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April 24th, 2015, 07:21 AM   #1
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cube flux

calculating the flux of cube length's S, centered at the origin, with the field being:

F=x^2(i)+y^2(j)+z^2(k)

So the flux of the face of the cube in the (i) direction :

F dot dA(i)= x^2*dA...but x at this surface is just S/2( cube's center is at origin) , so it (S^2)/4*dA, and dA is S^2...so its " (S^4)/4

now the negative (i) direction is -x^2*dA, but x is -S/2..So -(S^2)/4*dA...

the addition of the flux of two surfaces is 0..if I do the same for the (k) and (j) components. I end of with a grand total of 0 flux.. is the correct?????????
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April 24th, 2015, 01:19 PM   #2
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seems the flux of a cube whos center is at origin is 0 for a lot of vector fields
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April 25th, 2015, 06:56 AM   #3
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As long as there is no source or sink the flux through any region will be 0. Rather than integrating over each face, it is simpler to use the "Divergence Theorem":

Essentially that says that the net flow through the region is equal to the change in the quantitiy in in the region- if more flows in than out, the amount increases,

Here, so .

Find
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April 26th, 2015, 09:41 AM   #4
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thanks. I am starrting to look at greens theorem. But Id like to practice the math of finding the flux without it


Im looking at another flux now of a cylinder with its base on the x,y plane

with F=<x,y>*ln(x^2+y^2)...there is only F in radial direction so i dont have to consider the flux of the top and bottom of the cylinder

The normal of a cylinder is the radial vector..<x,y>/a, where a is the radius length

So F dot N, is (x^2+y^2)*ln(x^2+y^2)/a

want to integrate using cylindrical coords, so I want x=acos& , y=asin&

pluging this in to the dot product I end up with:

a*ln(a^2)

dS in cylindrical coords = a*d&*dz

so double integral: a^2*ln(a^2)*d&*dz

which is 2*Pi*a^2*ln(a^2)*height

But according to the text I am looking at the answer is :


4*Pi*a^2*ln(a)....

can someone point out my problem?

Last edited by Kinroh; April 26th, 2015 at 09:51 AM.
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April 26th, 2015, 09:51 AM   #5
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Never mind the last post.....


The text simplified the ln(a^2) to 2ln(a)... which would yield the same answer I got...
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