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

Question

Quizplus · Accepted Answer

Using Green's Theorem, we have
$$\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint\limits_D\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}ight)dA$$
where $D$ is the region enclosed by $C$, $P=-y-e^y\cos x$, and $Q=y-e^y\sin x$. Note that $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1$, so we need to find the area of $D$. We can do this by converting to polar coordinates and noting that the bounds are $0 \leq r\leq|\cos	heta|$ and $0\leq	heta\leq\frac{\pi}{4}$. Thus, we have:
\begin{align*}
\iint\limits_D dA &= \int_0^{\frac{\pi}{4}}\int_0^{|\cos	heta|}rdrd	heta\
&= \int_0^{\frac{\pi}{4}}\frac{1}{2}|\cos	heta|^2d	heta\
&= \frac{1}{2}\int_0^{\frac{\pi}{4}}\cos^2	heta d	heta\
&= \frac{1}{2}\cdot\frac{\pi}{4}\
&= \frac{\pi}{8}
\end{align*}
Therefore, the outward flux is:
$$\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint\limits_D\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}ight)dA = \frac{\pi}{8} \Rightarrow \boxed{	extbf{(A)}}$$

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