Line Integrals: Vector Fields - Multivariable Calculus

1. A vector field $\vec{F}$ satisfies $\nabla \times \vec{F} = \vec{0}$ in a region $R$, and $\int_{C_1} \vec{F} \cdot d\vec{r} = 8$ where $C_1$ is a curve from point $P$ to point $Q$ lying entirely in $R$. If $C_2$ is another curve from $P$ to $Q$ that passes through a small hole in region $R$ (where $\vec{F}$ is undefined), which statement is necessarily true?
   Answer: $\int_{C_2} \vec{F} \cdot d\vec{r}$ may not equal 8 because path independence requires the field to be defined on a simply connected domain While $\nabla \times \vec{F} = \vec{0}$ is a necessary condition for a conservative field, it is not sufficient unless the domain is simply connected. The presence of a hole makes the domain non-simply connected, which can break path independence even when the curl is zero. This is analogous to the vector field $\vec{F} = \frac{(-y, x)}{x^2 + y^2}$ around the origin, which has zero curl everywhere it's defined but is not conservative due to the singularity. Choice A incorrectly assumes curl-free implies conservative everywhere. Choice B ignores the topological requirement for path independence. Choice D is wrong because $C_2$ passes through the hole but $\vec{F}$ can still be evaluated along the parts of $C_2$ where it is defined.
2. A particle moves in the force field $\vec{F}(x,y) = \langle xy, x-y \rangle$. Calculate the work done as the particle moves from $(0,0)$ to $(2,4)$ along the parabola $y=x^2$, and then from $(2,4)$ to $(0,4)$ along a horizontal line segment.
   Answer: $-\frac{20}{3}$ When you encounter a work problem in a force field, you're calculating the line integral $W = \int_C \vec{F} \cdot d\vec{r}$. This requires parameterizing each path segment and integrating along the entire journey. For the first segment from $(0,0)$ to $(2,4)$ along $y = x^2$, parameterize as $\vec{r}_1(t) = \langle t, t^2 \rangle$ where $t$ goes from 0 to 2. Then $d\vec{r}_1 = \langle 1, 2t \rangle dt$. Along this path, $\vec{F} = \langle t \cdot t^2, t - t^2 \rangle = \langle t^3, t - t^2 \rangle$. The dot product gives: $\vec{F} \cdot d\vec{r}_1 = t^3(1) + (t - t^2)(2t) = t^3 + 2t^2 - 2t^3 = 2t^2 - t^3$. Integrating: $\int_0^2 (2t^2 - t^3) dt = \left[\frac{2t^3}{3} - \frac{t^4}{4}\right]_0^2 = \frac{16}{3} - 4 = \frac{4}{3}$ For the second segment from $(2,4)$ to $(0,4)$ along the horizontal line, parameterize as $\vec{r}_2(t) = \langle 2-t, 4 \rangle$ where $t$ goes from 0 to 2. Then $d\vec{r}_2 = \langle -1, 0 \rangle dt$. Here $\vec{F} = \langle (2-t)(4), (2-t) - 4 \rangle = \langle 8-4t, -2-t \rangle$. The work is: $\int_0^2 (8-4t)(-1) dt = \int_0^2 (-8+4t) dt = [-8t + 2t^2]_0^2 = -16 + 8 = -8$ Total work: $\frac{4}{3} + (-8) = \frac{4}{3} - \frac{24}{3} = -\frac{20}{3}$, confirming answer D. Answer A gives only the second segment's work. Answer B represents the first segment multiplied by 7. Answer C gives only the first segment's work. Always break complex paths into manageable segments and carefully track signs in your parameterization—the direction matters for line integrals.
3. Evaluate the line integral $\oint_C \vec{F} \cdot d\vec{r}$ where $\vec{F}(x,y) = \left\langle \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} \right\rangle$ and $C$ is the boundary of the square with vertices $(2,2)$, $(-2,2)$, $(-2,-2)$, and $(2,-2)$, oriented counter-clockwise.
   Answer: $2\pi$ When evaluating line integrals of vector fields, you should first check if the field is conservative or if Green's theorem applies. This particular vector field $\vec{F}(x,y) = \left\langle \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} \right\rangle$ is special—it's the classic "circulation field" that's not defined at the origin. Since the origin $(0,0)$ lies inside the square, we can't directly apply Green's theorem (which requires the field to be defined everywhere inside the region). However, we can recognize this field geometrically: it represents pure counterclockwise rotation around the origin with magnitude $\frac{1}{x^2+y^2}$. For any simple closed curve enclosing the origin once, this field produces a circulation of $2\pi$. You can verify this by parameterizing any circle around the origin—the integral always yields $2\pi$ regardless of the radius. Since our square encloses the origin once and is oriented counterclockwise, the answer is $2\pi$. Choice (A) $4\pi$ would result from incorrectly thinking the square somehow "wraps twice" around the origin. Choice (B) $0$ represents the misconception that you can apply Green's theorem directly—if the field were conservative, this would be correct, but this field has non-zero curl everywhere except the origin. Choice (C) $-2\pi$ would occur if you mistakenly used clockwise orientation instead of the given counterclockwise orientation. Study tip: When you see the field $\left\langle \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} \right\rangle$, immediately check if the origin is inside your curve. If yes, the answer is $\pm 2\pi$ depending on orientation.
4. Calculate the work done by the force field $\vec{F}(x,y) = \langle y^3, -x^3 \rangle$ on a particle that moves along the circle $x^2 + y^2 = 4$, oriented clockwise.
   Answer: $24\pi$ We use Green's Theorem, $\oint_C P\,dx + Q\,dy = \iint_D (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) \, dA$. Here $P=y^3$ and $Q=-x^3$, so $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = -3x^2 - 3y^2 = -3(x^2+y^2)$. The region $D$ is the disk $x^2+y^2 \le 4$. Green's theorem assumes counter-clockwise orientation. Since the path is oriented clockwise, we must negate the result. Work $W = -\iint_D -3(x^2+y^2) \, dA = 3 \iint_D (x^2+y^2) \, dA$. Converting to polar coordinates, $x^2+y^2 = r^2$ and $dA = r\,dr\,d\theta$. $W = 3 \int_0^{2\pi} \int_0^2 (r^2) r \, dr \, d\theta = 3 \int_0^{2\pi} \left[ \frac{r^4}{4} \right]_0^2 \, d\theta = 3 \int_0^{2\pi} 4 \, d\theta = 12(2\pi) = 24\pi$.
5. A vector field $\vec{F}(x,y) = (P(x,y), Q(x,y))$ is given where $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ throughout a simply connected domain. If the line integral $\int_{C_1} \vec{F} \cdot d\vec{r} = 5$ for a curve $C_1$ from point $A$ to point $B$, and $\int_{C_2} \vec{F} \cdot d\vec{r} = -3$ for a curve $C_2$ from point $B$ to point $C$, what is $\int_{C_3} \vec{F} \cdot d\vec{r}$ where $C_3$ is any curve from point $C$ to point $A$?
   Answer: $-2$ Since $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$, the vector field is conservative in the simply connected domain. For conservative fields, line integrals are path-independent and $\oint_C \vec{F} \cdot d\vec{r} = 0$ for any closed curve. The path from $A$ to $B$ to $C$ and back to $A$ forms a closed curve, so $\int_{C_1} \vec{F} \cdot d\vec{r} + \int_{C_2} \vec{F} \cdot d\vec{r} + \int_{C_3} \vec{F} \cdot d\vec{r} = 0$. Therefore, $5 + (-3) + \int_{C_3} \vec{F} \cdot d\vec{r} = 0$, which gives $\int_{C_3} \vec{F} \cdot d\vec{r} = -2$. Choice B assumes the wrong sign. Choice C adds the values instead of using the closed curve property. Choice D incorrectly computes $-(5+3)$.
6. Let $\vec{F}(x,y,z) = \langle yz e^{xyz} + 2x, xz e^{xyz} - 3y^2, xy e^{xyz} + 1 \rangle$. What is the work done by $\vec{F}$ along the curve $C$ parameterized by $\vec{r}(t) = \langle t, 1-t^2, t+1 \rangle$ for $t$ from $0$ to $1$?
   Answer: 3 Directly integrating $\int_C \vec{F} \cdot d\vec{r}$ would be extremely complicated. The vector field $\vec{F}$ is conservative because its domain is all of $\mathbb{R}^3$ and its curl is $\vec{0}$. We can find a potential function $f$ such that $\nabla f = \vec{F}$. By integration, we find $f(x,y,z) = e^{xyz} + x^2 - y^3 + z$. According to the Fundamental Theorem of Line Integrals, the work done is $f(B) - f(A)$, where $A$ and $B$ are the endpoints of the path. Here, $A = \vec{r}(0) = \langle 0, 1, 1 \rangle$ and $B = \vec{r}(1) = \langle 1, 0, 2 \rangle$. Thus, the work is $f(1,0,2) - f(0,1,1) = (e^0 + 1^2 - 0^3 + 2) - (e^0 + 0^2 - 1^3 + 1) = (1+1+2) - (1-1+1) = 4 - 1 = 3$.
7. Consider the vector field $\vec{F}(x,y) = \left(\frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2}\right)$ and two curves: $C_1$ is the upper semicircle of radius 2 centered at the origin from $(2,0)$ to $(-2,0)$, and $C_2$ is the line segment from $(2,0)$ to $(-2,0)$. Which statement about the line integrals $\int_{C_1} \vec{F} \cdot d\vec{r}$ and $\int_{C_2} \vec{F} \cdot d\vec{r}$ is correct?
   Answer: $\int_{C_1} \vec{F} \cdot d\vec{r} = \pi$ and $\int_{C_2} \vec{F} \cdot d\vec{r} = 0$ This vector field has the form $\vec{F} = \frac{1}{r^2}(-\sin\theta, \cos\theta)$ in polar coordinates, which is not conservative everywhere because it has a singularity at the origin. For $C_1$ (upper semicircle), we can parameterize as $\vec{r}(t) = (2\cos t, 2\sin t)$ for $t \in [0,\pi]$. The calculation yields $\int_{C_1} \vec{F} \cdot d\vec{r} = \pi$. For $C_2$ (line segment along x-axis), $y = 0$, so $\vec{F} = (-0/x^2, x/x^2) = (0, 1/x)$. Since $d\vec{r} = (dx, 0)$ along this path, $\vec{F} \cdot d\vec{r} = 0$, giving $\int_{C_2} \vec{F} \cdot d\vec{r} = 0$. Choice A incorrectly assumes conservativity. Choice C ignores the singularity at origin. Choice D has the wrong sign for the semicircle integral.
8. Calculate the work done by the force field $\vec{F} = \langle y^2, 2xy+1 \rangle$ along the path $C$, which is a segment of the parabola $y=x^2$ from $(0,0)$ to $(2,4)$. The problem notes that the path can be parameterized by either $\vec{r}_1(t) = \langle t, t^2 \rangle$ for $t \in [0,2]$ or $\vec{r}_2(u) = \langle \sqrt{u}, u \rangle$ for $u \in [0,4]$.
   Answer: 36 When you encounter a work problem with a vector force field, you're calculating the line integral $\int_C \vec{F} \cdot d\vec{r}$. This measures how much the force contributes along your path of motion. Let's use the first parameterization: $\vec{r}_1(t) = \langle t, t^2 \rangle$ for $t \in [0,2]$. First, find $d\vec{r} = \vec{r}'(t)dt = \langle 1, 2t \rangle dt$. Along this path, $x = t$ and $y = t^2$, so $\vec{F} = \langle (t^2)^2, 2t(t^2)+1 \rangle = \langle t^4, 2t^3+1 \rangle$. The dot product gives us: $\vec{F} \cdot d\vec{r} = \langle t^4, 2t^3+1 \rangle \cdot \langle 1, 2t \rangle = t^4 + (2t^3+1)(2t) = t^4 + 4t^4 + 2t = 5t^4 + 2t$ Now integrate: $\int_0^2 (5t^4 + 2t) dt = \left[t^5 + t^2\right]_0^2 = 32 + 4 = 36$ Answer (A) 34 likely comes from arithmetic errors in the integration. Answer (B) 32 results from forgetting the $2t$ term and only computing $\int_0^2 5t^4 dt = 32$. Answer (D) 4 might come from only evaluating $t^2$ at the bounds, missing the dominant $t^5$ term entirely. Study tip: Always double-check your parameterization and dot product calculation—these are the most common error sources. You can verify your answer using the alternate parameterization; both should give the same result since work depends only on the path, not how you parameterize it.
9. Let $\vec{F} = \langle 5y + x^2, 11x - \sin(y) \rangle$. Let $C$ be the boundary of a region $D$ in the xy-plane that has an area of 10. If $C$ is oriented counter-clockwise, what is the work done by $\vec{F}$ around $C$?
   Answer: 60 When you encounter a vector field and need to find work done around a closed curve, think about whether Green's theorem applies. This connects line integrals around closed curves to double integrals over the enclosed region. For work done by $\vec{F} = \langle P, Q \rangle$ around a counter-clockwise oriented curve $C$, Green's theorem states: $\oint_C \vec{F} \cdot d\vec{r} = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$ Here, $P = 5y + x^2$ and $Q = 11x - \sin(y)$. Computing the partial derivatives: - $\frac{\partial Q}{\partial x} = 11$ - $\frac{\partial P}{\partial y} = 5$ So $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 11 - 5 = 6$ Since this difference is constant, the double integral becomes: $\iint_D 6 \, dA = 6 \iint_D dA = 6 \times \text{Area}(D) = 6 \times 10 = 60$ The answer is B) 60. Choice A) 160 likely comes from miscalculating the partial derivatives or incorrectly using $11 + 5 = 16$ instead of the difference. Choice C) -60 represents the same calculation but with the wrong sign, perhaps from confusing clockwise versus counter-clockwise orientation. Choice D) Cannot be determined suggests missing that Green's theorem provides exactly the tools needed with the given information. Study tip: When you see work problems with closed curves and area given, immediately check if Green's theorem applies. The key insight is that constant differences in partial derivatives make the calculation straightforward regardless of the curve's specific shape.
10. Let $\vec{F}(x,y,z) = \langle y, -x, z^2 \rangle$. Calculate the work done by the vector field $\vec{F}$ on a particle that moves along the helical path $C$ parameterized by $\vec{r}(t) = \langle \cos(t), \sin(t), t \rangle$ for $0 \le t \le 2\pi$.
    Answer: $\frac{8\pi^3}{3} - 2\pi$ The work done is given by the line integral $W = \int_C \vec{F} \cdot d\vec{r}$. First, find the derivative of the parameterization: $\vec{r}'(t) = \langle -\sin(t), \cos(t), 1 \rangle$. Next, evaluate the vector field along the path: $\vec{F}(\vec{r}(t)) = \langle \sin(t), -\cos(t), t^2 \rangle$. The dot product is $\vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) = (\sin t)(-\sin t) + (-\cos t)(\cos t) + (t^2)(1) = -\sin^2(t) - \cos^2(t) + t^2 = -1 + t^2$. The integral for the work is then $\int_0^{2\pi} (t^2 - 1) \, dt = \left[ \frac{t^3}{3} - t \right]_0^{2\pi} = \frac{(2\pi)^3}{3} - 2\pi = \frac{8\pi^3}{3} - 2\pi$.
11. Let $\vec{F} = \langle 2xy^3 + \cos(x), 3x^2y^2 + e^y \rangle$. Calculate the work done by $\vec{F}$ on a particle that moves from $(0,0)$ to $(\pi, 1)$ along the curve $y = \sin^2(x/2)$.
    Answer: $\pi^2 + e - 1$ When you encounter a vector field and need to calculate work along a specific path, you have two main approaches: compute the line integral directly, or check if the field is conservative and use the fundamental theorem for line integrals. Let's check if $\vec{F} = \langle 2xy^3 + \cos(x), 3x^2y^2 + e^y \rangle$ is conservative by testing if $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$, where $P = 2xy^3 + \cos(x)$ and $Q = 3x^2y^2 + e^y$. We have $\frac{\partial P}{\partial y} = 6xy^2$ and $\frac{\partial Q}{\partial x} = 6xy^2$. Since these are equal, the field is conservative! This means the work depends only on the endpoints, not the specific path. For conservative fields, we need to find a potential function $f$ where $\nabla f = \vec{F}$. Integrating $\frac{\partial f}{\partial x} = 2xy^3 + \cos(x)$ gives $f = x^2y^3 + \sin(x) + g(y)$. Using $\frac{\partial f}{\partial y} = 3x^2y^2 + e^y$, we find $g'(y) = e^y$, so $g(y) = e^y + C$. Therefore, $f(x,y) = x^2y^3 + \sin(x) + e^y$, and the work is: $f(\pi,1) - f(0,0) = [\pi^2(1) + \sin(\pi) + e^1] - [0 + \sin(0) + e^0] = \pi^2 + 0 + e - 0 - 0 - 1 = \pi^2 + e - 1$ This matches answer D. Choices A, B, and C represent common arithmetic errors or incorrectly applying the fundamental theorem (like forgetting to subtract the initial point value or sign errors). Strategy tip: Always check for conservative vector fields first—it saves significant computation time compared to parametric line integrals.
12. Consider the vector field $\vec{F}(x,y,z) = (yz, xz, xy)$ and the curve $C$ which is the intersection of the cylinder $x^2 + y^2 = 1$ and the plane $z = x + y$, traversed once counterclockwise when viewed from above. To evaluate $\int_C \vec{F} \cdot d\vec{r}$, which parameterization and setup is most appropriate?
    Answer: $\vec{r}(t) = (\cos t, \sin t, \cos t + \sin t)$ for $t \in [0, 2\pi]$ with $\vec{F} \cdot d\vec{r} = (\sin t(\cos t + \sin t))(-\sin t) + (\cos t(\cos t + \sin t))(\cos t) + (\cos t \sin t)(\cos t - \sin t)$ The curve is the intersection of cylinder $x^2 + y^2 = 1$ and plane $z = x + y$, so the parameterization $\vec{r}(t) = (\cos t, \sin t, \cos t + \sin t)$ for $t \in [0, 2\pi]$ is correct (one full revolution). We have $\vec{r}'(t) = (-\sin t, \cos t, \cos t - \sin t)$. The vector field evaluated on the curve is $\vec{F}(\vec{r}(t)) = (\sin t(\cos t + \sin t), \cos t(\cos t + \sin t), \cos t \sin t)$. The dot product is $\vec{F} \cdot \vec{r}' = (\sin t(\cos t + \sin t))(-\sin t) + (\cos t(\cos t + \sin t))(\cos t) + (\cos t \sin t)(\cos t - \sin t)$. Choice B incorrectly uses $(\cos t + \sin t)$ instead of $(\cos t - \sin t)$ in the third component. Choice C has incorrect signs in the first two components. Choice D uses the wrong interval $[0,\pi]$ instead of $[0,2\pi]$.
13. Find the work done by the force field $\vec{F} = \langle y^2\cos(x), 2y\sin(x)+e^z, ye^z+3z^2 \rangle$ on a particle that moves along the path $\vec{r}(t) = \langle \pi t^2, \sin(\pi t/2), t \rangle$ for $t \in [1,2]$.
    Answer: $7 - e$ When you encounter a work problem with a vector field and parametric path, you're calculating $W = \int_C \vec{F} \cdot d\vec{r} = \int_a^b \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) \, dt$. This requires evaluating the force field along the path and taking the dot product with the velocity vector. First, find $\vec{r}'(t) = \langle 2\pi t, \frac{\pi}{2}\cos(\pi t/2), 1 \rangle$. Next, substitute $\vec{r}(t)$ into $\vec{F}$: $\vec{F}(\vec{r}(t)) = \langle \sin^2(\pi t/2)\cos(\pi t^2), 2\sin(\pi t/2)\sin(\pi t^2) + e^t, \sin(\pi t/2)e^t + 3t^2 \rangle$. The dot product $\vec{F}(\vec{r}(t)) \cdot \vec{r}'(t)$ gives: $2\pi t \sin^2(\pi t/2)\cos(\pi t^2) + \pi\cos(\pi t/2)\sin(\pi t/2)\sin(\pi t^2) + \pi\cos(\pi t/2)e^t + \sin(\pi t/2)e^t + 3t^2$ Using the identity $\sin(\pi t/2)\cos(\pi t/2) = \frac{1}{2}\sin(\pi t)$, this simplifies to: $\frac{\pi}{2}\sin(\pi t)\sin(\pi t^2) + e^t(\pi\cos(\pi t/2) + \sin(\pi t/2)) + 3t^2 + 2\pi t\sin^2(\pi t/2)\cos(\pi t^2)$ Integrating from 1 to 2 yields $7 - e$. Choice A ($e - 7$) has the wrong sign. Choice C (8) ignores the exponential terms entirely. Choice D ($e + 1$) incorrectly handles the trigonometric integrations. Remember: work calculations require careful substitution and often involve complex trigonometric identities. Always double-check your parametric substitutions and use trigonometric identities to simplify before integrating.
14. Let $\vec{F}(x,y) = (2xy + y^2, x^2 + 2xy - 1)$ be a vector field. If $C$ is the boundary of the region bounded by $y = x^2$ and $y = 2x$, oriented counterclockwise, and we want to evaluate $\int_C \vec{F} \cdot d\vec{r}$ using Green's theorem, what is the value of $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$ where $P = 2xy + y^2$ and $Q = x^2 + 2xy - 1$?
    Answer: $0$ (the vector field is conservative) To apply Green's theorem, we need $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$. We have $P(x,y) = 2xy + y^2$ and $Q(x,y) = x^2 + 2xy - 1$. Computing the partial derivatives: $\frac{\partial P}{\partial y} = 2x + 2y$ and $\frac{\partial Q}{\partial x} = 2x + 2y$. Therefore, $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = (2x + 2y) - (2x + 2y) = 0$. This means the vector field is conservative, so the line integral around any closed curve is zero. Choices B, C, and D represent common computational errors in taking partial derivatives.
15. A vector field $\vec{F}(x,y) = \left(\frac{y}{x^2 + y^2} + P(x,y), \frac{-x}{x^2 + y^2} + Q(x,y)\right)$ is given, where $P(x,y)$ and $Q(x,y)$ are smooth functions. If the line integral $\int_C \vec{F} \cdot d\vec{r}$ around any simple closed curve $C$ not enclosing the origin equals zero, but $\int_{C_0} \vec{F} \cdot d\vec{r} = 2\pi$ where $C_0$ is the unit circle centered at the origin oriented counterclockwise, what can be concluded about $P$ and $Q$?
    Answer: $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0$ everywhere, but the field cannot be conservative due to the singularity at the origin The vector field $\left(\frac{y}{x^2 + y^2}, \frac{-x}{x^2 + y^2}\right)$ has the property that its line integral around the unit circle is $-2\pi$ (this is the standard angular field). Since the total field gives $+2\pi$ around $C_0$, the contribution from $(P,Q)$ around $C_0$ must be $4\pi$. However, the condition that integrals around curves not enclosing the origin are zero, combined with the fact that such curves can be continuously deformed to points without crossing the origin, implies that $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0$ in any simply connected region not containing the origin. But the field cannot be globally conservative because the line integral around $C_0$ is non-zero. Choice A incorrectly suggests the field can be conservative. Choices C and D propose incorrect relationships for the curl of $P$ and $Q$.
16. Let $D$ be the region in the plane between the circles $x^2+y^2=1$ and $x^2+y^2=9$. Let $\vec{F}(x,y) = \langle -y + \sin(x^2), x + e^{-y^2} \rangle$. Calculate $\oint_C \vec{F} \cdot d\vec{r}$, where $C$ is the boundary of $D$ oriented such that the region $D$ is always on the left.
    Answer: $16\pi$ When you encounter a line integral around the boundary of a region, immediately consider Green's theorem, which converts the line integral into a double integral over the region's interior. The boundary of region $D$ consists of two circles: the outer circle $x^2+y^2=9$ (traversed counterclockwise) and the inner circle $x^2+y^2=1$ (traversed clockwise to keep $D$ on the left). For $\vec{F}(x,y) = \langle -y + \sin(x^2), x + e^{-y^2} \rangle$, Green's theorem gives us: $\oint_C \vec{F} \cdot d\vec{r} = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$ where $P = -y + \sin(x^2)$ and $Q = x + e^{-y^2}$. Computing the partial derivatives: $\frac{\partial Q}{\partial x} = 1$ and $\frac{\partial P}{\partial y} = -1$. Therefore: $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 - (-1) = 2$ The integral becomes: $\iint_D 2 \, dA = 2 \cdot \text{Area}(D)$ The area of the annular region is: $\pi(3^2) - \pi(1^2) = 9\pi - \pi = 8\pi$ So the line integral equals $2 \cdot 8\pi = 16\pi$. Choice A ($18\pi$) likely comes from miscalculating the area as $9\pi$. Choice B ($8\pi$) forgets the factor of 2 from the curl calculation. Choice D ($20\pi$) might result from using radius values incorrectly in the area formula. Study tip: Always check if Green's theorem applies before computing line integrals directly. When the curl is constant (as here), the problem reduces to finding the area of the region.
17. Consider the vector field $\vec{F}(x,y) = (e^{x+y}, xe^{x+y} + \cos y)$ and the curve $C$ parameterized by $\vec{r}(t) = (t, \sin t)$ for $0 \leq t \leq \pi$. To evaluate $\int_C \vec{F} \cdot d\vec{r}$, which expression correctly represents the integrand after substitution?
    Answer: $e^{t + \sin t} + te^{t + \sin t} \cos t + \cos(\sin t) \cos t$ We have $\vec{r}(t) = (t, \sin t)$, so $\vec{r}'(t) = (1, \cos t)$. The vector field along the curve is $\vec{F}(\vec{r}(t)) = \vec{F}(t, \sin t) = (e^{t + \sin t}, te^{t + \sin t} + \cos(\sin t))$. The dot product is $\vec{F} \cdot \vec{r}' = (e^{t + \sin t}, te^{t + \sin t} + \cos(\sin t)) \cdot (1, \cos t) = e^{t + \sin t} \cdot 1 + (te^{t + \sin t} + \cos(\sin t)) \cdot \cos t = e^{t + \sin t} + te^{t + \sin t} \cos t + \cos(\sin t) \cos t$. Choice A is missing the multiplication by the components of $\vec{r}'(t)$. Choice B shows the structure but doesn't distribute the terms. Choice C has the components of the dot product in the wrong order.
18. A conservative vector field $\vec{F}$ exists throughout $\mathbb{R}^3$. The work done by $\vec{F}$ along a path from point $A$ to point $B$ is 10. The work done by $\vec{F}$ along a path from point $C$ to point $B$ is $-3$. What is the work done by $\vec{F}$ along a path from point $A$ to point $C$?
    Answer: 13 When you encounter a conservative vector field problem, remember that the fundamental property is path independence—the work done depends only on the starting and ending points, not the specific path taken. Since $\vec{F}$ is conservative, you can use the additive property of work. Think of this like connecting destinations: if you know the "cost" to travel from A to B and from C to B, you can find the cost from A to C by considering how these paths relate. The work from A to C can be found by going from A to B, then from B to C. Since work from A to B is 10, and work from C to B is -3, the work from B to C must be +3 (reversing direction changes the sign). Therefore, work from A to C = work from A to B + work from B to C = 10 + 3 = 13. Choice A (7) represents the error of subtracting the magnitudes: 10 - 3 = 7. This ignores the proper vector addition needed when combining path segments. Choice C (-7) comes from incorrectly calculating 10 + (-3) - 10 = -7, showing confusion about how to combine the given information. Choice D suggests the answer cannot be determined, which would be true for non-conservative fields where work depends on the specific path, but conservative fields eliminate this path dependence. Remember: for conservative vector fields, work is path-independent, so you can always "connect the dots" between any three points using vector addition of work values, paying careful attention to direction.
19. Let $C$ be the upper semi-circle of radius 1 centered at the origin, traversed from $(1,0)$ to $(-1,0)$. Let $W_1$ be the work done by the vector field $\vec{F}_1 = \langle -y, x \rangle$ along $C$, and $W_2$ be the work done by $\vec{F}_2 = \langle x, y \rangle$ along $C$. Which of the following statements is true?
    Answer: $W_1 > W_2$ When you encounter work problems with vector fields, the key insight is recognizing whether the field is conservative or has circulation properties. Work is calculated as $W = \int_C \vec{F} \cdot d\vec{r}$, which measures how much the field "helps" the motion along the curve. For $\vec{F}_1 = \langle -y, x \rangle$, this field rotates counterclockwise around the origin. Since our upper semicircle also travels counterclockwise from $(1,0)$ to $(-1,0)$, the field consistently points in the same general direction as our path. Using the standard parameterization $\vec{r}(t) = \langle \cos t, \sin t \rangle$ for $t \in [0,\pi]$, we get $W_1 = \int_0^{\pi} \pi dt = \pi > 0$. For $\vec{F}_2 = \langle x, y \rangle$, this is a radial field pointing outward from the origin. Since our semicircular path moves perpendicular to the radial direction at each point, the dot product $\vec{F}_2 \cdot d\vec{r} = 0$ everywhere along the curve. Therefore, $W_2 = 0$. Since $W_1 = \pi > 0 = W_2$, we have $W_1 > W_2$, making C correct. Looking at the wrong answers: A suggests they're equal, but we showed $W_1 = \pi$ and $W_2 = 0$. B claims $W_1 < W_2$, which contradicts $\pi > 0$. D suggests $W_1 = -W_2$, but this would require $W_2 = -\pi$, not zero. Study tip: Always visualize vector fields geometrically. Rotational fields create circulation (nonzero work around closed or nearly closed paths), while radial fields typically produce zero work along circular paths due to perpendicularity.

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