Csdrhrt

Using the gaussian law I want to determine the field strength of a charged stick on another load.

The base formula for that is
$$epsilon_0 oint {vec{E} cdot d vec{A}} = q$$
A second formula for q is:
$$ q = lambda cdot h $$
$$vec{E} = frac{1}{4cdot pi cdot epsilon_0} cdot frac{q_1}{r^2} cdot hat{r}$$
In this case we use a cylinder to solve our problem and we only need the lateral surface.
$$A = 2cdot pi cdot d cdot h$$

My questions

1. I have a strong problem with a surface written as vector.
   
   Assuming we have a situation like in the first picture of this question where the lateral surface of our cylinder touches the load.
   
   How would I properly write the surface as vector (how can I determine the direction vector)?
   
   Usually my approach would be $vec{r} = (x_2 - x_1)hat{x} + (y_2 - y_1)hat{y}$. But as we have a solid here I have problems finding the start and end coordinates.




2. Until now I only solved integrals in the form of $int f(x) dx$ but now I have an equation like the one above. What would be my "x" here? How would I solve this integral?

As you can imagine this is a task from my electronics class. It's not a homework but it was an example.

Unfortunately we did explicitly not pay attention to the integral and we did not pay attention to the surface written as vector.

What we did is the following:
$$epsilon_0 oint {vec{E} cdot d vec{A}} = q$$
$$epsilon_0 cdot |vec{E}| cdot 2 cdot pi cdot h cdot d = q$$
Because $lambda$ is given in this particular task we can do
$$epsilon_0 cdot |vec{E}| cdot 2 cdot pi cdot h cdot d = lambda cdot h$$
$$|vec{E}| = frac{lambda}{2cdot pi cdot d cdot epsilon_0}$$

Due to symmetry considerations, $vec E$ at a given point is always pointing along the line from that point that is perpendicular to the wire. We can use cylindrical coordinates. With $d$ the distance from the wire, we can write $vec E=E(d)hat r$. For a surface, the vector $dvec A$ is always perpendicular to that surface. So in cylindrical coordinates we can write the surface element that is part of the cylinder with radius $r$ as $dvec A=r dtheta dh hat r$. Notice that $dvec A$ and $vec E$ are parallel (or antiparallel), so $$vec Ecdot dvec A=EdA=E d dtheta dh$$
Integrating $theta$ from $0$ to $2pi$ and $h$ from $0$ to $h$, you get your formula. In conclusion, you get the answer due to the fact that $vec E$ is always perpendicular to the surface of the cylinder of radius $d$, and is only dependent on the magnitude of $d$.