Determining correct orientation for Stokes' In general, how do you determine whether you have the correct orientation to apply Stokes'? E.g in the example: Use Stokes' to evaluate $$\iint_S \text{curl}\,\underline{F}\,\cdot \hat{n} dS,$$ where $\underline{F} = \langle -yz^2, xz^2, 2xyz \rangle$ and $S$ is the curved surface of the cylinder $x^2 + y^2 = 4$ for $ 0 \leq z \leq 3,$ and $\hat{n}$ points outwards.
I have the correct answer to this question by doing the actual surface integral. However, the question wants it to be done via line integral. I know we will have two boundary curves, one at $z = 3, z = 0$, but at $z=0$ the contribution will be $0$. So in parametrising the top circle, I get $\underline{r} = 2\cos t \hat{i} + 2\sin t \hat{j} + 3 \hat{k}$ How do I know whether the direction around the circle is in the right direction for Stokes'? 
 A: Given a surface $S\subset{\mathbb R}^3$ with boundary cycle (a "sum" of closed curves) $\partial S$ you are free to chose the orientation of $S$, i.e., the direction of the normal, but you have to orient $S$ and $\partial S$ coherently. This means that when you  look from the tip of the normal vector onto the surface the boundary $\partial S$ should be oriented counterclockwise, or, what is the same thing: The interior of $S$ should be to the left of $\partial S$.
As an exercice consider the annulus
$$A:=\{(x,y,0)\ |\ a^2\leq x^2+y^2\leq b^2\}$$
in the $(x,y)$-plane, and from the two possible normal unit vectors $(0,0,\pm 1)$ choose $n:=(0,0,1)$. Looking from the tip of $n$ means looking from high up on the $z$-axis. Its obvious that the outer boundary circle  of $A$ should be oriented counterclockwise. Staring at the figure you can convince yourself that the inner boundary circle has to be oriented clockwise to make the interior of $A$ lie to the left of $\partial A$. One might write
$$\partial A=\partial D_b-\partial D_a\ ,$$
where $D_r$ is the disk of radius $r$ centered at the origin, and its boundary circle $\partial D_r$ is oriented counterclockwise.
Now in the case of your cylindrical surface $S$, oriented such that $n$ points outwards, it is a similar thing. Convince yourself that the boundary circle at $z=0$ has to be oriented counterclockwise, i.e., parametrized as
$$t\mapsto(2\cos t,2\sin t,0)\qquad(0\leq t\leq 2\pi)\ ,$$
and the boundary circle at $z=3$ clockwise, i.e., should be parametrized as
$$t\mapsto(2\cos t,-2\sin t,3)\qquad(0\leq t\leq 2\pi)$$
to ensure that viewed from the tip of $n$ the interior of $S$ is to the left of $\partial S$, and to make Stokes' theorem work.
