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



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

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 $$\begin{align}\oint_C (F_x\,dx+F_y\,dy)&=\iint_S \left(\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\right)\,dx\,dy\\\\&=\iint_S (1)\,dx\,dy\\\\&=\int_{-\pi/4}^{\pi/4}\int_{0}^{\sqrt{\cos(2\theta)}} r\,dr\,d\theta\\\\&=\frac12 \int_{-\pi/4}^{\pi/4} \cos(2\theta)\,d\theta\\\\&=\frac12\end{align}$$

  • $\begingroup$ Can you explain that sorry I'm still a little lost $\endgroup$ – EconDude May 11 '16 at 20:41
  • $\begingroup$ The theorem states the equivalence of the line integral $\oint_C (F_xdx+F_ydy)$ and the integral $\int_S \left(\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\right)\,dx\,dy$. $\endgroup$ – Mark Viola May 11 '16 at 20:48
  • $\begingroup$ Oh okay I see, how does it equal 1? When I differentiate I get something off. $\endgroup$ – EconDude May 11 '16 at 20:49
  • $\begingroup$ What results do you have for each partial? The difference is $1$. $\endgroup$ – Mark Viola May 11 '16 at 20:51
  • $\begingroup$ With respect to y: -1-e^ycosx + 1-e^ysinx $\endgroup$ – EconDude May 11 '16 at 20:57

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

  • $\begingroup$ How did you get those endpoints? $\endgroup$ – EconDude May 11 '16 at 20:43
  • $\begingroup$ Using polar coordinates and the fact that the lemniscate is symmetric, maybe if you draw it then it will become clearer. Actually the area of the interior of this curve is well known so the result of the integral should be 1/2. $\endgroup$ – Joss May 11 '16 at 20:51
  • $\begingroup$ Where did the 2 come from? $\endgroup$ – EconDude May 11 '16 at 20:54
  • 1
    $\begingroup$ $A(C)=\int_{-\pi/4}^{\pi/4}\int_0^{\sqrt{cos2\theta}}rdrd\theta = 2\int_0^{\pi/4}\int_0^{\sqrt{cos2\theta}}rdrd\theta.$ Because the lemniscata has a symmetry with respect to the polar axis. $\endgroup$ – Joss May 11 '16 at 20:58
  • $\begingroup$ Ohhhh okay thanks I see now $\endgroup$ – EconDude May 11 '16 at 20:59

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