# Thread: Direction of normal vector stokes theorem

1. ## Direction of normal vector stokes theorem

Hi I am still a bit confused in determining the correct direction for a normal vector to a surface.

In the given example, I am ment to calculate the flux using stokes theorem with the open surface being a paraboloid: z= x^2 + y^2 bounded by plane z=4 and vector field F = (3y,-xz,yz)

My tutor said something about the normal vector always having to point outwards from the surface. Does this mean (using the right hand rule) that i have to travel in the clockwise direction around path c (the boundary between paraboloid and plane) so that my normal vector points outwards? it does not state orientation or direction of normal in the question. I have also seen that people use the convention that the surface is always on your left, but that would mean an inwards pointing vector.

so this is why iam confused.

can anyone enlighten me?

2. Yes, changing the direction of the normal vector to a surface would change the sign on the integral just as changing the direction the boundary is traversed changes the sign so the two sides will be equal as long as you are consistent. The way I learned it was- imaging walking around the boundary with you right side to the interior. Then the normal vector should be point "up" with respect to you.

The ideas of "inward" and "outward" do not apply here- they only apply to a closed surface which has no boundary.

3. So it doesnt really matter as long as I am consistent in my choice when verifying stokes theorem?

4. Yes.