Let $D$ denote the torus of revolution in $\Bbb{R}^3$ by revolving the circle
$$(y-2)^2+z^2=1,$$
along the $z-$axis.
If $\omega = z \, dx \wedge dy$, compute $\int_D \omega$.
Attempt:
So I am appealing to Stokes' theorem by computing
$$\int_M d \omega$$
where $M$ is the filled in torus (including the boundary) with one radius $2$ and one radius $1$. However, a quick computation shows
$$d \omega = dx \wedge dy \wedge dz.$$
Thus
$$\int_M d \omega$$
is just the volume of this torus which is
$$2\pi^2Rr^2$$
where $R=2$, $r=1$ so is the answer
$$4\pi^2??$$
Thanks in advance for any help!! so for Stokes' theorem, for any given exercise its always the case that one of the two sides of the equation will always be much more trivial to compute than the other side?