Finding the unit normal of a vector field

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robertjford80
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I'm on the last chapter of a 1200 page calc book, I'm really psyched.

Homework Statement



Textbook page titled Using Stokes' Theorem to Calculate Circulation listing exercises 1 to 5 with vector fields F and curves C, such as the ellipse 4x^2 + y^2 = 4 in the xy-plane and the square bounded by x = ±1 and y = ±1.


Handwritten-style worked solution: curl F equals the determinant of i, j, k with partials, giving 0i + 0j + (3 - 2)k = k, so n = k and curl F dot n = 1; then the surface integral of dσ = dx dy over region R equals the area of a circle, 9π.




The Attempt at a Solution



The method I learned for finding the unit normal of a vector field, n, is take the derivative of the equation and divide that by the magnitude of the derivative. This technique works for questions 3 and 4 above, but for 1,2 and 5, the solution manual just says n = k. I can't figure out why n = k. What are they talking about?
 
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I also need to know what d sigma = dxdy means.
 
robertjford80 said:
I also need to know what d sigma = dxdy means.

dσ=dx dy simply means that the differential of area is equal to the differential of x multiplied by the differential of y.
 
robertjford80 said:
I'm on the last chapter of a 1200 page calc book, I'm really psyched.

Homework Statement



Textbook page titled 'Using Stokes' Theorem to Calculate Circulation' listing six exercises, each giving a vector field F and a curve C in a plane, with exercises 1 to 5 shown in full.


Handwritten-style math derivation: curl F = del x F with a 3x3 determinant of i, j, k, partial derivatives and the components 2y, 3x, -z^2, giving 0i + 0j + (3 - 2)k = k; then n = k, curl F dot n = 1, dσ = dx dy, and the line integral of F dot dr equals the double integral over R of dx dy = Area of circle = 9π.




The Attempt at a Solution



The method I learned for finding the unit normal of a vector field, n, is take the derivative of the equation and divide that by the magnitude of the derivative. This technique works for questions 3 and 4 above, but for 1,2 and 5, the solution manual just says n = k. I can't figure out why n = k. What are they talking about?

Because the curve is parallel to the xy plane.
 
robertjford80 said:
The method I learned for finding the unit normal of a vector field, n, is take the derivative of the equation and divide that by the magnitude of the derivative. This technique works for questions 3 and 4 above, but for 1,2 and 5, the solution manual just says n = k. I can't figure out why n = k. What are they talking about?

[itex]\hat n = \vec k[/itex] means that the unit normal vector is parallel to the z-axis, in this case, it is in the vertically upward direction.
 
what about d sigma?

how do you find n if the curve is parallel to the xy plane, does it just equal k?
 
robertjford80 said:
how do you find n if the curve is parallel to the xy plane, does it just equal k?

Yes.
 
robertjford80 said:
what about d sigma?

d sigma is dx dy.
 
sharks said:
[itex]\hat n = \vec k[/itex] means that the unit normal vector is parallel to the z-axis, in this case, it is in the vertically upward direction.

Or in other words, the curve is parallel to the xy plane.
 
robertjford80 said:
how do you find n if the curve is parallel to the xy plane, does it just equal k?

n could be k or -k, depending on the orientation of the surface. n is the outward unit normal vector, and if the problem states "counterclockwise from above" then n is pointing upward, otherwise if it states ""counterclockwise from below"" then n = -k. If the surface lies in any plane (x=0, y=0 or z=0), then there are only 2 possible orientations for n, without any need for calculations, as you can deduce n directly.

If the surface lies in:
plane z=0, n is either k or -k
plane y=0, n is either j or -j
plane x=0, n is either i or -i

In all 3 cases, the positive value denotes moving along the positive direction parallel the respective axis, and similarly the negative value denotes moving in the negative direction parallel to the axis.
 
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sharks said:
n could be k or -k, depending on the orientation of the surface. n is the outward unit normal vector, and if the problem states "counterclockwise from above" then n is pointing upward, otherwise if it states ""counterclockwise from below"" then n = -k. If the surface lies in any plane (x=0, y=0 or z=0), then there are only 2 possible orientations for n, without any need for calculations, as you can deduce n directly.

If the surface lies in:
plane z=0, n is either k or -k
plane y=0, n is either j or -j
plane x=0, n is either i or -i

In all 3 cases, the positive value denotes moving along the positive direction parallel the respective axis, and similarly the negative value denotes moving in the negative direction parallel to the axis.

The OP said that the problem says n=k, so n is just k.
 
dimension10 said:
d sigma is dx dy.

but what does d sigma refer to? how do you calculate it? what does it mean?
 
##d\sigma## refers to the differential area, which when expressed in terms of Cartesian coordinates, converts to dxdy, as you are projecting the surface onto the x-y plane.
 
robertjford80 said:
but what does d sigma refer to? how do you calculate it? what does it mean?

As sharks has said, it is the differential of area.

[tex]{\rm{d}}\sigma = \left\| {\frac{{\partial \vec r}}{{\partial s}} \times \frac{{\partial \vec r}}{{\partial t}}} \right\|{\rm{d}}s{\rm{ d}}t[/tex]

In your case, [itex]\left\| {\frac{{\partial \vec r}}{{\partial s}} \times \frac{{\partial \vec r}}{{\partial t}}} \right\| = 1[/itex] and ##s=x,t=y##