Thread: flux through each side of cube

Consider the vector field .

Find the flux of the given vector field through a cube in the first octant with edge length c, one corner at the origin and edges along the axes.


I need to find the flux at each sides of the cube, can someone help me how I can do this?

Originally Posted by TheRekz

Consider the vector field .

Find the flux of the given vector field through a cube in the first octant with edge length c, one corner at the origin and edges along the axes.


I need to find the flux at each sides of the cube, can someone help me how I can do this?

I hope you know Gauss' theorem otherwise you will need to evaluate 6 surface integrals.

Use the divergence thm --beats 6 flux integrals

Flux = volume integral( divF) in this case divF =1

so you have simply c^3 for the flux

If you absolutely have to do 6 Flux calculations

since the x and z components are constant the flux through the bottom
cancels the flux through the top. Similarly for the front and back since the x component is 0

On the left side y= 0 so the flux is 0

the right face then is the only contribution to the flux

where y =c and you integrate over the square with sides c in the xzplane

Originally Posted by Calculus26

Use the divergence thm --beats 6 flux integrals

Flux = volume integral( divF) in this case divF =1

so you have simply c^3 for the flux

If you absolutely have to do 6 Flux calculations

since the x and z components are constant the flux through the bottom
cancels the flux through the top. Similarly for the front and back since the x component is 0

On the left side y= 0 so the flux is 0

the right face then is the only contribution to the flux

where y =c and you integrate over the square with sides c in the xzplane

yes.. I actually need to find the flux through each surface.. how can I do that?

Originally Posted by TheRekz

yes.. I actually need to find the flux through each surface.. how can I do that?

The definiton of Flux is


Since each of the sides of the cube are parallel to a coordinate axisis the normal vectors will be the unit vectors i, j k.

I will set one up for you

So on the top of the cube the normal vector point up in the Z direction.



You will need to do this with the other 5 sides and then add all of them up.

Originally Posted by TheEmptySet

The definiton of Flux is


Since each of the sides of the cube are parallel to a coordinate axisis the normal vectors will be the unit vectors i, j k.

I will set one up for you

So on the top of the cube the normal vector point up in the Z direction.



You will need to do this with the other 5 sides and then add all of them up.

for the bottom would it be just the inverse of the top?

Originally Posted by TheRekz

for the bottom would it be just the inverse of the top?

yes becuase your unit vector is

the negative not the inverse -- yes since n = -k (outward normal)

I am just confuse on how to get the left side of the cube... can someone enlighten me

so far I have:

First n = -j also y = 0 on the left face so F*n = 0 --on the right face
y = c

Originally Posted by Calculus26

First n = -j also y = 0 on the left face so F*n = 0 --on the right face
y = c

sorry I meant to ask for the right face as the left face is 0

when y = c, so what do I set the integral as?

For

simply put y= c in above integral

thanks!