Trying to understand why circle area is not $2 \pi r^2$ I understand the reasoning behind $\pi r^2$ for a circle area however I'd like to know what is wrong with the reasoning below:

The area of a square is like a line, the height (one dimension, length) placed several times next to each other up to the square all the way until the square length thus we have height x length for the area.

The area of a circle could be thought of a line (The radius) placed next to each other several times enough to make up a circle. Given that circumference of a circle is $2 \pi r$ we would, by the same reasoning as above, have $2 \pi r^2$. Where is the problem with this reasoning?

Lines placed next to each other would only go straight like a rectangle so you'd have to spread them apart in one of the ends to be able to make up a circle so I believe the problem is there somewhere. Could anybody explain the issue in the reasoning above?
The main issue is that you don't form an area by placing lines next to each other--you need to place strips next to each other.  As you say, to form an $a\times a$ square, you can place $n$ strips of dimension $a\times w$ next to each other, where $w=a/n$, giving total area $naw=a^2$.



Your suggestion amounts to forming the area of a circle of radius $r$ by placing $n$ strips of dimension $w\times r$ next to each other radially, where $w=2\pi r/n$, giving area $nrw=2\pi r^2$.  But look what happens if you do this:



The problem is that the strips overlap, so the total area of the $n$ strips is greater than the area of the circle.  If you run the animation (by reloading the page, if necessary) you can convince yourself that in the $n\rightarrow\infty$ limit, half of each strip contributes to the final area.  (Observe that as strips are added in the counterclockwise direction, roughly half of each strip gets covered by subsequent strips.)

We can fix the problem of overlapping strips more easily by using triangles of base $w$ and height $r$ instead of rectangles:



This gives area $\frac{1}{2}nrw=\pi r^2$.
Actually, you can use your method, but you have to be a bit careful.  Start with a circle of radius $r$ centered at the origin.  We can think of the total area of the circle as adding the "lengths" of all the circumferences of all the circles centered at the origin which have a radius $x \leq r$:

$$\text{Area of Circle} = \int_0^{r} 2 \pi x~\textrm{d}x = \pi r^2$$

In reality, we are approximating the circumference of each circle by an annulus of small width, and then letting that width tend to zero.

This method works with finding the area of the square too.  Suppose you have a square centered at the origin.  Then, the area of the square is the perimeter of each square centered at the origin which ha a side length $x &lt; r$.

$$
\text{Area of Square} = \int_{-r/2}^{r/2} 4x dx = r^2.
$$

The reason this doesn't work the way you originally thought was in order to add "fattened" radii all around the circle, the fattened radii overlap, and so you are over counting.