Let $\omega\in\Omega^1(\mathbb{R^2})$ be a 1-form such that $d\omega=dx\wedge dy$. Let $i:S^1\to \mathbb{R^2}$ be the inclusion and $S^1$ be the unit circle. I want to compute $\displaystyle\int_{S^1}i^*\omega$. By Stoke's we have that $$\displaystyle\int_{S^1}i^*\omega=\int_{B}d\omega\,\,\,\,\,,B\text{ is the unit disc}.$$
We can parametrize $B$ by $Y(r,\theta)=(r\cos\theta,r\sin\theta)$ where $0\leq r\leq1,$ $0\leq\theta\leq 2\pi.$ Now, I need to check if this parametrization is coherent with the orientation on $B$. I am not sure how to do this and would like some help on that. Assuming this is coherent, $$\int_B d\omega=\int_0^1 \int_0^{2\pi}\begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix}drd\theta=2\pi$$. Is this correct?
To check that the orientation induced by this parameterization is consistent with the orientation on $B$ it suffices to show that the determinant of the Jacobian of the transition matrix between these two charts is positive, which you already did!
I think there is a small mistake in your computation though since the $d\omega$ is the usual volume form on $\mathbb{R}^2$, and hence on $B$. So the integral should be the area of the circle, which is $\pi$. You have the limits of your integrals switched ($r$ doesn't go from $0$ to $2\pi$, $\theta$ does!), what you should have is $$\int_{B}d\omega=\int_{0}^{2\pi}\int_{0}^{1}\begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix}dr~d\theta=\int_{0}^{2\pi}\frac{1}{2}~d\theta=\pi.$$