We have set $A=\{(x,y,z):x^2+y^2=1,x+z=1\}$. We define manifold which contains all open line segments connecting $(0,0,0)$ with points included in $A$. Let's call this manifold $M$
Now we need to integrate $$\int_M|y|dS$$
I have problem with defining $M$ in a workable way before doing integration, so possibly help with this step would be enough to let me carry on.
The set $A$ is an ellipse in ${\mathbb R}^3$ with parametric representation $$\phi\mapsto (\cos\phi,\sin\phi, 1-\cos\phi)\qquad(0\leq\phi\leq 2\pi)\ .$$ Drawing the segments from $(0,0,0)$ to the points of this ellipse produces the surface $$M:\quad(t,\phi)\mapsto{\bf x}(t,\phi)=\bigl(t\cos\phi,t\sin\phi,t(1-\cos\phi)\bigr)\qquad(0\leq t\leq 1,\ 0\leq \phi\leq 2\pi)\ .$$ From $${\bf x}_t=(\cos\phi,\sin\phi,1-\cos\phi),\quad{\bf x}_\phi=(-t\sin\phi,t\cos\phi,t\sin\phi)$$ we obtain $${\bf x}_t\times{\bf x}_\phi=(t-t\cos\phi, t\sin\phi ,t)\ ,$$ so that the area element ${\rm d}S$ on $M$ is given by $${\rm d}S=\bigl|{\bf x}_t\times{\bf x}_\phi\bigr|\>{\rm d}(t,\phi)=t\sqrt{3-2\cos\phi}\ {\rm d}(t,\phi)\ .$$ Symmetry considerations then show that the integral ($=:J$) in question is given by $$\eqalign{J&=2\int_0^\pi\int_0^1 y(t,\phi)\> t\sqrt{3-2\cos\phi}\>dt\>d\phi\cr &=2\int_0^\pi\int_0^1 t^2\sqrt{3-2\cos\phi}\ \sin\phi\>dt\>d\phi\cr &={2\over3}\int_0^\pi\sqrt{3-2\cos\phi}\>\sin\phi\>d\phi\cr &={2\over9}\bigl(5\sqrt{5}-1\bigr)\ .\cr}$$