz))$ by computing: $\frac{\partial}{\partial x}[\sin(xyz)] = yz \cos(xyz)$, $\frac{\partial}{\partial y}[\sin(xyz)] = xz \cos(xyz)$, and $\frac{\partial}{\partial z}[\sin(xyz)] = xy \cos(xyz)$. Since $\mathbf{F}$ is conservative (being the gradient of a scalar function), we have $\nabla \times \mathbf{F} = 0$. By Stokes' theorem, $\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS = 0$. Alternatively, since $\mathbf{F}$ 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.
5. Let $S$ be the portion of the sphere $x^2 + y^2 + z^2 = 4$ for which $z \ge \sqrt{3}$, oriented with an upward-pointing normal vector. Let $\vec{F}(x,y,z) = \langle -y, x, zx^3y^2 \rangle$. Evaluate $\iint_S (\nabla \times \vec{F}) \cdot d\vec{S}$.
   Answer: $2\pi$ By Stokes' theorem, this surface integral is equal to the line integral of $\vec{F}$ over the boundary curve $C$ of $S$. The boundary $C$ is the intersection of the sphere $x^2 + y^2 + z^2 = 4$ and the plane $z = \sqrt{3}$. Substituting $z=\sqrt{3}$ into the sphere equation gives $x^2 + y^2 + (\sqrt{3})^2 = 4$, which simplifies to $x^2+y^2=1$. So, $C$ is a circle of radius 1 in the plane $z=\sqrt{3}$. The upward orientation of $S$ means the boundary $C$ must be traversed counter-clockwise when viewed from above. We can parameterize $C$ as $\vec{r}(t) = \langle \cos t, \sin t, \sqrt{3} \rangle$ for $t \in [0, 2\pi]$. Then $d\vec{r} = \langle -\sin t, \cos t, 0 \rangle dt$. On the curve $C$, the vector field is $\vec{F}(\vec{r}(t)) = \langle -\sin t, \cos t, \sqrt{3}(\cos t)^3(\sin t)^2 \rangle$. The line integral is: $\oint_C \vec{F} \cdot d\vec{r} = \int_0^{2\pi} \langle -\sin t, \cos t, \dots \rangle \cdot \langle -\sin t, \cos t, 0 \rangle dt$ $= \int_0^{2\pi} (\sin^2 t + \cos^2 t) dt = \int_0^{2\pi} 1 dt = 2\pi$ The direct computation of the surface integral would be much more difficult.
6. Let $\vec{F} = \langle 2y, 3x, 5z \rangle$. Let $C$ be the curve of intersection of the cylinder $x^2+y^2=4$ and the plane $x+z=3$, oriented counter-clockwise when viewed from above. Which of the following integrals correctly represents the value of $\oint_C \vec{F} \cdot d\vec{r}$ according to Stokes' theorem?
   Answer: $\iint_D 1 \, dA$, where $D$ is the disk $x^2+y^2 \le 4$ in the $xy$-plane. 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 $\oint_C \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot \hat{n} \, dS$, where $S$ is any surface bounded by curve $C$. First, calculate the curl of $\vec{F} = \langle 2y, 3x, 5z \rangle$: $\nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 2y & 3x & 5z \end{vmatrix} = \langle 0, 0, 3-2 \rangle = \langle 0, 0, 1 \rangle$ The curve $C$ is where the cylinder $x^2+y^2=4$ meets the plane $x+z=3$. The smartest choice for surface $S$ is the flat disk $D: x^2+y^2 \leq 4$ in the $xy$-plane (at $z=0$), since it's bounded by the projection of $C$. This disk has upward normal $\hat{n} = \langle 0, 0, 1 \rangle$ (matching the counter-clockwise orientation). Therefore: $\iint_S (\nabla \times \vec{F}) \cdot \hat{n} \, dS = \iint_D \langle 0, 0, 1 \rangle \cdot \langle 0, 0, 1 \rangle \, dA = \iint_D 1 \, dA$ Option A uses $5z$ instead of the curl. Option B incorrectly uses just the $z$-component value $5$ 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.
7. Consider the vector field $\mathbf{F} = (x^2 - z, y^2 + x, z^2 - y)$ and the curve $C$ consisting of the intersection of $x^2 + y^2 + z^2 = 9$ and $x + y + z = 3$, oriented counterclockwise when viewed from the point $(10, 10, 10)$. To apply Stokes' theorem efficiently, which surface choice would minimize computational complexity?
   Answer: The planar disk $x + y + z = 3$ bounded by $C$, using the parameterization $\mathbf{r}(u,v) = (u, v, 3-u-v)$ For Stokes' theorem $\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS$, we need to compute $\nabla \times \mathbf{F} = (-1-2z, -1-2x, 1-2y)$. The planar surface $x + y + z = 3$ has constant normal $\mathbf{n} = \frac{(1,1,1)}{\sqrt{3}}$ and simple surface element $dS = \sqrt{3} \, du \, dv$. The dot product becomes $\frac{1}{\sqrt{3}}[(-1-2z) + (-1-2x) + (1-2y)] = \frac{-1-2(x+y+z)}{\sqrt{3}} = \frac{-1-6}{\sqrt{3}} = \frac{-7}{\sqrt{3}}$, 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.
8. Let $S$ be the portion of the cone $z = \sqrt{x^2+y^2}$ that lies between the planes $z=1$ and $z=2$, with an upward-pointing normal. Let $\vec{F}$ be a vector field with $\nabla \times \vec{F} = \langle 0, 0, 2 \rangle$. Evaluate $\iint_S (\nabla \times \vec{F}) \cdot d\vec{S}$.
   Answer: $6\pi$ 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 $\nabla \times \vec{F} = \langle 0, 0, 2 \rangle$ is constant, you can factor it out of the surface integral: $\iint_S (\nabla \times \vec{F}) \cdot d\vec{S} = \iint_S \langle 0, 0, 2 \rangle \cdot d\vec{S} = 2 \iint_S \hat{k} \cdot d\vec{S}$. The remaining integral $\iint_S \hat{k} \cdot d\vec{S}$ represents the flux of the unit vector $\hat{k}$ through surface $S$. Since the cone has an upward-pointing normal and opens upward, this flux equals the area of the projection of $S$ onto the $xy$-plane. The surface $S$ projects to an annulus (ring) with inner radius 1 and outer radius 2, so the projected area is $\pi(2^2) - \pi(1^2) = 3\pi$. Therefore, $\iint_S (\nabla \times \vec{F}) \cdot d\vec{S} = 2 \cdot 3\pi = 6\pi$, which is answer C. Answer A ($3\pi$) represents just the projected area without the factor of 2 from the curl's $z$-component. Answer B ($4\pi$) might come from incorrectly using the full disk of radius 2 instead of the annulus. Answer D ($2\pi$) could result from using only the outer boundary area $\pi(2^2) - \pi(\sqrt{2})^2$ or other geometric errors. Remember: when curl is constant, factor it out and focus on the geometric flux calculation through the surface.
9. Let $S$ be the surface of the cylinder $x^2+y^2=1$ for $0 \le z \le 2$, with an outward-pointing normal. Note that $S$ does not include the top or bottom disks. For the vector field $\vec{F} = \langle z, x, y \rangle$, evaluate $\iint_S (\nabla \times \vec{F}) \cdot d\vec{S}$.
   Answer: $0$ By Stokes' theorem, the integral of the curl over the surface $S$ is equal to the line integral of $\vec{F}$ over the boundary of $S$. The boundary of this open cylinder consists of two curves: the circle $C_1$ at the top ($z=2$) and the circle $C_2$ at the bottom ($z=0$). The outward normal on the cylinder induces an orientation on the boundary curves. By the right-hand rule, the top curve $C_1$ is oriented counter-clockwise, while the bottom curve $C_2$ is oriented clockwise. Let's calculate the line integrals: For $C_1$: $\vec{r}_1(t) = \langle \cos t, \sin t, 2 \rangle$, $d\vec{r}_1 = \langle -\sin t, \cos t, 0 \rangle dt$. $\vec{F}$ on $C_1$ is $\langle 2, \cos t, \sin t \rangle$. The integral is $\int_0^{2\pi} (-2\sin t + \cos^2 t) dt = \pi$. For $C_2$ (clockwise): We can parameterize it as $\vec{r}_2(t) = \langle \cos t, -\sin t, 0 \rangle$, so $d\vec{r}_2 = \langle -\sin t, -\cos t, 0 \rangle dt$. $\vec{F}$ on $C_2$ is $\langle 0, \cos t, -\sin t \rangle$. The integral is $\int_0^{2\pi} (0 - \cos^2 t) dt = -\pi$. The total value is the sum of the integrals over the boundary components: $\pi + (-\pi) = 0$. Alternatively, one can compute the surface integral directly. $\nabla \times \vec{F} = \langle 1,1,1 \rangle$. The outward normal to the cylinder is $\vec{n} = \langle x, y, 0 \rangle = \langle \cos\theta, \sin\theta, 0 \rangle$. The dot product is $\cos\theta + \sin\theta$. Integrating $\int_0^2 \int_0^{2\pi} (\cos\theta + \sin\theta) d\theta dz = 0$.
10. Let $S_1$ be the disk $x^2 + y^2 \le 1$ in the $xy$-plane with normal $\vec{n}_1 = \hat{k}$. Let $S_2$ be the surface of the paraboloid $z = 1 - x^2 - y^2$ for $z \ge 0$ with normal $\vec{n}_2$ pointing upwards. Let $\vec{F}$ be a continuously differentiable vector field. What is the value of the expression $\iint_{S_1} (\nabla \times \vec{F}) \cdot d\vec{S} - \iint_{S_2} (\nabla \times \vec{F}) \cdot d\vec{S}$?
    Answer: 0 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 $S_1$ (the disk) and $S_2$ (the paraboloid) have the same boundary: the unit circle $x^2 + y^2 = 1$ in the $xy$-plane. By Stokes' theorem, $\iint_{S_1} (\nabla \times \vec{F}) \cdot d\vec{S} = \oint_C \vec{F} \cdot d\vec{r}$ and $\iint_{S_2} (\nabla \times \vec{F}) \cdot d\vec{S} = \oint_C \vec{F} \cdot d\vec{r}$, where $C$ 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: $\iint_{S_1} (\nabla \times \vec{F}) \cdot d\vec{S} = \iint_{S_2} (\nabla \times \vec{F}) \cdot d\vec{S}$ Therefore: $\iint_{S_1} (\nabla \times \vec{F}) \cdot d\vec{S} - \iint_{S_2} (\nabla \times \vec{F}) \cdot d\vec{S} = 0$ Choice A incorrectly suggests the result relates to flux through $S_2$. Choice B incorrectly assumes some doubling relationship. Choice D is wrong because this result holds for any continuously differentiable vector field $\vec{F}$. 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.
11. Let $C$ be the curve of intersection of the cylinder $x^2+y^2=1$ and the plane $z=y+2$. Evaluate $\oint_C \vec{F} \cdot d\vec{r}$ for the vector field $\vec{F} = \langle e^x \sin y, e^x \cos y, z^2 \rangle$.
    Answer: $0$ 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 $\vec{F} = \langle e^x \sin y, e^x \cos y, z^2 \rangle$. For a conservative field in 3D, we need $\nabla \times \vec{F} = \vec{0}$. Computing the curl: $\frac{\partial}{\partial y}(z^2) - \frac{\partial}{\partial z}(e^x \cos y) = 0 - 0 = 0$, $\frac{\partial}{\partial z}(e^x \sin y) - \frac{\partial}{\partial x}(z^2) = 0 - 0 = 0$, and $\frac{\partial}{\partial x}(e^x \cos y) - \frac{\partial}{\partial y}(e^x \sin y) = e^x \cos y - e^x \cos y = 0$. Since the curl is zero, $\vec{F}$ 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 $\nabla \times \vec{F} = \vec{0}$, then any closed curve integral immediately equals zero.
12. A student claims that for any closed curve $C$ and vector field $\mathbf{F}$, Stokes' theorem guarantees that $\oint_C \mathbf{F} \cdot d\mathbf{r} = 0$ because a closed curve bounds no surface. Which of the following best identifies the error in this reasoning?
    Answer: The student confuses 'closed curve' with 'boundary of a surface'; every simple closed curve in $\mathbb{R}^3$ is the boundary of some surface 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 $\mathbb{R}^3$ 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 $\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS$ for any surface $S$ with boundary $C$. The line integral is generally nonzero unless $\nabla \times \mathbf{F} = 0$ (i.e., $\mathbf{F}$ 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.
13. Let $S$ be the portion of the parabolic cylinder $z=y^2$ that lies inside the cylinder $x^2+y^2=1$. Let $C$ be the boundary of $S$, oriented counter-clockwise when viewed from above. Evaluate $\oint_C \langle -y, x, z \rangle \cdot d\vec{r}$.
    Answer: $2\pi$ We use Stokes' theorem to convert the line integral into a surface integral over $S$: $\oint_C \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot d\vec{S}$. First, calculate the curl of $\vec{F} = \langle -y, x, z \rangle$: $\nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ -y & x & z \end{vmatrix} = \hat{i}(0-0) - \hat{j}(0-0) + \hat{k}(1 - (-1)) = \langle 0, 0, 2 \rangle$ The surface $S$ is defined by $z=y^2$ over the domain $D$ which is the unit disk $x^2+y^2 \le 1$ in the $xy$-plane. We can parameterize $S$ as $\vec{r}(x,y) = \langle x, y, y^2 \rangle$. The normal vector is $\vec{r}_x \times \vec{r}_y = \langle 1,0,0 \rangle \times \langle 0,1,2y \rangle = \langle 0, -2y, 1 \rangle$. The $z$-component is positive, so this corresponds to the required upward orientation. The surface integral is: $\iint_S (\nabla \times \vec{F}) \cdot d\vec{S} = \iint_D \langle 0,0,2 \rangle \cdot \langle 0,-2y,1 \rangle dA = \iint_D 2 dA$ This is 2 times the area of the unit disk $D$. The area of $D$ is $\pi(1^2) = \pi$. So the integral is $2\pi$. Alternatively, we could use a simpler surface with the same boundary, namely the disk $x^2+y^2 \le 1$ itself. The integral would be $\iint_D \langle 0,0,2 \rangle \cdot \langle 0,0,1 \rangle dA = \iint_D 2 dA = 2\pi$.
14. Consider two surfaces $S_1$ and $S_2$ with the same boundary curve $C$, where $S_1$ is a flat disk and $S_2$ is a portion of a sphere. Both surfaces are oriented consistently with the same orientation of $C$. For a vector field $\mathbf{F}$ such that $\nabla \times \mathbf{F}$ is not constant, under what condition will $\iint_{S_1} (\nabla \times \mathbf{F}) \cdot \mathbf{n}_1 \, dS_1 = \iint_{S_2} (\nabla \times \mathbf{F}) \cdot \mathbf{n}_2 \, dS_2$?
    Answer: Always, since both integrals equal $\oint_C \mathbf{F} \cdot d\mathbf{r}$ by Stokes' theorem By Stokes' theorem, for any surface $S$ with boundary $C$ (oriented consistently), we have $\iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS = \oint_C \mathbf{F} \cdot d\mathbf{r}$. Since the line integral depends only on the vector field $\mathbf{F}$ and the boundary curve $C$ (with its orientation), and not on the choice of surface $S$, both surface integrals must be equal regardless of the specific properties of $\mathbf{F}$ or $\nabla \times \mathbf{F}$. This is a fundamental consequence of Stokes' theorem. Option A is wrong because $\mathbf{F}$ being conservative would make $\nabla \times \mathbf{F} = 0$ 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.

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