Let $S^{1} \times S^{1}$ be the torus embedded in $\mathbb{R}^{4}$. I want to compute
$\int_{S^{1} \times S^{1}} d\theta_{1} \wedge d\theta_{2}$
I believe this should be "essentially" $\int_{0}^{2\pi} \int_{0}^{2\pi} d\theta_{1} d\theta_{2}= 4 \pi^{2}$. I mainly just wanted to see justification.
I've been reading about differential forms in Guillemin and Pollack, where I did a similar exercise, where I saw that for $$\omega = \frac{x dy - y dx}{x^2+y^2}$$, the pull-back allows under the map $h(t)=(\cos(t), \sin(t))$ allows us to see that (see here):
$$\int_{S^{1}} \omega = \int^{2\pi}_{0} h^{*}\omega=2\pi$$
I feel like in this case, we have something similar with $$\omega_{i} = \frac{x_{i} dy_{i} - y_{i} dx_{i}}{x_{i}^2+y_{i}^2}$$ and considering $$\int_{S^{1} \times S^{1}} \omega_{1} \wedge \omega_{2}$$ in similar fashion change coordinates say by considering a map $H:(s,t) \mapsto (h(s), h(t))$ and considering the pull-back. Would this be a way to justify it?