Why is the flux on top and bottom of a cylinder ignored in Gauss's law?

[Acuben]

Homework Statement

(first of all, let me know if I am wrong in any parts)
gauss's law can be used on cylinder.
Why is it that flux on top and flux on bottom of the cylinder, with area pi*r^2 ignored? (aka, equal to 0)
such that
Flux= Integral(E*da)=E2(pi)rl=q/(Eo)
where (Eo) is permitivity of/in vacuum.

Homework Equations

Cylinder
Flux= Integral(E*da)=E2(pi)rl=q/(Eo)
Sphere
Flux= Integral(E*da)=E2(pi)r^2=q/(Eo)

The Attempt at a Solution

 

[zhermes]

What are you uses gauss' law ON? a wire? a uniform sheet? constant charge density? a banana?

 

[Mindscrape]

Mmm, gaussian banana.

 

[Acuben]

What are you uses gauss' law ON? a wire? a uniform sheet? constant charge density? a banana?

nothing in particular. I meant making an gaussian cylinder in general did not define what I am using it on.
Although if it's a long wire, I suppose there is no Electric field towards top and bottom and it cancels out.

 

[zhermes]

in that case the answer to your question is: 'it depends on the situation.'

 

[stevenb]

Mmm, gaussian banana.

That can only be solved using a bananal coordinate system.