Approximation of areas with lines Let $$\textbf{r}(t)=(\frac{1}{2}\sin t+\frac {1}{2};\frac{t}{a}) $$
With $t\in[0,a]$ and $a$ is a constant.The $y$ component of this will go from 0 to 1 and the $x$ component will go back and fourth between 0 and 1. Now imagine that we let $a\rightarrow \infty$. In this case the $x$ component will continue to go back and forth between 0 and 1, but the $y$ will increase very very slowly. in the limit this curve will equal the square $[0,1]\times [0,1]$. Is it possible to calculate the area of this square using some kind of integration with the "length" of $\textbf{r}(t)$?
 A: The question you are asking is very much in the realm of the philosophy of mathematics so I will answer it as such. Integrals are foundationally not equipped to handle the computation as you intend.
The reason for this is intuitively simple - integrals can only sum up objects of the same dimensions, but this computation would need us to add a bunch of lengths to get an area. This is baked into the laws and assumptions behind the math we use to make integrals, we can't add things with different dimensions normally (think adding a meter to a kilogram) from the Riemann sum or Darboux sums. We can set up integrals to integrate a function to hold the extra dimensions, but then we are not summing lengths anymore.
Torricelli, however, used indivisbles (today known as Cavilieri's principle) to prove that Gabriel's horn had a finite volume. In effect, it added up discs of a certain area to get a volume, which is the procedure we want to do.
Consider the unit square to the right of $[0,1]^2$. Freeze $a$ and draw the lines $y=\frac{n\pi}{a}$. Each half period of the sine wave (which always connects the left side of the square to right) intersects exactly one of these lines, which goes on to be a set of parallel line segments in the next square with length $1$. 
In the limit the arc length of each branch goes to $1$, so since each branch corresponds to a perfectly parallel line in the next square over with the same length, the two areas must be equal.
While not rigorous this sketch can get you the proof you need with some squeeze theorem manipulations. This is a reasonable interpretation  of "lengths adding up to areas".
But in the usual sense of adding up lengths to get an area, this is impossible. You can prove that the normal arc length integral has to diverge.
