Problem with the orientation of a surface

[Amaelle]

Homework Statement

circuitation and surface orientation

Relevant Equations

circuitation and surface orientation

Greetings All, I have confusion to establish the orientation of a surface so please bear with me

σ

Those two surfaces are of course intersecting in z=10 on a circle .
if I take the surface σ and I want it to be oriented so that it forms an acute angle with the axe Z , the orientation of the outer border (the circle at z=10 because if I imagine a little turtle moving the border the surface should remain at it right ) will be negative? am´I correct?

in the other hand if I take the surface τ and want it to be oriented so that it forms an acute angle with the axe Z , the orientation of the outer border will be positive.

Thank you so much!

 

[Mark44]

Those two surfaces are of course intersecting in z=10 on a circle .

This is not true. K is the disk in the plane of radius 1, centered at (0, 0). $\sigma$ is a map from this disk to $\mathbb R^3$ that represents a cup-shaped surface whose perimeter is at a height of 4, and whose center is at (0, 0, 3). $\tau$ is a portion of a sphere whose radius is $\sqrt{17}$ and whose center is at the origin.

The two surfaces intersect in a circle of radius 1, centered at (0, 0, 4).

I don't understand what you are saying that you want $\sigma$ and $\tau$ oriented so that they make acute angles with the z axis (note spelling - an axe is a tool for chopping wood).

 

[Delta2]

Yes it is axis or axle not axe e hehe.

 

[Amaelle]

This is not true. K is the disk in the plane of radius 1, centered at (0, 0). $\sigma$ is a map from this disk to $\mathbb R^3$ that represents a cup-shaped surface whose perimeter is at a height of 4, and whose center is at (0, 0, 3). $\tau$ is a portion of a sphere whose radius is $\sqrt{17}$ and whose center is at the origin.

The two surfaces intersect in a circle of radius 1, centered at (0, 0, 4).

I don't understand what you are saying that you want $\sigma$ and $\tau$ oriented so that they make acute angles with the z axis (note spelling - an axe is a tool for chopping wood).

Thank you
if I take the paraboloid σ and I want it to be orientated in a way it forms an acute angle with the Z axis, and imagine I have a vector field F(x,y,z) , and I want to compute the circuitation of F around the upper limit by keeping the orientation (an acute angle with the Z axis) it sould be equal to =-∫CURL(F(x,y,z)dA (A is the the circle of intersection)

 

[Mark44]

if I take the surface σ and I want it to be oriented so that it forms an acute angle with the axe Z ,

What does this have to do with Quiz 4? It seems that there is some information missing in your post. For example, there is no mention of how the vector field (campo vettoriale) F is defined, other than it maps R³ to R³.

What is the full statement of this problem?

 

[Amaelle]

What does this have to do with Quiz 4? It seems that there is some information missing in your post. For example, there is no mention of how the vector field (campo vettoriale) F is defined, other than it maps R³ to R³.

What is the full statement of this problem?

Here is it and thank you so much!

 

[pasmith]

The normal $\mathbf{n}$ of $\sigma$ either has positive $z$ component or negative $z$ component. Which of those will yield an acute angle ($\mathbf{n} \cdot \mathbf{e}_z > 0$) with the unit vector parallel to the $z$ axis, $\mathbf{e}_z = (0,0,1)$?

 

[Amaelle]

The normal $\mathbf{n}$ of $\sigma$ either has positive $z$ component or negative $z$ component. Which of those will yield an acute angle ($\mathbf{n} \cdot \mathbf{e}_z > 0$) with the unit vector parallel to the $z$ axis, $\mathbf{e}_z = (0,0,1)$?

both of them, if i´m not wrong

 

[Mark44]

The normal $\mathbf{n}$ of $\sigma$ either has positive $z$ component or negative $z$ component. Which of those will yield an acute angle ($\mathbf{n} \cdot \mathbf{e}_z > 0$) with the unit vector parallel to the $z$ axis, $\mathbf{e}_z = (0,0,1)$?

both of them, if i´m not wrong

No, not true. The dot product has an alternate definition: $\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos(\theta)$, where $\theta$ is the angle between the two vectors. If the dot product is negative, then the angle is greater than 90 degrees.

 

[Amaelle]

No, not true. The dot product has an alternate definition: $\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos(\theta)$, where $\theta$ is the angle between the two vectors. If the dot product is negative, then the angle is greater than 90 degrees.

Sorry, I think I´m lost here

• normal vector of the surface σ=<-2x;-2y;1>
• vector of the surface τ

< x/sqrt(17-x^2-y^2); y/sqrt(17-x^2-y^2; 1>
those two have positive orientation regarding the axis Z