Using Green's Theorem to compute counterclockwise circulation

Question

Using Green's Theorem, compute the counterclockwise circulation of $\mathbf F$ around the closed curve C. $$\mathbf F = (-y - e^y \cos x)\mathbf i + (y - e^y \sin x)\mathbf j$$ C is the right lobe of the lemniscate $r^2 = \cos 2\theta$

I need help starting this question. I already know the formula for Green's Theorem, but how do I set this up so that I can apply that formula.Thanks

Answer 1

HINT:

$$\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}=1 \tag 1$$

Now, what is the area enclosed by $C$?

SPOILER ALERT: Scroll over the highlighted area to reveal the solution

Note that we have $$\frac{\partial F_y}{\partial x}=-e^y\cos(x)$$and $$\frac{\partial F_x}{\partial y}=-1-e^y\cos(x)$$Thus, taking the difference, we obtain the result in $(1)$. Then, from Green's Theorem $$\begin{align}\oint_C (F_x\,dx+F_y\,dy)&=\iint_S \left(\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\right)\,dx\,dy\\\\&=\iint_S (1)\,dx\,dy\\\\&=\int_{-\pi/4}^{\pi/4}\int_{0}^{\sqrt{\cos(2\theta)}} r\,dr\,d\theta\\\\&=\frac12 \int_{-\pi/4}^{\pi/4} \cos(2\theta)\,d\theta\\\\&=\frac12\end{align}$$

Answer 2

$A(C)=2\int_0^{\pi/4}\int_0^{\sqrt{cos2\theta}}rdrd\theta.$ This integral is easy to calculate