Stokes' Theorem & Curl Integrals - Calculus 3

1. Consider the vector field $\vec{F} = (z, x, y)$ and the surface $S$ defined by the paraboloid $z = 4 - x^2 - y^2$ for $z \geq 0$, with outward normal orientation. The boundary curve $C$ is traversed counterclockwise when viewed from above. Which statement about applying Stokes' Theorem is correct?
   Answer: Stokes' Theorem gives $\oint_C \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot \hat{n} \, dS = 4\pi$ Stokes' Theorem applies to any orientable surface with boundary, so option B is wrong. We compute $\nabla \times \vec{F} = \nabla \times (z, x, y) = (1-0, 0-1, 1-0) = (1, -1, 1)$. The surface normal vector is $\vec{N} = \nabla(z - 4 + x^2 + y^2) = (2x, 2y, 1)$, but we need the outward normal. Since the paraboloid opens downward, the outward normal points away from the origin, so $\hat{n} = \frac{(2x, 2y, 1)}{\sqrt{4x^2 + 4y^2 + 1}}$. However, it's easier to use the projection method: $(\nabla \times \vec{F}) \cdot \hat{n} \, dS = (1, -1, 1) \cdot (2x, 2y, 1) \, dx \, dy = (2x - 2y + 1) \, dx \, dy$. Over the disk $x^2 + y^2 \leq 4$: $\iint_{x^2+y^2 \leq 4} (2x - 2y + 1) \, dx \, dy = \iint_{x^2+y^2 \leq 4} 1 \, dx \, dy = 4\pi$ since the $2x$ and $-2y$ terms integrate to zero by symmetry.
2. Let $\vec{F}(x,y,z) = (2y + z^2, 2x + z^3, 2xz + 3yz^2)$ and let $S$ be the portion of the sphere $x^2 + y^2 + z^2 = 9$ with $z \geq 0$, oriented with upward normal. If we apply Stokes' Theorem to find $\oint_{\partial S} \vec{F} \cdot d\vec{r}$, what computational approach is most efficient?
   Answer: Use the flat disk $z = 0$, $x^2 + y^2 \leq 9$ as the surface since $\nabla \times \vec{F} = \vec{0}$ First, we compute $\nabla \times \vec{F}$: $\nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 2y + z^2 & 2x + z^3 & 2xz + 3yz^2 \end{vmatrix} = (6yz - 3z^2)\hat{i} - (2z - 2z)\hat{j} + (2 - 2)\hat{k} = (6yz - 3z^2, 0, 0)$. Wait, let me recalculate: $\frac{\partial}{\partial y}(2xz + 3yz^2) - \frac{\partial}{\partial z}(2x + z^3) = 3z^2 - 3z^2 = 0$, $\frac{\partial}{\partial z}(2y + z^2) - \frac{\partial}{\partial x}(2xz + 3yz^2) = 2z - 2z = 0$, $\frac{\partial}{\partial x}(2x + z^3) - \frac{\partial}{\partial y}(2y + z^2) = 2 - 2 = 0$. So $\nabla \times \vec{F} = \vec{0}$. By Stokes' Theorem, the line integral equals zero regardless of which surface we choose, but using the flat disk is computationally simplest. Option A would work but is unnecessarily complex, option B avoids Stokes' Theorem entirely, and option D is incorrect since divergence theorem applies to closed surfaces and volume integrals.
3. Let $\vec{F} = (P, Q, R)$ where $P = yz + x^2$, $Q = xz + y^2$, and $R = xy + z^2$. Consider the curve $C$ that is the boundary of the surface $S$ defined by $z = 1 - x^2 - y^2$ for $z \geq 0$, with $C$ oriented counterclockwise when viewed from above. What is $\oint_C \vec{F} \cdot d\vec{r}$?
   Answer: $0$ We apply Stokes' Theorem: $\oint_C \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot \hat{n} \, dS$. First, we compute the curl: $\nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ yz + x^2 & xz + y^2 & xy + z^2 \end{vmatrix}$. Computing each component: $i$-component: $\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} = \frac{\partial (xy + z^2)}{\partial y} - \frac{\partial (xz + y^2)}{\partial z} = x - x = 0$. $j$-component: $\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} = \frac{\partial (yz + x^2)}{\partial z} - \frac{\partial (xy + z^2)}{\partial x} = y - y = 0$. $k$-component: $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \frac{\partial (xz + y^2)}{\partial x} - \frac{\partial (yz + x^2)}{\partial y} = z - z = 0$. Therefore, $\nabla \times \vec{F} = \vec{0}$, which means $\iint_S (\nabla \times \vec{F}) \cdot \hat{n} \, dS = 0$. Hence $\oint_C \vec{F} \cdot d\vec{r} = 0$.
4. Consider the vector field $\vec{F} = (y^2, -x^2, xyz)$ and the helical curve $C$ parameterized by $\vec{r}(t) = (\cos t, \sin t, t)$ for $0 \leq t \leq 2\pi$. To use Stokes' Theorem to evaluate $\oint_C \vec{F} \cdot d\vec{r}$, which statement about choosing an appropriate surface $S$ is correct?
   Answer: Stokes' Theorem cannot be applied because $C$ is not a simple closed curve Stokes' Theorem applies to surface integrals over orientable surfaces whose boundary is a simple closed curve. The given curve $C: \vec{r}(t) = (\cos t, \sin t, t)$ for $0 \leq t \leq 2\pi$ is a helix that starts at $(1,0,0)$ and ends at $(1,0,2\pi)$. Since the starting and ending points are different, $C$ is not a closed curve at all, let alone a simple closed curve. Therefore, Stokes' Theorem does not apply. Option A incorrectly suggests using a surface that contains $C$, but the boundary of such a surface would not be $C$ alone. Option B correctly identifies that $C$ is not closed but gives the wrong reason. Option C misunderstands the requirements for Stokes' Theorem - we need a surface whose boundary is exactly $C$, and since $C$ has endpoints, no such surface exists in the context of Stokes' Theorem.
5. For the vector field $\vec{F} = (z^2 \sin y, x^2 \cos z, y^2 \sin x)$, consider two different surfaces $S_1$ and $S_2$ that have the same boundary curve $C$. If $\iint_{S_1} (\nabla \times \vec{F}) \cdot \hat{n}_1 \, dS = 5$ where $\hat{n}_1$ gives the correct orientation for Stokes' Theorem, what is $\iint_{S_2} (\nabla \times \vec{F}) \cdot \hat{n}_2 \, dS$ where $\hat{n}_2$ also gives the correct orientation?
   Answer: $5$, because Stokes' Theorem guarantees the surface integrals are equal This is a direct application of Stokes' Theorem: $\oint_C \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot \hat{n} \, dS$ for any orientable surface $S$ with boundary $C$, where $\hat{n}$ is chosen to be consistent with the orientation of $C$. Since both $S_1$ and $S_2$ have the same boundary $C$, and both normal vectors $\hat{n}_1$ and $\hat{n}_2$ are chosen to give the correct orientation for Stokes' Theorem (meaning they both correspond to the same orientation of $C$), the surface integrals must be equal. This is the fundamental content of Stokes' Theorem - the surface integral depends only on the boundary curve and its orientation, not on the specific surface chosen. Option B is incorrect because the orientations are both correct, not opposite. Option C misses the point of Stokes' Theorem. Option D is false; there's no reason to expect zero circulation for this particular vector field.
6. Consider the vector field $\vec{F} = (yz, xz + z^2, xy + 2yz)$ and let $C$ be the triangle with vertices at $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$, oriented counterclockwise when viewed from the origin. To apply Stokes' Theorem most efficiently, which surface $S$ with boundary $C$ should be used?
   Answer: The triangular surface with vertices $(1,0,0)$, $(0,1,0)$, $(0,0,1)$ since any surface choice gives the same result but this minimizes area By Stokes' Theorem, $\oint_C \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot \hat{n} \, dS$ for any surface $S$ with boundary $C$. We compute $\nabla \times \vec{F} = (x + 2z - x, y - z, z - z) = (2z, y - z, 0)$. The triangular surface lies in the plane $x + y + z = 1$, so $\hat{n} = \frac{(1,1,1)}{\sqrt{3}}$. We have $(\nabla \times \vec{F}) \cdot \hat{n} = \frac{1}{\sqrt{3}}(2z + y - z) = \frac{1}{\sqrt{3}}(y + z)$. Since the triangle has the smallest area among reasonable surface choices and the curl is not particularly simple, using the triangle minimizes computation. Option B is wrong because $\nabla \times \vec{F}$ is not constant and its normal component to the plane $x + y + z = 1$ is not zero. Options C and D would work but involve more complex parameterizations without computational benefit.
7. Let $\vec{F}(x,y,z) = (x^2 - y, xy + z, yz - x)$ and consider the closed curve $C$ formed by the intersection of the cylinder $x^2 + z^2 = 1$ and the plane $y = 2$, oriented so that $\oint_C \vec{F} \cdot d\vec{r} > 0$. Using Stokes' Theorem with the disk $S: y = 2, x^2 + z^2 \leq 1$, what is the value of this line integral?
   Answer: $\pi$ First, we compute $\nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x^2-y & xy+z & yz-x \end{vmatrix} = (y-1)\hat{i} - (-1-(-1))\hat{j} + (y-(-1))\hat{k} = (y-1, 0, y+1)$. For the disk $S: y = 2, x^2 + z^2 \leq 1$, we need to determine the correct orientation. The curve $C$ lies in the plane $y = 2$ and forms a circle $x^2 + z^2 = 1$. If we want $\oint_C \vec{F} \cdot d\vec{r} > 0$, we need to choose the normal direction appropriately. With $\hat{n} = \hat{j}$ (positive $y$-direction), we get $(\nabla \times \vec{F}) \cdot \hat{n} = 0$. With $\hat{n} = -\hat{j}$ (negative $y$-direction), we still get $(\nabla \times \vec{F}) \cdot \hat{n} = 0$. Wait, let me recalculate the curl more carefully. We have $\nabla \times \vec{F} = \left(\frac{\partial (yz-x)}{\partial y} - \frac{\partial (xy+z)}{\partial z}, \frac{\partial (x^2-y)}{\partial z} - \frac{\partial (yz-x)}{\partial x}, \frac{\partial (xy+z)}{\partial x} - \frac{\partial (x^2-y)}{\partial y}\right) = (z - 1, 0 - (-1), y - (-1)) = (z-1, 1, y+1)$. Now $(\nabla \times \vec{F}) \cdot \hat{j} = 1$, so $\iint_S (\nabla \times \vec{F}) \cdot \hat{n} \, dS = \iint_{x^2+z^2 \leq 1} 1 \, dx \, dz = \pi$.

Start this quiz