I'm learning about surface integrals, and while I know how to find the curl of a vector field and how to simplify the dS, I have trouble finding and understanding how the normal vector of a surface is found.
Example question:
Use Stokes' Theorem to find the circulation of the vector field $\overrightarrow{F}=2xz\overrightarrow{i} +(x+5yz)\overrightarrow{j} +2x^2\overrightarrow{k}$ around the circle $x^2+y^2=4$, $z=4$, oriented counterclockwise when viewed from above.
I know that curl $\overrightarrow{F} = \langle-5y,-2x,1\rangle$, and in the answer that I looked at, it says that the normal vector is $\langle0,0,1\rangle$, but I'm not sure how this was determined. In simpler problems, I've been finding the normal by computing the gradient of a function and taking the coefficients, but that idea doesn't seem to work here.
First thing you should note is that you are not given a surface, you are given a curve, and you can choose any surface which has that curve as its boundary. Whichever surface you choose, the integral will be the same (by Stokes' Theorem).
So you should try to choose the surface which will make your calculations as easy as possible. In this case the boundary is a horizontal circle $4$ units above the $(x,y)$ plane, so the obvious choice is a circular disc. To find the normal, the best way is to visualise it (as in the comment by Andrew Li). However, if you really want to use the gradient, you should note that the disc is part of the plane $z=4$, which is a level curve of $$f(x,y,z)=z$$ with gradient $$\nabla f=(0,0,1)\ .$$ So the normal is $\pm(0,0,1)$, and to determine the sign...
...you probably need a diagram after all :)