Using Green's Theorem, find the outward flux of F across the closed curve C.
-$$\mathbf { F } = \ln \left( x ^ { 2 } + y ^ { 2 } \right) i + \tan ^ { - 1 } \left( \frac { x } { y } \right) ; \quad C \text { is the region defined by the polar coordinate inequalities } 3 \leq r \leq 6 \text { and } 0 \leq \theta \leq \pi$$
A) 27
B) 45
C) 0
D) 54

Question

Quizplus · Accepted Answer

Green's Theorem relates a line integral around a simple closed curve C to a double integral over the plane region D bounded by C. The theorem is given by $\oint_C (L dx + M dy) = \iint_D \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right) dA$, where $F = L i + M j$. For the given vector field $\mathbf{F} = \ln(x^2 + y^2) i + \tan^{-1}\left(\frac{x}{y}\right) j$, we have $L = \ln(x^2 + y^2)$ and $M = \tan^{-1}\left(\frac{x}{y}\right)$. The partial derivatives are $\frac{\partial M}{\partial x} = \frac{1}{1 + (x/y)^2} \cdot \frac{1}{y} = \frac{1}{x^2 + y^2}$ and $\frac{\partial L}{\partial y} = \frac{1}{x^2 + y^2} \cdot 2y = \frac{2y}{x^2 + y^2}$. The difference $\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}$ is $0$, indicating that the outward flux of $\mathbf{F}$ across the closed curve $C$ is $0$, according to Green's Theorem.

Using Green's Theorem, Find the Outward Flux of F Across