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Using Greenʹs Theorem, compute the counterclockwise circulation of F around the closed curve C:

$F = (x - e^x \cos y)\vec{i} + (x + e^x \sin y)\vec{j}$; $C$ is the lobe of the lemniscate $r^2 = \sin 2θ$ that lies in the first quadrant.

$$\frac{\partial{Q}}{dx} = 1 + e^x\sin y,\quad \frac{\partial{P}}{dy} = e^x\sin y \implies \frac{\partial{Q}}{dx} - \frac{\partial{P}}{dy} = 1,\qquad \iint 1 \; dA$$

How do I calculate the bounds? I considered converting to polar and thought I could put $0 <\theta < 2\pi$ but I couldn't figure out what $r$ would be. Any advice on how to figure out the bounds as I seem to have problems with this a lot?

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So when we have a double integral like this we are just finding the area so we just need the area of the lemniscate. So to determine the bounds you have to seen where r=0 and that will tell you the angles you need. So we get that it equals 0 first from 0 to $\frac{\pi}{2}$ as earlier mentioned this double integral is just the area in polar so we get $\int_{0}^{\frac{\pi}{2}} \frac{r^2}{2}\text{d}\theta=\int_{0}^{\frac{\pi}{2}}\frac{\sin (2\theta)}{2}\text{d}\theta$

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  • $\begingroup$ If I graph $r = \sqrt{sin(2\theta)}$, then $r = 0$ when $\theta$ is at $0$ and $2\pi$? $\endgroup$ – asdfghjkl Dec 10 '16 at 19:09
  • $\begingroup$ Yeah so finding the area is just r^2/2d\theta where r is the graph $\endgroup$ – Teh Rod Dec 10 '16 at 19:11
  • $\begingroup$ I don't understand isn't it a double integral, wouldn't I need bounds for both $\theta $ and $r$? $\endgroup$ – asdfghjkl Dec 10 '16 at 19:13
  • $\begingroup$ You would but since this integral is just with respect to area (1dA) then we just need to find the area of the lemniscate. So we don't necessarily need a double integral $\endgroup$ – Teh Rod Dec 10 '16 at 19:14
  • $\begingroup$ So we can choose a single integral and integrate with respect to either $\theta$ or $r$? Does this work: $\int_{\theta=0}^{\pi/2} 1 \space d\theta$? $\endgroup$ – asdfghjkl Dec 10 '16 at 19:17

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