# Computing integral of $2$ - form on a torus

I am looking at problem 16-2 of Lee's Smooth Manifolds, second edition.

Problem 16 - 2: Let $\Bbb{T}^2 \subseteq \Bbb{R}^4$ be the two torus defined as the set of points $(w,x,y,z)$ such that $w^2 + x^2 = y^2 + z^2 = 1$, with the product orientation determined by the standard orientation on $\Bbb{S}^1$. Compute $\int_{\Bbb{T}^2} \omega$ where $\omega$ is the following two form on $\Bbb{R}^4$: $$\omega = xyz \hspace{1mm} dw \wedge dy.$$

Now if I want to evaluate such a two form do I need to care about the orientation? I am tempted to set up a map $F : \Bbb{R}^2 \to \Bbb{R}^4$ that sends $\varphi, \theta$ to $(\cos \varphi, \sin \varphi, \cos \theta, \sin \theta)$ and then taking the integral of $\omega$ on the torus to be $$\int_{[0,2\pi]^2} F^\ast \omega .$$

Is my reasoning correct, or do I need to care about the product orientation?

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To answer your question: yes you have to pay attention to the orientation. First of all you have to observe that $F:[0,2\pi)^2\to\mathbb{T}^2\subset\mathbb{R}^4$ is an isometry (the way you defined it), because of that then you can conclude as you wanted:
$$\int_{[0,2\pi)}F^\star\omega$$