Calculate $\int_CPdx+Qdy+Rdz$ where $\vec F=(P,Q,R)=(y,z,x)$ and $C$ it's the intersection of $x+y=2$ and $x^2+y²+z^2=2x+2y$ going on the counter-clockwise direction when saw from the origin.
Answer: $-2\sqrt2\pi$
My attempt:
We isolate $y=2-x$ on the first equation to substitute on the other one and have the ellipsis equation, we get:
$\vec r(t)=\frac12(1+\cos t)\vec i+\frac12(3-\cos t)\vec j+\frac{1}{\sqrt2}\sin t\vec k$
$\vec r'(t)=-\frac12\sin t \vec i+\frac12\sin t\vec j+\frac{1}{\sqrt2}\cos t\vec k$
$\vec F(t)=\frac12(3-\cos t)\vec i+\frac{1}{\sqrt2}\sin t\vec j+\frac12(1+\cos t)\vec k$
$\int_0^{2\pi}\left(-\frac34\sin t+\frac14\sin t\cos t\right)dt+\int_0^{2\pi}\frac{1}{2\sqrt2}\sin^2 tdt+\int_0^{2\pi}\left(\frac{1}{2\sqrt2}\cos t+\frac{cos^2t}{2\sqrt2}\right)dt$
$=\frac{\pi}{2\sqrt2}+\frac{\pi}{2\sqrt2}$
What was my mistake here?