Electric Flux through a square

[Cappy9000]

The electric field in a certain region of space pints in the z-direction and has magnitude E =5zx, where x and z are measured from some origin. Calculate the flux of that filed through a square perpendicular to the z-axis; the corners of the square are at (x,y,z) = (-1, -1, 1), (-1, 2, 1), (2, 2, 1), and (2, -1, 1). (All fields are measured in N/C, all distances in m).

I used Guass' law Flux = ∯E*dA

After integrating and such I ended up with:


(5zy)|*(.5x^2)| (lower boundary being -1 and upper boundary being 2).

My question is, in the solution z=3; I cannot for the life of me figure out why.

 

[jbsteele]

Can someone explain why this is?The z value in the electric field equation represents the distance from the origin. In this case, the corners of the square are all at coordinates (x, y, z) = (1, -1, 1), (-1, 2, 1), (2, 2, 1) and (2, -1, 1). So the z value for all of these coordinates is 1, which means the distance from the origin to each corner is 1m. Therefore, the z value used in the equation should be 1.