Use Stoke's theorem to solve the following integral (each time the curve is oriented counterclockwise when viewed from above): $$\int \limits_C xdx +(x-2yz)dy+(x^2+z)dz$$ where $C$ is the intersection of the sphere $x^2+y^2+z^2=1$, the cylinder $x^2+y^2=x$, and the halfspace $z>0$.
I let $y=r\cos\theta$, and let $z=r\sin\theta$ which makes $x=1-r^2\sin^2\theta$ so $$S(r,\theta)=(1-r^2\sin^2\theta, r\cos\theta, r\sin\theta)$$ with $r\in[0,1]$ and $\theta \in [0,2\pi]$
I got the normal $n=(r,0, 2r^2 \sin\theta)$ and
curl($F$)$=(2y,-2x,1)=(2r\cos\theta, -2-2r^2 \sin^2 \theta, 1)$
but when I take the dot product of these two and double integrate, I end up with zero which seems wrong. Please help.