Conservative Fields & Potentials - Calculus 3

1. Let $f(x,y,z) = x^2yz + xy^2z + xyz^2$. The gradient field $\mathbf{F} = \nabla f$ is used to compute $\int_C \mathbf{F} \cdot d\mathbf{r}$ where $C$ is the intersection of the sphere $x^2 + y^2 + z^2 = 9$ with the plane $x + y + z = 3$, oriented counterclockwise when viewed from the origin. What is the value of this line integral?
   Answer: The integral equals $0$ since the curve $C$ is a closed loop and $\mathbf{F}$ is conservative When you encounter a line integral of a gradient field over a closed curve, immediately think about conservative vector fields and the fundamental theorem for line integrals. The key insight is recognizing that $\mathbf{F} = \nabla f$ means $\mathbf{F}$ is conservative by definition. Since $f(x,y,z) = x^2yz + xy^2z + xyz^2$ is a polynomial function, it's smooth everywhere, making $\mathbf{F} = \nabla f$ conservative throughout its domain. For any conservative vector field, the fundamental theorem for line integrals states that $\int_C \mathbf{F} \cdot d\mathbf{r} = f(\text{endpoint}) - f(\text{startpoint})$. However, since $C$ is a closed curve (the intersection of a sphere and plane forms a circle), the starting and ending points are identical. Therefore, $\int_C \mathbf{F} \cdot d\mathbf{r} = f(\text{point}) - f(\text{point}) = 0$. Option A is wrong because recognizing conservative fields eliminates the need for explicit parameterization. Option B incorrectly suggests the answer is 27 and misapplies the fundamental theorem—the theorem actually guarantees zero for closed loops with conservative fields. Option C proposes $9\pi$, which might tempt students thinking about the geometric properties of the intersection, but geometric relationships don't determine line integrals of conservative fields over closed curves. Remember this pattern: whenever you see $\int_C \nabla f \cdot d\mathbf{r}$ over a closed curve, the answer is always zero due to the conservative nature of gradient fields. This saves enormous computation time on exams.
2. Consider the vector field $\mathbf{F}(x,y) = (P(x,y), Q(x,y))$ where $P$ and $Q$ are continuously differentiable on $\mathbb{R}^2$. If $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ and $\int_C \mathbf{F} \cdot d\mathbf{r} = 8$ for some path $C$ from $(1,1)$ to $(3,2)$, what can be concluded about $\int_{C'} \mathbf{F} \cdot d\mathbf{r}$ where $C'$ is the straight line from $(1,1)$ to $(3,2)$?
   Answer: The integral along $C'$ equals $8$ because conservative fields are path-independent on simply connected domains Since $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ and $P$, $Q$ are continuously differentiable on $\mathbb{R}^2$ (which is simply connected), the vector field $\mathbf{F}$ is conservative. For conservative fields, line integrals are path-independent, meaning they depend only on the endpoints, not the specific path taken. Since both $C$ and $C'$ connect the same endpoints $(1,1)$ and $(3,2)$, we have $\int_{C'} \mathbf{F} \cdot d\mathbf{r} = \int_C \mathbf{F} \cdot d\mathbf{r} = 8$. Choice B is incorrect because conservative fields don't have a 'preferred' path that minimizes the integral - all paths between the same endpoints give the same value. Choice C is wrong because path independence means we don't need the explicit forms of $P$ and $Q$. Choice D incorrectly suggests that geometric complexity affects the integral value for conservative fields.
3. The vector field $\mathbf{F}(x,y) = \left(\frac{-y}{x^2+y^2}\right)\mathbf{i} + \left(\frac{x}{x^2+y^2}\right)\mathbf{j}$ is defined on the domain $D = \{(x,y) : x^2 + y^2 > 0\}$. Which statement correctly characterizes this field?
   Answer: The field is conservative on any simply connected subdomain of $D$ but not on $D$ itself First, let's check if $\nabla \times \mathbf{F} = 0$. We have $\frac{\partial}{\partial x}\left(\frac{x}{x^2+y^2}\right) = \frac{(x^2+y^2) - x(2x)}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2}$ and $\frac{\partial}{\partial y}\left(\frac{-y}{x^2+y^2}\right) = \frac{-(x^2+y^2) + y(2y)}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2}$. So $\nabla \times \mathbf{F} = 0$ everywhere in $D$. However, $D$ is not simply connected because it has a hole at the origin. The line integral $\oint_C \mathbf{F} \cdot d\mathbf{r}$ around a circle $x^2+y^2=r^2$ equals $2\pi$ (this is related to the angle function $\theta$). Since this integral is nonzero, the field is not conservative on $D$. However, any simply connected subdomain of $D$ (such as $D$ with a ray removed from the origin) would make the field conservative there. Choice A is wrong because zero curl doesn't guarantee conservativeness in non-simply connected domains. Choice B is correct about non-conservativeness but wrong about the reason. Choice D is wrong because the partial derivatives exist and are equal everywhere in $D$.
4. A vector field $\mathbf{F}(x,y) = (ax^2 + bxy)\mathbf{i} + (cxy + dy^2)\mathbf{j}$ is conservative for all values of the constants $a$, $b$, $c$, and $d$ except when a certain relationship between these constants fails to hold. Additionally, if the potential function $f(x,y)$ satisfies $f(0,0) = 0$ and $f(1,1) = 2$, what constraint must be satisfied by the constants?
   Answer: $b = c$ and $\frac{a}{3} + \frac{b}{2} + \frac{d}{3} = 2$ for conservativeness and the boundary conditions When you encounter a vector field and need to determine when it's conservative, remember that a vector field $\mathbf{F} = P\mathbf{i} + Q\mathbf{j}$ is conservative if and only if $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$. For $\mathbf{F}(x,y) = (ax^2 + bxy)\mathbf{i} + (cxy + dy^2)\mathbf{j}$, we have $P = ax^2 + bxy$ and $Q = cxy + dy^2$. Taking partial derivatives: $\frac{\partial P}{\partial y} = bx$ and $\frac{\partial Q}{\partial x} = cy$. For these to be equal for all points, we need $bx = cy$ for all $x$ and $y$, which is only possible if $b = c = 0$. However, since the problem asks for a general relationship, we need $b = c$. When $b = c$, the potential function $f(x,y)$ satisfies $\frac{\partial f}{\partial x} = ax^2 + bxy$ and $\frac{\partial f}{\partial y} = bxy + dy^2$. Integrating, we get $f(x,y) = \frac{a}{3}x^3 + \frac{b}{2}x^2y + \frac{b}{2}xy^2 + \frac{d}{3}y^3 + C$. Using $f(0,0) = 0$, we find $C = 0$. Using $f(1,1) = 2$: $\frac{a}{3} + \frac{b}{2} + \frac{b}{2} + \frac{d}{3} = 2$, which simplifies to $\frac{a}{3} + b + \frac{d}{3} = 2$. Answer D correctly identifies both conditions. Answer A uses an incorrect symmetric form assumption. Answer B has the wrong coefficient relationship from the boundary condition. Answer C incorrectly assumes $b + c = 0$ rather than $b = c$. Remember: always check the conservative condition first, then use the potential function to satisfy boundary conditions.
5. Consider two potential functions $f_1(x,y) = x^2y + xy^2 + 3$ and $f_2(x,y) = x^2y + xy^2 - 5$. Their corresponding gradient fields $\mathbf{F}_1 = \nabla f_1$ and $\mathbf{F}_2 = \nabla f_2$ are used to compute line integrals along the same path $C$ from $(0,0)$ to $(2,3)$. If $\int_C \mathbf{F}_1 \cdot d\mathbf{r} = 30$, what is $\int_C \mathbf{F}_2 \cdot d\mathbf{r}$?
   Answer: $\int_C \mathbf{F}_2 \cdot d\mathbf{r} = 30$ because the constant difference between $f_1$ and $f_2$ vanishes in the gradient operation Since $f_1(x,y) = x^2y + xy^2 + 3$ and $f_2(x,y) = x^2y + xy^2 - 5$, we have $f_1(x,y) = f_2(x,y) + 8$. Taking gradients: $\mathbf{F}_1 = \nabla f_1 = \nabla(f_2 + 8) = \nabla f_2 + \nabla(8) = \nabla f_2 + \mathbf{0} = \nabla f_2 = \mathbf{F}_2$. Since the gradient of a constant is zero, $\mathbf{F}_1 = \mathbf{F}_2$. Therefore, $\int_C \mathbf{F}_2 \cdot d\mathbf{r} = \int_C \mathbf{F}_1 \cdot d\mathbf{r} = 30$. Choice A gives the right answer with correct reasoning but slightly imprecise language. Choice B incorrectly suggests the integral values differ. Choice D incorrectly suggests an additive effect from the constant.
6. Consider the vector field $\mathbf{F}(x,y,z) = (yz + 2x)\mathbf{i} + (xz + 3y^2)\mathbf{j} + (xy + z^2)\mathbf{k}$. A student claims that since $\nabla \times \mathbf{F} = \mathbf{0}$, the field is conservative. Which statement best describes this conclusion?
   Answer: The conclusion is correct for this field since the domain is all of $\mathbb{R}^3$ which is simply connected First, let's verify that $\nabla \times \mathbf{F} = \mathbf{0}$: \nabla \times \mathbf{F} = \left(\frac{\partial}\partial y}(xy + z^2) - \frac{\partial}{\partial z}(xz + 3y^2)\right)\mathbf{i} + \left(\frac{\partial}{\partial z}(yz + 2x) - \frac{\partial}{\partial x}(xy + z^2)\right)\mathbf{j} + \left(\frac{\partial}{\partial x}(xz + 3y^2) - \frac{\partial}{\partial y}(yz + 2x)\right)\mathbf{k} = (x - x)\mathbf{i} + (y - y)\mathbf{j} + (z - z)\mathbf{k} = \mathbf{0}. Since the domain is $\mathbb{R}^3$, which is simply connected, zero curl is sufficient for the field to be conservative. Choice A is wrong because zero curl is sufficient only in simply connected domains. Choice C is wrong because the curl calculation shows all mixed partials are equal. Choice D is wrong because in simply connected domains, zero curl implies global conservativeness.

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