Answer: $2\pi$

Parametrize the unit circle: Let $\mathbf{r}(t) = \langle \cos t, \sin t \rangle$ for $t \in [0, 2\pi]$. Then $d\mathbf{r} = \langle -\sin t, \cos t \rangle\, dt$. Evaluate the integrand: $\mathbf{F}(\cos t, \sin t) = \langle -\sin t, \cos t \rangle$. The dot product is $\mathbf{F} \cdot d\mathbf{r} = (-\sin t)(-\sin t) + (\cos t)(\cos t) = \sin^2 t + \cos^2 t = 1$. Compute the integral: $\oint_C \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} 1\, dt = 2\pi$. Why distractors fail: Option A ($0$) would be the result for a conservative field, but $\mathbf{F} = \langle -y, x \rangle$ is not conservative. Option B ($\pi$) uses incorrect limits or a half-circle. Option D ($4\pi$) doubles the correct answer, perhaps from a radius error.

Answer: (i) $\oint_C \mathbf{F} \cdot d\mathbf{r}$ and (ii) $\iint_S \mathbf{F} \cdot d\mathbf{S}$

Identify the physics of each task: Task (i) asks about energy gained along a path, which is work/circulation: $\oint_C \mathbf{F} \cdot d\mathbf{r}$. Task (ii) asks about net volume flow through a surface, which is 3D flux: $\iint_S \mathbf{F} \cdot d\mathbf{S}$. Why the correct answer works: Option B correctly pairs the circulation integral for energy along the loop with the surface flux integral for volumetric flow rate through $S$. Why distractors fail: Option A reverses the two integrals. Option C uses the 2D flux integral $\oint_C \mathbf{F} \cdot \mathbf{n}\, ds$ for task (i), but that measures flow across a curve, not energy along a path. Option D uses only line integrals and omits the required surface integral for task (ii).

Answer: $\iiint_V 3\, dV$

Apply the Divergence Theorem: The Divergence Theorem states $\iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_V \nabla \cdot \mathbf{F}\, dV$. Compute the divergence: $\nabla \cdot \mathbf{F} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 1 + 1 + 1 = 3$. Why the correct answer works: Since $\nabla \cdot \mathbf{F} = 3$, the flux equals $\iiint_V 3\, dV = 3\, \text{Vol}(V)$. Why distractors fail: Option A uses divergence $= 1$, computing only one partial derivative. Option C confuses the divergence (sum of partial derivatives) with the sum of the components. Option D uses $\|\mathbf{F}\|^2$, which has no role in the Divergence Theorem.

Answer: $2\pi$

Use the 2D flux formula: For $\mathbf{F} = \langle P, Q \rangle$ and a counterclockwise curve, the outward flux is $\oint_C \mathbf{F} \cdot \mathbf{n}\, ds = \oint_C (P\, dy - Q\, dx)$. Parametrize and compute: With $\mathbf{r}(t) = \langle \cos t, \sin t \rangle$, we have $dx = -\sin t\, dt$ and $dy = \cos t\, dt$. So the integrand is $\cos t \cdot \cos t - \sin t \cdot (-\sin t) = \cos^2 t + \sin^2 t = 1$. Thus the flux is $\int_0^{2\pi} 1\, dt = 2\pi$. Alternative via divergence: Since $\text{div}\, \mathbf{F} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} = 2$, by Green's theorem (divergence form), $\oint_C \mathbf{F} \cdot \mathbf{n}\, ds = \iint_D 2\, dA = 2\pi$. Why distractors fail: Option A ($0$) would hold if the divergence were zero. Option B ($\pi$) comes from forgetting the factor of 2 in the divergence or computing half the area. Option D ($4\pi$) doubles the result, possibly using a radius of $\sqrt{2}$.

Answer: $1$

Determine the normal vector: The plane $z = 1$ has upward unit normal $\mathbf{n} = \langle 0, 0, 1 \rangle$, so $d\mathbf{S} = \langle 0, 0, 1 \rangle\, dA$. Compute the dot product: On $S$, $z = 1$, so $\mathbf{F} = \langle 1, 0, x \rangle$. Then $\mathbf{F} \cdot d\mathbf{S} = \langle 1, 0, x \rangle \cdot \langle 0, 0, 1 \rangle\, dA = x\, dA$. Evaluate the integral: $\iint_S \mathbf{F} \cdot d\mathbf{S} = \int_0^1 \int_0^1 x\, dx\, dy = \int_0^1 \frac{1}{2}\, dy = \frac{1}{2}$. Wait — let me recheck. $\int_0^1 x\, dx = \frac{1}{2}$ and $\int_0^1 dy = 1$, so the result is $\frac{1}{2}$. Actually, upon further review, we need to recompute. $\mathbf{F} \cdot \mathbf{n} = x$, so $\iint_S x\, dA = \frac{1}{2}$. Let me reconsider the problem. With $\mathbf{F} = \langle z, 0, x \rangle$ on $z=1$: $\mathbf{F} = \langle 1, 0, x \rangle$, $\mathbf{F} \cdot \langle 0,0,1 \rangle = x$. The integral is $\int_0^1\int_0^1 x\,dx\,dy = \frac{1}{2}$. So the correct answer is $\frac{1}{2}$. Why distractors fail: Option A ($0$) would occur if $\mathbf{F} \cdot \mathbf{n} = 0$ everywhere. Option C ($1$) results from incorrectly using the first component of $\mathbf{F}$ (which is $z = 1$) instead of the third. Option D ($2$) comes from integrating the wrong component or doubling.

Answer: $7$

Parametrize the path: The straight line from $(0,0)$ to $(1,2)$ can be parametrized as $\mathbf{r}(t) = \langle t, 2t \rangle$ for $t \in [0, 1]$. Then $d\mathbf{r} = \langle 1, 2 \rangle\, dt$. Set up and evaluate the integral: Along this path, $\mathbf{F}(t, 2t) = \langle 2t, 6t \rangle$. So $W = \int_0^1 \langle 2t, 6t \rangle \cdot \langle 1, 2 \rangle\, dt = \int_0^1 (2t + 12t)\, dt = \int_0^1 14t\, dt = 7$. Why distractors fail: Option A ($5$) results from computing $\int_0^1 (2t + 6t)\, dt$ by forgetting the factor of 2 in the $dy$ component. Option C ($8$) comes from a parametrization error. Option D ($10$) results from using $\int_0^1 (2t + 6t) \cdot 2\, dt$ incorrectly.

Answer: The outward-pointing unit normal vector to $C$

Identify the components of the 2D flux integral: The 2D flux integral $\oint_C \mathbf{F} \cdot \mathbf{n}\, ds$ measures how much of the field crosses the curve. Here $\mathbf{n}$ is the outward-pointing unit normal to $C$ and $ds$ is the arc length element. Why the correct answer works: Option C correctly identifies $\mathbf{n}$ as the outward unit normal, which is what gives the integral its interpretation as net flow across the boundary. Why distractors fail: Option A (unit tangent) would yield a circulation integral, not a flux integral. Option B is nonsensical since $\mathbf{F}$ is a vector field and the gradient of a vector field is not standard in this context. Option D (curvature vector) is unrelated to the flux integrand.

Answer: The field has a net component in the direction of traversal around the curve

Understand circulation: Circulation $\oint_C \mathbf{F} \cdot d\mathbf{r}$ sums the tangential component of $\mathbf{F}$ along the closed curve. A positive value means the field, on net, aligns with the direction of traversal. Why the correct answer works: Option B correctly describes that positive circulation means the tangential component of $\mathbf{F}$ agrees with the curve's orientation on average. Why distractors fail: Option A describes flux (the normal component), not circulation (the tangential component). Option C is incorrect because a conservative field has zero circulation around every closed curve, contradicting a positive value. Option D confuses magnitude change with tangential alignment.

Answer: The work done by $\mathbf{F}$ on a particle moving along $C$

Recall the definition of work: In vector calculus, the work done by a force field $\mathbf{F}$ on a particle moving along a curve $C$ is defined as $W = \int_C \mathbf{F} \cdot d\mathbf{r}$. The dot product projects the force onto the direction of motion at each point. Why the correct answer works: Option B directly states the physical interpretation of this integral: it computes the work done by the force along the path. Why distractors fail: Option A describes flux, which uses $\int_C \mathbf{F} \cdot \mathbf{n}\, ds$ in 2D. Option C describes a scalar line integral $\int_C \|\mathbf{F}\|\, ds$, not a vector line integral. Option D describes circulation, which specifically requires a closed curve; the general work integral does not require $C$ to be closed.

Answer: No, because zero circulation implies zero curl, not zero divergence, so flux can be nonzero

Relate circulation to curl via Green's theorem: By the circulation form of Green's theorem, $\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$. Zero circulation for all closed curves implies zero curl ($\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0$). Flux depends on divergence, not curl: By the divergence form of Green's theorem, $\oint_C \mathbf{F} \cdot \mathbf{n}\, ds = \iint_D \left(\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}\right) dA$. Zero curl does not imply zero divergence. Counterexample: The field $\mathbf{F} = \langle x, y \rangle$ is conservative (zero curl everywhere), so circulation is always zero. But its divergence is $2$, so flux through any closed curve enclosing a region of area $A$ is $2A \neq 0$. Why distractors fail: Option A incorrectly equates curl and divergence. Option B is false; circulation and flux measure fundamentally different things (tangential vs. normal flow). Option D is wrong because 2D flux is a well-defined concept.

Answer: Compute $\nabla \times \mathbf{F}$ and integrate over the enclosed region for rotation; compute $\nabla \cdot \mathbf{F}$ and integrate over the same region for expansion

Identify the two quantities needed: Rotational tendency corresponds to circulation, related to curl via the circulation form of Green's theorem: $\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_D (\partial Q/\partial x - \partial P/\partial y)\, dA$. Expansion rate corresponds to flux, related to divergence: $\oint_C \mathbf{F} \cdot \mathbf{n}\, ds = \iint_D (\partial P/\partial x + \partial Q/\partial y)\, dA$. Why Option B is most efficient: By computing curl and divergence of $\mathbf{F}$ (both are simple partial derivative computations) and integrating each over the same domain $D$, both quantities are obtained efficiently without parametrizing the boundary twice. Why Option A is less efficient: Option A is correct in principle but requires two separate boundary parametrizations, which is less efficient than using area integrals of curl and divergence over the same region. Why Options C and D fail: Option C is wrong because divergence measures only expansion, not rotation. Option D is wrong because circulation measures only rotational tendency, not expansion.

Answer: Both use $\int \mathbf{F} \cdot d\mathbf{r}$, but circulation specifically requires a closed curve while work does not

Compare the definitions: Work along a path is $W = \int_C \mathbf{F} \cdot d\mathbf{r}$, and circulation is $\oint_C \mathbf{F} \cdot d\mathbf{r}$. Both use the same integrand $\mathbf{F} \cdot d\mathbf{r}$. Key distinction: The critical difference is that circulation requires $C$ to be a closed curve (indicated by $\oint$), measuring net rotational tendency. Work can be computed along any curve, open or closed. Why distractors fail: Option A incorrectly attributes the flux integrand $\mathbf{F} \cdot \mathbf{n}\, ds$ to circulation. Option C is wrong because circulation is defined for any vector field; for conservative fields it happens to be zero. Option D is wrong because work is a line integral, not a surface integral.

Answer: No, the engineer should compute the outward flux $\oint_C \mathbf{F} \cdot \mathbf{n}\, ds$ instead, since flux measures net flow out of the region

Distinguish source/sink from rotation: A source or sink is characterized by net outward or inward flow, which is measured by flux (normal component), not circulation (tangential component). Why the correct answer works: Option C correctly identifies that the engineer needs flux $\oint_C \mathbf{F} \cdot \mathbf{n}\, ds$. Positive flux indicates a net source; negative flux indicates a net sink; zero flux indicates neither (or balanced sources and sinks). Why distractors fail: Options A and B incorrectly claim circulation measures source/sink behavior; circulation instead measures rotational tendency. Option D is wrong because 2D flux is perfectly valid for detecting sources and sinks in the plane.