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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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    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?2012-12-20

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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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    Hi. Can any of you suggest books where I can read about this stuff, specifically about rigorous proofs of Stokes' theorem and Gauss's theorem and about issues related with orientation of surfaces?2014-03-15