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### Course: Multivariable calculus > Unit 5

Lesson 5: Stokes' theorem- Stokes' theorem intuition
- Green's and Stokes' theorem relationship
- Orienting boundary with surface
- Orientation and stokes
- Orientations and boundaries
- Conditions for stokes theorem
- Stokes example part 1
- Stokes example part 2
- Stokes example part 3
- Stokes example part 4
- Stokes' theorem
- Evaluating line integral directly - part 1
- Evaluating line integral directly - part 2

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# Orienting boundary with surface

Determining the proper orientation of the boundary given the orientation of the surface. Created by Sal Khan.

## Want to join the conversation?

- Isn't it easier to use a right-hand rule? Curling four fingers across the direction of the curve with your thumb pointing to the positive normal vector direction? Or are there cases where this does not apply?(20 votes)
- You can do that for every contour. The right-hand rule its just a more intuitive way to know how a cross product will look like. Positive it your thumb its on the positive direction of z or x or y, depending on which axis you're working and negative if it points to the negative direction.Some call it the Screwdriver rule, if the screwdriver is turning a screw and it would take it out of the surface or point or whatever the resultant vector is positive and vice-versa for the negative situation.

Did i made it more clear to you?(3 votes)

- Hello,

I am struggling with following question for so many days now. It would be great if you could help.

If I have a Normal at a point on a surface on the surface boundary and a tangent at the same point on the curve, how do I check whether the curve is in the positive orientation with surface? In short, how can I verify that the surface is really on the left or not from the curve tangent and surface normal information?

In case of planar surfaces, it is quite easier to solve because surface normal is constant all over the surface. On a non planar surface it is more difficult because the normal changes. e.g. in case of cylinder which is cut by two circular curves (one at z=0 and one at z = h) the curve orientations are required to be opposite of each other.

Thank you very much in advance.

Regards,

Anup(4 votes)- If the concern is making sure the curve is positively orientated I believe you could find the "curl" with a line integral. If the line integral turned out positive the surface path would be counterclockwise which is positive by convention and tell you the surface was on your left. If the value was negative then the path of the line integral is clockwise making the surface on your right.(2 votes)

- So, it's just the right hand rule? Using the thumb as N and wrapping fingers is where the circulation is going? N positive = counterclockwise. N negative = clockwise?(3 votes)
- For the simple case of a "normal" surface it's just the right hand rule right.

Image for example a cylinder with open end on both ends.

So now you can't you the right hand rule. The normal vector of the sides of the cylinder is pointing radial outwards. On you have to orientate the TWO boundary curves (because in this case you have 2 boundary Curves) according to the rule Sal explained. And the two curved will have opposite orientation. Hope this makes sense to you.(2 votes)

- This should be in the Tips & Thanks section, but since people rarely view that section I will leave this in the Questions.

No need to watch this video

Just watch the video after this :

https://www.khanacademy.org/math/multivariable-calculus/greens-theorem-and-stokes-theorem/stokes-theorem/v/orientation-and-stokes

It's basically the identical video but it has an added insight at the end (using the "corkscrew" method to visualize direction). Personally, I prefer the right hand rule but any of these methods work(2 votes) - When he says that if you orient your head in the direction of the normal vector and you were to walk along the path, the surface would be on your left (1:15-1:26), does that imply that you are outside of the surface?(1 vote)
- Yes, it is like you are walking around a small hill in a counterclockwise direction, with the hill always to your left.(2 votes)

- a question about direction. in a 3d cordinate system, if we have a curve, what is the normal at a point going to be like. is it a plane perpendicular to the tangent.

well i know for a surface we can find a normal at a point p, and than its tangent plane by simple formula <p-x>.<n> = 0.(1 vote)- The normal vector, like any other vector, is
**always**a one dimensional ray. When doing Stoke's theorem problems, you will be dealing with the normal*vector*,**not**the normal*plane*. The normal vector of a surface S at point P is the vector perpendicular to the tangent plane of surface S at point P.

http://math.etsu.edu/multicalc/prealpha/Chap3/Chap3-6/printversion.pdf

You don't have to read the whole thing. Just look at the picture on page 1. The green arrow is the normal vector.

also reference1:18-- n^ is a normal vector to the surface(1 vote)

- this may be in the context of geometry, does integrating ∯1.n̂ds just gives the area enclosed by the curve.(1 vote)

## Video transcript

I've restated Stokes' theorem. And what I want to
do in this video is make sure that we get
our orientation right. Because when we think about
a normal vector to a surface there are actually
two normal vectors. There-- based on the
way I've drawn it right over here, there could be
the one that might pop outward, like this. Or there might be the one that
pops inward, just like that. Both of those would be normal
to the surface right there. And also, when we
think about a path that goes around the
boundary of a surface, there's two ways to
think about that path. We could be going-- based on
how I've oriented it right now, we could go in a
counterclockwise orientation, or direction. Or we could go in a clockwise
orientation, or direction. So in order to make sure
we're using Stokes' theorem correctly, we need to
make sure we understand which each convention
it is using. And the way we think
about it is, whatever the normal direction
we pick-- and so let's say we pick this normal
direction right over here, the one I am drawing in yellow. So if we pick this
as our normal vector. So we're essentially
saying maybe that's the top, one way
of doing it is, that's the top of our surface, then
the positive orientation that we need to traverse the path in is
the one that if your head was pointed in the direction
of the normal vector, and you were to walk along
that path, the inside, or the surface itself
would be to your left. And so, if my head is
pointed in the direction of the normal vector-- so
this is me right over here-- my head is pointed
in the direction of the normal vector-- I'm
wearing a big arrow hat right over there-- and if I'm
walking around the boundary, the actual surface
needs to be to my left. So I need to be-- this is me
walking right over here-- I need to be walking in the
counterclockwise direction just like that. Then that's the
convention that we use when we're thinking
about Stokes' theorem. If oriented this
thing differently, or if we said that
no, no, no, no, no, this is not the normal vector. This is not the, essentially,
the top that we want to pick. If we wanted to pick
it the other way, if we wanted to go
in that direction, If we wanted that to be
our normal vector, in order to be consistent, we would
have to now do the opposite. I would now have to have my
head going in that direction. And then I would have
to walk, once again, and this might be a little
bit harder to visualize. I would have to walk
in the direction that the surface is to my left. And now, in this
situation, instead of the surface looking
like a hill to me, the surface would look
like some type of a bowl, or some type of a valley or
something like that to me. And the way that I
would have to do it now, and it's a little
bit hard to visualize the upside down Sal,
but the upside down Sal would have to walk in
this direction in order for the bowl, or the
dip, to be to my left. So that's just important
to keep this in mind in order for this to be
consistent with this right over here. Put your head in the direction
of the normal vector. Or you can kind of view that
as the top of the direction that the top of the
surface is going in. And then the contour,
or the direction that you would have to traverse
the boundary in order for this to be true, is the
direction with which the surface is to your left.