Using Green's Theorem, find the outward flux of F across the closed curve C.
-$\mathbf { F } = \left( \mathrm { x } - \mathrm { e } ^ { \mathrm { x } } \cos \mathrm { y } \right) \mathbf { i } + \left( \mathrm { x } + \mathrm { e } ^ { \mathrm { x } } \sin \mathrm { y } \right) \mathbf { j } ; \mathrm { C }$ is the lobe of the lemniscate $\mathrm { r } ^ { 2 } = \sin 2 \theta$ that lies in the first quadrant
A) $\frac { 1 } { 4 }$
B) 0
C) 1
D) $\frac { 1 } { 2 }$

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 outward flux of $\mathbf{F}$ across C is given by the circulation form of Green's Theorem: $\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA$, where $\mathbf{F} = P\mathbf{i} + Q\mathbf{j}$. For $\mathbf{F} = (x - e^x\cos y)\mathbf{i} + (x + e^x\sin y)\mathbf{j}$, we have $P = x - e^x\cos y$ and $Q = x + e^x\sin y$. Thus, $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 + e^x\sin y + e^x\cos y$. The region D is the lobe of the lemniscate $r^2 = \sin 2\theta$ in the first quadrant, which is symmetric and bounded. Converting to polar coordinates and integrating over the appropriate bounds for the first quadrant lobe of the lemniscate gives the result $\frac{1}{2}$.

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