This will work for all six sides of the cube.

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SUMMARY

The discussion focuses on calculating the total flux of the vector field F(x,y,z) = 8x²y i + 6yz² j + y³z k through a cube defined by its vertices. The surface integral is set up using the general flux formula Φ = ∫∫ F ⋅ n dS, where n is the unit normal vector. The user initially attempts to compute the flux using a specific surface setup but encounters issues with the normal vectors on certain cube faces. The recommendation is to apply the general flux formula for accurate results across all six sides of the cube.

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Homework Statement


Calculate the total flux of vectorF(x,y,z)=8x^2y i + 6yz^2 j + y^3z k outward through the cube whose verticies are(0,0,0), (1,0,0), (1,1,0), (0,1,0), (0,0,1), (1,0,1),(1,1,1), (0,1,1).


Homework Equations



\int\int \widehat{}F \bullet (-partial z/dx i -partial z/dy j + k) dxdy



The Attempt at a Solution


I set up the surface S: xyz\leq 1
so z \leq 1/xy

dz/dx= 1/y lnx
dz/dy= 1/x lny

so F (dot) (-dz/dz i -dz/dy j + k)
=-8x^2lnx - (6yz^2lny)/x + y^3z

I then plugged in z

=-8x^2lnx - 6lny/x^3y + y^2/x

\int\int =-8x^2lnx - 6lny/x^3y + y^2/x dxdy

0 \leq x \leq 1
0 \leq y \leq 1

Does this look correct?
 
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Your formula there should work for the top and bottom faces of the cube. But your front/back, left/right sides might pose problems, because their normal vectors don't have a k component.

I think you might want to consider using the general flux formula:
\Phi=\int\int \vec{F}\circ\hat{n}dS, where \hat{n}dS is the unit normal vector of the surface times the differential area (i.e. dx dy, dy dz, etc).
 

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