I am starting to study calculus III and I came across the following situation
Given the following form $$ydx - xdy$$ Why the integral along the semicircle $$P(t) = cos(t)\vec{i} + sin(t)\vec{j},\:0 \leq t \leq \pi$$ It is the same when using the following parameterization $$y = \sqrt{1-x^2}, -1 \leq x \leq 1$$ Since the parameterizations are reversed?
When computing the first integral, I got $ - \pi $.
$$\int_c ydx - xdy$$ $$x = cos(t) \Rightarrow \frac{dx}{dt} = -sin(t)$$ $$y = sin(t) \Rightarrow \frac{dt}{dt} = cos(t)$$ $$\int_0^\pi ( sin(t)(-sin(t)) - cos(t)cos(t)) dt = -\pi$$
How do I compute the second one that has an inverted sense of integration than the first one?
Thanks in advance!
You must know the orientation of the curve. In this case whether it is anti-clockwise or clockwise. That decides whether you go from $0$ to $\pi$ or $\pi$ to $0$ (similarly $-1$ to $1$ or from $1$ to $-1$).
$y = \sqrt{1-x^2}$
$dy = -\frac{x}{\sqrt{1-x^2}} \ dx$
Please note that if we have anti-clockwise orientation as in your first integral then we go from $x = 1$ to $x = -1$. So the line integral is
$\displaystyle \int_{1}^{-1} (\sqrt{1-x^2} + \frac{x^2}{\sqrt{1-x^2}}) \ dx = -\pi$