Calculute the flux resulting from a certain vector field in a cube.

[mahrap]

Let's say there is a cube sitting in the first octant. Our F(x,y,z): <ax , by, cz> and Each face of the cube is oriented to outward pointing normal. Can I just calculate the the flux of one face and then multiply this by the number of faces to get the total flux? Will flux in a cube always be symmetric?


I think you can parametrize a face by letting G(x,y,z)=<x,y,1>. Then use this to find normal and dot this with F(x,y,z) and integrate with dx dy with x and y ranging from where the sides start to where they end?


I'm not sure. Please Help.

 

[SammyS]

Let's say there is a cube sitting in the first octant. Our F(x,y,z): <ax , by, cz> and Each face of the cube is oriented to outward pointing normal. Can I just calculate the the flux of one face and then multiply this by the number of faces to get the total flux? Will flux in a cube always be symmetric?

No. You cannot calculate the flux by considering only one face. The field is not the same in all directions.

I think you can parametrize a face by letting G(x,y,z)=<x,y,1>. Then use this to find normal and dot this with F(x,y,z) and integrate with dx dy with x and y ranging from where the sides start to where they end?

I'm not sure. Please Help.

If it's a unit cube, then yes, the face at z = 1, can be parametrized as <x,y,1>. Furthermore, the normal vector to this face, pointing outwards from the cube is <0, 0, 1> .

To find the flux through this face, take the dot product (inner product) of <0, 0, 1> and F(x,y,z) and integrate over the face.

etc.


BTW: The normal to the opposite face is <0, 0, -1> . This is the face at z = 0 .

 

[HallsofIvy]

It might be simpler to use the "divergence theorem":
\int\int\int \nabla\cdot \vec{F} dV= \int\int \vec{F}\cdot d\vec{S}
in other words, instead of integrating the vector function over the faces of the cube, integrate its divergence over the cube itself.

(Unless, of course, this exercise asks you to do both and compare!)