Let $C$ be the curve which obtains from the intersection of the plane $z=x$ and the cylinder $x^{2}+y^{2}=1$, oriented counterclockwise. If $F$ is a vector field $F=(x,z,2y) \in \mathbb{R}^{3}$, compute $$I= \oint_{C} F \cdot dr $$
Hint: Use Stoke's Theorem.
I have that, $\nabla \times F= i$, then $$I= \int_{S} (\nabla \times F) \cdot dS= \int \int_{D} 1dA $$ where $D= \{(x,y)| \quad x^{2} + y^{2} \le 1 \}$.
Is this fine?, I don't know if I am using correctly the theorem.
Let's call the surface enclosed by the curve $C$ as $D$. Then, as you said, $$I= \oint_{C} F \cdot dr = \int \int_{D} 1dA$$
One of the possible ways to parameterize the surface $D$ is:
$$\vec{\omega}(r,\theta) = <r\cos\theta,r\sin\theta, r\cos\theta >\quad 0 \leq r \leq 1, 0\leq \theta\leq 2\pi$$
Like you said, $\nabla \times F= <1,0,0>$ and now we need to get the normal vector of the surface (pointing upward) to evaluate the surface integral, which can be obtained by: $\vec{\omega_r}\times\vec{\omega_\theta}$ or $-\vec{\omega_r}\times\vec{\omega_\theta}$, choosing whichever having the positive $z$ value.
According to my calculation of the cross product, I get $\vec{\omega_r}\times\vec{\omega_\theta} = <-r, 0, r>$ .
Therefore the integral becomes $$\int_0^{2\pi} \int_0^1 <1,0,0>\cdot (\vec{\omega_r}\times\vec{\omega_\theta}) drd\theta$$
$$= \int_0^{2\pi} \int_0^1 <1,0,0>\cdot <-r, 0, r>drd\theta$$
$$= -\pi$$