Evaluate $\int_Mxdy\wedge dz$ where $M$ is the torus obtained by rotating the circle $(x-2)^2+z^2=1$ around the $y$ axis. I've parameterized $M$ using $\alpha:(0,2\pi)\times (0,2\pi)\rightarrow M$ where $$\alpha(\theta,\phi)=((2+\cos\theta)\cos\phi,\sin\theta,(2+\cos\theta)\sin\phi)$$ But since $\int_Mxdy\wedge dz=\int_U\alpha^*(xdy\wedge dz)$ we get $$\int_U\alpha^*(xdy\wedge dz)$$ $$=\int_U(2+\cos\theta)\cos\phi((\cos\theta d\theta)\wedge((2+\cos\theta)\cos\phi d\phi-\sin\theta\sin\phi d\theta))$$ $$=\int_U(2+\cos\theta)\cos\phi((2+\cos\theta)\cos\theta\cos\phi d\theta d\phi)$$ $$=\int_U(2+\cos\theta)^2\cos^2\phi\cos\theta d\phi d\theta$$
We integrate from $0$ to $2pi$ in the $\phi$ direction and get
$$\pi\int_0^{2\pi}(2+\cos\theta)^2cos\theta=\pi\int_0^{2\pi}4\cos\theta+4\cos^2\theta+cos^3\theta$$ $$=4\pi^2$$
which I believe is correct. I have checked my calculations a few times but I may have missed something.
The question also asks to consider the orientation on $M$ induced by the normal field $v$ where $v(3,0,0)=(1,0,0)$ but I have no idea what kind of orientation this is or how that will effect my integral.