I am approaching line integrals for the first time. To understand how they work and, most importantly, how they could be used, I tried to come up with simple examples.
One such example is to use line integrals in order to recover the area of the unit square $[0,1]^2$ on $R^2$, which is trivially equal to 1x1=1.
The canonical way to do that is via Riemann integrals:
$\int_0^1 \int_0^1 1\ d \tilde x \ d \tilde y =1$
That is, integrating a measure of 1 over the unit square.
A different approach could use a line integral. Intuitively we could use the linear function $y=ax$ as a `windshield wiper' to get the area of the square.
Below an attempt to compute the first half of the area by having the slope go from $a=0$ to $a=1$. Let $x(t)=t$ and $y(t)=at$. I get that the area is the "sum" of the integrals of a family of linear functions $y=ax$:
$\int_{0}^1 \int_{c(a)} 1 \ ds \ da $
where $c(a)={\{ (\tilde x , \tilde y \}:\tilde y=a\cdot\tilde x \text { and } (\tilde y,\tilde x) \in [0,1]^2 }$. Soving leads to
$\int_0^1 \int_0^1 1 \sqrt{1^2+a^2}dt \ da = \int_{0}^1 \frac{1}{2} \left(\sqrt{a^2+1} a+\sinh ^{-1}(a)\right)da = \frac{1}{2} \left(\sqrt{2}+\sinh ^{-1}(1)\right) \not = 1/2.$
What am I missing? Thanks for helping me understand.