$$\oint \vec{E}^{\,} \cdot \vec{dA}^{\,}$$
This is Gauss' law from physics but my question is more maths related.
Say I have $\vec{E}^{\,}=3.5\times 10^3N/C\times e_x$ and $A = 0,35\times0,70m^2$. The plane that has that area, where the electric field $E$ is applied, is parallel to $YZ$. So it has a coordinate in $e_x$. Right?
What does the circle in the integral mean in practice?
What is the value of $dA$? Why is it a vector?
How do I solve this integral?
On
The circle in the integral specifies that integration is being performed over a closed surface (i.e. one without a boundary). The surface of a sphere would be considered a closed surface, but the surface of a plane would not.
Gauss's law states that the net electric flux through any arbitrary closed surface enclosing charge is proportional to the total charge.
We obtain the electric flux through an infinitesimal element of a surface by taking the dot product $\overrightarrow{E} \cdot d\overrightarrow{A}$. The vector d$\overrightarrow{A}$ refers to the unit normal vector of the infinitesimal surface element. The net electric flux is obtained by integration over the entire surface.
If we were to consider your surface to be closed (which a plane is not) and that the electric field was constant and always normal to that surface, then $\oint \overrightarrow{E} \cdot d\overrightarrow{A}= |\overrightarrow{E}|A$.
On
Gauss' law involves a surface integral. The answers to your questions come from understanding how this surface integral works.
In a vector integral over some surface $S$ like $$ \int_S \vec{F} \cdot \mathrm{d}\vec{A} $$ we have $2$ parts at play:
P1) $\vec{F}$ is a vector field, which is just a function that gives you a vector as an output.
P2) $\mathrm{d}\vec{A}$ is the area differential vector. Explicitly, it represents the unit normal vector of the surface $\hat{n}$ being multiplied by the area differential $\mathrm{d}S$ (which you can think of as an infinitesimally small portion of the surface area). In practice, you constantly have to use $\mathrm{d}\vec{A} =\hat{n}\, \mathrm{d}S$ to solve the integrals.
These 2 quantities are being multiplied using the dot product, which by definition will transform our integrand from vectors being multiplied into some (scalar) number.
Here's a diagram to better understand these components:
Here $S$ is the surface we're integrating over, the vector field $\color{red}{\vec{F}}$ are all the arrows in red (of which there are one for each point in space), the normal unit vectors $\color{blue}{\hat{n}}$ are the length-one vectors perpendicular to (each point on) the surface, and each $\mathrm{d}S$ is one of the small patches of surface area (separated by the grid lines). Now to answer the questions:
What does the circle in the integral mean in practice?
The circle just means that your surface $S$ of integration is closed. This just means that Gauss' law doesn't apply to any blanket-like surface you want, it HAS to completely enclose the charge (which is the physical reason for this). This is why you'll find that you usually take spheres or cylinders as the surface $S$ in these problems. In practice, it's no different as to solving any other surface integral, it only affects the limits of integration.
What is the value of $\vec{dA}$? Why is it a vector?
By definition, it's the area vector we described previously.
How do I solve this integral?
I'll use your data to try and show the general approach. Our vector field $\vec{E}$ is given by $$ \vec{E} = (3500,0,0) $$ Now, the way I understand your question as written is that our surface of integration $S$ in this case is a segment of a plane parallel to the $y-z$ plane. This means that the normal vector to this plane is the unitary vector in the $x$ direction, or in other words: $$ \hat{n} = (1,0,0) $$ Note here that in this case our surface is not closed, since a plane does not enclose some volume like a sphere, for example. This is why it wouldn't be correct to use $\oint$ in this case, we should just use $\int_S$. Having said this, the integral we want becomes \begin{align*} \int_S \vec{E} \cdot \vec{dA} &\overset{\color{blue}{\text{P}2}}{=} \int_S\left( \vec{E} \cdot \hat{n}\right) \ dS \\ &= \int_S\left( (3500,0,0) \cdot(1,0,0)\right) \ dS \\ &= \int_S 3500\ dS \\ & = 3500 \int_S \ dS \end{align*} but by definition of the surface integral, if we're integrating the area differential over the whole surface we're interested in, we just get the area of the entire surface $S$ as a result. In other words: $$ \int_S \ dS=\text{Area of }S = 0.35 \times 0.70 = 0.245 $$ which is the area of our plane segment. And so \begin{align*} \int_S \vec{E} \cdot \vec{dA} & = 3500 \int_S \ dS\\ & = 3500 (0.245)\\ & = 857.5 \end{align*} I've dropped the units in the proceadure, but this last result would have units of $\frac{N m^2}{C}$.
As a final note, this kind of vector-surface integral is also known as the flux. It's called this way because it essentially tells you "how much field" passes through the surface we're integrating over. You can see this by the use of the dot product between $\vec{F} \cdot \vec{dA}= \vec{F} \cdot \hat{n} \ dS$ in the diagram at the beginning since when $\vec{F}$ and the normal vector to the surface are parallel the dot product will give a bigger number, whereas if they don't point in the same direction we get a smaller dot product.
Lastly, in a general approach to solve the integral we usually can't easily calculate the area or factor out the integrand. In these cases you would proceed to parametrize the surface (in cartesian coordinates, for example) and then use that $dS = dy \ dx$ to solve the integrals explicitly.
It looks like you're overcomplicating things massively for this simple example.
In this case, the answer is just multiply the two numbers together. No integral solving, nothing sophisticated. This is because you just have a constant electric field passing perpendicular through an area. I am not sure how your course defines electric flux, but for a uniform field passing through a flat surface, you multiply the area by the component of your field that is perpendicular to your surface. In your question, field and surface are perpendicular, so you don't even need to worry about any potential $\cos\theta$ factors.
For your other questions, these are all things from vector calculus, which I'm guessing you'll learn about sooner or later. The above result is a very special case of the general Gauss law with vector integrals etc. For the purpose of this question, you don't need to worry about all that. I can expand more on the mathematics/vector calculus if you wish.
OK so the following is a super rapid overview of vector integrals.
You no doubt know how differentiation/integration for functions $f:\mathbb R\to\mathbb R$ works. The idea of vector calculus is to generalise things to functions $\mathbb R^3\to\mathbb R$, $\mathbb R\to\mathbb R^3$, $\mathbb R^3\to\mathbb R^3$ etc. There's nothing special about $\mathbb R^3$, it just so happens that the world we live in has three spatial dimensions, so physics people care an awful lot about $\mathbb R^3$.
Let's consider a concrete example. Suppose we have some material contained in a volume $V$, with surface $S$. We are interested in finding how much heat flows out of this block. We can find this by computing the heat flow out of the surface $S$.
To do this, we somehow need to integrate the heat flow over the surface. We write $\mathop{}\!\mathrm{d}S$ for the differential corresponding to a tiny chunk of this surface (similar to how $\mathop{}\!\mathrm{d}x$ represents a small change in $x$ in single variable calculus). If our surface happened to be the $yz$-plane, then we would have $\mathop{}\!\mathrm{d}S=\mathop{}\!\mathrm{d}y\mathop{}\!\mathrm{d}z$; I hope it's intuitively clear why.
Suppose $\mathbf H(\mathbf v)$ represents the heat flow at some point $\mathbf v$ in space. Note that $\mathbf H$ takes a vector for the position as input and spits out a vector for the heat flow at that position (per unit area and time) as output. The heat flow out of a chunk of surface $\mathop{}\!\mathrm{d}S$ is the area times the component of $\mathbf H$ perpendicular to $\mathop{}\!\mathrm{d}S$. The reason we only take the perpendicular component is that the component of $\mathbf H$ that flows parallel to the surface doesn't flow in or out of the material, so doesn't contribute to the quantity we care about.
If $\mathbf n$ represents the unit length vector normal to $S$ (so $\mathbf n$ may vary depending on where on $S$ you are standing), then the component of $\mathbf H$ perpendicular to the surface is just $\mathbf H\cdot\mathbf n$, so the heat flow through that little chunk is $\mathbf H\cdot\mathbf n\mathop{}\!\mathrm{d}S$.
To find the total heat flow, we need to sum the contributions from all chunks of our surface, i.e. we integrate over the surface: $$\int\mathbf H\cdot\mathbf n\mathop{}\!\mathrm{d}S.$$
Since $\mathbf n$ is intrinsic to the surface $S$, we often write as an abbreviation $\mathop{}\!\mathrm{d}\mathbf S=\mathbf n\mathop{}\!\mathrm{d}S$. This is the vector area element. With this notation, the total heat flow is $$\int\mathbf H\cdot\mathop{}\!\mathrm{d}\mathbf S.$$ There's nothing special about heat of course: I just picked it because it's easy to visualise heat flowing out of stuff. The same expression above, applied to the electric field $\mathbf E$, just tells you how much of the electric field is flowing through the surface. We call this quantity the (electric) flux through the surface.
Let's apply all this to the question at hand. Your electric field is uniform $\mathbf E=3500\mathbf e_x$. Let's say the surface is a rectangle with vertices $(0,0,0)$, $(0,0,0.35)$, $(0,0.7,0.35)$, $(0,0.7,0)$. The normal vector is clearly just $\mathbf e_x$, anywhere on the surface. So $\mathop{}\!\mathrm{d}\mathbf S=\mathbf e_x\mathop{}\!\mathrm{d}S=\mathbf e_x\mathop{}\!\mathrm{d}y\mathop{}\!\mathrm{d}z$, and applying the above, \begin{align} \int\mathbf E\cdot\mathop{}\!\mathrm{d}\mathbf S&=\int_0^{0.35}\int_0^{0.7}3500\mathbf e_x\cdot\mathbf e_x\mathop{}\!\mathrm{d}y\mathop{}\!\mathrm{d}z\\ &=3500\cdot\int_0^{0.35}\int_0^{0.7}1\mathop{}\!\mathrm{d}y\mathop{}\!\mathrm{d}z\\ &=3500\cdot\int_0^{0.35}0.7\mathop{}\!\mathrm{d}z\\ &=3500\times0.35\times0.7 \\ &=857.5. \end{align} So this integral is the long way of saying "just multiply everything". Of course, when the field $\mathbf E$ or the surface $S$ are more complicated, the integrals won't be nearly so trivial. But the basic method is the same: