# Green's Theorem $F = (x - e^x cos y)i + (x + e^x sin y)j$; $C$ is the lobe of the lemniscate $r^2 = sin 2θ$ that lies in the first quadrant.

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?

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 $\int_{0}^{\frac{\pi}{2}} \frac{r^2}{2}\text{d}\theta=\int_{0}^{\frac{\pi}{2}}\frac{\sin (2\theta)}{2}\text{d}\theta$
• If I graph $r = \sqrt{sin(2\theta)}$, then $r = 0$ when $\theta$ is at $0$ and $2\pi$? – asdfghjkl Dec 10 '16 at 19:09
• I don't understand isn't it a double integral, wouldn't I need bounds for both $\theta$ and $r$? – asdfghjkl Dec 10 '16 at 19:13
• So we can choose a single integral and integrate with respect to either $\theta$ or $r$? Does this work: $\int_{\theta=0}^{\pi/2} 1 \space d\theta$? – asdfghjkl Dec 10 '16 at 19:17