Compute the integral: $ \int_{S} F \cdot N d \sigma$ where $S$ is the intersection of $2 x^{2}+z^{2} \leq 1$ and the plane $x=y,$ with $F=(x,-y, \sin y)$ and $S$ oriented with normal pointing in $+x$ direction.
Attempt: The boundary of $S$ is given by $2 x^{2}+z^{2}=1, y=x,$ oriented in the counterclockwise direction when viewed from far out the $x$ axis. It can be parameterized by $\phi(t)=(\cos t / \sqrt{2}, \cos t / \sqrt{2}, \sin t), t \in[0,2 \pi].$
Let curl $(P, Q, R)=(x,-y, \sin y) .$ If $R_{y}=x$ and $Q_{z}=0,$ then $R=x y+f(x, z)$ and $Q=g(x, y) .$ Thus $R_{x}=y+f_{x}$ and we can set $f=0, P=0,$ and $g=0 .$ Hence $P=\cos y, Q=0, R=x y,$ and $(P, Q, R) \circ \phi=$ $\left(\cos (\cos t / \sqrt{2}), 0, \cos ^{2} t / 2\right) .$ Therefore, it follows from Stokes' Theorem that
$$\iint_{S} \omega=\int_{0}^{2 \pi}\left(-\frac{\sin t}{\sqrt{2}} \cos \left(\frac{\cos t}{\sqrt{2}}\right)+\frac{\cos ^{2} t}{2} \cos t\right) d t=\int_{1 / \sqrt{2}}^{1 / \sqrt{2}} \cos u d u+0=0.$$
Can you check the proof, if there is a mistake in the proof or if it is false, then can you help? Thanks...