Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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'?

share|improve this question
    
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
add comment

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.

share|improve this answer
    
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
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.