Let $F: \Bbb{R}^2 \to \Bbb{R}^3$ be given as $$(\theta,\phi) \mapsto ((2+\cos \phi)\cos \theta,(2+\cos\phi)\sin\theta,\sin\phi).$$ Let $M$ be the two torus obtained as the image of $F$. Consider the $2-$form $\omega=x \, dy \wedge dz$, compute $F^*\omega$ and use this to compute $\int_M \omega$.
Attempt:
As $M$ is orientation preserving, by a proposition in Lee,
$$\int_M \omega = \int_0^{2\pi}\int_0^{2\pi}F^* \omega.$$
I calculated $F^* \omega$ to be
$$F^* \omega = (2 \cos \phi + \cos^2 \phi)\cos^2 \theta \, d \theta \wedge d \phi$$
thus
\begin{align} \int_M \omega&=\int_0^{2\pi}\int_0^{2\pi} (2\cos \phi + \cos^2 \phi)\cos^2 \theta d \theta d \phi\\ &=5\pi^2. \end{align}
Is my work correct?
For the pullback I did this:
\begin{align} F^*\omega&=F^*(x \, dy \wedge dz)\\ &=((2+\cos \phi)\cos \theta)\, d((2+\cos\phi)\sin\theta) \wedge d(\sin\phi)\\ &=(2+\cos\phi)\cos\theta(-\sin\phi\sin\theta\, d \phi + \cos \theta (2+\cos\phi)\, d \theta) \wedge \cos \phi \, d \phi \end{align}
and the $d \phi \wedge d \phi$ cancels so its
$$(2+\cos\phi)^2 \cos^2 \theta \cos \phi\, d \theta \wedge d \phi$$