Green's Theorem: Circulation & Flux - Calculus 3

1. A vector field $\mathbf{F}(x,y) = (e^x \cos y - y, -e^x \sin y + x)$ is given in a simply connected domain. To determine whether the circulation around any simple closed curve in this domain is zero, which condition should be verified using Green's theorem?
   Answer: Check if $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0$ everywhere in the domain For circulation to be zero around any simple closed curve in a simply connected domain, the vector field must be conservative. By Green's theorem, $\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$. For this to equal zero for any region $D$, we need $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0$ everywhere, which is equivalent to $\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}$. Choice A tests the condition for flux, not circulation. Choice C reverses the order ($\frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x} = 0$ is equivalent to choice B). Choice D uses the wrong combination of partial derivatives.
2. A vector field $\mathbf{F}(x,y) = (P(x,y), Q(x,y))$ has the property that $\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} = 6xy$ throughout a simply connected region $D$. If the outward flux of $\mathbf{F}$ through the boundary of the square with vertices at $(0,0)$, $(2,0)$, $(2,2)$, and $(0,2)$ is computed using Green's theorem, what is the result?
   Answer: $24$ For outward flux through a closed curve, Green's theorem gives $\oint_C \mathbf{F} \cdot \mathbf{n} \, ds = \iint_D \left(\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}\right) dA$. We're given that $\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} = 6xy$. The region $D$ is the square $[0,2] \times [0,2]$. Therefore, the flux is $\iint_D 6xy \, dA = \int_0^2 \int_0^2 6xy \, dx \, dy = \int_0^2 6y \left[\frac{x^2}{2}\right]_0^2 dy = \int_0^2 6y \cdot 2 \, dy = \int_0^2 12y \, dy = 12 \left[\frac{y^2}{2}\right]_0^2 = 12 \cdot 2 = 24$.
3. The vector field $\mathbf{F}(x,y) = (2x - y, x + 3y)$ is defined in the plane. A student wants to compute the circulation of $\mathbf{F}$ around a triangle with vertices at $(1,0)$, $(4,0)$, and $(2,3)$, oriented counterclockwise. Using Green's theorem, which of the following expressions correctly represents this circulation?
   Answer: $\iint_D (1 - (-1)) \, dA = \iint_D 2 \, dA$ For circulation using Green's theorem: $\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$. Here $P(x,y) = 2x - y$ and $Q(x,y) = x + 3y$. Computing the partial derivatives: $\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x + 3y) = 1$ and $\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(2x - y) = -1$. Therefore $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 - (-1) = 2$. Choice B incorrectly adds the partial derivatives instead of subtracting. Choice C uses the wrong partial derivatives (3 and 2). Choice D also uses wrong values and computes $2-3$ instead of the correct expression.
4. Consider the region $D$ bounded by $x^2 + y^2 = 4$ and the vector field $\mathbf{F}(x,y) = \left(\frac{y}{x^2+y^2}, \frac{-x}{x^2+y^2}\right)$. A student attempts to use Green's theorem to compute $\oint_C \mathbf{F} \cdot d\mathbf{r}$ where $C$ is the circle $x^2 + y^2 = 4$ oriented counterclockwise. What is the fundamental issue with this approach?
   Answer: Green's theorem doesn't apply because the vector field is not continuously differentiable throughout the region $D$ When applying Green's theorem to compute a line integral, you must verify that all conditions are met. Green's theorem states that $\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$ when the vector field $\mathbf{F} = (P, Q)$ is continuously differentiable on a simply connected region $D$. Here, $P(x,y) = \frac{y}{x^2+y^2}$ and $Q(x,y) = \frac{-x}{x^2+y^2}$. The critical issue is that both $P$ and $Q$ have denominators of $x^2+y^2$, making them undefined at the origin $(0,0)$. Since the region $D$ is the disk $x^2 + y^2 \leq 4$, it includes the origin where the vector field is not continuously differentiable. This violates a fundamental requirement for Green's theorem. Option A is incorrect because counterclockwise orientation is actually the correct orientation for Green's theorem. Option B is wrong since the circle $x^2 + y^2 = 4$ is indeed a simple, closed curve. Option C misidentifies the issue—the region $D$ (a disk) is simply connected, but that's not the problem here. The correct answer is D because the vector field fails to be continuously differentiable throughout the region due to the singularity at the origin. Study tip: Always check that vector fields are well-defined and differentiable throughout the entire region before applying Green's theorem. Look for denominators that could equal zero within your domain—this is a common trap in circulation problems.
5. A vector field $\mathbf{F}(x,y) = (P(x,y), Q(x,y))$ satisfies $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = x + y$ in a region $D$. If the boundary $\partial D$ consists of two simple closed curves: an outer curve $C_1$ oriented counterclockwise and an inner curve $C_2$ oriented clockwise, what is $\oint_{C_1} \mathbf{F} \cdot d\mathbf{r} + \oint_{C_2} \mathbf{F} \cdot d\mathbf{r}$ in terms of the given information?
   Answer: $\iint_D (x + y) \, dA$ where $D$ is the region between $C_1$ and $C_2$ When you encounter a vector field with given curl and multiple boundary curves, you're dealing with Green's theorem in a region with holes. The key insight is understanding how orientations work with the divergence theorem. Green's theorem states that $\oint_{\partial D} \mathbf{F} \cdot d\mathbf{r} = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$, where $\partial D$ represents the complete boundary of region $D$ with proper orientation. For a region with holes, the "positive orientation" means the outer boundary goes counterclockwise while inner boundaries go clockwise - both orientations keep the region to your left as you traverse the boundary. Since $C_1$ is oriented counterclockwise (correct for outer boundary) and $C_2$ is oriented clockwise (correct for inner boundary), both curves have the proper orientation for the region between them. Therefore, $\oint_{C_1} \mathbf{F} \cdot d\mathbf{r} + \oint_{C_2} \mathbf{F} \cdot d\mathbf{r} = \iint_D (x + y) dA$ where $D$ is the region between the curves. Answer A incorrectly uses the entire region bounded by $C_1$, ignoring the hole created by $C_2$. Answer B has the wrong sign, which would occur if one of the orientations were reversed. Answer D reflects the misconception that opposite orientations automatically cancel - this ignores that the curves bound different parts of the region's boundary. Remember: for regions with holes, Green's theorem applies to the actual region (between boundaries), and "positive orientation" means outer curves go counterclockwise while inner curves go clockwise.

Start this quiz