...for the curve $C$, which goes along the circle of radius $3$, from the point $(3,0)$ to the point $(-3,0)$, and then in a straight line along the $x$-axis back to $(3,0$).
So I set the half-circle part as $C_1$ and the line part as $C_2$.
For $C_1$, I used $x=rcos(t)$, $y=rsin(t)$, $0≤t≤π$, to get $(3cos(\pi t), 3sin(πt)$, $0≤t≤1$.
Now what I'm stuck on is $C_2$. I thought I could just set $x=$t, $y=0$ with $-3≤t≤3$, but the book says it's $(6t-9, 0)$, $1≤t≤2$. Is there really a difference or would both of these be correct?
I haven't done parametrizations before now, when we're using them for path integrals, so I'm a bit lost. I'm not even sure how you'd use something like this for a path integral. Any help would be much appreciated.
The essential reason why the book has a different parametrization is because if you have already used $t \in [0,1]$ to describe the semicircular arc, you have to continue from $t \ge 1$ to describe the diameter. The way you've defined it, $$(x(t), y(t)) = \begin{cases} (3 \cos \pi t, 3 \sin \pi t), & 0 \le t \le 1, \\ (t, 0), & -3 \le t \le 3, \end{cases}$$ does not give a well-defined ordered pair $(x,y)$ for $t \in [0,1]$, but two different ordered pairs. You can't do that. The way the book defined it, $$(x(t), y(t)) = \begin{cases} (3 \cos \pi t, 3 \sin \pi t), & 0 \le t < 1, \\ (6t-9, 0), & 1 \le t \le 2, \end{cases}$$ clearly gives only one choice for any $t \in [0,2]$; moreover, at $t = 1$, the curve is continuous: $$(3 \cos \pi, 3 \sin \pi) = (-3,0) = (6(1)-9, 0).$$
The book's parametrization is not a uniquely legitimate one, however. For example, you can choose to parametrize the diameter as $(x,y) = (6t^2-9,0)$, $1 \le t \le \sqrt{2}$. But what you absolutely cannot do is define a parametrization that does not specify a unique coordinate for each value of the parameter in its domain.