z = x Let be the curve of intersection of the cylinder x2+y2=1 and the...

Step 1
Stock's Theorem:-
${\displaystyle {\int }_{C}F.dr\int \int curl\left(F\right)ds}$
Given that a curve say D (as name not mention in problem) of intersection of the cylinder ${\displaystyle x^2+y^2=1}$ and the plane z=x and $F=yi+zj+2xk$
Also, S be inside of this curve, oriented with upward-pointing normal, parameterize of this curve is
${\displaystyle r=<x,y,z\ge <x,y,x>}$ as z = z
${\displaystyle r=\cos 0i+\sin 0j+\cos 0k}$
since, we know ${\displaystyle x=1.\cos 0=\cos 0,y=1.\sin 0=\sin 0,z=x=\cos 0}$ as radius =1
Here, ${\displaystyle 0\le 0\le 2\pi (as{x}^2+y^2=1^2)}$
then,
${\displaystyle r=<\cos 0,\sin 0,\cos 0>}$
${\displaystyle \frac{dr}{d0}=r^{\prime }=<-\sin 0,\cos 0,-\sin 0>}$
$F(x,y,z)=yi+zj+2xk$
${\displaystyle F\left(r\left(0\right)\right)=\sin 0i+\cos 0j+2\cos 0k}$
Step 2
Now,
${\displaystyle {\int }_{D}Fdr={\int }_0^{2\pi }F\left(r\left(0\right)\right)\frac{dr}{d0}d0}$
${\displaystyle {\int }_{D}Fdr={\int }_0^{2\pi }<\sin 0,\cos 0,2\cos 0><-\sin 0,\cos 0,-\sin 0>d0}$
${\displaystyle {\int }_{D}Fdr={\int }_0^{2\pi }(-{\sin }^{2}0+{\cos }^{2}0-2\sin 0\cos 0)d0}$
Note: ${\displaystyle -{\cos }^{2}0-{\sin }^{2}0=\cos 20,\sin 20=2\sin 0\cos 0}$
${\displaystyle {\int }_{D}Fdr={\int }_0^{2\pi }(\cos 20-\sin 20)d0={[\frac{\sin 20}{2}+\frac{\cos 20}{2}]}_0^{2\pi }}$
Note: