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I have this problem

Use Green’s Theorem to find the counter-clockwise circulation for the field $\mathbf{F}$ and curve $C$.

with this image

enter image description here

Green's Theorem says that the counter-clockwise circulation is $\oint_C \mathbf{F} \cdot \mathbf{T} \ ds = \oint_C M \ dx + N \ dy$. I will use the latter formula.

We can see from the vector field $\mathbf{F}$ that $M = x + 3y$ and $N = 2x - y$.

The parameterization I used is $x = \cos(t)$, $y = \sin(t)$ $\forall \ 0 \le t \le 2\pi$. Therefore, the parameterized curve is $r(t) = 2\cos(t) \mathbf{i} + \sin(t) \mathbf{j}$.

Taking the derivatives, we get $dx = -\sin(t) \ dt$ and $dy = \cos(t) \ dt$.

So we get

$\int_C M \ dx + N \ dy = \int^{2\pi}_0 (\cos(t) + 3(\sin(t))(-\sin(t)) + (2\cos(t) - \sin(t))(\cos(t)) \ dt = -\pi$ by my calculations. I confirmed my calculations by using this calculator.

But I'm unsure if this is correct. Can someone please check my work?

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  • $\begingroup$ Green's theorem converts the line integral to a double integral. You just calculated the line integral by parametrization, which is a valid approach as well, but not what the question asks. However, you parametrized the elipse wrong. $\endgroup$ – Sorfosh Jul 22 '18 at 22:23
  • $\begingroup$ @Sorfosh Ok thanks. What is the correct parameterization? $\endgroup$ – Wyuw Jul 22 '18 at 22:29
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As I mentioned in my comment Green's theorem draws a connection between the double integral over the region enclosed by the curve, and the line integral of a vector field over that curve. In symbolic terms:

$$\oint_{\partial D} (P\, dx+Q\, dy) = \iint_D dx\,dy \: \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)$$

Now in our case, $$\frac{\partial Q}{\partial x}=2,\frac{\partial P}{\partial y}=3$$

Hence, We get, by the formula: $$\iint_D dx\,dy \: \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)=\iint_D 2-3dx\,dy=-\iint_D 1dx\,dy$$ This is simply $-1$ times the area of the enlosed region, which we know is simply $\sqrt{2}\pi$. So:

$$\oint_{\partial D} (P\, dx+Q\, dy) =-\sqrt{2}\pi$$

Regarding the parametrization, it should be $x=\sqrt{2}\cos(t)$ and $y=\sin(t)$ and it should give you the same answer.

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  • $\begingroup$ Thank you for clearing this up. Can you please demonstrate how you arrived at that parameterization? I'm unsure why we use $\sqrt{2}$. Also, doesn't a negative value for circulation mean that it is circulating clockwise instead of anti-clockwise? Is that ok in this context? $\endgroup$ – Wyuw Jul 22 '18 at 22:59
  • $\begingroup$ @Wyuw The problem says counterclockwise, and if you look at the picture, it is indeed counterclockwise. So we got the direction right. Regarding parametrization, it is something you should memorize, it is a convienient tool. Simply note that when we plug in our parametrization into the equation of the elipse, we get $x^2+2y^2=2\cos^2(t)+2\sin^2(t)=2(\cos^2(t)+\sin^2(t))=2$. So our curve lays on the elipse. $\endgroup$ – Sorfosh Jul 23 '18 at 8:52
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    $\begingroup$ Your ${\partial Q\over\partial x}$ and ${\partial P\over\partial y}$ are wrong. They should be $3$ and $2$. $\endgroup$ – Christian Blatter Jul 23 '18 at 14:59
  • $\begingroup$ @ChristianBlatter oh boy, my bad. Corrected, thank you! $\endgroup$ – Sorfosh Jul 23 '18 at 15:11
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    $\begingroup$ @Wyuw the drawing of the ellipse $x^2+2y^2=2$ is incorrect. I used the formula rather than the picture. Regarding positive negative, it is not the case that the line integral will always be positive with positive orientation. What textbook are you using? This is plainly false. $\endgroup$ – Sorfosh Jul 23 '18 at 16:16

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