Flux of Vector Field F out of Cube: 1

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The flux of the vector field F = xi + zj out of the specified cube is confirmed to be 1. The divergence of the vector field is calculated as del dot F, which equals 1. A triple integral of the divergence over the cube yields a result of 1. Multiple participants in the discussion agree on this solution. The final answer is indeed 1.
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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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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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