Answer: A vector field is conservative if and only if line integrals of the field depend only on the endpoints, not on the particular path taken

Path independence explained: A line integral $\int_C \mathbf{F} \cdot d\mathbf{r}$ is path-independent if its value depends only on the starting and ending points of $C$, not on the specific route. This is equivalent to $\mathbf{F}$ being conservative. Why the correct answer works: Option B correctly captures the biconditional relationship: $\mathbf{F}$ is conservative if and only if $\int_C \mathbf{F} \cdot d\mathbf{r}$ is path-independent. Why distractors fail: Option A conflates arc-length integrals with vector line integrals. Option C is false; most continuous fields are not path-independent. Option D incorrectly claims the value must be positive — path-independent integrals can be zero or negative.

Answer: $f(B) - f(A)$

Fundamental Theorem for Line Integrals: If $\mathbf{F} = \nabla f$ and $C$ goes from $A$ to $B$, then $\int_C \mathbf{F} \cdot d\mathbf{r} = f(B) - f(A)$. This is the line-integral analogue of the Fundamental Theorem of Calculus. Why the correct answer works: Option B gives $f(B) - f(A)$, which is the correct application of the theorem: evaluate the potential at the terminal point minus the potential at the initial point. Why distractors fail: Option A reverses the order, giving $f(A) - f(B)$, a common sign error. Option C adds the values instead of subtracting. Option D gives a vector quantity, but the line integral of a vector field produces a scalar.

Answer: $\mathbf{F} = \langle 2x + y, x + 2y \rangle$

Apply the component test: For $\mathbf{F} = \langle P, Q \rangle$ on $\mathbb{R}^2$ (simply connected), $\mathbf{F}$ is conservative if and only if $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$. Test each option: Option A: $P_y = -1$, $Q_x = 1$. Not equal — not conservative. Option B: $P_y = 2y$, $Q_x = 3y$. Not equal — not conservative. Option C: $P_y = 1$, $Q_x = 1$. Equal — conservative! Option D: $P = x + y^2 \Rightarrow P_y = 2y$; $Q = 2xy + x \Rightarrow Q_x = 2y + 1$. Not equal — not conservative. Why distractors fail: Option A represents a rotation field, which has nonzero curl. Option B fails because $2y \neq 3y$. Option D fails because $P_y = 2y$ while $Q_x = 2y + 1$; they differ by a constant 1.

Answer: $f(x,y) = x^2 y + 3x - 2y^2 + C$

Integrate the first component with respect to $x$: We need $f_x = 2xy + 3$. Integrating with respect to $x$: $f(x,y) = x^2 y + 3x + g(y)$ for some function $g(y)$. Determine $g(y)$ using the second component: We need $f_y = x^2 + g'(y) = x^2 - 4y$. So $g'(y) = -4y$, giving $g(y) = -2y^2 + C$. Assemble the potential function: Therefore $f(x,y) = x^2 y + 3x - 2y^2 + C$, which is Option A. Why distractors fail: Option B uses $+4y$ instead of $-2y^2$, failing to integrate $-4y$ correctly. Option C does not integrate $2xy$ properly (should get $x^2 y$, not $2xy$). Option D has $-3x$ instead of $+3x$, a sign error.

Answer: $10$

Parameterize the line segment: Let $\mathbf{r}(t) = \langle 3t, 4t \rangle$ for $t \in [0,1]$. Then $\mathbf{r}'(t) = \langle 3, 4 \rangle$ and $\|\mathbf{r}'(t)\| = 5$, so $ds = 5\,dt$. Evaluate the integral: On $C$, $y = 4t$. So $\int_C y\, ds = \int_0^1 4t \cdot 5\, dt = 20 \int_0^1 t\, dt = 20 \cdot \frac{1}{2} = 10$. Why distractors fail: Option A ($8$) might result from forgetting the arc-length factor and computing $\int_0^1 4t \cdot 4\,dt = 8$. Option C ($12$) could result from a wrong speed calculation. Option D ($20$) omits the $\frac{1}{2}$ from integrating $t$.

Answer: The integral of $f$ along $C$ weighted by arc length

Meaning of the scalar line integral: The scalar line integral $\int_C f\, ds$ integrates the scalar function $f$ along the curve $C$, where $ds = \|\mathbf{r}'(t)\| \, dt$ represents the arc-length element. It accumulates the values of $f$ weighted by arc length. Why the correct answer works: Option B correctly identifies the integral as the integral of $f$ weighted by arc length along the curve. Why distractors fail: Option A describes a vector line integral (work), which involves a dot product $\mathbf{F} \cdot d\mathbf{r}$, not a scalar function times $ds$. Option C describes a flux integral, which is a different type of integral. Option D would be $\frac{1}{b-a}\int_a^b f(\mathbf{r}(t))\,dt$, which lacks the arc-length weighting $\|\mathbf{r}'(t)\|$.

Answer: $\operatorname{curl} \mathbf{F} = \mathbf{0}$

Curl test for conservative fields: On a simply connected domain, a smooth vector field $\mathbf{F}$ is conservative if and only if $\operatorname{curl} \mathbf{F} = \mathbf{0}$. This follows because if $\mathbf{F} = \nabla f$, then $\operatorname{curl}(\nabla f) = \mathbf{0}$ always holds, and on a simply connected domain the converse is also true. Why the correct answer works: Option B states $\operatorname{curl} \mathbf{F} = \mathbf{0}$, which is the standard equivalent condition for conservativeness on a simply connected domain. Why distractors fail: Option A describes a solenoidal (divergence-free) field, which is a different property. Option C has no connection to conservative fields. Option D imposes an orthogonality condition that is not related to conservativeness.

Answer: $1$

Check if $\mathbf{F}$ is conservative: We verify: $\frac{\partial}{\partial y}(yz) = z = \frac{\partial}{\partial x}(xz)$, $\frac{\partial}{\partial z}(yz) = y = \frac{\partial}{\partial x}(xy)$, and $\frac{\partial}{\partial z}(xz) = x = \frac{\partial}{\partial y}(xy)$. So $\operatorname{curl} \mathbf{F} = \mathbf{0}$ and $\mathbf{F}$ is conservative. Find the potential function: Integrating $f_x = yz$ gives $f = xyz + g(y,z)$. Then $f_y = xz + g_y = xz$ implies $g_y = 0$, and $f_z = xy + g_z = xy$ implies $g_z = 0$. So $f(x,y,z) = xyz + C$. Apply the Fundamental Theorem: $\int_C \mathbf{F} \cdot d\mathbf{r} = f(1,1,1) - f(1,0,0) = (1)(1)(1) - (1)(0)(0) = 1 - 0 = 1$. Why distractors fail: Option A ($0$) would be correct for a closed path. Option C ($2$) and Option D ($3$) arise from arithmetic errors or incorrect potential functions.

Answer: $\mathbf{F} = \langle 2x + y, x + 2y \rangle$

Apply the component test for conservative fields in 2D: A field $\mathbf{F} = \langle P, Q \rangle$ is conservative on a simply connected domain if and only if $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$. Test each option: Option A: $P = -y \Rightarrow P_y = -1$; $Q = x \Rightarrow Q_x = 1$. Not equal — not conservative. Option B: $P = 2x+y \Rightarrow P_y = 1$; $Q = x+2y \Rightarrow Q_x = 1$. Equal — conservative! Option C: $P = y^2 \Rightarrow P_y = 2y$; $Q = 3xy \Rightarrow Q_x = 3y$. Not equal — not conservative. Option D: $P = 2xy \Rightarrow P_y = 2x$; $Q = x^2 + x \Rightarrow Q_x = 2x + 1$. Not equal — not conservative. Why distractors fail: Option A represents a rotation field with $P_y = -1 \neq 1 = Q_x$. Option C fails because $2y \neq 3y$. Option D fails because $P_y = 2x$ while $Q_x = 2x + 1$; they differ by a constant 1.

Answer: $\sqrt{2}$

Parameterize the line segment: Let $\mathbf{r}(t) = \langle t, t \rangle$ for $t \in [0, 1]$. Then $\mathbf{r}'(t) = \langle 1, 1 \rangle$ and $\|\mathbf{r}'(t)\| = \sqrt{2}$, so $ds = \sqrt{2}\,dt$. Substitute and evaluate: On $C$, $x = t$ and $y = t$, so $x + y = 2t$. The integral becomes $\int_0^1 2t \cdot \sqrt{2}\,dt = 2\sqrt{2} \int_0^1 t\,dt = 2\sqrt{2} \cdot \frac{1}{2} = \sqrt{2}$. Why distractors fail: Option A ($1$) omits the arc-length factor $\sqrt{2}$ and also makes an arithmetic error. Option C ($2$) omits the arc-length factor $\sqrt{2}$, computing only $2 \int_0^1 t\,dt = 1$ — actually closer to Option A; $2$ results from forgetting to evaluate $\int_0^1 t\,dt$ and treating it as $1$. Option D ($2\sqrt{2}$) correctly applies the arc-length factor but forgets to multiply by $\frac{1}{2}$ when integrating $t$.

Answer: The integral equals zero

Closed-loop property of conservative fields: For a conservative field $\mathbf{F} = \nabla f$, any closed curve $C$ has the same start and end point, say $A$. Then $\oint_C \mathbf{F} \cdot d\mathbf{r} = f(A) - f(A) = 0$. Why the correct answer works: Option B is correct because the integral over any closed path in a conservative field is always zero, a direct consequence of path independence. Why distractors fail: Option A confuses the vector line integral with the scalar line integral $\int_C 1\, ds$. Option C confuses the line integral with area, which relates to Green's theorem for specific fields. Option D is incorrect because the value is zero regardless of orientation.

Answer: Because $ds$ is always positive regardless of orientation, while $d\mathbf{r}$ reverses direction when the curve is traversed in the opposite sense

Arc-length element vs. displacement element: The element $ds = \|\mathbf{r}'(t)\|\,dt$ is always non-negative because it measures arc length, which is independent of traversal direction. In contrast, $d\mathbf{r} = \mathbf{r}'(t)\,dt$ is a vector that reverses when the direction of travel reverses. Why the correct answer works: Option B correctly identifies that $ds > 0$ always, so $\int_C f\,ds$ is orientation-independent, while $d\mathbf{r}$ flips sign under orientation reversal, causing $\int_C \mathbf{F} \cdot d\mathbf{r}$ to negate. Why distractors fail: Option A has it backwards: $ds$ does not reverse sign. Option C is wrong because $f$ can be negative; the invariance comes from $ds$, not $f$. Option D is wrong because the vector line integral uses a dot product, not a cross product.

Answer: $\mathbf{F} = \nabla f$ for some scalar potential function $f$

Definition of a conservative field: A vector field $\mathbf{F}$ is called conservative if there exists a scalar function $f$ such that $\mathbf{F} = \nabla f$. The function $f$ is called the potential function of $\mathbf{F}$. Why the correct answer works: Option B directly states the definition: $\mathbf{F} = \nabla f$ for some scalar potential function $f$. This is the standard characterization of a conservative vector field. Why distractors fail: Option A confuses conservative with constant-magnitude fields, which is unrelated. Option C describes no standard property of conservative fields. Option D describes an incompressible (solenoidal) field, not a conservative one; conservative fields have zero curl, not zero divergence.

Answer: $16\pi$

Parameterize the curve: The circle $x^2+y^2=4$ has radius 2. Parameterize as $\mathbf{r}(t) = \langle 2\cos t, 2\sin t \rangle$ for $t \in [0, 2\pi]$. Then $\|\mathbf{r}'(t)\| = 2$, so $ds = 2\,dt$. Evaluate the integrand on the curve: On this circle, $x^2 + y^2 = 4$ everywhere. So the integral becomes $\int_0^{2\pi} 4 \cdot 2\, dt = 8(2\pi) = 16\pi$. Why distractors fail: Option A ($0$) would occur for an odd function integrated over a symmetric path, not this case. Option B ($8\pi$) forgets to include the arc-length factor of 2, computing $4 \cdot 2\pi$ instead. Option C ($4\pi$) uses the integrand value of 4 times only $\pi$ instead of $2\pi$, or omits the arc-length factor entirely.