Verify Stokes' Theorem I need to verify Stokes' Theorem for the vector field $F(x,y,z)=(x^2,xy,z^2)$, for the surface, $S$, given by the part of the plane $x+y+z=1$ that is inside the cylinder $x^2+y^2=x$. 
For this I first parametrized the surface using cylindrical coordinates:
$$x=r \cos\theta +1/2 $$
$$y=r \sin\theta $$
$$z= 1- r\cos\theta - r\sin\theta$$
For $0<r<1/2$ and $0<\theta<2\pi$. This gives $||T_r \times T_{\theta}||=(r,r,r)$. And $rotF=(0,-z,y)=(0,-1+r\cos\theta +r\sin\theta,r \sin\theta)$. Then by the Stokes' Theorem:
$$\oint_{\delta S}F=\int_0^{2\pi} \int_0^{1/2}(0,-1+r\cos\theta +r\sin\theta,r \sin\theta)(r,r,r) \ dr \ d \theta$$
$$=-\pi /8$$
However, when I calculate this using line integral I'm getting, letting $x,y,z$ be:
$$x=1/2 \cos\theta +1/2 $$
$$y=1/2 \sin\theta $$
$$z= 1- 1/2\cos\theta - 1/2\sin\theta$$
And hence $\delta S '(\theta)=(-1/2 \sin\theta, 1/2\cos\theta,1/2 \sin\theta - 1/2 \cos \theta )$
And so:
$$\oint_{\delta S}F=\int_0^{2\pi} (1/4 \cos ^2 \theta +1/2 \cos\theta + 1/4,1/4 \cos \theta \sin \theta +1/4 \sin\theta ,1- 2 (\cos\theta + \sin\theta) +\sin^2 \theta+ 2 \sin\theta \cos\theta +cos^2\theta)(-1/2 \sin\theta, 1/2\cos\theta,1/2 \sin\theta - 1/2 \cos \theta )  \ d \theta$$
$$=\pi $$
Where am going wrong? Did I parametrize correctly? I just can't figure out why I can't verify the theorem.
 A: You are incorrectly assuming the circular arc  $\{(x,y):0 \leqslant x \leqslant 1,x^2+y^2 = x\}$ is entirely within the region  $\{(x,y):0 \leqslant x \leqslant 1,0 \leqslant y \leqslant 1-x\}.$
The surface is $S = S_1 \cup S_2$ where 
$$S_1 = \{(x,y,z):1/2 \leqslant x \leqslant 1, 0 \leqslant y \leqslant 1-x, z = 1 - x - y\},$$
and
$$S_2 = \{(x,y,z):0 \leqslant x \leqslant 1/2, 0 \leqslant y \leqslant \sqrt{(x(1-x)}, z = 1 - x - y\}.$$
We have 
$$\nabla \times F = \langle0,0,y\rangle,\\\mathbf{n}= \frac1{\sqrt{3}}\langle 1,1,1 \rangle,\\dS= \sqrt{3}\,dx\,dy.$$
Hence,
$$\int_S (\nabla \times F) \cdot \mathbf{n} \, dS=\int_{S_1} y/\sqrt{3}\, dS+\int_{S_2} y/\sqrt{3} \, dS\\=\int_{1/2}^1 \int_0^{1-x}y \,dy\,dx + \int_{0}^{1/2} \int_0^{\sqrt{x(1-x)}}y \,dy\,dx\\=\frac1{2}\int_{1/2}^1 (1-x)^2 \,dx + \frac1{2}\int_{0}^{1/2} x(1-x) \,dx\\=\frac1{16}.$$
Now you should be able to calculate the line integral over $\partial S= C_1 \cup C_2 \cup C_3$ where
$$C_1 = \{(x,y,z):1/2 \leqslant x \leqslant 1, 0 \leqslant y \leqslant 1-x, z=0\},\\C_2 = \{(x,y,z):0 \leqslant x \leqslant 1, y =0 , z=1-x\},\\C_3 = \{(x,y,z):0 \leqslant x \leqslant 1/2, 0 \leqslant y \leqslant \sqrt{x(1-x)} , z=1-x-y\}.$$
