Let L be a line going through (0,0) in $\mathbb{R}^2$
Let $\alpha$ be some $C^1$ curve belonging to L
Let $F(x,y)=(y,-x)$
How one goes to prove that $\int_{\alpha}F\cdot d\alpha=0$
I am not asking for proof per se. Line " Let $\alpha$ be some $C^1$ curve belonging to L" is crude translation from finnish, which I have trouble to grasp.
Decided to ask stackexchange, because if $F(x,y)=(y,-x)$, $\int_{\alpha}F\cdot d\alpha=0$ concepts ring a bell - please give me a pointer, what does it mean "some curve belonging to a line going through (0,0)"?
Got it.
In this context it means that $\alpha(t) \in L$, ie subset of L
as example $\alpha(t)=(t^3,t^3)$ for $t \in \mathbb{R}$, or $\alpha(t)=(sin(t),sin(t))$ for $t \in [-1,1]$
And original question regarding proof is then
$int_{\alpha} F \cdot d\alpha=\int_{\alpha} (-y,x)\cdot (\alpha_x'(t),\alpha_y'(t))$
Since $\alpha(t) \in L$, we can say it is of a form $(g(t),kg(t))$ and $\alpha'(t)=(g'(t),kg'(t))$
dot product is then $g(t)kg'(t)-kg(t)g't=0$
and $\int_a^b 0dt=0$ for any a,b
sorry for bothering