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Help compute integral (electric potential)

  • Thread starter Carl140
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  • #1
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Homework Statement



How to calculate the following integral:

Integral( 1/ |x-y| dS) where x and y are vectors in R^2, || represents norm.
So say x= (x1,x2), y=(y1,y2). Then the integral is [ dS/ sqrt ((y1-x1)^2 + (y2-x2)^2)].

The problem is: I need to integrate this over any plane in R^3.

The Attempt at a Solution



So I assume the plane is of the form ax+by+cz+d =0. Then I really have no clue how to proceed.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
tiny-tim
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Hi Carl140! :smile:

(have an integral: ∫ and a square-root: √ and try using the X2 tag just above the Reply box :wink:)
Integral( 1/ |x-y| dS) where x and y are vectors in R^2, || represents norm.
So say x= (x1,x2), y=(y1,y2). Then the integral is [ dS/ sqrt ((y1-x1)^2 + (y2-x2)^2)].

The problem is: I need to integrate this over any plane in R^3.
Probably easiest if you let P be the closest point on the plane to the origin, and use coordinates based on P. :wink:
 
  • #3
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Hi Carl140! :smile:

(have an integral: ∫ and a square-root: √ and try using the X2 tag just above the Reply box :wink:)


Probably easiest if you let P be the closest point on the plane to the origin, and use coordinates based on P. :wink:
Thanks but how should I choose the limits of integration and how to calculate the area element? I'm very confused with this part.
 
  • #4
Born2bwire
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By itself, the integral doesn't make any sense. What are the variables of integration? Are you integrating over the vector x or y or are both vectors a function of the position on the plane? This would help determine the area element. As for the limits of integration, that's up to what you are trying to calculate.

You say the vectors are in R^2 space but you want to integrate over R^3. How do these vectors lie in the R^3 space?
 
  • #5
chiro
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What you want to do is use Stokes Theorem by first projecting the plane to a two dimensional plane (that is the XY plane for example) and then you use what you call the surface integral relationship which is given by

What you need to picture is the projection of a three dimensional plane onto a 2D space.

I'll just say a few things about projections and how they work. There are two major types of projections:

The first one is the perspective projection and the second major class is the orthographic projection.

The perspective projection essentially transforms the x,y,z coordinate system to the xy plane in the following manner:

x' = x/z
y' = y/z

Note here that we can't use z = 0 because the result will be infinite but essentially it scales the coordinates so that we get a feeling of "depth".

If we have say the points of a cube and we transform these eight points into the XY plane, then we get the following result:

Initial points:

Front face of cube
-1,+1,+1
-1,-1,+1
+1,-1,+1
+1,+1,+1

Back face of cube
-1,+1,+2
-1,-1,+2
+1,-1,+2
+1,+1,+2

Now we take the perspective transformation on all points to the x,y plane by using the transformations listed above.

Transforming Front face of cube

(-1,+1,+1) - > (-1,1)
(-1,-1,+1) - > (-1,-1)
(+1,-1,+1) - > (1,-1)
(+1,+1,+1) - > (1,1)

Transform Back face of cube
(-1,+1,+2) -> (-1/2,1/2)
(-1,-1,+2) -> (-1/2,-1/2)
(+1,-1,+2) -> (1/2,-1/2)
(+1,+1,+2) -> (1/2,1/2)

If you draw these points on a paper with a cartesian coordinate system you will see that the points corresponding to the larger z coordinate are smaller which means that as
distance increases you get the illusion of further distance. Take note however this is
a simple projective transformation. Other transformations can exist that are nonlinear and not of this type but i've just decicided to illustrate a simple example.

As for orthographic transformations these potentially should preserve length in a coordinate system. So what you want to do is use orthographic projections in place of perspective when talking about projecting the surface to the XY plane.

Now we have to explain how the surface integral works.

This is a rough sketch of how to explain the surface integral.

If we divide the surface up into extremely small "bits" then we get a normal vector at each point. Now what we do is if we picture the surface to be a collection of "points" with a normal vector at each point, then what we are picturing is basically a relative
measure (ie another form of saying the angular measure) between the standard normal (ie 0,0,1) and the surface normal.

The contribution is given by the inner product (or just dot product) of the surface normals with the standard base normal ie (0,0,1). If we sum these components and multiply by the projected area, then we have what the call the surface area.

Hope that helps.
 

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