Consider the $2$-torus $\mathbb{T}^2 = S^1 \times S^1$, viewed as a subset of $\mathbb{C}^2$, endowed with the product manifold smooth structure. Define the following vector fields on $\mathbb{T}^2:$ $$\left(\frac{\partial}{\partial \theta_1}\right)_{(z_1, z_2)} = \left. \frac{d}{dt} \right\rvert_{t = 0} (e^{it} \cdot z_1, z_2) $$ and $$\left(\frac{\partial}{\partial \theta_2}\right)_{(z_1, z_2)} = \left. \frac{d}{dt} \right\rvert_{t = 0} (z_1, e^{it} \cdot z_2), $$ for every $(z_1, z_2) \in \mathbb{T}^2$.
Now, we can see that these two vector fields are smooth. Moreover, these two vector fields form a basis for the tangent space $T_{(z_1, z_2)}(\mathbb{T}^2)$, for every $(z_1, z_2) \in \mathbb{T}^2$. Hence, we can talk about their dual basis in $T^*(\mathbb{T}^2)$. Call those differential $1$-forms $d\theta_1$ and $d\theta_2$. We immediately see that for every $f \in C^\infty(\mathbb{T}^2)$, we have that $$df = \frac{\partial f}{\partial \theta_1} d\theta_1 + \frac{\partial f}{\partial \theta_2} d\theta_2. $$
It is not difficult to show that the $1$-forms $d\theta_1$ and $d\theta_2$ are not exact, since the differential of every smooth real-valued function defined on a compact manifold must vanish. It is also easy to see that those two $1$-forms are closed, and that $d\theta_2 \wedge d\theta_1$ is a volume form on $\mathbb{T}^2$. My question is the following:
How to we compute the volume of $\mathbb{T}^2$, with respect to the volume form $d\theta_2 \wedge d\theta_1$, i.e. how do we compute $$\int_{\mathbb{T}^2} d\theta_2 \wedge d\theta_1. $$
I've always seen this integration on manifolds (without the use of the theorem of Stokes) as only theoretical, because in order to compute the above integral, we firstly need a finite cover of $\mathbb{T}^2$ with coordinate charts, then we need to integrate the the pullback of $d\theta_2 \wedge d\theta_1$ through every chart and so on.
How do we find the answer in this case? I don't know how to apply the theorem of Stokes here, since I would firstly need to determine what submanifold of $\mathbb{R}^4$ has $S^1 \times S^1$ as its boundary. But even if I would be able to do that, the differential of $d\theta_2 \wedge d\theta_1$ is $0$, but the volume cannot be zero.