Flux of Vector Field F out of Cube: 1

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SUMMARY

The flux of the vector field F = xi + zj out of the cube defined by the corners (0,0,0), (1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1), and (1,1,1) is confirmed to be 1. The divergence of the vector field is calculated as del · F = 1. The triple integral of the divergence over the cube results in a value of 1, validating the solution.

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Homework Statement


What is the flux of vector field F = xi + zj out of the cube that has corners in (0,0,0),(1,0,0),(0,1,0),(0,0,1),(1,1,0),(1,0,1),(0,1,1),(1,1,1).

The Attempt at a Solution


Is it 1?

Divergence = del dod F = 1
Triple integral of divergence= SSS 1 dx dy dz each from 0 to 1
= 1

I don't have a solution for this one, so I'd like to know if I got it right. :smile:
 
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Yes, it's 1.
 
Tom Mattson said:
Yes, it's 1.

Thanks!
 

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