Rotating cycloid We have cycloid $(x=t-\sin t,y=0,z=1-\cos t)$ and $t\in(0,\pi]$.
Now I need do integrate some differential form over manifold $M$ which is created by rotating above cycloid around line $x=\pi,y=0$.
Could someone explain to me how exactly obtain such Manifold from above data? This should be easy, but I have hard time with understanding how exactly I have to modify above formula in order to obtain relevant figure.
Then we need to find $\int_M\omega=\int_M(y^2-x^2)dy\wedge dz+(z-x)dz\wedge dx+(2z-y)dx\wedge dy$ assuming that vector $[0,0,1]$ in point $(\pi,0,2)$ is determining orientation of $M$.
EDIT:
I changed my approach a bit, and my idea now looks like this:
First we set $t=0$, so we are in point $(0,0,0)$. We rotate this point around point $(\pi,0,0)$ on plane $z=0$. So our circle satisfies equation $(x-\pi)^2+y^2=\pi^2$. We parametrize this circle and obtain $x=\pi\cos\alpha+\pi,y=\pi\sin\alpha$. Now from every point of this circle start cycloid. So final parametrization looks like this:$$x=\pi\cos\alpha+\pi+t+\cos t,y=\pi\sin\alpha,z=1-\cos t$$
Still I do not know if it's correct and it lead to solution of integral. It seem too complicated to my mind.
 A: To rotate a point $(x,y,z)^T$ around the $z$-axis ($x = 0, y = 0$) by angle $\theta$, we apply the linear transformation
$$ \begin{pmatrix} x \\ y \\ z \end{pmatrix} \mapsto \begin{pmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix}. $$
To rotate a point $(x,y,z)^T$ around the axis $x = \pi, y = 0$, we can first translate everything so that the rotation axis becomes the $z$-axis, then perform a rotation and then translate the result back. Explicitly, the resulting map $T_{\theta}$ is given by
$$ \begin{pmatrix} x \\ y \\ z \end{pmatrix} \mapsto \begin{pmatrix} x - \pi \\ y \\ z \end{pmatrix} \mapsto
\begin{pmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} x - \pi \\ y \\ z \end{pmatrix} = \begin{pmatrix} (x - \pi) \cos \theta + y \sin(\theta) \\ -(x - y) \sin \theta + y \cos \theta \\ z \end{pmatrix} \mapsto  \begin{pmatrix} (x - \pi) \cos \theta + y \sin(\theta) + \pi \\ -(x - y) \sin \theta + y \cos \theta \\ z \end{pmatrix}.$$
Your manifold $M$ is the result of rotating the curve $\gamma(t) = (t - \sin t, 0, 1 - \cos t)$ around the axis $x = \pi, y = 0$ so the resulting manifold can be described as the image of the map 
$$ \varphi(t,\theta) = T_{\theta}(\gamma(t)) = ((t - \sin t - \pi) \cos \theta + \pi, -(t - \sin t - \pi) \sin \theta, 1 - \cos t) $$
where $t \in (0, \pi]$ and $\theta \in [0,2\pi]$. The resulting surface $M$ looks like 

where the black curve on the surface is $\gamma$. Set $D = (0,\pi) \times (0,2\pi)$. You can also verify that this parametrization induces the correct orientation on $M$ (in the sense that it is orientation preserving with respect to the the natural orientation on $D$) and so
$$ \int_{M} \omega = \int_D \varphi^{*}(\omega). $$
To compute the integral using the definitions, compute the pullback $\varphi^{*}(\omega)$ (it will be of the form $F(t,\theta) dt \wedge d\theta$) and then the integral you are looking for is
$$ \int_{M} \omega = \int_D \varphi^{*}(\omega) = \int_{(0,\pi) \times (0, 2\pi)} F(t,\theta) \, dt \, d\theta. $$
Unless there are some unexpected cancellations, the calculation of the pullback and the integral doesn't look very pleasant.
