Flux of a Vector field of a square on a plane x+y+z=20

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SUMMARY

The discussion centers on calculating the flux of the vector field F(vector) = 5i + 8j through a square of side 2 in the plane defined by x + y + z = 20, oriented away from the origin. The user attempted to find the flux by using the integral of the vector field multiplied by the normal vector of the square's area but arrived at an incorrect answer of 20800. The correct approach involves determining the unit normal vector, which should be (-1,-1,-1)/sqrt(3) and ensuring it points away from the origin, leading to a flux calculation of -52/sqrt(3).

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I have been trying this problem for multiple hours now, and cannot figure out what I am doing wrong.

--Calculate the flux of the vector field F(vector)= 5i + 8j through a square of side 2 lying in the plane x + y + z = 20 oriented away from the origin.


I realize that I need the integral of the Vector field (F) multiplied by the normal vector of the area of the square. To do this, I assigned values for the points of the triangle made, and made 2 new vectors out of it. I crossed these vectors to get the normal vector of the plane given. Once I had that, I dotted the Vector field by the normal vector, then multiplied that by the area of the square.

The answer I got was 20800. This was the incorrect answer!

Can anyone help me!?


Thanks
 
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You are probably working too hard to find the normal, but you can do it that way. Did you find the UNIT normal or just any normal? Show the details of your solution if you are still confused.
 
If you got (-1,-1,-1)/sqrt(3) for the unit normal and -52/sqrt(3) for the flux, then are you sure you've got the unit normal pointed in the right direction? The problem says it's oriented 'away from the origin'.
 
I tried it both ways. The normal ends up being the same magnitude either way, and neither answer works.

I cannot figure out what I am doing wrong :(
 
If you tried 4*13/sqrt(3) with both signs, I'm not sure what the problem is either.
 

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