We have to points. $P$ is moving on circle $x^2+y^2=1,z=1$ counter-clockwise and in time $t=0$ it is $P_t=(1,0,1)$.
Second point $Q$ is moving on circle $x^2+y^2=1,z=0$ with the same angular speed (also counter-clockwise) as point $P$ but when $t=0$ it is $Q_t=(0,1,0)$.
Now manifold $M$ consists of open line segments connecting $P_t$ with $Q_t$.
Find $\int_M|2z-1|dS$
I know that I should use polar coordinates to this one. Perhaps I should do something like that: $P_t=(\cos t,\sin t,1), t\in[0,2\pi)$ and $Q_t=(\sin t,\cos t,0)$. Now $M$ is graph of function $f(t,a)=aP_t+(1-a)Q_t,a\in(0,1)$, therefore it is honest smooth manifold. But Gram determinant looks terrible so something must have gone wrong.
EDIT: I still hope someone will help. I checked with wolfram this above parametrization and it is hopeless, but I suspect I made mistake somewhere, but I cannot find it.
EDIT2: Now I see that this approach cannot be right because it does not represent proper manifold. I guess I shouldn't interchange $\sin$ with $\cos$, but modify its arguments, so $P_t=(\cos t,\sin t,1)$ and $Q_t=(\sin (t+q),\cos(t+q),0)$ for some $q$, but I am to tired to work this out, so I'll come tommorow to this.
$(x,y,z)\to(a(\cos t-\cos(t+\pi/2))+\cos(t+\pi/2),a(\sin t-\sin(t+\pi/2))+\sin(t+\pi/2),a)=(a(\cos t+\sin t)-\sin t,a(\sin t-\cos t)+\cos t,a)$
Then $G=6a^2-6a+2$ and our domain of integration is what we obtain under above transformation for $t\in(0,2\pi),a\in(0,1)$, so out integral looks like: $$\int_0^1\int_0^{2\pi}\sqrt2\sqrt{3a^2-3a+1}|2a-1|dtda$$