Sorry I‘m on the way home right now but would you have some time to discuss it afterwords, so I could upload one computation from the lecture where we used this „trick“ maybe for you it is clearer then. for me it would be helpful.

sure please do...

Okey perfect thank you very much! I appreciate your help. so I think I should be able to upload it in the next 40 minutes.

I added it above, I hope you understand what I mean by the the manifold is defined when the first coordinate is smaller than $0$, because then with $=0$ then we have direclty the boundary.

If something isn't clear you clearly can ask so that i can rewrite something

ok I was just going through your edit. I do not follow why you are taking the first component negative. May be that is just a general practice in your lectures of choosing orientation or may be I am missing something. Also the way I do it is by first seeing the direction in which the normal vector to the surface is pointing and then orient the boundary accordingly.

So maybe it could have something to do integration on the standart halfspace $H_k=\{(x_1,...,x_k)\in \Bbb{R}^k: x_1\leq 0\}$

sorry could you explain this with the normal vector a bit more, because i have never heard about that

So for me it would be nice if I get a understandable way how to find a parametrization of the boundary with the right orientation. Because this one is okey but I don't really see why it workes.

On my point about normal vector - If $M$ is a two-dimensional submanifold on $\mathbb{R}^3$, then at $P \in M$, you have two normal vectors. Take an example of $x^2 + y^2 + z^2 = 1, z \geq 0$. If you choose outward normal vector to the surface (pointing in positive z-direction)], boundary $C: x^2 + y^2 = 1, z = 0$ is oriented counter-clockwise.

so Is there a generalisation for arbitrary spaces and arbitrary manifolds in them? And why is this true that then if it points outwards the orientation is counterclockwise?

If it points inside is it then always clockwise orientation

No this may not generalize for arbitrary spaces and arbitrary manifolds.

As to how we find the orientation of the curve, just follow the right hand rule. Have you heard of it?

Or would you have another way how to find the orientation in an arbitrary case?

No unfortunately I would not know or I would not remember... I studied these long long back

where it would work - for a surface in Euclidean space R^3

so if I have a submanifold of dimension 1 or 2 in R^3 I can always say that if the normal vector on a point on the boundary points outwards then the boundary is orientated counterclockwise, and if it points inside then it is orientated clockwise?

@MathLover okei no problem but can you tell me maybe if I get the orientation of the boundary of your exapmle what does it help me? so I know that the orientation of the boundary is counterclockwise and then also my parametrization needs to go in that direction

Not necessarily. Take surface z = x^2 + y^2, 0 \leq z \leq 1

the outward normal vector to the surface is pointing downward

so that will correspond to boundary curve $x^2 + y^2 = 1, z = 1$ oriented clockwise

aha so It only depends on if the outer normal vector points upwards or downwards

but on which point do we consider this normal vector?

point the thumb in the direction of the normal vector and see which way your fingers curl, that gives the orientation of the boundary curve

so but I mean I can take a point on the left or on the right so for example (1,1,1)

then it would not point downward?

say surface is z = x^2 + y^2 and boundary is x^2 + y^1 = 1, z = 1. So I take my right hand near the edge of the surface (near the boundary). I point my thumb downward as the outer normal vector is pointing down. Now I see how my fingers curl and that shows clockwise direction. So I will parametrize the curve as (cos theta, - sin theta, 1), 0 < theta < 2pi

sorry my thump point upwards and therefore counterclockwise direction

I thinka about this surface as a cup and then if I hold the cup to drink my thumb points upwards and my fingers curls counterclockwise

z = x^2 + y^2 is a paraboloid that opens up... out of the surface normal would mean generally downward

think a tangent plane to this surface at any point and a vector perpendicular to the tangent plane pointing away from the paraboloid surface

aha and what if I have the same surface as I gave in my example so this sphere which is cut from above and below, then the normal vector of the above curve points also downwards and the one of the lower curve also?

for the boundary at z = 1/2, you know the outward normal vector is pointing generally upward so curve has to be oriented counter-clockwise

I think it was z = 1/sqrt2.. but that doesnt matter

but isn't it the same as with the parabloid

sorry I really don't get this at the moment

paraboloid is opening up... sphere above z = 0 is not opening up... it is rather folding and at z = 1, x^2 + y^2 = 0

but I cut the above part away from the sphere so it is opened up not?

or do I consider the whole sphere

yes but that does not matter. do you agree that as z increases above 0, the radius of the sphere decreases?

cross sections

yess I agree with the radius

so then the curve at z=-1/2 goes in counterclockwise direction

below z = 0, the outward normal (take a vector from origin to a point on the sphere below z = 0) is pointing generally downward (negative z direction) so it should be clockwise

but the circle is also closed below, as you said above. Sorry now I don't understand nothing, i thought I got it but I'm far away from everithing with this thumb rule. it seems nice but I don't get it

when you point the thumb downward, your fingers curl in clockwise direction

but do I always collect a point of the boundery with the center? so i mean there is no center in a parabloid

take a cup that is opening up.. that is radius of cross section increases as you go up. place a coaster as tangent plane at a point and now take a line perpendicular to the plane at the point of tangency. Now there are two directions... one that takes you inside the surface and one that takes you away from it. the one that takes you away is pointing in which direction? down or up?

it points down but not vertically only if the coaster is at (0,0,0). as highter your point on the surface is as less vertically the vector shows

yes but that is why we call it generally downward...

it cannot point horizontally as that would be a cylinder then

aha okey now I think it makes sence and then on a sphere for all points above the x-y- plane it points generally upwards and for all below it points generally downwards

ok we are getting somewhere now

so isn't it correct?

since you said somewhere

yes it is correct

ah okey but then I only know the orientation, so now i need to find a parametrization with the same orientation right? (is this parametrization then clalled orientation faithful)

yes and also to mention, we typically seek orientation given in the question and then accordingly decide orientation of the boundary curve... many a times questions do not mention then we assume outward normal vector and then accordingly orientation of the boundary curve.

now for your question on x + y + z = 1 if I assume normal pointing in positive z direction then the boundary will go from (1, 0, 0) to (0, 1, 0) to (0, 0, 1) and to (1, 0, 0)

right because the normal vector points upwards?

ah this makes sence, but again to the parabloid. at z=1 we have x^2+y^2=1 which is orientated clockwise. Then I think we should parametrize x,y with polar coordinates and z=1. But then how do I know where I need to change the polar coordinates to go clockwise since thex normally go counterclockwise right

I need to go now... if there is more we can discuss later... I am really late and need to wake up in a few hours. I hope someone answers your question for arbitrary manifolds

on your last question - (cos theta, - sin theta, 1) works right?

ah okey no problem thank you for your help. sure I also hope someone can answer my question.

@MathLover ah yes I saw it. thanks