The exercise that generates me a doubt is the following:
Let $F$ be the vector field given by $F= (xy+\sin(x^2), zy^2, xy)$
Calculate using a surface integral: $\int_c \vec F \ d\vec r$
$c$ is the intersection curve between $5x^2+y^2+z^2=1$ and the plane $z+ \dfrac{y}{2}=1$
My doubt is that I got a negative flow and a normal vector that I interpret that points outward from the surface. I understand that this is not possible so I leave my resolution below and I thank if someone can correct me and help me to understand this concept if I wrong about it.
The intersection between $5x^2+y^2+z^2=1$ and the plane $z+ \dfrac{y}{2}=1$ is $(2x)^2 +(y-2)^2 = \left( \dfrac{2}{5} \right)^2$ that is the domain of the parameterization
$$\varphi(r,\theta) = \left( \frac{r}{2} \cos \theta ,r \sin\theta + \frac{2}{5}, 1- \frac{r}{2} \sin \theta - \frac{1}{5}\right)$$
Then, the normal is: $ \vec N= \varphi_r \times \varphi_\theta = \left(0,\dfrac{r}{8}, \dfrac{r}{4}\right)$ this normal vector points out of the surface
On the other hand $ \nabla \times \vec F= (x-y^2,-y,-x$
Then, by Stokes Theorem $$\int_C \vec F \cdot \vec {dr} = \iint_S (\nabla \times \vec F) \cdot \vec {dS}=\iint_S (\nabla \times \vec F)(\varphi(r,\theta)) \cdot (\varphi_r \times \varphi_\theta) \, dr \, d \theta = \dfrac{-2 \pi}{125}$$ This is a negative flux and and the normal vector I got points out of the surface.