Line Integrals of Vector Fields - Calculus 3

1. A vector field $\mathbf{F}(x,y) = (P(x,y), Q(x,y))$ satisfies $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ in a simply connected domain $D$. For any closed curve $C$ lying entirely within $D$, what can be concluded about $\oint_C \mathbf{F} \cdot d\mathbf{r}$?
   Answer: The integral equals zero because the field is conservative in $D$, and circulation around closed paths in conservative fields vanishes The condition $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ means $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0$, so the field has zero curl. In a simply connected domain, this implies the field is conservative, meaning there exists a potential function $f$ such that $\mathbf{F} = \nabla f$. For any closed curve $C$, $\oint_C \mathbf{F} \cdot d\mathbf{r} = \oint_C \nabla f \cdot d\mathbf{r} = 0$ by the fundamental theorem for line integrals. Choice B is incorrect because Green's theorem applies to any piecewise smooth closed curve, not just simple ones. Choice C is wrong because conservative fields always have zero circulation around closed curves. Choice D misapplies Green's theorem - when the curl is zero, the double integral gives zero, not the area.
2. A vector field $\mathbf{F}(x,y,z) = (P, Q, R)$ satisfies $\nabla \times \mathbf{F} = \mathbf{0}$ throughout a simply connected region. If $\int_{C_1} \mathbf{F} \cdot d\mathbf{r} = 5$ where $C_1$ goes from point $A$ to point $B$, and $\int_{C_2} \mathbf{F} \cdot d\mathbf{r} = -3$ where $C_2$ goes from point $B$ to point $C$, what is $\int_{C_3} \mathbf{F} \cdot d\mathbf{r}$ where $C_3$ goes from point $C$ to point $A$?
   Answer: $-2$ because the sum of line integrals around any closed loop in a conservative field must equal zero Since $\nabla \times \mathbf{F} = \mathbf{0}$ in a simply connected region, the field is conservative. This means line integrals are path-independent and depend only on endpoints. For any closed loop, the circulation must be zero. The closed path $A \to B \to C \to A$ gives us: $\int_{C_1} \mathbf{F} \cdot d\mathbf{r} + \int_{C_2} \mathbf{F} \cdot d\mathbf{r} + \int_{C_3} \mathbf{F} \cdot d\mathbf{r} = 0$. Substituting known values: $5 + (-3) + \int_{C_3} \mathbf{F} \cdot d\mathbf{r} = 0$, so $\int_{C_3} \mathbf{F} \cdot d\mathbf{r} = -2$. Choice B incorrectly assumes energy is always positive and adds magnitudes. Choice C makes an error in the relationship between the integrals. Choice D incorrectly suggests the answer depends on the specific path, contradicting the path-independence of conservative fields.
3. Consider the work integral $W = \int_C \mathbf{F} \cdot d\mathbf{r}$ where $\mathbf{F}(x,y,z) = (yz \cos(xyz), xz \cos(xyz), xy \cos(xyz))$ and $C$ is the curve from $(0,1,1)$ to $(\pi,1,1)$ parameterized by $\mathbf{r}(t) = (t, 1, 1)$ for $t \in [0,\pi]$. What is the value of $W$?
   Answer: $W = \sin(\pi) - \sin(0) = 0$ by recognizing that $\mathbf{F} = \nabla(\sin(xyz))$ and applying the fundamental theorem When you encounter a work integral with a vector field that looks like it might be conservative, your first instinct should be to check if the field is the gradient of some scalar function. This can dramatically simplify the calculation. Looking at $\mathbf{F}(x,y,z) = (yz \cos(xyz), xz \cos(xyz), xy \cos(xyz))$, notice that each component involves the product of two variables multiplied by $\cos(xyz)$. This pattern suggests the field might be $\nabla(\sin(xyz))$. Let's verify: if $f(x,y,z) = \sin(xyz)$, then $\frac{\partial f}{\partial x} = yz\cos(xyz)$, $\frac{\partial f}{\partial y} = xz\cos(xyz)$, and $\frac{\partial f}{\partial z} = xy\cos(xyz)$. Indeed, $\mathbf{F} = \nabla(\sin(xyz))$. Since the field is conservative, we can apply the Fundamental Theorem for Line Integrals: $W = f(\pi,1,1) - f(0,1,1) = \sin(\pi \cdot 1 \cdot 1) - \sin(0 \cdot 1 \cdot 1) = \sin(\pi) - \sin(0) = 0 - 0 = 0$. This confirms answer C. Answer A is wrong because while it correctly identifies that $W = 0$, the reasoning about "level surfaces" is irrelevant—the result follows from the Fundamental Theorem. Answer B incorrectly suggests direct computation without recognizing the conservative nature. Answer D shows a confused calculation that doesn't properly relate to the given vector field. Strategy tip: Always check if a vector field is conservative before computing line integrals directly. Look for patterns suggesting $\mathbf{F} = \nabla f$ for some function $f$—this can save significant computation time.
4. Consider the line integral $\oint_C \mathbf{F} \cdot d\mathbf{r}$ where $\mathbf{F}(x,y) = (x^2y, xy^2 - x)$ and $C$ is the boundary of the region bounded by $y = x^2$ and $y = 2x$, traversed counterclockwise. Which method would be most efficient for evaluating this integral?
   Answer: Green's theorem since $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = y^2 - 1 - x^2 \neq 0$, making the double integral straightforward To determine the best method, we check if the field is conservative by computing $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$ where $P = x^2y$ and $Q = xy^2 - x$. We get $\frac{\partial Q}{\partial x} = y^2 - 1$ and $\frac{\partial P}{\partial y} = x^2$, so $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = y^2 - 1 - x^2 \neq 0$ in general. Since the field is not conservative, Green's theorem is appropriate and often more efficient than parameterizing multiple curve segments. Choice A is wrong because the field is not conservative. Choice C incorrectly states that Green's theorem doesn't apply when curl is nonzero - it actually applies specifically in such cases. Choice D is wrong because the field is not conservative, so circulation need not be zero.
5. The vector field $\mathbf{F}(x,y) = \left(\frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2}\right)$ is defined for all $(x,y) \neq (0,0)$. If $C_1$ is the unit circle centered at the origin traversed counterclockwise and $C_2$ is the circle of radius 2 centered at $(3,0)$ traversed counterclockwise, what is the relationship between $\oint_{C_1} \mathbf{F} \cdot d\mathbf{r}$ and $\oint_{C_2} \mathbf{F} \cdot d\mathbf{r}$?
   Answer: $\oint_{C_1} \mathbf{F} \cdot d\mathbf{r} = 2\pi$ and $\oint_{C_2} \mathbf{F} \cdot d\mathbf{r} = 0$ since $C_2$ doesn't enclose the singularity at the origin When you encounter a vector field with a singularity (undefined point) and need to compute line integrals around different closed curves, think about whether each curve encloses the singularity and whether the field is conservative in the relevant regions. This vector field $\mathbf{F}(x,y) = \left(\frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2}\right)$ has a singularity at the origin and is actually the gradient of the angle function $\theta = \arctan(y/x)$. For $C_1$ (unit circle around origin), as you traverse counterclockwise, the angle $\theta$ increases by $2\pi$, so $\oint_{C_1} \mathbf{F} \cdot d\mathbf{r} = 2\pi$. For $C_2$ (circle centered at $(3,0)$ with radius 2), this curve doesn't enclose the origin. Since $\mathbf{F}$ is conservative on any simply connected region that doesn't contain the origin, $\oint_{C_2} \mathbf{F} \cdot d\mathbf{r} = 0$. Answer A incorrectly claims both integrals equal $2\pi$. While the field does have zero curl where defined, $C_2$ doesn't enclose the singularity, so its integral is zero, not $2\pi$. Answer B suggests the circulation depends on radius, but what matters is enclosing the singularity, not the size of the circle. Answer C incorrectly states both integrals are zero—this misses that $C_1$ encloses the singularity where the field isn't conservative. Remember: for vector fields with singularities, line integrals around closed curves depend on whether the curve encloses the singularity, not the size or shape of the curve. Check enclosure first, then apply Green's theorem or conservation properties accordingly.

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