My question is very simple yet crucial to the understanding of many fields of mathematics. What is a line integral? If I choose some arbitrary line segment $\mathbb{A}$ to integrate a function $\Gamma$ over, what does $$\oint_{\mathbb{A}} \Gamma (s) ds $$ mean? For example: View page three of paper. What exactly does integrating over a line segment tell us? How is this integration performed?
For example:
We consider the cube $L=[0,r_0]^3$ where $r_0 > 0$ is a fixed real number.
The lattice is generated by the level set $L_r$
We define the level function $\lambda (x,y,z) \to x+y+z =r$
For $ r \leq r_0$
$$L_r = \{x \in L : \lambda(r) = r\}$$ is an equilateral triangle with vertices $(r,0,0), (0,r,0),(0,0,r)$.
For $r_0 < r < 2r_0$ the level set $L_r$ is a hexagon with vertices $$(r_0, r-r_0,0), (0,r_0,r-r_o), (0,r - r_0, r_0),(r-r_0,r_0,0), (r-r_0, 0, r_0)\;.$$
Define $\mu_L (r)$ to represent the surface area of each level set.
Therefore $$\mu_L (r) = \begin{cases} \frac{\sqrt3}{2}r^2, & \text{if } 0\leq r \leq r_0 \\\\ \frac{\sqrt3}{2}(r^2-3(r-r_0)^2), & \text{if } r_0 < r \leq 2r_0 \\\\ \frac{\sqrt3}{2}(3r_0-r^2), & \text{if }2r_0 \leq r \leq 3r_0 \\ \end{cases} $$
I'm a little confused on the second premise of $\mu_L (r)$.
It follows that $\mu_L(r)$ is continuously differentiable and $\mu_L^{'}(r_o) = \sqrt3 r_0$
If $x$ is on the line segment from $(r_0,0,0)$ to $(0,r_0,0)$, then the outer normal to $A=L_{r_0}$ is given by
$$\frac{ (-r_0,r_0,0) \times (1,1,1)}{|| (-r_0,r_0,0) \times (1,1,1) ||} = \frac{1}{\sqrt6} (1,1,-2)$$
and it follows that
$\Phi(n(x)) = \frac{1}{\sqrt6}$ Integrating along the aforementioned line segment gives $$\int_{(r_0,0,0)}^{(0,r_0,0)} \Phi (n(s)) ds = \frac{1}{\sqrt3} r_0$$ therefore $$\oint_{\partial A} \Phi(n(s)) ds = \sqrt3r_0$$.
Hence at least in this example, $$\oint_{\partial L_{r_0}} \Phi (n(s)) ds = \mu^{'}_L (r_0) $$
I have literally no idea what is going on here. Can someone explain? What is the integration for?