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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'?

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When you look from the tip of the normal vector onto the surface the boundary curve should be oriented counterclockwise. This means, now looking from high above on the $z$-axis, that the circle $z=0$ has to be oriented counterclockwise and the circle $z=3$ clockwise. –  Christian Blatter Dec 20 '12 at 10:00
For the circle in the $xy$ plane, it's normal points in the direction $-\hat{k}$. So this means from the tip of this vector, looking up, the circle has to be oreinted counterclockwise. (as you said). For the circle at $z=3$, the normal points $\hat{k}$, so looking down, shouldn't it be oreinted counterclockwise too? –  CAF Dec 20 '12 at 10:15
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1 Answer

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.

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Thanks for this answer. Does the above not contradict the right hand rule though? If I consider the normal vector pointing in negative $z$, then my thumb, representing the normal, points towards the floor. So from the top of my thumb's perspective, we have a counterclockwise oreintation. This is what we wanted. But for the normal pointing towards the ceiling (i.e from $z = 3$), my thumb points upwards, and from the top of the thumb's perspective, it is also counterclockwise? –  CAF Dec 20 '12 at 11:48
@CAF: The normal of your cylindrical surface $S$ is parallel to the $(x,y)$-plane and points outward. Place yourself at the point $(1,0,2.9)\in S$ and verify, in which direction the upper boundary circle should go, if (seen from the outside) you should be to the left of the oriented circle. –  Christian Blatter Dec 20 '12 at 12:00
Am I not considering the normals to the circles at $z=0$ and $z=3$? –  CAF Dec 20 '12 at 12:21
@CAF: No. We are considering the normals to $S$ and the tangential vectors to $\partial S$. –  Christian Blatter Dec 20 '12 at 12:26
Sorry, I still don't see it. Looking from above, the normal vector points to the right say. (So looking down from high above, I see a circle with an arrow attached to it pointing right). Now what should I do? Thanks. –  CAF Dec 20 '12 at 12:42
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