This will work for 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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