In a proof for Green's theorem where we first assume that the area bounded by a closed $C^1$ curve $\gamma$ is of the following form: $$D = \{ (x,y) \ | \ x \in [a,b], \ \mu(x) \le y \le v(x) \}$$ Then we split $\gamma$ into four pieces in the following way, and I am not sure what's going on here and why we can do this?
$$ r_1(t) = \begin{pmatrix}t \\ \mu(t) \end{pmatrix} \ \ \ \ \text{when} \ t \in [a,b] $$ $$ r_2(t) = \begin{pmatrix} b \\(1-t)\mu(b) + tv(b) \end{pmatrix} \ \ \ \ \text{when} \ t \in [0,1]$$ $$ r_3(t) = \begin{pmatrix} b-t \\ v(b-t) \end{pmatrix} \ \ \ \ \text{when} \ t \in [0,b-a]$$ $$r_4(t) = \begin{pmatrix} a \\t\mu(a) + (1-t)v(a) \end{pmatrix} \ \ \ \ \text{when} \ t \in [0,1]$$ What's going on? Is this just some common known fact? There's no explanation given in the book, and my experience with parametric equations is very limited.
The parametric equation $r(t) = (f(t),g(t))$ gives equations $x=f(t)$ and $y=g(t)$ for the coordinates $x$ and $y$ in terms of a variable $t$.
The region $D$ is bounded (above and below) by $\mu(x)$ and $\nu(x)$, and the $r_1$ and $r_3$ are a parametrisation of these bits ($r_3$ goes backwards because we are going anticlockwise around the region): notice that the $x$ coordinate is just given by $t$ in the case of $r_1$, so that $r_1$ satisfies $y=\mu(x)$ (second coordinate is $\mu$ applied to the first one).
$r_2$ parametrises the segment of the line $x=b$ (notice the first coordinate again) between $\mu(b)$ and $\nu(b)$: notice that $t=0$ gives $y=\mu(b)$ and $t=1$ gives $y=\nu(b)$. $r_4$ works in the same way, but on $x=a$, with $t=0$ giving $y=\nu(a)$ and $t=1$ giving $y=\mu(a)$.