• Support PF! Buy your school textbooks, materials and every day products Here!

Area vector

  • Thread starter Ozzyman
  • Start date
1
0
1. Homework Statement
I am wondering what the set of area vectors for a surface would be. For a plane on the xy-plane, I know the set of area vectors is <0,0,dx*dy>.


2. Homework Equations
So, for a set of points (x,y,z) that make a paraboloid, if F(x,y,z)=0 then [grad(F)•<dy*dz, dx*dz, dx*dy>]/mag(grad(F)) would be the set of area vectors?
(i'm not sure if I'm correct in assuming the area vector needs to be a unit vector)
grad(F) is the gradient of F
mag(grad(F)) is the magnitude of the gradient field of F

3. The Attempt at a Solution
So for paraboloid x^2+y^2+z=0, the area vectors would be <(2x/sqrt(4x^2+4y^2+1))*dydz, (2y/sqrt(4x^2+4y^2+1))*dxdz, (1/sqrt(4x^2+4y^2+1))*dxdy>
yea, I am very confused. I would like to know this for integrating electric vector fields for flux.
 

Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
41,732
893
First, there is no such thing as an "area vector" (except for planes). You are talking about the "vector differential of area". Second, it can't be a unit vector- all the information about the area is contained in its length.
One thing you haven't said is how you are reducing this to two dimensions.
If you are given the surface as a function z= f(x,y), then it would make sense to reduce to the xy-plane. Write the function as F(x,y)= z- F(x,y)= 0 and find the gradient: here Fxi- Fyj+ k. That, times dxdy is the "vector differential of area"- its length is the "scalar differential of area" and it is perpendicular to the surface at each point.

In the paraboloid case, z= x2+ y2, F(x,y)= z- x2- y2 and the gradient is -2xi- 2yj+ k. The "vector differential of area" is (-2xi- 2yj+ k)dxdy and the "scalar differential of area" is its length, [itex]\sqrt{4x^2+ 4y^2+ 1}dxdy[/itex].

If the surface is given in a more general form, F(x,y,z)= constant, then, again, take the gradient of F but now divide by the k component in order to integrate in the xy-plane, by the j component in order to integrate in the xz-plane, or by the i component in order to integrate in the yz-plane.

A more general method is to find the "fundamental vector product". Suppose you are given the surface as a vector equation depending on two parameters, u, and v: r(u,v)= x(u,v)i+ y(u,v)j+ z(u,v)k. (In z= F(x,y), you could use x and y themselves as parameters- x= u, y= v, z= F(u,v)). r(u,v)= ui+ vj+ F(u,v)k. Differentiate with respect to u getting on vector in the tangent plane and wth respect to v to get another vector in the plane. Their dot product is perpendicular to the plane, multiplied by dudv, is the "vector differential of area".

Again for the paraboloid, z= x2+ y[/sup], we could take x and y as parameters and write r(x,y)= xi+ yj+ (x2+ y[/sup]2[/sup] as its vector equation. Differentiating with respect to x gives rx[/sup]= i+ 2xk and with respect to y gives ry= j+ 2yk. The cross product of those two vectors is -2xi- 2yj+ k just as before.
 
Last edited by a moderator:

Related Threads for: Area vector

  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
2
Views
29K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
5
Views
693
  • Last Post
Replies
14
Views
1K
  • Last Post
Replies
3
Views
867
  • Last Post
Replies
3
Views
986
Top