This will work for all six sides of the cube.

[hils0005]

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?

 

[PingPong]

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).