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I calculated the circulation of the vector field :

$$\vec{v} = -y\omega \, \vec{i} + x\omega \, \vec{j}$$

over the ellipse : $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

I found $2 \pi \omega a b$.

Now I'm supposed to find the same result using Green's theorem:

$$\int\limits_C{(-y\omega \, dx + x\omega \, dy}) = \iint\limits_D \left(\frac{\partial(x\omega)}{\partial x} - \frac{\partial (-y\omega)}{\partial y} \right)dxdy = 2\omega\iint\limits_D \, dx \, dy$$

with D: $$\frac{x^2}{a^2} + \frac{y^2}{b^2} < 1$$

and I don't understand what I'm supposed to do with this...

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Remember that $$\iint\limits_D \, dx \, dy = \text{Area}(D).$$ What is the area of an ellipse?

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  • $\begingroup$ Ok but I tried to use parametrization and it didn't work: $2\omega\int_0^{2\pi}{(\int_0^{2\pi}{-absin(t)cos(t)dt})}dt=0$ $\endgroup$ Oct 3, 2014 at 9:06
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    $\begingroup$ @user1234161 You use $ x = a r \cos (\theta) $ and $ y = b r \sin (\theta) $. The integral becomes $$\iint\limits_D \, dx \, dy = \int_0^{2 \pi} \hspace{-5pt} \int_0^1 ab \, r \, d r \, d \theta = \pi ab.$$ You parameterized the boundary only. You don't substitute things like a line integral. $\endgroup$ Oct 3, 2014 at 12:10

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