# Flux of a vector field through a cube

• Apr 18th 2012, 06:04 PM
OneMileCrash
Flux of a vector field through a cube
Calculate the flux of the vector field F = (4-x)i through the cube with verticies (0,0,0),(5,0,0),(0,5,0),(0,0,5), oriented outwards.

First, I drew the cube. Since the vector field is only in the i direction, I eliminated all but two sides. Namely, the sides parallel to the x-axis. Is that a correct thing to do?

Secondly, I noted that the only sides that are perpendicular to the direction of the vector field at any point have the following attributes:

Surface #1: x = 0 at all points. Surface area = 25. This side is oriented towards -i.
Surface #2: x = 5 at all points, surface area also 25. This side is oriented towards +i.

Thus I figured it would be a simple problem of adding the fluxes through these two surfaces.

Surface #1:
Since x = 0 at all points, vector field F = 4i at all points on the surface. Since the orientation is -i, A vector = -25i.

Dotting these two vectors is just -100.

Surface #2:

Since x = 5 at all points, vector field F = -i at all points on the surface. Since the orientation is +i, Area vector = 25i.

Dotting these two vectors is -25.

And,

-100 + -25 = -125

So, what part of this is incorrect thinking? Care to shed some light?

Thank you!!
• Apr 19th 2012, 05:07 PM
OneMileCrash
Re: Flux of a vector field through a cube