# flux through each side of cube

• April 22nd 2009, 05:58 PM
TheRekz
flux through each side of cube
Consider the vector field $\vec{F} = 10\vec{i} + y\vec{j} + 5\vec{k}$ .

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?
• April 22nd 2009, 06:07 PM
TheEmptySet
Quote:

Originally Posted by TheRekz
Consider the vector field $\vec{F} = 10\vec{i} + y\vec{j} + 5\vec{k}$ .

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.

$\iiint_V \nabla \cdot F dV$

$\int_{0}^{c}\int_{0}^{c}\int_{0}^{c}1dV=c^3$
• April 22nd 2009, 06:08 PM
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
• April 22nd 2009, 07:01 PM
TheRekz
Quote:

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?
• April 22nd 2009, 07:11 PM
TheEmptySet
Quote:

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
$\int F \cdot \vec n dS$

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.

$\int_{0}^{c}\int_{0}^{c} (10 \vec i +y \vec y + 5 \vec k)\cdot \vec k dxdy=\int_{0}^{c}\int_{0}^{c}5dxdy=5c^2$

You will need to do this with the other 5 sides and then add all of them up.
• April 22nd 2009, 08:13 PM
TheRekz
Quote:

Originally Posted by TheEmptySet
The definiton of Flux is
$\int F \cdot \vec n dS$

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.

$\int_{0}^{c}\int_{0}^{c} (10 \vec i +y \vec y + 5 \vec k)\cdot \vec k dxdy=\int_{0}^{c}\int_{0}^{c}5dxdy=5c^2$

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?
• April 22nd 2009, 08:15 PM
TheEmptySet
Quote:

Originally Posted by TheRekz
for the bottom would it be just the inverse of the top?

yes becuase your unit vector is $-\vec k$
• April 22nd 2009, 08:16 PM
Calculus26
the negative not the inverse -- yes since n = -k (outward normal)
• April 23rd 2009, 07:43 AM
TheRekz
I am just confuse on how to get the left side of the cube... can someone enlighten me

so far I have:

$\int_{0}^{c}\int_{0}^{c} y dx dz$
• April 23rd 2009, 09:43 AM
Calculus26
First n = -j also y = 0 on the left face so F*n = 0 --on the right face
y = c
• April 23rd 2009, 11:33 AM
TheRekz
Quote:

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
• April 23rd 2009, 08:56 PM
TheRekz
when y = c, so what do I set the integral as?
• April 23rd 2009, 09:01 PM
Calculus26
For http://www.mathhelpforum.com/math-he...e246cfe4-1.gif

simply put y= c in above integral
• April 23rd 2009, 09:03 PM
TheRekz
thanks! :)