Let $D$ be the disk \begin{equation} D=\{(x,y)\in\mathbb{R}^{2}\:|\:x^{2}+y^{2}\leq 1\}, \end{equation} which is easily verified to be a compact $2$-differentiable manifold with boundary.
Let $d\omega$ be the $2$-differential form on $D$ given by \begin{equation} d\omega=(1-x^{2})\:dx\wedge dy. \end{equation}
I want to compute the following integral \begin{equation} \int_{D}d\omega \end{equation}
How can I do that? I don't really understand how to parametrize $D$ in order to keep the standard orientation of $\mathbb{R}^{2}$, which is given by the canonical basis ordered as $[e_{1},\:e_{2}]$.
Can someone help me? Thanks.
We use the polar transform $\phi(x,y) =\big(r\cos(\theta), r\sin(\theta)\big)$ and find the pullback of $d\omega$ under this mapping.
$$\phi^*dx = d(r\cos(\theta)) = \frac{\partial r\cos(\theta)}{\partial r}dr + \frac{\partial r\cos(\theta)}{\partial \theta}d\theta = \cos(\theta)dr-r\sin(\theta)d\theta \\ \phi^*dy = d(r\sin(\theta)) = \sin(\theta)dr + r\cos(\theta)d\theta \\ \begin{align}\implies \phi^*d\omega &= \phi^*\left[(1-x^{2})\ dx\wedge dy\right] \\ &= \big[1-r^2\cos^2(\theta)\big]\ \big(\cos(\theta)dr-r\sin(\theta)d\theta\big)\wedge \big(\sin(\theta)dr + r\cos(\theta)d\theta\big) \\ &= \big[1-r^2\cos^2(\theta)\big]r\ dr\wedge d\theta\end{align}$$
Then $$\int_D d\omega = \int_D (1-x^2)\ dx\wedge dy = \int_{\phi^{-1}(D)} \big[1-r^2\cos^2(\theta)\big]r\ dr\wedge d\theta = \int_{\phi^{-1}(D)} \phi^*d\omega$$
The only thing left is to find $\phi^{-1}(D)$. In this case it's pretty obvious that it's $\phi^{-1}(D) = \{(r,\theta) \mid 0 \le r \le 1, 0 \le \theta \lt 2\pi\}$. This is a simple region (actually it's just a rectangle) in the $r$-$\theta$ plane. So using Fubini's theorem we can see that this integral can be written as an iterated integral:
$$\int_{\phi^{-1}(D)} \big[1-r^2\cos^2(\theta)\big]r\ dr\wedge d\theta = \int_0^{2\pi} \int_0^1 \big[1-r^2\cos^2(\theta)\big]r\ drd\theta$$
This integral is then rather simple to evaluate.