How Does Electric Flux Through a Cylinder Change with a Central Point Charge?

[villani]

Homework Statement

A point charge is on the axis of a cylinder at its center. The electric field is given by:

$$\vec{E} = q \vec{r} / \| \vec{r} \| ^3$$

Compute the flux of $$\vec{E}$$ outward through the cylinder: $$x^2 + y^2 = R^2$$, for $$0 \leq z \leq h$$

Homework Equations

$$\int \vec{E}d\vec{A} = \int \vec{E}\vec{n}dA$$

The Attempt at a Solution

We consider the three surfaces: side of the cylinder, top and bottom of the cylinder.
The unit normal vector to the side is: $$\frac{x \vec{i} + y \vec{j}}{\sqrt{x^2 + y^2}}$$
The unit vector to the top is k
The unit vector to the bottom is -k

If we consider the three surfaces and three integrals. We sum them to get the result.
Should the result be (Gauss Law):
$$4q\pi$$ ? How do we prove it?
Are the integrals for the top and bottom equals to zero?

How do we calculate the integral for the side of the cylinder?

$$\int \vec{E}\vec{n}dA = \int q \vec{r} / \| \vec{r} \| ^3 \vec{n}dA$$

 

[SEngstrom]

The field described is that of a point charge (q) at the origin, so the flux out of a volume that encloses the charge is $$4\pi*q$$. Now, for the given cylinder the charge lies in the plane of the bottom lid $$(z=0)$$, reducing the flow to half of that you'd get if the charge was enclosed in the volume. To convince yourself of this you can crank through the integrals using cylindrical coordinates $$(R,\phi,z)$$. The $$z=0$$ integral is zero, and the other two add up to $$2\pi*q$$.