Here is the problem:
"A plane, whose normal (directed away from the origin) is at an angle α to the z axis, and whose perpendicular distance to the origin is a, is intersected by the sphere $ x^2 + y^2 + z^2 = R^2$ along the circle C. The direction of C is positive relative the outward normal of the sphere on the part of the sphere with the greater surface area. Calculate the integral.
$$\oint_c \bar{A} • d\bar r$$
Where $A = (3xz + a(z + x))\bar x+ (3xz + a(z-x))\bar y + (3xy + a(x+y))\bar z$ "
I know that for this we use stokes theorem so we get:
$$\nabla \times \bar A = -2a \bar z$$
So we get $$\oint_c \bar{A} • d\bar r = \iint \nabla \times \bar A • d\bar S = \iint -2a \bar z•\hat n dS $$
So in the solution is says $$ \bar z•\hat n =-cos\alpha $$
So if I understood this part I would be able come to the solution my self. Lets say we instead have $$ \bar y •\hat n = ? $$ $$ \bar x •\hat n = ? $$ What would we get instead of questionmark?
I am simply not understanding the dot product of the vector with the normal. I wish someone can explain this to me so that I may understand it.