Evaluate the path integral of $f(x, y) = y$ over the graph of the semicircle
$y = \sqrt{1-x^2}, -1 \leq x \leq 1$
solution:
$\vec{C}(t) = (cos(t), sin(t))$
$\vec{C}'(t) = (-sin(t), cos(t))$
$||\vec{C}(t)'|| = 1$
$\int_{C} f ds = \int_{0}^{\pi} sin(t)dt = -cos(t)|_{0}^{\pi} = -(-1-1) = 2$
Could someone explain the solution please. How they get "$\vec{C}(t) = (cos(t), sin(t))$"
what's the use of "$||\vec{C}(t)'|| = 1$"
where does $0 \leq t \leq \pi$ come from?