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Show that the integral of the form

$$\omega = \frac{xdy - ydx}{x^2 + y^2}$$

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

does not depend on a or b.

My solution:

$ d\omega = 0$ (tedious algebra)

Take $A = \{ \frac{x^2}{a^2} + \frac{y^2}{b^2} \leqslant1\}$ then $\partial A = \{ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\}$

by Stokes $$\oint_{\partial A} \omega = \int_{A} d\omega = \int_{A}0 =0$$

Conclusion: Integral independent of choice for $a$ and $b$.

Now this is dubious to me because if a=b=1 then I would have $$\int_{A} d\theta = \int_{0}^{2\pi}d\theta =2\pi$$ - or I am missing something?

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    $w$ is not a smooth form since $(0,0)$ is in the region. – 2017-01-14
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    Add a small circle about $(0,0)$ oriented so that together with the ellipse they bound the region between. Now the integral of $\omega$ over the union of the two curves is zero. Meaning the integral over the ellipse is the negative of the integral over the circle. The restriction of $\omega$ to the circle is just $d\theta$. – 2017-01-14

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The error in your argument is that $\omega$ is not defined at then origin. To fix the error apply your argument to the region between the ellipse and the small circle $$x^2+y^2=\epsilon$$ for small enough $\epsilon$.