Answer: $2x + z + 3z^2$

Recall the divergence formula: The divergence of $\mathbf{F} = P\,\mathbf{i} + Q\,\mathbf{j} + R\,\mathbf{k}$ is $\nabla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}$. Compute each partial derivative: $\frac{\partial}{\partial x}(x^2) = 2x$, $\frac{\partial}{\partial y}(yz) = z$, $\frac{\partial}{\partial z}(z^3) = 3z^2$. Summing gives $2x + z + 3z^2$. Why distractors fail: Option B replaces $z$ with $y$, confusing which variable survives the partial of $yz$ with respect to $y$. Option C simply lists the components without differentiating. Option D keeps the product $yz$ instead of differentiating $Q = yz$ with respect to $y$.

Answer: The claim is true; it follows from the equality of mixed partial derivatives (Clairaut's theorem), causing all terms to cancel in pairs

Verify the identity: $\nabla \cdot (\nabla \times \mathbf{F})$ expands to a sum of six mixed second partial derivatives. By Clairaut's theorem ($f_{xy} = f_{yx}$ for smooth functions), these terms cancel pairwise, giving zero. This is a universal vector identity: The identity $\nabla \cdot (\nabla \times \mathbf{F}) = 0$ holds for any sufficiently smooth $\mathbf{F}$. It does not depend on the field being conservative or on any particular theorem of integration. Why distractors fail: Option A is simply false — the identity always holds for smooth fields. Option B gives an incorrect justification; the Divergence Theorem is an integration result, not a pointwise differential identity. Option D is wrong because the identity holds for all smooth fields, not just conservative ones.

Answer: The flux through any closed surface is zero

Apply the Divergence Theorem: By the Divergence Theorem, $\iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_E (\nabla \cdot \mathbf{F})\,dV$. If $\nabla \cdot \mathbf{F} = 0$ everywhere, then the triple integral is zero for any region $E$. Conclusion: Therefore the net flux through any closed surface is zero. This is the hallmark of an incompressible or solenoidal field. Why distractors fail: Option A is wrong because zero divergence guarantees zero flux, not positive flux. Option B is wrong because the result holds regardless of the surface shape. Option D confuses divergence-free (solenoidal) with curl-free (conservative) — these are different conditions.

Answer: $\langle x - x, y - y, z - z \rangle = \langle 0, 0, 0 \rangle$

Apply the curl formula: With $P = yz$, $Q = xz$, $R = xy$: the $\mathbf{i}$-component is $\frac{\partial(xy)}{\partial y} - \frac{\partial(xz)}{\partial z} = x - x = 0$. Compute remaining components: The $\mathbf{j}$-component is $\frac{\partial(yz)}{\partial z} - \frac{\partial(xy)}{\partial x} = y - y = 0$. The $\mathbf{k}$-component is $\frac{\partial(xz)}{\partial x} - \frac{\partial(yz)}{\partial y} = z - z = 0$. Significance: A zero curl indicates that $\mathbf{F}$ is conservative (irrotational). Indeed, $\mathbf{F} = \nabla(xyz)$. Why distractors fail: Option B gives the gradient of $x^2 + y^2 + z^2$ incorrectly. Option C results from sign errors in the partial derivatives. Option D simply restates $\mathbf{F}$ itself.

Answer: $3\pi$

Apply Green's Theorem: Green's Theorem gives $\oint_C P\,dx + Q\,dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$ where $P = 2y$ and $Q = 5x$. Compute the integrand: $\frac{\partial Q}{\partial x} = 5$ and $\frac{\partial P}{\partial y} = 2$, so the integrand is $5 - 2 = 3$. Evaluate the double integral: $\iint_D 3\,dA = 3 \cdot (\text{area of unit disk}) = 3\pi$. Why distractors fail: Option A ($7\pi$) results from adding the partial derivatives instead of subtracting. Option C ($5\pi$) uses only $\frac{\partial Q}{\partial x} = 5$. Option D ($-3\pi$) reverses the subtraction order.

Answer: Use the Divergence Theorem to convert to a triple integral of $3(x^2 + y^2 + z^2)$ over the cube

Identify the setting: We need the flux through a closed surface. The Divergence Theorem is specifically designed for this: $\iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_E (\nabla \cdot \mathbf{F})\,dV$. Compute the divergence: $\nabla \cdot \mathbf{F} = 3x^2 + 3y^2 + 3z^2 = 3(x^2 + y^2 + z^2)$. The triple integral over $[0,1]^3$ is straightforward. Why distractors fail: Option A is correct in principle but far more tedious (six separate integrals). Option C is wrong because Stokes' Theorem relates a surface integral of curl to a line integral, not a flux integral to a line integral. Option D misapplies Green's Theorem, which is a 2D result and does not directly apply to individual faces of a 3D flux computation.

Answer: The Divergence Theorem does not apply because $\mathbf{F}$ is not defined at the origin, which lies inside the sphere

Identify the issue: The field $\mathbf{F} = \mathbf{r}/|\mathbf{r}|^3$ has a singularity at the origin. The Divergence Theorem requires $\mathbf{F}$ to be continuously differentiable throughout the entire enclosed volume $E$. The actual divergence: Away from the origin, $\nabla \cdot (\mathbf{r}/|\mathbf{r}|^3) = 0$ is indeed correct. However, at the origin the field is undefined, so the hypothesis of the Divergence Theorem is violated. The correct flux: Direct computation (or using a limiting argument with a small sphere) shows that the flux of this field through any closed surface containing the origin is $4\pi$, not zero. Why distractors fail: Option A is wrong because the divergence computation away from the origin is correct. Option C is wrong because the Divergence Theorem applies to any piecewise smooth closed surface. Option D is wrong because Stokes' Theorem relates line and surface integrals, and is not relevant to flux through a closed surface.

Answer: Stokes' Theorem

Apply Stokes' Theorem: Stokes' Theorem states $\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$. If $\nabla \times \mathbf{F} = \mathbf{0}$, the surface integral is zero, so the line integral is zero. Role of simple connectivity: Simple connectivity ensures that every closed curve $C$ bounds a surface $S$ within the domain, so Stokes' Theorem can be applied. Why distractors fail: Option A relates flux to volume integrals, not circulation to surface integrals. Option B misstates Green's Theorem (it involves curl, not divergence, in 2D). Option D is a consequence of the field being conservative, but the direct link from $\nabla \times \mathbf{F} = \mathbf{0}$ to the vanishing integral is via Stokes' Theorem.

Answer: The claim is true because $\nabla \times \mathbf{F} = \mathbf{0}$, and by Stokes' Theorem every closed line integral vanishes

Compute the curl: $\mathbf{i}$: $\frac{\partial(2yz)}{\partial y} - \frac{\partial(xe^y + z^2)}{\partial z} = 2z - 2z = 0$. $\mathbf{j}$: $\frac{\partial(e^y)}{\partial z} - \frac{\partial(2yz)}{\partial x} = 0 - 0 = 0$. $\mathbf{k}$: $\frac{\partial(xe^y + z^2)}{\partial x} - \frac{\partial(e^y)}{\partial y} = e^y - e^y = 0$. Apply Stokes' Theorem: Since $\nabla \times \mathbf{F} = \mathbf{0}$ and the domain is simply connected, Stokes' Theorem guarantees $\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S \mathbf{0} \cdot d\mathbf{S} = 0$ for any closed $C$. Why distractors fail: Option A incorrectly uses divergence to draw conclusions about circulation; divergence relates to flux, not circulation. Option C incorrectly states that zero divergence implies conservative — zero curl is the relevant condition. Option D is baseless; the presence of a $z^2$ term does not prevent path independence.

Answer: A double integral over the region enclosed by $C$

State Green's Theorem: Green's Theorem states $\oint_C (P\,dx + Q\,dy) = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$, where $D$ is the planar region enclosed by $C$. Why the correct answer works: The right-hand side is a double integral over the 2D region $D$ bounded by $C$, making Option B correct. Why distractors fail: Option A describes something like the Divergence Theorem (triple integral). Option C describes Stokes' Theorem (surface integral in 3D). Option D describes a property of conservative fields and path independence, not Green's Theorem.

Answer: Both approaches yield the same answer because Stokes' Theorem holds for any surface bounded by $C$

Key insight from Stokes' Theorem: Stokes' Theorem states $\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$. The left side depends only on $C$, not on the choice of $S$. Therefore any surface $S$ bounded by $C$ with consistent orientation gives the same value. Why both surfaces work: Both the hemisphere and the flat disk share the same boundary $C$ (the unit circle in the $xy$-plane). As long as the orientation is consistent with the same traversal of $C$, the surface integral of curl is the same. Why distractors fail: Option A is wrong because Stokes' Theorem imposes no requirement that the surface match the curve's shape. Option B is wrong because Stokes' Theorem works for any smooth (or piecewise smooth) oriented surface. Option D is wrong because consistent orientation with the same $C$ is guaranteed by hypothesis, yielding the same result.

Answer: When the surface $S$ in Stokes' Theorem is a flat region in the $xy$-plane, Stokes' Theorem reduces to Green's Theorem

Connect the two theorems: Stokes' Theorem states $\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$. If we choose $S$ to be a flat region $D$ in the $xy$-plane with unit normal $\mathbf{k}$, then the surface integral reduces to $\iint_D (\nabla \times \mathbf{F}) \cdot \mathbf{k}\,dA$. Show the reduction: For $\mathbf{F} = P\,\mathbf{i} + Q\,\mathbf{j}$ (a 2D field), $(\nabla \times \mathbf{F}) \cdot \mathbf{k} = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$, recovering exactly the integrand in Green's Theorem. Why distractors fail: Option B is incorrect because zero divergence is unrelated to this reduction. Option C is incorrect because a curve on the $z$-axis does not bound a planar region. Option D reverses the relationship — Green's Theorem is the 2D special case, not a 3D generalization.

Answer: $6\pi$

Recall the area formula from Green's Theorem: Green's Theorem gives the area of a region as $A = \frac{1}{2}\oint_C (x\,dy - y\,dx)$. For standard shapes we can directly use the known area formula: $A = \pi ab$ for an ellipse with semi-axes $a$ and $b$. Apply to the given ellipse: Here $a = 3$ and $b = 2$, so $A = \pi(3)(2) = 6\pi$. Why distractors fail: Option B ($12\pi$) doubles the area (perhaps forgetting the $\frac{1}{2}$ factor). Option C ($36\pi$) uses $a^2 b^2$ instead of $ab$. Option D has no natural derivation from this problem.

Answer: Apply Stokes' Theorem to replace the line integral with a surface integral of $\nabla \times \mathbf{F}$ over a surface bounded by $C$

Identify the theorem: The work integral $\oint_C \mathbf{F} \cdot d\mathbf{r}$ around a closed curve is related to a surface integral by Stokes' Theorem: $\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$. Why this helps: Even though $C$ is hard to parametrize, we can choose a surface $S$ bounded by $C$ (such as the portion of the paraboloid) and evaluate the surface integral of curl, which may be simpler. Why distractors fail: Option A is wrong because the Divergence Theorem converts a surface flux integral to a volume integral, not a line integral to a volume integral. Option C fails because $\nabla \times \mathbf{F} \neq \mathbf{0}$ (verifiable by computation), so $\mathbf{F}$ is not conservative and has no potential function. Option D is incorrect because Green's Theorem applies only to curves in the $xy$-plane, not curves on a paraboloid in 3D.

Answer: $\left(\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}\right)\mathbf{i} + \left(\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}\right)\mathbf{j} + \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\mathbf{k}$

Recall the determinant formula for curl: The curl is computed as $\nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \partial/\partial x & \partial/\partial y & \partial/\partial z \\ P & Q & R \end{vmatrix}$. Expand the determinant: The $\mathbf{i}$-component is $\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}$, the $\mathbf{j}$-component is $\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}$ (note the sign from cofactor expansion), and the $\mathbf{k}$-component is $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$. Why distractors fail: Option B reverses the sign of every component (a common error from mishandling the determinant signs). Option C gives the components of the gradient applied component-wise, which is related to divergence, not curl. Option D has an incorrect $\mathbf{j}$-component — it omits the sign change needed for the middle cofactor.

Answer: A triple integral of $\nabla \cdot \mathbf{F}$ over the enclosed volume

State the Divergence Theorem: The Divergence Theorem says $\iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_E (\nabla \cdot \mathbf{F})\,dV$, where $E$ is the solid region enclosed by the closed surface $S$. Why the correct answer works: The theorem converts a flux integral over a closed surface into a volume (triple) integral of the divergence, which is Option C. Why distractors fail: Option A describes the conversion done by Stokes' Theorem (surface to line). Option B describes a 2D integral, but the Divergence Theorem works in 3D. Option D describes the integrand in Stokes' Theorem, not the Divergence Theorem.

Answer: $32\pi$

Compute the divergence: $\nabla \cdot \mathbf{F} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 1 + 1 + 1 = 3$. Apply the Divergence Theorem: $\iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_E 3\,dV = 3 \cdot V$, where $V$ is the volume of the sphere of radius $2$. Compute the volume: $V = \frac{4}{3}\pi(2)^3 = \frac{32\pi}{3}$. Therefore the flux is $3 \cdot \frac{32\pi}{3} = 32\pi$. Why distractors fail: Option A ($16\pi$) is the surface area of the sphere, not the flux. Option C ($8\pi$) uses the wrong radius or divergence. Option D ($\frac{32\pi}{3}$) is just the volume, without multiplying by the divergence.

Answer: Invoke the Divergence Theorem: since $\nabla \cdot \mathbf{v} = 0$ throughout $E$, the theorem immediately gives $\iint_{\partial E} \mathbf{v} \cdot d\mathbf{S} = \iiint_E 0\,dV = 0$

Recognize the Divergence Theorem shortcut: The Divergence Theorem states $\iint_{\partial E} \mathbf{v} \cdot d\mathbf{S} = \iiint_E (\nabla \cdot \mathbf{v})\,dV$. If $\nabla \cdot \mathbf{v} = 0$ everywhere in $E$, the volume integral is trivially zero. Why this is the most efficient approach: No parametrization of the boundary is needed. The conclusion follows immediately from knowing the divergence is zero, regardless of how complex the boundary geometry is. Why distractors fail: Option A is correct in principle but extremely inefficient for complex boundaries. Option C misapplies Stokes' Theorem, which converts a surface integral of curl to a line integral, not a flux integral. Option D confuses conservative (curl-free) fields with incompressible (divergence-free) fields; being conservative does not guarantee zero flux.

Answer: The field has a constant rotational tendency about the $z$-axis at every point

Interpret the curl physically: The curl of a vector field measures the local rotational tendency (circulation density) of the field. A constant curl $\langle 0, 0, 2 \rangle$ means the field induces the same amount of counterclockwise rotation about the $z$-axis at every point. Why distractors fail: Option A describes divergence (expansion), not curl (rotation). Option C is wrong because a nonzero curl means the field is not conservative. Option D is wrong because $\nabla \cdot \mathbf{F} = \frac{\partial(-y)}{\partial x} + \frac{\partial(x)}{\partial y} + \frac{\partial(0)}{\partial z} = 0 + 1 + 0 = 1$, not $2$.